An Introduction to Symbolic Logic

Version Aug 2013 of An Exposition of Symbolic Logic is a lightly revised version of the August 2012 version of An Introd

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Chapter Two
Sentential Logic with 'and', 'or', if-and-only-if'
1 SYMBOLIC NOTATION
2 ENGLISH EQUIVALENTS OF THE CONNECTIVES
3 COMPLEX SENTENCES
8 DERIVED RULES
9 OFFICIAL CONDITIONS FOR DERIVATIONS
10 TRUTH TABLES AND TAUTOLOGIES
11 TAUTOLOGICAL VALIDITY
Chapter Five
Identity and Operation Symbols
1 IDENTITY
2 AT LEAST AND AT MOST, EXACTLY, AND ONLY
3 DERIVATIONAL RULES FOR IDENTITY
44c23ac2-d876-40cb-9427-16641034ed00.pdf
Chapter Two
Sentential Logic with 'and', 'or', if-and-only-if'
1 SYMBOLIC NOTATION
P(Q ( R
P ( Q(R
P ( (Q(R)
2 |
T
2 2
T
2 2
EXERCISES
1. For each of the following state whether it is a sentence in official notation, or a sentence in informal notation, or not a sentence at all. If it is a sentence, parse it as indicated above.
2 ENGLISH EQUIVALENTS OF THE CONNECTIVES
The book is short, and it is interesting
The book is short, but it is interesting
The book is short; it is interesting
The game will be called off just in case it rains: Q ( R
EXERCISES
1. For each of the following sentences say which symbolic sentence it is equivalent to.
R ( P
R ( P
R ( P
W ( R
W ( R
R ( S
R ( S
Q ( R
Q ( R
2. Symbolize each of the following using this translation scheme:
S Sally will walk
V Veronica will give Sally a ride
R It will rain
Q Barbara will come with Quincy
T Barbara will come with Tom
3 COMPLEX SENTENCES
This is a complex sentence, with at least two different but equivalent symbolizations.
Neither Polk nor Quincy was president.
If neither Wilma nor Sally attends, either Robert or Peter will be bored.
If neither Wilma [attends] nor Sally attends, either Robert [will be bored] or Peter will be bored.
If neither W nor S, either R or P
~(W(S) ( (R(P)
If neither Wilma nor Sally attends, either Robert or Peter, but not Tom, will be bored.
If neither Wilma [attends] nor Sally attends, either Robert [will be bored] or Peter [will be bored], but Tom will not be bored.
If neither W nor S, either R or P, but not T
~(W(S) ( (R(P) & ~T
~(W(S) ( (R ( (P&~T))
(~(W(S) ( (R(P)) & ~T
Either Robert or Tom will attend, but not both
Either Robert [will attend] or Tom will attend, but not both [will attend]
Either R or T, but not R and T
(R(T) (~(R(T)
Robert will attend if Sally does, but she won't attend if neither Tom nor Wilma attend.
Robert will attend if Sally does [attend], but she won't attend if neither Tom [attends] nor Wilma attends.
R if S, but not S if neither T nor W
(S(R) ((~(T(W) ( ~S)
Neither Sally nor Robert will run, but if either Tom or Quincy run, Veronica will win.
Neither S nor R, but if either T or Q, V
~(S(R) ((T(Q ( V).
Given that Sally and Robert won't both run, Tom will run exactly if Q does.
Given that not both S and R, T exactly if Q.
~(S(R) ( (T(Q)
A variety of English expressions that we have not mentioned affect how a sentence is to be symbolized. Examples:
Quincy will whistle if Reggie sings without Susan singing or Susan sings without Reggie, but he won't whistle if they both sing
Q if R and not S or S and not R, but not Q if S and R
((R(~S) ( (S(~R) ( Q) ((S(R ( ~Q)
EXERCISES
This derivation illustrates how the conjunction rules are used:
P ( Q
( Q ( P
1. Show Q ( P
Addition indicates that from any sentence you may infer its disjunction with any other sentence.
For mtp you need the negation of a disjunct. In the case given, if 'T' and 'W' were both true, then the argument would have true premises and a false conclusion.
S ( P
P(Q ( ~R
R ( ~P
EXERCISES
These strategy hints will be put to use below, as we extend our list of Theorems from Chapter 1.
Notice that T26 and T4 from the previous chapter are both called "hypothetical syllogism".
1. Show (P(Q ( R) ( (P( (Q(R))
2. Show (P(Q ( R) ( (P( (Q(R))
4. Show P( (Q(R)
The rest of the work is filling in the remaining subderivations. It is often useful to develop a derivation as we did here by first sketching its overall structure, and then flesh it out with details afterwards.
EXERCISES
P
1. Show R
Here is a more highly abbreviated derivation.
P ( Q
R ( ~Q
Rule r also gives you a sentence -- the sentence on the line cited.
EXERCISES
1. Use the method of abbreviating derivations to produce shortened derivations for T38, T40-43.
Here are two arguments, and derivations, that use some theorems from Chapter 1 as rules.
S ( T
S ( Q
EXERCISES
Some additional theorems are given here for reference.
8 DERIVED RULES
EXERCISES
9 OFFICIAL CONDITIONS FOR DERIVATIONS
Let us summarize here what we can now use in constructing an unabbreviated derivation.
EXERCISES
10 TRUTH TABLES AND TAUTOLOGIES
EXERCISES
11 TAUTOLOGICAL VALIDITY
P ( P
EXERCISES

Citation preview

An Exposition of Symbolic Logic with Kalish-Montague derivations

Preface The system of logic used here is essentially that of Kalish & Montague 1964 and Kalish, Montague and Mar, Harcourt Brace Jovanovich, 1992. The principle difference is that written justifications are required for boxing and canceling: 'dd' for a direct derivation, 'id' for an indirect derivation, etc. This text is written to be used along with the UCLA Logic 2010 software program, but that program is not mentioned, and the text can be used independently (although you would want to supplement the exercises). The system of notation is almost the same as KK&M; major differences are that the signs '∀' and '∃' are used for the quantifiers, name and operation symbols are the small letters between ‘a’ and ‘h’, and variables are the small letters between ‘i’ and ‘z’. The exercises are new. Chapters 1-3 cover pretty much the same material as KM&M except that the rule allowing for the use of previously proved theorems is now in chapter 2, immediately following the section on theorems. (Previous versions of this text used the terminology ‘tautological implication’ in section 2.11. This has been changed to ‘tautological validity’ to agree with the logic program.) Chapters 4-6 include invalidity problems with infinite universes, where one specifies the interpretation of notation "by description"; e.g. "R(): ≤". These are discussed in the final section of each chapter, so they may easily be avoided. (They are not currently implemented in the logic program.) Chapter 4 covers material from KK&M chapter IV, but without operation symbols. Chapter 4 also includes material from KK&M chapter VII, namely interchange of equivalents, biconditional derivations, monadic sentences without quantifier overlay, and prenex form. Chapter 5 covers identity and operation symbols. Chapter 6 covers Fregean definite descriptions, as in KK&M chapter VI.

Version Aug 2013 of An Exposition of Symbolic Logic is a lightly revised version of the August 2012 version of An Introduction to Symbolic Logic (also known as Terry-Text).

Introduction -- 2

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CONTENTS Chapter One Sentential Logic with 'if' and 'not' 1 SYMBOLIC NOTATION 2 MEANINGS OF THE SYMBOLIC NOTATION 3 SYMBOLIZATION: TRANSLATING COMPLEX SENTENCES INTO SYMBOLIC NOTATION 4 RULES 5 DIRECT DERIVATIONS 6 CONDITIONAL DERIVATIONS 7 INDIRECT DERIVATIONS 8 SUBDERIVATIONS 9 SHORTCUTS 10 STRATEGY HINTS FOR DERIVATIONS 11 THEOREMS 12 USING PREVIOUSLY PROVED THEOREMS IN DERIVATIONS

Chapter Two Sentential Logic with 'and', 'or', if-and-only-if' 1 SYMBOLIC NOTATION 2 ENGLISH EQUIVALENTS OF THE CONNECTIVES 3 COMPLEX SENTENCES 4 RULES 5 SOME DERIVATIONS USING RULES S, ADJ, CB 6 ABBREVIATING DERIVATIONS 7 USING THEOREMS AS RULES 8 DERIVED RULES 9 OFFICIAL CONDITIONS FOR DERIVATIONS 10 TRUTH TABLES AND TAUTOLOGIES 11 TAUTOLOGICAL VALIDITY

Chapter Three Individual constants, Predicates, Variables and Quantifiers 1 INDIVIDUAL CONSTANTS AND PREDICATES 2 QUANTIFIERS, VARIABLES, AND FORMULAS 3 SCOPE AND BINDING 4 MEANINGS OF THE QUANTIFIERS 5 SYMBOLIZING SENTENCES WITH QUANTIFIERS 6 DERIVATIONS WITH QUANTIFIERS 7 UNIVERSAL DERIVATIONS 8 SOME DERIVATIONS 9 DERIVED RULES 10 INVALIDITIES 11 EXPANSIONS

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Chapter Four Many-Place Predicates 1 MANY-PLACE PREDICATES 2 SYMBOLIZING SENTENCES USING MANY-PLACE PREDICATES 3 DERIVATIONS 4 THE RULE "INTERCHANGE OF EQUIVALENTS" 5 BICONDITIONAL DERIVATIONS 6 SENTENCES WITHOUT OVERLAY OF QUANTIFIERS 7 PRENEX NORMAL FORMS 8 SOME THEOREMS 9 SHOWING INVALIDITY 10 COUNTER-EXAMPLES WITH INFINITE UNIVERSES

Chapter Five Identity and Operation Symbols 1 IDENTITY 2 AT LEAST AND AT MOST, EXACTLY, AND ONLY 3 DERIVATIONAL RULES FOR IDENTITY 4 INVALIDITIES WITH IDENTITY 5 OPERATION SYMBOLS 6 DERIVATIONS WITH COMPLEX TERMS 7 INVALID ARGUMENTS WITH OPERATION SYMBOLS 8 COUNTER-EXAMPLES WITH INFINITE UNIVERSES

Chapter Six Definite Descriptions 1 DEFINITE DESCRIPTIONS 2 SYMBOLIZING SENTENCES WITH DEFINITE DESCRIPTIONS 3 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: PROPER DESCRIPTIONS 4 SYMBOLIZING ORDINARY LANGUAGE 5 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: IMPROPER DESCRIPTIONS 6 INVALIDITIES WITH DEFINITE DESCRIPTIONS 7 UNIVERSAL DERIVATIONS 8 COUNTER-EXAMPLES WITH INFINITE UNIVERSES

Introduction -- 4

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Introduction

Introduction Logic is concerned with arguments, good and bad. With the docile and the reasonable, arguments are sometimes useful in settling disputes. With the reasonable, this utility attaches only to good arguments. It is the logician's business to serve the reasonable. Therefore, in the realm of arguments, it is the logician who distinguishes good from bad. Kalish & Montague 1964 p. 1

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2 TRUTH & VALIDITY A principle unit of investigation in logic is called an argument. An "argument", in its technical sense, consists of two parts: a set of sentences, called the premises, and a sentence called the conclusion. The term "argument" may suggest a dispute, but in logic something is called an argument whether or not any people ever have or ever will disagree about it. Likewise, the "premises" of such an argument may or may not have been believed or asserted by somebody, and it is sometimes useful to examine arguments whose "premises" would never be believed by any rational person. Likewise, by calling something a "conclusion" we do not suggest that anyone ever has or even should "conclude" this thing on the basis of the premises given. The point of the terminology is this: a major topic in the study of deductive logic is validity. This is a relationship between a set of sentences and another sentence; this relationship holds whenever it is logically impossible for there to be a situation in which all the sentences in the first set are true and the other sentence false. It turns out to be very useful to study this relationship in complete generality. That is, it is useful to have a theory which tells us when this relationship holds between any set of sentences and any other sentence. Since a major practical application of such a theory is to pieces of reasoning that people actually use, the tradition has arisen of calling the first set of sentences the "premises", and the other sentence the "conclusion". And since a practical application of logic is to situations in which people disagree, it is perhaps appropriate to call the whole thing an "argument". But these are now technical terms. An argument is simply something that has two parts: a set of sentences called the premises, and another sentence called the conclusion. For logical purposes, any such combination counts as an argument. In displaying arguments it is customary to write their premises first, and to indicate the conclusion by the word like 'so' or a symbol such as '∴' Either Polk was a president or Whitney was a president. Whitney was not a president. ∴ Polk was a president. The triangle made of three dots is an abbreviation of the word `therefore', and is a way of identifying the conclusion of an argument. In order to save on writing, and also to begin displaying the form of the arguments under discussion, we will start abbreviating simple sentences by capital letters. For the time being we will abbreviate Polk was a president by `P',

Introduction -- 6

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Introduction and Whitney was a president by `W'. We will abbreviate Whitney was not a president by `not W'. So the argument can be shortened to: P or W not W ∴P A major point of this book is to explore the notion of deductive validity. Since the deductive kind is the only one considered here, we simply refer to it as "validity". In this section we will go over certain consequences of the following definition of validity: •

An argument is valid if, and only if, there is no logically possible situation in which all of its premises are true and its conclusion false.

When we talk about "truth" here we do not have anything deep or mysterious in mind. For example, we say that the sentence 'There is beer in the refrigerator' is true if there is beer in the refrigerator, and false if there isn't beer in the refrigerator. That's all there is to it. We have already seen one case of a valid argument which has all of its premises true and its conclusion true as well: P or W not W ∴P

True True True

What other possibilities are there? Well, as we noted above, it is possible to have some of the premises false and the conclusion false too. (This is sometimes referred to as a case of the "garbage in, garbage out" principle.) Suppose we use `R' to abbreviate Robert E. Lee was a president. Then this argument does not have all of its premises true, nor is its conclusion true: R or W not W ∴R

False True False

Yet this argument is just as good, as far as its validity is concerned, as the first one. If its premises were true, then that would guarantee that its conclusion would be true too. There is no logically possible situation in which the premises are all true and the conclusion false. This argument, though it starts with a false premise and ends up with a false conclusion, has exactly the same logical form as the first one. This sameness of logical form lies at the foundation of the theory in this book; it is discussed in the following section. Although false inputs can lead to false outputs, there is no guarantee that this will happen, for you can reason validly from false information and accidentally end up with a conclusion that is true. Here is an example of that: P or not W W ∴P

True False True

In this example, one of the premises is false, but the conclusion happens to be true anyway. Mistaken assumptions can sometimes lead to a true conclusion by chance. The one combination that we cannot have is a valid argument which has all true premises and a false conclusion. This is in keeping with the definition given above: a deductively valid argument is one for which it is logically impossible for its conclusion to be false if its premises are all true. We have seen that there are valid arguments of each of these sorts: PREMISES CONCLUSION

all true true

not all true false

not all true true

What about invalid arguments? (That is, what about arguments that are not deductively valid?) What combination of truth-values can the parts of invalid arguments have? The answer is that they can have any combination of truth-values whatsoever. Here are some examples:

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Introduction P or W P ∴W

True True False

PREMISES ALL TRUE CONCLUSION FALSE

P not W ∴ not R

True True True

PREMISES ALL TRUE CONCLUSION TRUE

P or W W ∴P

True False True

PREMISES NOT ALL TRUE CONCLUSION TRUE

W or R P ∴R

False True False

PREMISES NOT ALL TRUE CONCLUSION FALSE

The moral of the story so far is that if you know that an argument is invalid, that fact alone tells you nothing at all about the actual truth-values possessed by its parts. And if you know that it is valid, all that that fact tells you about the actual truth-values of its parts is that it does not have all of its premises true plus its conclusion false. However, there is more to be said. Suppose that you want to show that an argument is invalid, but the argument does not already have all true premises and a false conclusion. How can you do this? One approach is to appeal directly to the characterization of validity, and describe a possible situation in which the premises are all true and the conclusion false. For example, suppose someone has given this (invalid) argument: Either Roosevelt or Truman (or perhaps both) was a president. Truman was a president. ∴ Roosevelt was a president. There is no mistake of fact involved here, but the argument is a bad one, and you would like to establish this. You could do so as follows. You say: "Suppose that Truman had been a president, but not Roosevelt. In that situation the premises would have been true, but the conclusion false." This is enough to show the reasoning bad, that is, to show the argument invalid. We can do even more than this, as we will see in the next section.

EXERCISES This book provides a stock of exercises as an aid to learning. They were written in the belief that the "hands on" approach to modern logical theory is the best way to master it. You will also be supplied with answers to many of the exercises. You should attempt every exercise on your own, and then check your efforts against the answers that are given. If you do not understand one or more of the exercises, ask for help! Several of the exercises contain material that supplements the explanations in the body of the text. None of the exercises presuppose material that is not provided in the text or in the exercise itself. 1. Decide whether each of the following arguments is valid or invalid. If the argument is invalid then describe a possible situation in which its premises are all true and its conclusion false. a. Either Polk or Lee was a president. Either Lee or Whitney was a president. ∴ Either Polk or Whitney was a president. b. Lee wasn't a president, and Polk was. Either Polk or Whitney was a president. ∴ Whitney was a president. Copyrighted material

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Introduction c. Polk was a president and so was Lee. Whitney was a president. ∴ Polk was a president and so was Whitney. d. Either Polk or Whitney was a president. Lee was not a president. ∴ Lee wasn't a president and Polk was. 2. Which of these are true, and which are false: a. b. c. d. e. f. g.

Some valid arguments have false conclusions. No invalid argument has all true premises and a false conclusion. If an argument is valid, and you produce a new argument from it by adding one or more premises to it, the resulting argument will still be valid. If an argument is invalid, and you produce a new argument from it by adding one or more premises to it, the resulting argument must still be invalid. If an argument has an impossible premise, it is valid. (An example of an impossible sentence is `Some giraffes aren't giraffes'.) If an argument has a necessarily true conclusion, it is valid. (An example of a necessarily true sentence is `Every giraffe is a giraffe'.) If an argument has a false premise, it is valid.

3. An argument which is valid and which also has all of its premises true is called sound. Based on this definition, which of the following are true, and which false: a. b. c. d. e. f.

All valid arguments are sound. All invalid arguments are unsound. All sound arguments have true conclusions. If an argument is sound, and you produce a new argument from it by adding one or more premises to it, the resulting argument will still be sound. All unsound arguments are invalid. If an argument has a necessarily true conclusion, it is sound.

4. Suppose that we have two arguments which are both valid: W ∴Q

Q ∴S

what do we know about this argument? W ∴S 5. Suppose that the following two simple arguments are both sound: W ∴Q

Q ∴S

what do we know about this argument? W ∴S 6. Suppose that the following argument is invalid: W ∴S what do we know about these arguments? W ∴R

R ∴S

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Introduction 7. (a) Give an example of a "reversing" argument, that is, one which is guaranteed to have a false conclusion if its premises are true, and is guaranteed to have a true conclusion if any of its premises are false. (b) Give an example of an argument that must have a true conclusion no matter what the truth-values of its premises. Is this argument valid?

3 LOGICAL FORM If you want to show that an argument is invalid, you can describe a possible situation in which the premises are all true and the conclusion false. We illustrated this above with the argument: Either Roosevelt or Truman (or perhaps both) was a president. Truman was a president. ∴ Roosevelt was a president. But this direct appeal to possible situations is sometimes difficult to articulate, and judgments of possibility can differ. Fortunately, there is another technique that is often more useful. You could challenge the above reasoning by saying: "That reasoning is no good. If that reasoning were good, we could prove that McGovern was a president! For we know that: Either McGovern or Truman was a president, and we know: Truman was a president. So, by your reasoning we should be able to conclude that McGovern was a president too!" This challenge, like the first one, also shows that the argument given above is invalid. But whereas the first type of challenge focuses on how the ORIGINAL argument works in some POSSIBLE situation, this second challenge is based on how some OTHER argument works in the ACTUAL situation. What we do in this second technique is to give an argument that is different than the first, but closely related to it. In the case in question, the new argument is: Either McGovern or Truman was a president. Truman was a president. ∴ McGovern was a president. We know the new argument is invalid because it actually has all true premises and a false conclusion (we chose it on purpose to be this way). Since the new argument is invalid, so is the original one. But why should the original argument be invalid just because this second argument is invalid? The answer is that, intuitively speaking, they both employ the same reasoning, and it is the reasoning that is being assessed when we make a judgment about validity. But how can we tell that they employ the same reasoning? The answer is that they both have the same form. Each argument is one in which one of the premises is an "or" statement, with the other premise being one of the parts of the "or" statement and the conclusion being the other part. This sameness of structure or form indicates a sameness of the reasoning involved. A key assumption on which all of modern logical theory is based is that goodness of deductive reasoning is a matter of form. Any argument which has just the same form as the argument we were just discussing is invalid, no matter whether its subject matter is religion, politics, mathematics, or baseball. Likewise, any argument which has this form: P or W not W ∴P

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Introduction is valid, regardless of its subject matter. With this in mind, we can give a modern account of validity due to form: •

An argument is formally valid if and only if every argument with exactly the same form is valid.

It follows from this definition that if an argument is formally valid, so is any argument with exactly that form, if an argument is not formally valid, neither is any argument with exactly that form. A central preoccupation of modern logic, then, is the investigation and classification of logical forms. (That is why this logic is called "formal logic".) This will be our business throughout the chapters that follow.

EXERCISES 1. Decide whether each of the following arguments is valid or invalid. If the argument is invalid then give an argument which has the same form, and which actually has all true premises and a false conclusion. a. Either Polk or Lee was a president. Either Whitney or Lee was a president. ∴ Either Polk or Whitney was a president. b. Lee wasn't a president, and Polk was. Either Polk or Whitney was a president. ∴ Whitney was a president. c. Polk was a president and so was Lee. Whitney was a president. ∴ Polk was a president and so was Whitney. d. Either Polk or Whitney was a president. Lee was not a president. ∴ Lee wasn't a president and Polk was. 2. Which of these are true, and which are false: a. b.

Some invalid arguments have the same forms as valid ones. You can show an argument valid by producing another argument which has the same form and which has true premises and a true conclusion.

c.

If you are wondering whether an argument is valid or not, and you fail to find another argument which has the same form and all true premises and a false conclusion, that shows the original argument to be valid.

3. Here are some argument forms. For each, say whether every argument with that form is valid. If it is not valid, give an example of an argument with the given form that has true premises and a false conclusion. a.

If A then B A ∴B

b.

If A then B B ∴A

c.

not (A and B) not-B ∴ not A

d.

A or B B ∴A Copyrighted material

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Introduction e.

A and not-A ∴B

f.

A ∴ B or not-B

g.

A or B not-B or C ∴ A or C

4. Recall that an argument which is valid and which also has all of its premises true is called sound. a.

If you are wondering whether an argument is sound, and you manage to find another one with the same form and having all true premises and a false conclusion, does that show the original argument to be unsound? Why?

b.

If you are wondering whether an argument is sound, and you manage to find another one with the same form and having all true premises and a true conclusion, does that show the original argument to be sound? Why?

c.

If you are wondering whether an argument is sound, and you manage to find another one with the same form and having all false premises and a false conclusion, does that show the original argument to be sound? To be unsound? Why?

5. For each of the examples in 3, say whether or not every argument with that form is sound, and also say whether some argument with that form is sound. 6. {This question is speculative, and does not necessarily have a straightforward answer} Could there be an argument that is valid but not formally valid? Could there be an argument that is formally valid but not valid?

4 SYMBOLIC NOTATION Our investigation of logical forms will take an indirect route, but one that has proved to be worthwhile. Instead of attempting a direct classification of the logical forms of sentences of English, we will develop an artificial language that is considerably simpler than English. It will in some ways be like English without some of the logically irrelevant aspects of English. And it will lack some of the characteristics that make the use of English confusing when used in argumentation. For example, the artificial language will lack some of the structural ambiguity of English. Consider this English sentence: Mary teaches little girls and boys. Does this tell us that Mary teaches little girls and little boys, or that she teaches little girls and regular-size boys? If this sentence occurred in an argument, the validity of the argument might turn on how the sentence was read. In the artificial language to be developed, structural ambiguities of this sort will be absent. The artificial language will be especially designed to make logical form perspicuous. You are already familiar with this from arithmetic. Consider the partly symbolic sentence: For any two numbers x and y, x+y = y+x. It is clear what this says. The same thing can also be said without any symbols: Given any two numbers, the result of adding them together in one order is the same as the result of adding them together in the reverse order. It is apparent that the use of symbols makes the claim clearly and vividly. Our logical symbolism will be like this.

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Introduction In fact, it will be possible in the symbolic language to tell the logical forms of its sentences just by examining the shapes and arrangements of its symbols. And it will be possible to evaluate formal validity and invalidity of arguments expressed in it by a variety of techniques that appeal directly to the visible arrangements of the symbols in the sentences used. This does not mean that we will lose sight of reasoning that is expressed in our native tongue. One of the tasks of learning the artificial language will be to learn how to take sentences of English and re-express them in the artificial language we are learning. One of our goals, after all, is to learn about reasoning that we encounter every day, in its natural habitat.

5 IDEALIZATIONS The data that we have to deal with are incredibly complex, and this is only an introductory text. So we will idealize from time to time. This is no different from any other art or science. In physics you usually begin by studying the behavior of bodies falling in a perfectly uniform gravitational field, or sliding down frictionless planes. There are no perfectly uniform gravitational fields, and no frictionless planes, yet studying these things gives us clear and simple models that can be applied to real phenomena as approximations. And then in the advanced courses you can learn how friction affects the sliding, and how non-uniform fields affect the movement of things in them. Here are some of the idealizations that we will make in this book: We will look only at arguments with indicative sentences, not with imperative or interrogative. We will ignore any problems due to vagueness. For example, given a perfect understanding of the situation, you may still be unsure whether to say that Mary loves John, because of the vagueness of distinguishing between loving and liking. We will also totally ignore the fact that sentences may change truth-value over time and with differing situations. If I say today: I'm feeling great! this may be true, but the very same sentence may be false tomorrow. And it may be true when I say it, yet false when someone else utters it. This "context dependence" of truth has aroused a great deal of interest, and there are many theories about how it works. They all presuppose that their readers have already learned the material in this book. We will pretend in our investigations that sentences come with unique truth-values that do not change with context. The effects of context constitute an advanced study. We will also assume that each sentence is either true or false. Again, the question of whether, and which, sentences lack truth-value is interesting, but is not to be pursued at the beginning. Many other idealizations will become apparent as we proceed.

6 THE PLAN OF THE TEXT In this text we will develop a symbolic notation in a step-by-step process. In chapter 1 we consider simple sentences that are combined together with the "connectives" 'or' and 'if . . .then . . .'. Even with this very austere notation we can formulate and study a number of formal validities. In chapter 2 we expand this notation by introducing further connectives: 'and', 'or', and 'if and only if'. In chapter 3 we vastly expand our symbolic language with the introduction of variables and quantifiers. Each expansion of the notation builds on what has gone before, so we are continually increasing our ability to validate formally valid arguments and invalidate arguments that are not formally valid. Chapters 4-6 contain additional expansions.

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Introduction - Answers to the Exercises

Answers to the Exercises -- Introduction SECTION 2 1. a.

INVALID

Any possible situation in which Lee was a president but neither of the others was.

b.

INVALID

Any possible situation in which Polk was a president but neither of the others was.

c.

VALID

d.

INVALID

2. a. b. c. d. e. f. g. 3. a. b. c. d. e. f.

Any possible situation in which Whitney was a president and neither of the others was.

True. (Such an argument will always have at least one false premise.) False. Some do; some don't. True. False. Sometimes adding a premise converts an invalid argument into a valid one, and sometimes it does not. It depends on what you add. True. There can't be a possible situation in which it has all true premises and a false conclusion because there can't be a possible situation in which it has all true premises. True. There can't be a possible situation in which it has all true premises and a false conclusion because there can't be a possible situation in which it has a false conclusion. False. It might be valid, or it might be invalid. False. Valid arguments with false premises aren't sound. True. An invalid argument isn't sound because it isn't even valid. True. The premises are all true, and it's valid, so its conclusion must be true too. False. If you add a true premise it will remain sound, but if you add a false premise it will become unsound. False. A valid argument is unsound if it has a false premise. False. If the conclusion is necessarily true the argument will be valid, but it still might have a false premise, and thus be unsound.

4. It has to be valid. For suppose it were not. Then there would be a possible situation in which A is true and C is false. Since the first argument is valid, B is true in this situation; but then since the second argument is valid, C is also true in that situation, contradicting our supposition that there is a situation in which A is true and C is false. 5. It has to be sound. It has to be valid for the same reason as in the previous example. And since the first argument is sound, A is true. So its premise is true. 6. We know that at least one of them is invalid, but we don't know which. If they were both valid, the first argument would have to be valid, as in exercise 4. So they aren't both valid. But there are cases in which the first is valid and the second invalid, and cases in which the first is invalid and the second valid, and cases in which they are both invalid. First valid and second invalid: A B C

Polk was a president Polk or Lee was a president Lee was a president

First invalid and second valid: A B C

Polk was a president Polk and Lee were presidents Lee was a president

Both invalid:

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Introduction - Answers to the Exercises A B C 7. a.

b.

Polk was a president Nixon was a president Lee was a president

Polk was a president. ∴ Polk wasn't a president.

[Naturally, the argument is invalid.]

Polk was a president. ∴ Either Whitney was a president or he wasn't. This argument is valid; it cannot have all true premises and a false conclusion because it cannot have a false conclusion.

SECTION 3 1. a.

Either McGovern or Nixon was president. Either Nixon or Goldwater was president ∴ Either McGovern or Goldwater was president.

b.

The original argument will do; it already has all true premises and a false conclusion.

c.

VALID.

d.

Either Whitney or Polk was a president. Lee was not a president. ∴ Lee wasn't a president and Whitney was.

2. a. b. c.

3. a.

False. False. This does not show that no argument with that form has true premises and a false conclusion. False. You might not have looked hard enough.

VALID

b.

INVALID

If Polk and Lee were both presidents, Polk was a president. Polk was a president. ∴ Polk and Lee were both presidents.

c.

INVALID

not (Polk was a president and Lee was a president) not Lee was a president ∴ not Polk was a president

d.

INVALID

Lee or Polk was a president. Polk was a president. ∴ Lee was a president.

e.

VALID

f.

VALID

g.

VALID

4. a. b.

(This depends interpreting `or' inclusively; this is discussed in chapter 2 below.)

Yes. It shows the original argument invalid, and an invalid argument is not sound. No. The original argument could still be invalid, or have a false premise, or both. Example: Original argument: Lee was a president ∴ Whitney was a president

"Found" argument: Nixon was a president. ∴ Kennedy was a president.

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Introduction - Answers to the Exercises c.

It shows neither. Examples: Original unsound argument: Lee wasn't a president ∴ Whitney wasn't a president Original sound argument: Lee wasn't a president ∴ Lee wasn't a president

5. a. b. c. d. e. f. g.

6.

"Found" argument: Nixon wasn't a president. ∴ Kennedy wasn't a president. "Found" argument: Nixon wasn't a president. ∴ Nixon wasn't a president.

Some arguments with this form are sound: the ones with true premises. But not all; some of them have false premises. None are sound, since none are valid. None are sound, since none are valid. But not all; some of them have false premises. None are sound, since none are valid. None are sound, since none has a true premise. Some arguments with this form are sound: the ones with true premises. But not all; some of them have false premises. Some arguments with this form are sound: the ones with true premises. But not all; some of them have false premises. Many logicians think that there are arguments that are valid, but not formally valid. An example is: Herman is a bachelor ∴ Herman is unmarried The validity of this argument comes from the meaning of the word 'bachelor', and not from the form of the sentences in the argument. As we have defined 'formally valid', any argument that is formally valid is automatically valid.

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CHAPTER 1 SECTION 1

Chapter One Sentential Logic with 'if' and 'not' 1 SYMBOLIC NOTATION In this chapter we begin the study of sentential logic. We start by formulating the basic part of the symbolic notation mentioned in the Introduction. For purposes of this chapter and the next, our symbolic sentences will consist entirely of simple sentences, called atomic sentences, together with molecular sentences made by combining simpler ones with connectives. The simple sentences are capital letters, which can be thought of as abbreviating sentences of English, as in the Introduction. In this chapter, the connectives are the negation sign, '~', and the conditional sign, '→'. The negation sign, '~', is used much as the word 'not' is used in English, to state the opposite of what a given sentence says. For example, if 'P' abbreviates the sentence 'Polk was a president', then '~P' abbreviates the sentence 'Polk was not a president'. The conditional sign, '→', is used much as 'if . . , then . . .' is used in English. If 'P' abbreviates the sentence 'Polk was a president' and 'W' abbreviates 'Whitney was a president' then '(P→W)' abbreviates the sentence 'If Polk was a president, then Whitney was a president'. We need to be precise about exactly what the symbolic sentences of Chapter 1 are: • • •

Any capital letter between 'P' and 'Z' is a symbolic sentence. (Numerical subscripts may also be used, as in 'P3', 'Q24'.) If '□' is a symbolic sentence, so is '~□' If '□' and '○' are symbolic sentences, so is '(□→○)'.

Nothing is a symbolic sentence of Chapter 1 unless it can be constructed by means of these provisions. Some terminology: • A symbolic sentence containing no connectives at all is an atomic sentence. In this chapter and the next, only sentence letters are atomic. • Any symbolic sentence that contains one or more connectives is called a molecular sentence. • We call '~□' the negation of '□'. • We call any symbolic sentence of the form '(□ → ○)' a conditional sentence; we call '□' the antecedent of the conditional, and '○' the consequent of the conditional. Examples of symbolic sentences with minimal complexity are: U ~U (U→V) The first is an atomic sentence. The second is the negation of that atomic sentence. The last is a conditional whose antecedent is the atomic sentence 'U' and whose consequent is the atomic sentence 'V'. Once a molecular sentence is constructed, it can itself be combined with others to make more complex molecular sentences: ~(U→V)

it is not the case that if U then V

(~V → (U→V))

if it is not the case that V then if U then V

~~(V → (U→V))

it is not the case that it is not the case that if V then if U then V

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CHAPTER 1 SECTION 1

The formation rules determine when parentheses occur in a symbolic sentence. When adding a negation sign to a sentence you do not add any parentheses. These are not symbolic sentences because they contain extra (prohibited) parentheses : ~(U), ~(~U), ~((U→V)) Although '~(U→V)' has a parenthesis immediately following the negation sign, that parenthesis got into the sentence when constructing '(U→V)', and not because of the later addition of the negation sign. When combining sentences with the conditional sign, parentheses are required. For example, this is not a sentence: U→V→W There is one exception to the need for parentheses. If a sentence appears all by itself, not as part of a larger sentence, then its outer parentheses may be omitted. So these sentences are taken informally to be conditional symbolic sentences: U→V ~U → V (U→V) → ~U For purposes of Chapter 1: A sentence is in official notation if it can be constructed by using the processes given in the box above. It is in informal notation if it can be put into official notation by enclosing it in a single pair of parentheses. Anything that is not in either official notation or informal notation is not a sentence at all. Any well-formed sentence can be "parsed" into its constituents. You begin with the sentence itself, and you indicate below it how it is constructed out of its constituents. First you locate the main connective, which is the last connective introduced when constructing the sentence. If the sentence is a negation, the main connective is the negation sign; you draw a vertical line under it and write the part of the sentence to which the negation sign is applied. If it is a conditional, the main connective is the conditional sign; you draw branching lines below the main conditional sign and write the antecedent and consequent: ~P | P

P→Q 2 P Q

If the parts are themselves complex, the parsing may be continued: ~~(P → Q) | ~(P→ Q) | P→Q 2 P Q

P → ~(Q→R) 2 P ~(Q→R) | Q→R 2 Q R

(P→Q) → (~R→S) 2 P→Q ~R→S 2 2 P Q ~R S | R

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CHAPTER 1 SECTION 1

EXERCISES 1. For each of the following state whether it is a sentence in official notation, or a sentence in informal notation, or not a sentence at all. If it is a sentence, parse it as indicated above. a. b. c. d. e. f. g. h. i.

~~~P ~Q→~R ~(Q~→R) ~(~P)→~R (P→Q) → (R→~Q) P → (Q→R) → Q (P → (Q→R) → Q) (~S→R) → ((~R→S) → ~(~S→R)) P → (Q→P)

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CHAPTER 1 SECTION 2

2 MEANINGS OF THE SYMBOLIC NOTATION The negation sign: The logical import of the negation sign is this: it makes a sentence that is false if the sentence to which it is prefixed is true, and true if the sentence to which it is prefixed is false. It is common to summarize this behavior of the negation sign by means of what is called a truth table: □ T F

~□ F T

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CHAPTER 1 SECTION 2

this text is that 'if . . , then. . .' is sometimes used to express what is called the "material conditional". This is the use of 'if . . , then . . .' where a conditional sentence is false in case the antecedent is true and the consequent false, and it is true in every other case. This use is truth functional. It is described by means of this truth table: □ ○ □→○ T T T T F F F T T F F T The conditional is used in this way by mathematicians, and by others. We will assume in doing exercises and examples that the logical import of 'if . . , then' is intended to coincide with our symbolic '→'. There may be other uses of 'if . . , then' that convey more than '→', but we will not address them in this text. The word 'if' in English has many synonyms. In at least some contexts these are all interchangeable: if provided that assuming that given that in case on the condition that

If Maria sings, Xavier will leave Provided that Maria sings, Xavier will leave Assuming that Maria sings, Xavier will leave Given that Maria sings, Xavier will leave In case Maria sings, Xavier will leave On the condition that Maria sings, Xavier will leave

Using 'S' for 'Maria sings' and 'X' for 'Xavier will leave', these can all be symbolized as S→X 'If' clauses in English may also occur at the end of a sentence instead of at the beginning. So these also may be symbolized as 'S→X': Xavier will leave if Maria sings Xavier will leave provided that Maria sings Xavier will leave assuming that Maria sings Xavier will leave given that Maria sings Xavier will leave in case Maria sings Xavier will leave on the condition that Maria sings In either use, the word 'If' immediately precedes the antecedent of the conditional. Many of these "synonyms" of 'if' can be used to say more than what is said with a simple use of the word 'if'. For example, a person who says 'assuming that' may want to convey that s/he is indeed making a certain assumption, and not just saying 'if'. But in other contexts no assuming is indicated. A physicist who says 'Assuming that there are planets with orbits outside the orbit of Pluto, we will need to send space probes to investigate them' may simply be responding to the question 'What if there are planets beyond Pluto?', and not doing any assuming at all. In doing the exercises we will take for granted that the locutions identified above are being used in the most minimal sense of 'if', which we take to be that of the connective '→'. Only if: The word 'only' can be added to the word 'if', to make 'only if'. The 'only' has the effect of reversing antecedent and consequent. As a result, whereas 'if', when used alone, immediately precedes the antecedent of a conditional, 'only if' immediately precedes the consequent. So we have these equivalences: If P, Q Only if P, Q

P→Q Q→P

P if Q P only if Q

Q→P P→Q

Some will find it more natural to represent 'P only if Q' by 'If not Q then not P', or '~Q → ~P'. It will turn out that this is logically equivalent to 'P → Q'. We will generally use the latter form because it's simpler.

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When ‘only if' comes first, there are grammatical changes in the last clause, which converts into its interrogative word order: The game will be called off only if it rains = Only if it rains will the game be called off 'Only' may also precede any of the synonyms of 'if', so these all may be symbolized as 'X → S': Xavier will leave only if Maria sings Xavier will leave only provided that Maria sings Xavier will leave only assuming that Maria sings Xavier will leave only given that Maria sings Xavier will leave only in case Maria sings Xavier will leave only on the condition that Maria sings Only if Maria sings will Xavier leave Only provided that Maria sings will Xavier leave Only assuming that Maria sings will Xavier leave Only given that Maria sings will Xavier leave Only in case Maria sings will Xavier leave Only on the condition that Maria sings Xavier will leave

EXERCISES For these exercises assume that 'S' abbreviates 'Susan will be late' and 'R' abbreviates 'It will rain'. 1. For each of the following sentences say which symbolic sentence is equivalent to it. a. Only if it rains will Susan be late S→R R→S b. Susan will be late provided that it rains S→R R→S c. Susan won't be late ~S ~~S d. Susan will be late only if it rains S→R R→S e. Given that it rains, Susan will be late S→R R→S 2. Symbolize each of the following: a. b. c. d. e.

Susan will be late only provided that it rains Only on condition that it rains will Susan be late Susan will be late only in case it rains Susan will be late only if it rains It is not the case that Susan will be late

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3 SYMBOLIZATION: TRANSLATING COMPLEX SENTENCES INTO SYMBOLIC NOTATION In representing simple sentences of English by sentential letters you need to say which letter abbreviates which sentence. So far this has been done informally, by choosing sentential letters that are already used in a prominent place in the English sentence, as in using 'S' for 'Sally will be late'. But if different English sentences share their prominent letters, we can soon run out of natural sentential letters to choose. So instead we give a scheme of abbreviation, which pairs off sentence letters with the English sentences that they abbreviate. An example is: X Y

Susan will be late It will rain

With this scheme we would represent 'Given that it rains, Susan will be late' by: Y→X Any way of "symbolizing" English sentences in logical notation, or of "translating" English into logical notation, is done relative to a scheme of abbreviation. A translation or symbolization that is correct on one scheme may be incorrect on others. Complex sentences of English generally translate into complex sentences of the logical notation. Here it is important to be clear about the grouping of clauses in the English sentence. Consider, the sentence: If Roberta doesn't call, Susan will be distraught This is a conditional whose antecedent is a negation. Using 'P' for 'Roberta calls' and 'Q' for 'Susan will be distraught', this may be symbolized: ~P → Q It is not the negation of a conditional: ~(P→Q) To make the negation of a conditional, you need to say something like: It is not the case that if Roberta calls, Susan will be distraught which is symbolized as: ~(P → Q) There are a few fundamental principles that govern symbolizations of English sentences in the logical notation. SOURCES OF '~' The locution 'fail to' always yields a negation sign that applies to the symbolization of the smallest sentence that 'fail to' is part of. Likewise for the word 'not'. The expression 'it is not the case that' applies to a sentence immediately following it. Notice that in the sentence 'It is not the case that Willa will leave if Sam does' there are two grammatical sentences that immediately follow 'It is not the case that', namely, 'Willa will leave' and 'Willa will leave if Sam does'. So there are two ways to symbolize the sentence: S → ~W ~(S→W) The sentence is in fact ambiguous.

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SOURCES OF '→' If: The word 'if' always gives rise to a conditional, '□→○'. Wherever 'if' occurs (not as part of 'only if'), the antecedent of the conditional is the symbolization of a sentence immediately following 'if. The consequent of the conditional is either the symbolization of a sentence immediately preceding 'if' (with no comma in between) -- as in '○ if □' -- or it is the symbolization of a sentence immediately following the sentence that is symbolized as the antecedent -- as in 'if □ then ○'. Then: If 'then' occurs it must be paired with a preceding 'if'. The antecedent of the conditional introduced by 'if' is the symbolization of the sentence exactly between 'if' and 'then'. Its consequent is the symbolization of a sentence immediately following 'then'. Only if: The expression 'only if' always gives rise to a conditional, '□→○'. The consequent of '□→○' is the symbolization of a sentence immediately following 'only if' -as in '□ only if ○' -- or as in 'only if ○, □'. The antecedent of '□→○' is the symbolization of a sentence immediately preceding 'only if' (with no comma in between) -- '□ only if ○' -- or of a sentence immediately following the consequent -- as in 'only if ○, □'. In the latter case, that sentence is grammatically changed (to its interrogative word order). Illustration: These principles determine that the sentence: 'Pat won't call only if it is not the case that the quilt is dirty' is symbolized: ~P → ~Q The (contracted) 'not' in 'Pat won't call' yields a negation that applies directly to 'P'. The 'it is not the case that' yields a negation that applies directly to 'Q', since 'the quilt is dirty' is the only sentence immediately following 'it is not the case that'. The only sentence immediately to the left of the 'only if' is 'Pat won't call', so that is the antecedent of the conditional, and the only sentence immediately to the right of the 'only if' is 'it is not the case that the quilt is dirty', so that is the consequent. Illustration: In the sentence 'If Wilma leaves then Xavier stays if Yolanda sings' the first 'if . , then . . .' exactly encloses 'Wilma leaves', so W is the antecedent of the first conditional. There are two sentences immediately following 'then'; they are the whole 'Xavier stays if Yolanda sings' and just 'Xavier stays'. So the sentence must have the form: W → (Xavier stays if Yolanda sings) or (W → X) if Yolanda sings The second 'if' comes between its consequent and antecedent. It must give rise to a conditional that has 'Y' as its antecedent, since the only sentence following the 'if' is 'Yolanda sings'. The consequent of that conditional can be the symbolization of either just 'Xavier stays', or 'If Wilma leaves then Xavier stays', since each of these immediately precedes the 'if'. The first of these fits with the first partial symbolization above, giving: W → (Y→X) and the second fits with: Y → (W→X). Both of these symbolizations are possible, which agrees with the intuition that the original English sentence is ambiguous. (Some people find the first reading more natural than the second, but the second is a possible reading under some circumstances.)

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Illustration: In the sentence 'If Wilma leaves then Xavier stays only if Yolanda sings' the first 'if' is exactly as in the previous case, so the sentence has the form: W → (Xavier stays only if Yolanda sings) or (W → X) only if Yolanda sings The 'only if' comes between its antecedent and consequent. It must give rise to a conditional that has 'Y' as its consequent, since the only sentence following the 'only if' is 'Yolanda sings'. The antecedent of that conditional can be the symbolization of either just 'Xavier stays', or 'If Wilma leaves then Xavier stays', since each of these immediately precedes the 'only if'. The first of these fits with the first partial symbolization above, giving: W → (X→Y) and the second fits with: (W→X) → Y Again, the sentence is ambiguous. The principles above also apply when 'if' is replaced by one of its synonyms, such as 'given that'.

So 'If Wilma leaves then Xavier stays provided that Yolanda sings' has the same symbolization options as 'If Wilma leaves then Xavier stays if Yolanda sings': W → (Y→X) and Y → (W→X). Commas: We have seen that the fundamental principles governing words that yield negations and conditionals can permit a significant amount of ambiguity. A common way to eliminate such ambiguity from sentences is to use commas to indicate how parts of the symbolization are to be grouped. Commas are used for a wide variety of purposes, so the presence of a comma may be irrelevant to the symbolization. But sometimes they are used to indicate that sentences should be grouped together, that is, combined into a single sentence. When this happens, the comma appears right after the sentence that results from the grouping. Or, a comma may be used to indicate that sentences to the right should be grouped together. COMMAS A comma indicates that the symbolizations of sentences to its left should be combined into a single sentence, or that sentences to its right should be combined into a single sentence. For example, above we saw that the sentence: If Wilma leaves then Xavier stays if Yolanda sings has two possible symbolizations: W → (Y→X) and Y → (W→X). If a comma appears before the first 'then': If Wilma leaves, then Xavier stays if Yolanda sings this forces the first symbolization, since it requires that 'X' and 'Y' be grouped together. But if the comma appears later:

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If Wilma leaves then Xavier stays, if Yolanda sings this forces the second symbolization, since now 'W' and 'X' must be grouped together. Sometimes there are no phrases to be kept together, as in: If Wilma leaves, then Xavier stays There is only one sentence to the left of the comma, and only one to the right, so the comma is redundant; we just have: W→X

EXERCISES For these questions please use the following scheme of abbreviation: V W Y

Veronica will leave William will leave Yolanda will leave

1. For each of the following say which of the proposed translations are correct. a. If Veronica doesn’t leave William won’t either ~(V→W) ~V→~W V → ~~W b. William will leave if Yolanda does, provided that Veronica doesn’t (W→Y) → ~V V → (W→Y) ~V → (Y→W) c. If Yolanda doesn’t leave, then Veronica will leave only if William doesn’t ~Y → (~W → V) ~Y → (V → ~W) ~W → (~Y→V) d. If Yolanda doesn’t leave then Veronica will leave, given that William doesn’t ~Y → (~W → V) ~Y → (V → ~W) ~W → (~Y→V) 2. For each of the following produce a correct symbolization a. William will leave if Veronica does b. Veronica won't leave if William does c. If Veronica leaves, then if William doesn't leave, Yolanda will leave d. If Veronica doesn't leave if William doesn't, then Yolanda won't. e. William won’t leave provided that Veronica doesn’t leave f. If William leaves, then if Veronica leaves so will Yolanda g. William will leave only if if Veronica leaves then so will Yolanda h. William will leave only if Veronica leaves, only provided that Yolanda will leave

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CHAPTER 1 SECTION 4

4 RULES A "rule" is a particular valid form of argument which may be used in extended reasoning. Certain rules involving negations and conditionals have been recognized for centuries, and they have traditional names. The basic rules used in this chapter are: RULES Repetition:

□ ∴ □

Modus Ponens:

□→○ □ ∴○

Modus Tollens:

□→○ ~○ ∴ ~□

Double negation:

□ ∴ ~~□

or

~~□ ∴□

Repetition is the most trivial rule; it indicates that if you have any sentence you may validly infer it from itself. Although trivial, this rule will have an important use later when we construct a method of showing that an argument is valid. Modus ponens indicates that if you have any conditional sentence along with its antecedent you may infer its consequent. For example, this valid argument is an instance of modus ponens: If Polk was a president, so was Whitney Polk was a president ∴ Whitney was a president

P→W P ∴ W

This rule may be justified by noting that a conditional with a true antecedent and false consequent is false. Modus Tollens indicates that if you have any conditional sentence along with the negation of its consequent, you may infer the negation of its antecedent. For example, this valid argument is an instance of modus tollens: If Polk was a president, so was Whitney Whitney wasn't a president ∴ Polk wasn't a president

P→W ~W ∴ ~P

This rule may be justified in a similar way to that used in justifying modus ponens. Double negation indicates that from any sentence you may infer the result of putting two negation signs on the front, or vice versa. For example, both of these valid arguments are instances of double negation: Polk was a president

P ∴ ~~P

It is not the case that Polk wasn't a president

~~P ∴ P

∴ It is not the case that Polk wasn't a president ∴ Polk was a president

It should be obvious upon reflection that any argument whose conclusion follows from its premises by a single application of one of these rules is formally valid. That is, there cannot be a situation in which it has true premises and a false conclusion. These rules apply to anything that fits their pattern, even if it is complex. For example, this is an instance of modus ponens:

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If Polk was a president, Whitney wasn't Polk was a president ∴ Whitney wasn't a president

P→~W P ~W

And so is this: If Polk was a president then if Roosevelt was a president, so was Truman Polk was a president ∴ If Roosevelt was a president then so was Truman

P→(R→T) P R→T

The following is not an instance of modus ponens: If Whitney was a president, so was Truman Truman was a president ∴ Whitney was a president

W→T T ∴ W

This argument is in fact invalid. It is an instance of a famous fallacy called "affirming the consequent". Likewise, the following is not an instance of modus tollens: If Whitney was a president, so was Truman Whitney wasn't a president ∴ Truman wasn't a president

W→T ~W ∴ ~T

This too is a famous fallacy, called "denying the antecedent".

EXERCISES 1. For each of the following arguments, say whether it is an instance of modus ponens, or modus tollens, or double negation, or none of the above. a.

P→~Q Q ∴ ~P

b.

~P → Q ∴ ~(P → Q)

c.

~~(P→Q) ∴ P→Q

d.

~P → ~Q ~P ∴ ~Q

e.

~P → ~Q ~~Q ∴ ~~P

f.

P→Q ~R ∴ ~P

g.

P → (R→Q) P ∴ R→Q

h.

P → (R→Q) R → ~Q ∴ ~P

i.

~~(P → Q) ~Q ∴ P→Q

2. Given the sentences below, say what, if anything, can be inferred in one step by modus ponens, or modus tollens, or double negation. a.

~W → ~X ~W ∴ ?

b.

~W → ~X ~~X ∴ ?

c.

W→X ~W ∴ ?

d.

W → (R→X) W ∴ ?

e.

W → (R→X) ~R → X ∴ ?

f.

~~(W → X) ~X ∴ ?

g.

W→~X X ∴ ?

h.

~W → X ∴ ?

i.

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~~(W→X) ∴ ?

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CHAPTER 1 SECTION 5

5 DIRECT DERIVATIONS Complex reasoning often consists of stringing together simple inferences so as to show the validity of a complex argument. Here is an example. We are given the following premises and conclusion: If Polk was a president then if Whitney was a president so was Trump. Polk wasn't a president only if Trump was a president. Trump wasn't a president. ∴ Whitney wasn't a president. We may reason as follows: Since Trump wasn't a president (given) and Polk wasn't a president only if Trump was (given), it is not the case that Polk wasn't a president. So Polk was a president. But if Polk was a president, then if Whitney was a president, so was Trump (given); so if Whitney was a president so was Trump. But Trump wasn't a president (given), so Whitney wasn't a president. The pattern of reasoning is easier to follow if the sentences are given in symbolic form: First premise: Second premise: Third premise: Conclusion:

P→(W→T) ~P→T ~T ∴ ~W

The above reasoning can be systematized and explained using the following format, where each line contains a sentence followed by a “justification” – an explanation of why the sentence is there. 1. To show ~W: "Since Trump wasn't a president (given) and Polk wasn't a president only if Trump was (given), it is not the case that Polk wasn't a president." 2. 3. 4.

~T ~P→T ~~P

pr pr 2 3 mt

this line is a premise this line is a premise this line follows from 2 and 3 by modus tollens

"So Polk was a president" 5.

P

4 dn

this line follows from line 4 by double negation

"But if Polk was a president, then if Whitney was a president, so was Trump (given); so if Whitney was a president so was Trump." 6. 7.

P→(W→T) W→T

pr 5 6 mp

this line is a premise this line follows from 5 and 6 by modus ponens

"But Trump wasn't a president (given), so Whitney wasn't a president." 8. 9.

~T ~W

pr 7 8 mt

this line is a premise this line follows from 7 and 8 by modus tollens

Lines 2-4 indicate that the first part of the reasoning appeals to two of the premises of the argument, and it draws a conclusion from them by modus tollens. Line 5 indicates that the reasoning goes from line 4 to 5 by double negation. Lines 6 and 7 indicate that line 5 together with the first premise lead to the sentence on line 7 by modus ponens. Finally, lines 8 and 9 indicate that line 7 together with the third premise lead to the conclusion of the argument. This layout of premises and inferences constitute most of the ingredients of what we will call a derivation. Every line consists of a line number followed by a sentence followed by a justification. The sentence on each line either (i) occurs as a premise, and the line is justified by writing "pr", or (ii) follows from previous lines by a rule, and the line is justified by writing the number(s) of the line(s) from which it follows, along with a short name of the rule. The short names of the rules that we have so far are “r”, "mp", "mt", and "dn".

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The particular approach taken here is to see a derivation as carrying out a task. Each task is to show that a sentence follows from certain things. Our derivations begin with a special line stating the task; that is, stating what is to be shown. In the sequence of steps above it would come first, and would be of this form: 1.

Show ~W

A "show" line may be introduced at any time, and it does not need a justification, because it only states what it is we intend to derive. All other lines need justifications. Suppose a derivation is to be constructed, guided by the reasoning given above. Following the 'show' line we repeat two of the premises, justifying them with the notation 'pr': 1. 2. 3.

Show ~W ~T ~P→T

pr pr

From these two lines we infer the third by modus ponens: 1. 2. 3. 4.

Show ~W ~T ~P→T ~~P

pr pr 2 3 mt

Next, we write 'P', indicating that we are deriving it from line 4: 1. 2. 3. 4. 5.

Show ~W ~T ~P→T ~~P P

pr pr 2 3 mt 4 dn

Still following out the reasoning on the previous page, we repeat another premise, and then infer 'W → T' from it together with line 5 using modus ponens: 1. 2. 3. 4. 5. 6. 7.

Show ~W ~T ~P→T ~~P P P→(W→T) W→T

pr pr 2 3 mt 4 dn pr 5 6 mp

Again, we insert the second premise, and infer '~W' from it and line 7 by modus tollens: 1. 2. 3. 4. 5. 6. 7 8. 9.

Show ~W ~T ~P→T ~~P P P→(W→T) W→T ~T ~W

pr pr 2 3 mt 4 dn pr 5 6 mp pr 7 8 mt

The reasoning ends here, where we have completed the task of showing '~W'. At this point we need a way to indicate that the task of showing '~W' has been completed. The completion of the task is indicated by writing "dd" after that line (meaning "direct derivation"); then the "Show" on the show line is cancelled by drawing a line through it -- because the task has been completed -- and the steps used in that process are boxed off. Specifically, since the sentence to be shown occurs on line 9, you may write "dd" to its right, box all lines below the show line, and cancel the "Show":

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1.

Show ~W

2. 3. 4. 5. 6. 7. 8. 9.

~T ~P→T ~~P P P→(W→T) W→T ~T ~W

Cancel the "Show" pr pr 2 3 mt 4 dn pr 5 6 mp pr 7 8 mt dd

Box the lines

Write "dd"

The cancellation of the show line indicates that the task was successfully completed, and the boxing encloses the lines used in completing that task. It is also permissible to wait and write the "dd" on a later line. Such a line contains no sentence itself; its justification consists of the number of the line where the target sentence occurs, followed by "dd". Here is the same derivation with the dd justification on a later line: 1.

Show ~W

2. 3. 4. 5. 6. 7. 8. 9. 10.

~T ~P→T ~~P P P→(W→T) W→T ~T ~W

pr pr 2 3 mt 4 dn pr 5 6 mp pr 7 8 mt 9 dd

Empty line with "dd"

It is often a matter of taste which technique to use for indicating the completion of a direct derivation. Here is another illustration of a direct derivation, used to show this argument valid: Q→~S V→X ~V→S ~X ∴ ~Q The derivation begins with a line indicating that the task is to show the conclusion of the argument: 1.

Show ~Q

The next few lines give the reasoning steps: 2. 3. 4. 5. 6. 7. 8. 9.

V→X ~X ~V ~V→S S ~~S Q→~S ~Q

pr pr 2 3 mt pr 4 5 mp 6 dn pr 7 8 mt

On line 9 we have completed the task. So we write "dd" and box and cancel:

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1.

Show ~Q

2. 3. 4. 5. 6. 7. 8. 9.

V→X ~X ~V ~V→S S ~~S Q→~S ~Q

pr pr 2 3 mt pr 4 5 mp 6 dn pr 7 8 mt dd

(Line 7 is necessary before using modus tollens with line 8. This form of inference is indeed valid: □→~○ ○ ∴ ~□ however, it is not itself an instance of modus tollens. It is instead an inference that is easily justified using double negation along with modus tollens.) We indent all of the lines immediately following a show line; this is a device for keeping track (by indentation) of where the task that is initiated by the "show" is being carried out. The indentation also reserves a space for the box that will be drawn if the derivation is successful. In getting precise about how to construct a direct derivation, it will help to specify what previously occurring things can be appealed to when applying a rule. We will say that a previous line is available from a given line just in case it is an earlier line that is not an uncancelled show line and is not already in a box: In a derivation from a set P of premises, a line is available from a given line just in case it is a member of P or it is an earlier line that is not an uncancelled show line and is not already in a box. Whether a line is available or not depends on your perspective. A show line is not available from the line immediately below it -- because it is not yet cancelled. But once it is cancelled, it is available from all lines below the line from which it was cancelled. And a line may be available from a given line, but once it is boxed, it is not available from any line outside the box. A direct derivation from a set of sentences P consists of a sequence of lines (including justifications when appropriate) that is built up, step by step, where each step is in accordance with these provisions: • • • •

A show line consists of the word "Show" followed by a sentence. The first step of producing a derivation must be to introduce a show line. A show line also may be introduced at any later step. Show lines are not given a justification. At any step, any sentence from the set of sentences, P, may be introduced, justified with the notation "pr". At any step a line may be introduced if it follows by a rule from previous available lines in the derivation; it is justified by citing the numbers of those previous lines and the name of the rule. If a line is introduced whose sentence is the same as the sentence in the closest previous uncancelled show line, one may, as the next step, write "dd" at the end of that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. (Alternatively) At any step, if any previous available line contains a sentence that is the same as that in the closest previous uncancelled show line, one may introduce a line with no sentence on it, justifying it by citing the number of the earlier line followed by "dd"; one then draws a line through the word "Show", and draws a box around all the lines below that show line, including the current line.

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These instructions show how to construct a derivation in a step by step fashion. Any number of steps results in a derivation, but not necessarily one that completes any of the tasks set out by its show lines. For example, when constructing a derivation above, at a certain stage we had reached this far in the process: 1. 2. 3. 4.

Show ~W ~T ~P→T ~~P

pr pr 2 3 mt

This sequence of lines satisfies the conditions for being a derivation as defined above. But there is a sense in which it is not yet finished. For this purpose we define a "complete" derivation: A derivation is complete if every show line is cancelled and every line that is not a show line is boxed. The point of doing a derivation is often to show that a certain argument is formally valid. When a derivation shows that an argument is formally valid, we say that it "validates" the argument: A derivation validates an argument if and only if it is a complete derivation from the premises of that argument, and the conclusion of the argument appears on an unboxed cancelled show line in the derivation.

EXERCISES 1. Check through each line of the following direct derivations to determine whether it can be constructed by means of the provisions for direct derivations given above, where the set P is taken to be the premises of the displayed arguments. (When assessing a given line, assume that all previous lines are correct.) Argument:

1. 2. 3. 4. 5. 6. 7. 8. 9.

P → (Q→~R) ~P → ~Q Q ∴ ~R

Show ~R Q ~~Q ~P → ~Q ~~P P P → (Q→~R) Q→~R ~R

pr 2 dn pr 3 4 mt 6 dn pr 6 7 mp 2 8 mp dd

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Argument:

1. 2. 3. 4. 5. 6. 7. 8. 9.

Show ~R Q ~P→~Q P R→~Q ~~Q ~~R R ~Q

Argument:

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

P → (R→~Q) ~P → ~Q Q ∴ ~R

pr 2 pr 2 3 mt 4 mp 2 dn 5 6 mt 7 dn 5 8 mp dd

~O S → (W→~O) O→S W ∴ ~S

Show ~S W pr ~S pr O→S pr ~O 3 4 mt W pr ~(W→~O) 5 6 mt S → (W→~O) pr ~S 7 8 mt 9 dd

2. Construct direct derivations to validate each of the following arguments: P Q → ~P R→Q ∴ ~R W → ~(V→~Y) X → (V→~Y) V→Y (V→Y) → X ∴ ~W (W→Z) → (Z→W) (Z→W) → ~X P→X ~~P ∴ ~(W→Z)

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6 CONDITIONAL DERIVATIONS In this section we learn about one of the most powerful and useful procedures for constructing proofs in a natural way. The procedure is called conditional derivation. It is meant to reflect a natural reasoning process that involves hypothetical inference. Suppose that you wish to show that the following argument is valid: If Robert drives, Sam won't drive. If Sam doesn't drive, Teresa won't go. Willa will go only if Teresa does. ∴ If Robert drives, Willa won't go. If we try to reason as above using our rules mp, mt, and dn, we will not succeed; none of them apply to the premises we are given. What you would probably do on your own is to reason somewhat as follows: ASSUME that Robert drives Well, if he drives, Sam won't (given); so Sam won't drive But if Sam doesn't drive, Teresa won't go (given), so Teresa won't go. But Willa will go only if Teresa does (given), so Willa won't go. So, SUMMING UP, if Robert drives, Willa won't go. The middle three steps look familiar; they are inferences from premises and previously stated sentences, and they are all justifiable by rules that we have. But the first and last steps are new. What does it mean to "assume", as we have done in the first step, and what is this "summing up" in the last step? What role do these have as legitimate parts of a piece of reasoning? Here is what goes on in "conditional" reasoning. Our goal is to show that a certain conditional sentence follows from certain premises. (In the example above, the conditional sentence is 'If Robert drives, Willa won't go'.) We then "assume" the antecedent of the conditional. If we can use this to derive the consequent of the conditional, we conclude that this reasoning has shown the conditional itself to follow from the given premises. An example: R→~S ~S→~T W→T ∴ R→~W A derivation using the conditional derivation technique begins with a line specifying the task, which is to show the conclusion. This is followed by an assumption of the antecedent of the conditional to be shown: 1. 2.

Show R→~W R

ass cd

assumption for conditional derivation (the goal is now to derive the consequent: ~W)

The reasoning in the center of the derivation proceeds normally: 3. 4. 5. 6. 7. 8.

R→~S ~S ~S→~T ~T W→T ~W

pr 2 3 mp pr 4 5 mp pr 6 7 mt

Now that we have derived the consequent of the conditional on line 8, We cite "cd", and we box and cancel:

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1.

Show R→~W

2. 3. 4. 5. 6. 7. 8.

R R→~S ~S ~S→~T ~T W→T ~W

ass cd pr 2 3 mp pr 4 5 mp pr 6 7 mt cd

Cancel the "Show"

Box the lines

Write "cd"

Lines 2-8 show that given the premises, we may derive '~W' from 'R'. Our conditional derivation technique says that this amounts to showing that those premises validate 'R→~W', so we may box and cancel. Here is another example used to show that the following very short argument is valid: ∴

S (S→R)→R

1.

Show (S→R)→R

2. 3. 4.

S→R S R

ass cd pr 2 3 mp

We have assumed the antecedent of the conditional on the show line, and we have now succeeded in deriving the consequent of that conditional. So we may use the technique of conditional derivation: 1.

Show (S→R)→R

2. 3. 4.

S→R S R

ass cd pr 2 3 mp cd

EXERCISES 1. For each of the following derivations, determine which lines are correct and which incorrect. (In assessing a line, assume that previous lines are correct.) a.

P → (~Q→R) ~R ∴ P→Q 1. Show P → Q 2. 3. 4. 5. 6. 7.

b.

P P → (~Q→R) ~Q → R ~R ~~Q Q

ass cd pr 2 3 mp pr 4 5 mt 6 dn cd

P → (Q→~R) R ∴ P → ~Q 1. Show P → ~Q 2. 3. 4. 5. 6. 7.

P ~Q P → (Q→~R) Q → ~R R ~Q

ass cd 1 2 mp pr 2 4 mp pr 5 6 mt cd

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c.

~S → (Q→R) R → ~(Q→R) ~P → R ∴ P → ~S 1. Show P → ~S 2. 3. 4. 5. 6. 7. 8. 9.

~S Q→R ~~(Q→R) R → ~(Q→R) ~R ~P → R ~~P P

ass cd 2 mp 3 dn pr 4 5 mt pr 6 7 mt 8 dn cd

2. Construct correct derivations for each of the following arguments using conditional derivations. a.

P → (Q → (R→S)) ~Q → ~R R ∴ P→S

b.

Q → ~(R→S) P → (R→S) ~Q → R ∴ P→S

c.

U → (U→V) ~R → ~(U→V) R → ~S ∴ U → ~S

3. Symbolize the following arguments using the sentence letters given, and then give derivations to validate them. a.

If Seymour likes papayas he'll have them for tea. He won't have them for tea if we don't have any. If we didn't shop yesterday, we don't have any papayas. So if Seymour likes papayas, we shopped yesterday. (P: Seymour likes papayas; T: Seymour will have papayas for tea; X: We have some papayas; S: We shopped yesterday.)

b.

If today is Thursday, then Saturday is two days from now. If the party is on Saturday, then if Saturday is two days from now, then so is the party. I can't go to the party if it's two days from now. The party is on Saturday. So if today is Thursday, I can't go to the party. (T: Today is Thursday; S: Saturday is two days from now; P: The party is two days from now; Y: The party is on Saturday; X: I can go to the party.)

c.

If Samantha is at home, she won't cause any trouble. If she isn't at home, she can't be reached by telephone. If she can't be reached by telephone, it's too late to tell her about the party. If she comes to the party she will cause some trouble. So if it's not too late to tell her about the party, she won't come. (S: Samantha is at home' T: Samantha will cause trouble; R: Samantha can be reached by telephone; X: It's too late to tell Samantha about the party; Y: Samantha will come to the party.)

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7 INDIRECT DERIVATIONS There is a third technique for doing derivations, called indirect derivation. It is often used in cases where direct derivation and conditional derivation do not obviously apply. An example is an attempt to show that this inference is valid: Polk was a president Whitney wasn't a president ∴ It is not the case that if Polk was a president, so was Whitney The conclusion is not a conditional, so a straightforward application of conditional derivation does not seem possible. Nor is it clear how to derive the conclusion using mp, mt, or dn. To validate this argument you might reason as follows: We want to show that it is not the case that if Polk was a president, so was Whitney. Well, assume the opposite: assume that it is the case that if Polk was a president then so was Whitney. Then, since we are given that Polk was a president, so was Whitney. But we are given that Whitney wasn't. So we are led to absurd conclusions: Whitney was a president and Whitney was not a president. So the assumption we made, which lead to these inferences, must not be true (given the premises of the argument). The reasoning is called indirect because in order to show something, you assume the opposite and derive contradictory sentences from it (along with the premises). (Sentences are contradictory when one is the negation of the other.) If you succeed in doing this, you have shown that the negation of what you are trying to derive isn't true; it can’t be true because it entails contradictory sentences. So what you are trying to derive must itself be true. The technique of indirect derivation has two parts. First, there is a new kind of assumption: immediately following a show line you may assume the opposite of the sentence on the show line. (The opposite of the sentence is its negation, or its "unnegation" if it is already a negation.) Then when you have two sentences, one of which is the negation of the other, after the last one derived you add the line number of the other and write "id" for "indirect derivation"; then you box and cancel. (Alternatively, you may write a later line with no sentence on it, citing the line numbers of both of the contradictory sentences, write "id'; and then box and cancel.) Here is a formal derivation corresponding to the above reasoning: P ~W ∴ ~(P→W) 1. 2. 3. 4. 5.

Show ~(P→W) P→W P W ~W

ass id

assumption for indirect derivation

pr 2 3 mp pr

Having reached line 5 we may add to it the line number 4 and "id"; then box and cancel: 1. Show ~(P→W) 2. 3. 4. 5.

P→W P W ~W

ass id pr 2 3 mp pr 4 id

Cancel the "Show" Box the lines Write "id"

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The "4 id" at the end of line 5 indicates that the sentence on line 4 contradicts that on the current line, line 5, and thus the assumption (on line 2) which lead to them must be false (given the premises of the argument). Here is another example of a derivation for this short valid argument: U→S ~U→S ∴ S 1. 2. 3. 4. 5. 6.

Show S ~S U→S ~U ~U→S S

ass id pr 2 3 mt pr 4 5 mp

Since we have ~S on line 2 and S on line 6, we are in a position to box and cancel: 1. 2. 3. 4. 5. 6.

Show S ~S U→S ~U ~U→S S

ass id pr 2 3 mt pr 4 5 mp 2 id

Notice that we could have postponed the id step to a later line: 1. 2. 3. 4. 5. 6. 7.

Show S ~S U→S ~U ~U→S S

ass id pr 2 3 mt pr 4 5 mp 26 id

THINKING UP DERIVATIONS: Suppose that you have an argument and you are not sure whether or not it is valid. You can show it not to be valid by producing what is usually called a "counter-example" -- a logically possible situation in which its premises are all true and its conclusion false. If you get a counterexample, the argument is invalid. Suppose that you aren't able to find a counter-example. That might be because the argument is valid, or it might be because you haven't been lucky enough or clever enough to find a counter-example. So failing to find a counter-example, by itself, shows nothing. However, sometimes when you try to find a counter-example, and you fail, this is because any attempt to make the premises true and conclusion false leads you to assign opposite truth values to some sentence. If this happens to you, your failure can be used as a guide to producing a derivation that validates the argument. Typically, it is a guide to producing an indirect derivation. Here is an example. You are given the argument:

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If Pedro left early then if Taylor stayed, Zack left Taylor stayed If Zack left, Pedro didn't leave early ∴ Pedro didn't leave early Symbolized, it is: P → (T → Z) T Z → ~P ∴ ~P Consider an attempt to produce a counter-example. You need a case in which the conclusion, '~P', is false; that is, one in which 'P' is true. So assume that 'P' is true. By the first premise, that makes 'T→Z' be true, and then by the second premise, 'Z' must be true. Then by the third premise, '~P' must be true. Oops! We can't make both 'P' and '~P' true. So there must be no counter-example. Here is a derivation based on that reasoning. The assumption that 'P' is true is just like an assumption for purposes of indirect derivation: 1. Show ~P 2.

P

ass id

Now the next few steps of our attempt to get a counter-example are paralleled by familiar steps in a derivation: 1. Show ~P 2. 3. 4. 5. 6. 7. 8.

P ass id P → (T→Z) pr T→Z 2 3 mp T pr Z 4 5 mp Z → ~P pr ~P 6 7 mp

The "Oops" in the failed counter-example search parallels the fact that we now have contradictories on lines 2 and 8. So we may box and cancel: 1. Show ~P 2. 3. 4. 5. 6. 7. 8.

P P → (T→Z) T→Z T Z Z → ~P ~P

ass id pr 2 3 mp pr 4 5 mp pr 6 7 mp 2 id

This then is an indirect derivation that validates the argument.

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EXERCISES 1. For each of the following derivations, determine which lines are correct and which incorrect. (In assessing a line, assume that previous lines are correct.) a.

R → (S → T) S ~T ∴ ~R 1. Show ~R 2. 3. 4. 5. 6. 7.

b.

R R → (S → T) S→T S T ~T

ass id pr 2 3 mp pr 4 5 mp pr 6 id

~S → P R S → ~R ∴ P 1. Show P 2. 3. 4. 5. 6. 7.

c.

R ~S → ~R ~S ~P ~S → P ~~S

pr pr 2 3 mt ass id pr 5 6 mt 5 id

U → (V→~W) X → (U→V) (V → ~W) → X ∴ U→V 1. Show U → V 2. 3. 4. 5. 6. 7.

U → ~V U → (V→~W) ~V X → (U→V) ~X ~(U→V)

ass id pr 2 3 mt pr 2 5 mt 5 6 mt 2 id

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2. Construct correct derivations for each of the following arguments using indirect derivations. a.

~Q → R S → ~R ~S → Q ∴ Q

b.

(P→Q) → R S → (P→Q) ~S → R ∴ R

c.

~P → (R→S) (R→S) → T ~T Q → (R→S) ∴ ~(P → Q)

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8 SUBDERIVATIONS Reasoning can be intricate. One way in which this happens is when derivations occur within derivations. Consider the following argument in symbolic form: ~(R→Q) → P P → (~Q→Q) ~Q ∴ ~R 1.

Show ~R

It is not clear how to directly derive ~R using our rules mp, mt, and dn. A conditional derivation is not applicable because ~R is not a conditional. One could assume R for purposes of doing an indirect derivation, but it is not clear how to proceed from there. An alternative approach is to use a derivation within the main derivation. Here is one way to proceed: Try to derive the negation of ‘~Q→Q’ from the third premise, and then use modus tollens on the second, and then on the first, premise to get R→Q, which then leads to the desired conclusion using mt and the third premise. The first part of this strategy -- deriving ‘~(~Q→Q)’ from the third premise -- requires a derivation of its own. You can do this as a conditional derivation, leaving you with: 1. 2.

Show ~R Show ~(~Q→Q)

3. 4. 5.

~Q→Q ~Q Q

ass id pr 3 4 mp 4 id

You now may use line 2 just as you would use any other line; once the "Show" is cancelled, the line is no longer something you are trying to derive, it is something you have derived. (A cancelled 'show' means "shown".) You proceed: 1. 2.

Show ~R Show ~(~Q→Q)

3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

~Q→Q ~Q Q P → (~Q→Q) ~P ~(R→Q) → P ~~(R→Q) 7 8 R→Q ~Q ~R

ass id pr 3 4 mp 4 id pr 2 6 mt pr mt 9 dn pr 10 11 mt

This is a complete direct derivation, so you may box and cancel:

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1. 2.

Show ~R Show ~(~Q→Q)

3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

~Q→Q ~Q Q P → (~Q→Q) ~P ~(R→Q) → P ~~(R→Q) R→Q ~Q ~R

ass id pr 3 4 mp 4 id pr 2 6 mt pr 7 8 mt 9 dn pr 10 11 mt dd

The subderivation was just a way to derive '~(~Q→Q)'. Here is another example: (P→~Q) → (R →S) Q→~P T→R ∴ T→S We begin the derivation by stating what is to be shown: 1.

Show T→S

Generally, the easiest way to derive a conditional is to use conditional derivation. So we write this: 1. Show T→S 2.

T

ass cd

with the goal of deriving S, thereby completing the conditional derivation. At this point it is unclear what to do next. However, we notice that S occurs only once in the premises -- in the first premise. And if we could derive the antecedent of that premise, we could use mp to infer R→S, which would get us close to completing the derivation. So we try to derive the antecedent of the first premise: 1. 2. 3.

Show T→S T Show P→~Q

ass cd

Line 3 is a conditional, so we try to use a conditional derivation: 1. Show T→S 2. 3.

T Show P→~Q

4.

P

ass cd ass cd

Our immediate goal is now to derive ~Q. We can do that by appealing to the double negation of P along with the second premise: 1. 2. 3.

Show T→S T Show P→~Q

4. 5. 6. 7.

P ~~P Q→~P ~Q

ass cd ass cd 4 dn pr 5 6 mt

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This completes the conditional subderivation, so we box and cancel: 1. 2. 3.

Show T→S T Show P→~Q

4. 5. 6. 7.

P ~~P Q→~P ~Q

ass cd ass cd 4 dn pr 5 6 mt

We may now do the rest of the derivation: 1. 2. 3.

Show T→S T Show P→~Q

4. 5. 6. 7. 8. 9. 10. 11. 12.

P ~~P Q→~P ~Q

ass cd ass cd 4 dn pr 5 6 mt cd

(P→~Q) → (R→S) R→S T→R R S

pr 3 8 mp pr 2 10 mp 9 11 mp

Line 12 completes the main conditional derivation, so we box and cancel, and we are done: 1. 2. 3.

Show T→S T Show P→~Q

4. 5. 6. 7. 8. 9. 10. 11. 12.

P ~~P Q→~P ~Q

ass cd ass cd 4 dn pr 5 6 mt cd

(P→~Q) → (R→S) pr R→S 3 8 mp T→R pr R 2 10 mp S 9 11 mp cd

Now that we have derivations within derivations, the availability of previous lines for the purpose of applying rules can change from line to line; a line that is not available at one point can become available later, and one that is available may become unavailable. Examples: At line 4 above, line 3 is not available, because from the point of view of line 4, line 3 is an uncancelled show line. But the 'show' on line 3 is cancelled at line 7, so from the point of view of line 8, line 3 is available. On the other hand, at line 6, line 5 is available; but line 5 is no longer available at line 8, because it has been boxed at line 7. We can now state explicitly what may appear in a derivation which may contain subderivations:

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DERIVATIONS A derivation from a set of sentences P consists of a sequence of lines that is built up in order, step by step, where each step is in accordance with these provisions: • • • •

• •

• •

Show line: A show line consists of the word "Show" followed by a symbolic sentence. A show line may be introduced at any step. Show lines are not given a justification. Premise: At any step, any symbolic sentence from the set P may be introduced, justified with the notation "pr". Rule: At any step, a line may be introduced if it follows by a rule from sentences on previous available lines; it is justified by citing the numbers of those previous lines and the name of the rule. Direct derivation: When a line (which is not a show line) is introduced whose sentence is the same as the sentence on the closest previous uncancelled show line, one may, as the next step, write "dd" following the justification for that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. Assumption for conditional derivation: When a show line with a conditional sentence is introduced, as the next step one may introduce an immediately following line with the antecedent of the conditional on it; the justification is "ass cd". Conditional derivation: When a line (which is not a show line) is introduced whose sentence is the same as the consequent of the conditional sentence on the closest previous uncancelled show line, one may, as the next step, write "cd" at the end of that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. Assumption for indirect derivation: When a show line is introduced, as the next step one may introduce an immediately following line with the [un]negation of the sentence on the show line; the justification is "ass id". Indirect derivation: When a sentence is introduced on a line which is not a show line, if there is a previous available line containing the [un]negation of that sentence, and if there is no uncancelled show line between the two sentences, as the next step you may write the line number of the first sentence followed by "id" at the end of the line with the second sentence. Then you cancel the closest previous "show", and box all sentences below that show line, including the current line.

Except for steps that involve boxing and canceling, every step introduces a line. When writing out a derivation, every line that is introduced is written directly below previously introduced lines. Optional variant: When boxing and canceling with direct or conditional derivation, the "dd" or "cd" justification may be written on a later line which contains no sentence at all, and which is followed by the number of the line that satisfies the conditions for direct or conditional derivation. With indirect derivation, the "id" justification may be written on a later line which contains no sentence at all, and which is followed by the numbers of the two lines containing contradictory sentences. In both cases, the lines cited must be available from the later line.

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CHAPTER 1 SECTION 8

EXERCISES 1. For each of the following derivations, determine which lines are correct and which incorrect. (In assessing a line, assume that previous lines are correct.) Tip: When a box occurs in a correctly formed derivation, it is put there by the command ('dd', or 'cd' or 'id') that appears at the end of the last line within the box. When a "show" is cancelled, it is cancelled by the same command that puts the box immediately below the "show'. a.

P → (Q→R) ~Q → S ∴ P → (~S→R) 1. Show P→(~S→R) 2. 3.

P Show ~S → R

4. 5. 6. 7. 8. 9. 10.

~S ~Q → S ~~Q Q P → (Q→R) Q→R R

11.

b.

ass cd ass cd pr 4 5 mt 6 dn pr 2 8 mp 7 9 mp cd 3 cd

R→Q Q→P ∴ R→P 1. Show R → P 2. 3. 4. 5. 6. 7. 8.

R R→Q Q Show P ~P Q→P ~Q

9.

c.

ass cd pr 2 3 mp ass id pr 6 7 mt 4 id

error on this line!

5 cd

P→Q (R→Q) → S (U→S) → ~P ∴ ~P

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CHAPTER 1 SECTION 8

1. Show ~P 2. 3.

P Show R → Q

4. 5. 6.

R P→Q Q

7.

Show U→ S

8. 9. 10. 11. 12. 13.

U (R→Q) → S S (U→S) → ~P ~P P

ass id ass cd pr 2 6 mp cd ass cd pr 3 9 mp cd pr 7 11 mp 2 r 12 id

2. Construct correct derivations for each of the following arguments. a.

P → (Q→R) S→Q ∴ S → (P→R)

b.

(P→Q) → Q P→R Q → ~Q ∴ ~(R → Q)

c.

(U → V) → (W→X) U→Z ~V → ~Z X→Z ∴ W→Z

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9 SHORTCUTS Writing long derivations can be tedious. Here are two shortcuts. Citing premises without rewriting them Every time we have wanted to appeal to a premise when using a rule of inference, we have written the premise into the derivation, justifying it by 'pr', and then when we use the rule we cite the number of the line where the premise was written. We can skip writing down the premise entirely if the justification of the rule identifies the premise by name. For example, instead of having: 8. Q → R 9. Q 10. R

pr

(where 'Q' is the third premise) 8 9 mp

we just write: 8. 9.

Q→R R

8 pr3 mp

This is equivalent to just assuming that the premises all come with line numbers: pr1, pr2, pr3, . . ., and citing those line numbers when we use a rule of inference. (Our directions above for constructing derivations already include this option.) For example, here is a derivation we gave earlier: Premises:

Conclusion: 1.

P→(W→T) ~P→T ~T ∴ ~W

Show ~W

2. 3. 4. 5. 6. 7. 8. 9.

~T ~P→T ~~P P P→(W→T) W→T ~T ~W

pr pr 2 3 mt 4 dn pr 5 6 mp pr 7 8 mt dd

Using the shortcut, we can essentially skip lines 2, 3, 6, 8, to get this shortened derivation which has analogues of lines 1, 4, 5, 7, 9 in the original derivation: 1.

Show ~W

2. 3. 4. 5.

~~P P W→T ~W

pr2 pr3 mt 2 dn 3 pr1 mp 4 pr3 mt dd

Explicitly, the technique is: Citing Premises in Rules When citing a premise in applying a rule of inference, use 'pr1', 'pr2' 'pr3', . . . to identify the first, second, third, . . . premises.

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CHAPTER 1 SECTION 9

Mixed derivations: Our derivation rules are already formulated in a way that lets you use any one of dd, cd, id, in cases in which you expect to use another of them. For example, suppose you are trying to show 'P→Q' by cd, assuming 'P'. But you derive 'P→Q' instead of 'Q', Then you can use dd to box and cancel. The fact that you have assumed 'P' for purposes of producing a conditional derivation does not interfere with this use of dd. Given these lines: 1. Show □→○ 2. □ 3. …….. 7. 8.

ass cd

…….. □→○

you can box and cancel: 1. Show □→○ 2. 3. 7. 8.

□ ass cd …….. …….. □→○

dd

This is a "mixed" derivation: an assumption is made for constructing a conditional derivation, and then 'dd' is used instead of 'cd' to complete it. That's OK because this is just a shortcut. Whenever you are in the position described above, you could instead add a step to the end of the derivation and then conclude the derivation as a conditional derivation. Just add step 9: 1. 2. 3. 4. 7. 8. 9.

Show □→○ □ ass cd …….. …….. …….. □→○ ○

2 8 mp

and then box and cancel using cd: 1. Show □→○ 2. 3. 4.

□ …….. ……..

7. 8. 9.

…….. □→○ ○

ass cd

2 8 mp cd

So using dd at the end of line 8 is merely a way to save a step.

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CHAPTER 1 SECTION 9

Similarly, if you are trying to do an indirect derivation and you end up deriving the sentence on the show line, you may use dd. That is, if you have: 1. 2. 3. 4.

Show □ ~□ …….. ……..

7. 8.

…….. □

ass id

you may box and cancel with dd: 1.

Show □

2. 3. 4.

~□ …….. ……..

7. 8.

…….. □

ass id

dd

Here the shortcut is obvious; you are already in a position to use id since you already have the contradictory sentences that you need; you can instead cite the line number of the other contradictory and use id: 1.

Show □

2. 3. 4.

~□ …….. ……..

7. 8.

…….. □

ass id

2 id

Similarly you can use cd when you are set up for a direct derivation of a conditional, or when you are trying to derive a conditional using id; and you can use id when you have derived contradictories even if you are set up for a direct or conditional derivation. In allowing for mixed derivations, we are not actually changing anything. Our rules already allow for them. So this is a summary of things we can already do with the rules as stated: Mixed derivations You may use dd, cd, and id to complete a derivation by boxing and canceling whenever they apply, whether or not an assumption has been made, and regardless of the type of assumption if any.

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CHAPTER 1 SECTION 9

EXERCISES 1. Each of the following derivations is a mixed derivation. In each case produce another derivation which is not mixed. a.

P→R Q → ~R ~Q → Q ∴ P→Q 1. Show P → Q 2. 3. 4. 5. 6. 7. 8. 9.

b.

ass cd pr 2 3 mp 4 dn pr 5 6 mt pr 7 8 mp 7 id

Q→U Q → ~U R→Q R ∴ P 1. Show P 2. 3. 4. 5. 6. 7. 8.

c.

P P→R R ~~R Q → ~R ~Q ~Q → Q Q

R R→Q Q Q→U U Q → ~U ~U

pr pr 2 3 mp pr 4 5 mp pr 4 7 mp 6 id

U → (V→W) X→U ~X → W ∴ V→W 1. Show V → W 2. 3. 4. 5. 6. 7. 8.

~(V → W) U → (V→W) ~U X→U ~X ~X → W W

ass id pr 2 3 mt pr 4 5 mt pr 6 7 mp cd

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2. Do the derivations from 1a-c above using premise line numbers instead of rule pr. 3. Earlier we stipulated that a conditional sentence is false when its antecedent is true and its consequent false, and true in all other cases. This was not arbitrary. Given the rules and derivation procedures that we have adopted, these choices are forced on us. For, using our rules and procedures, we can produce derivations to show each of the following arguments to be valid: P Q ∴ P→Q

P ~Q ∴ ~(P→Q)

~P Q ∴ P→Q

~P ~Q ∴ P→Q

The first of these tells us that when the antecedent and consequent of a conditional are both true, so is the conditional. The second tells us that when the antecedent of a conditional is true and the consequent false, the conditional is false. The third and fourth tell us that in either case in which the antecedent is false, the conditional is true. Produce short derivations for each of those four arguments.

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CHAPTER 1 SECTION 10

10 STRATEGY HINTS FOR DERIVATIONS Some strategies are generally useful in thinking up how to construct derivations. They do not always work, but they are often the best way to begin the search for a successful derivation. 1. Try to reason out the argument for yourself. If you can do that, then write down the steps you went through in your own reasoning. These steps will often be an outline of a good derivation. This idea of reasoning things out and then turning the steps into a derivation has been illustrated above when introducing direct, conditional, and indirect derivations. This is by far the best approach to thinking up a derivation. 2. Begin with a sketch of an outline of a derivation, and then fill in the details. For example, in thinking up a derivation for this argument: P → (Q→R) Q ∴ P→R you might begin by saying: "I'll do a conditional derivation: I'll assume P and then use this together with the premises to derive R. Then I'll box and cancel by cd." This outline gives you this much of a derivation: 1.

Show P→ R

2.

P

ass cd ::::: ::::: :::::

13.

R

cd

. . . . then box and cancel

All you need to do then is to fill in lines 3 to 12. (I'm just guessing that it will take exactly eight more steps to finish the derivation. If it takes more or less, change the '13' to the appropriate number.) 3. Write down obvious consequences: Write down obvious consequences of premises or of sentences that have already been derived. If you write down the simple consequences of things that you already have, then you have lots of resources right in front of you. Of course, you can do this in your head instead of on paper, and sometimes that is sufficient. But if things are not completely clear to you, writing down obvious consequences can be useful. Suppose you are trying to construct a derivation to validate this argument: P → (Q→R) S→Q Q→P S ∴ R Let's say that you have written down the show line: 1. Show R but you are momentarily stuck. Write down some obvious consequences of the premises: 2. 3. 4.

Q P Q→R

pr4 pr2 mp pr3 2 mp pr1 3 mp

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CHAPTER 1 SECTION 10

At this point, if you look over what is available, it will be obvious that you can get the desired conclusion in one additional step: 5.

R

2 4 mp

4. If you are trying to derive a conditional, use conditional derivation. This is almost always the easiest way to derive a conditional. This has been amply illustrated above. 5. If you have a conditional, try to derive its antecedent (and then use modus ponens), or try to derive the negation of its consequent (and then use modus tollens). Here is an illustration: P → (Q → R) S→Q ~P → ~S S ∴ R The only premise in which R occurs is the first one, which is a conditional. You could apply modus ponens to this conditional if you could prove P. Our strategy rule suggests that you try to derive P. That is not difficult to do: 1. Show R 2. ~~S 3. ~~P 4. P

pr4 dn 2 pr3 mt 3 dn

This immediately gives us: 5.

Q→R

4 pr1 mp

It is then easy to complete the derivation: 6. 7.

Q R

pr2 pr4 mp 5 6 mp

and you are ready to box and cancel. 6. Try indirect derivation. When you reach a place where none of the other strategies clearly apply, assume the negation of what you are trying to derive and try to derive a contradiction. This too has been amply illustrated above. 7. When doing an indirect derivation, try to derive the negation of a premise or the negation of something that you already have derived. This is especially useful if you already have the negation of a conditional, for you can try to derive the conditional by using conditional derivation, and then you will have both the conditional and its negation, and you can box and cancel with id. An example. You have: R (Q → S) → ~R ∴ ~S Now begin a derivation, and follow strategy rule 1: write out obvious consequences of what you have: 1. Show ~S 2. S 3. ~~R 4. ~(Q→S)

ass id pr1 dn 3 pr2 mt

The moves so far are pretty straightforward. But it may not be clear what to do next. Strategy rule 7 suggests that you try to derive the conditional 'Q→S', in order to contradict '~(Q→S)'. You do this with a conditional subderivation:

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CHAPTER 1 SECTION 10

5.

Show Q → S

6. 7.

Q S

ass cd 2 r cd

(this step is obvious once you notice it)

Having derived the conditional on line 5 and its negation on line 4 the indirect derivation is just about complete. The complete derivation is: 1. Show ~S 2. 3. 4. 5.

S ~~R ~(Q→S) Show Q → S

6. 7.

Q S

8.

ass id pr1 dn 3 pr2 mt ass cd 2 r cd 4 5 id

EXERCISES Produce correct derivations to validate these arguments. 1.

S (R→S) → W ∴ W

2.

P → (S→R) P → (W→S) W→P ∴ W→R

3.

(P→Q) → S S→T ~T → Q ∴ T

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CHAPTER 1 SECTION 11

11 THEOREMS A truth of logic (a sentence that is logically true) is a sentence that is true in any logically possible situation. It must be true no matter what. Because of this, a truth of logic is like the conclusion of a valid argument which has no premises. If such an argument is valid, it does not have all true premises and a false conclusion in any logically possible situation. When there aren’t any premises, this is equivalent to saying that it does not have a false conclusion in any logically possible situation. That is, its conclusion is true in every logically possible situation. It is a truth of logic. Since derivations show arguments valid, if a derivation is used to show an argument with no premises to be valid, that amounts to showing that the conclusion is logically true. There is a special word, ‘theorem’, that refers to any sentence shown by a technique like a derivation when no premises at all are used. So the topic of this section is Theorems. It is customary to indicate a theorem by placing a "therefore" sign in front of it, as if it were an argument with its premises missing. So writing "∴□" indicates that □ is a theorem. Here is a derivation to show that ‘P→P’ is a theorem: ∴ P→P 1.

Show P→P

2. 3.

P

ass cd 2 cd

In simple cases, theorems are obviously trivial statements. Even when they are complex, they are still trivial in the sense that they say nothing beyond what is logically true. In this book we will list several theorems, giving them the names "T1", "T2", and so on. Here are some given in increasing degrees of complexity. Some of them have common names; these are indicated to the right. T1

P→P

T2

Q → (P→Q) 1. 2. 3.

Show Q → (P→Q) Q Show P→Q

4. 5.

P Q

6.

T3

ass cd ass cd 2 r cd 3 cd

P → ((P→Q) → Q) 1. 2. 3.

Show P → ((P→Q) → Q) P ass cd Show (P→Q) → Q

4. 5. 6.

P→Q Q

ass cd 2 4 mp cd 3 cd

T4

(P→Q) → ((Q→R) → (P→R))

"Syllogism"

T5

(Q→R) → ((P→Q) → (P→R))

"Syllogism"

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CHAPTER 1 SECTION 11

T6

(P → (Q→R)) → ((P→Q) → (P→R)) 1.

"Distribution of → over →"

Show (P → (Q→R)) → ((P→Q) → (P→R))

2. 3.

P → (Q→R) Show (P→Q) → (P→R)

4. 5.

P→Q Show P→R

6. 7. 8. 9.

P Q Q→R R

ass cd ass cd ass cd 4 6 mp 2 6 mp 7 8 mp cd

10.

5 cd

11.

3 cd

T7

((P→Q) → (P→R)) → (P→(Q → R))

"Distribution of → over →"

T8

(P→ (Q→R)) → (Q → (P→R))

Commutation

T9

(P → (P→Q)) → (P→Q)

T10

((P→Q) → Q) → ((Q→P) → P)

T11

~~P → P

Double negation

T12

P → ~~P

Double negation

T13

(P→Q) → (~Q → ~P)

Transposition

T14

(P → ~Q) → (Q → ~P)

Transposition

T15

(~P → Q) → (~Q → P)

Transposition

T16

(~P → ~Q) → (Q → P)

Transposition

T17

P → (~P→Q)

T18

~P → (P→Q)

T19

(~P→P) → P

T20

(P→~P) → ~P

T21

~(P→Q) → P

T22

~(P→Q) → ~Q

T23

((P→Q) → P) → P

Peirce's law

EXERCISES 1. Prove theorems T8, T11, T13, T17, T19, T21, T22.

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12 USING PREVIOUSLY PROVED THEOREMS IN DERIVATIONS One major use of theorems is to avoid repeating derivations that you have done earlier. For example, suppose you have previously derived 'P → (P → (Q→P))' from no premises. Now you are doing a new derivation where you have derived 'R', and you want to derive 'S→R'. By following out the reasoning from earlier, but using 'R' instead of 'P', you could derive this: 'R → (R → (S→R))', and then use modus ponens twice to get 'S → R'. But why write out the new derivation of 'R → (R → (S→R))' when it may be lengthy, and it just involves repeating moves that you made earlier? It would be nice to have a way to "reuse" old derivations for new purposes when the reasoning is the same. One way to do this is with this new procedure that we adopt hereafter: Theorems: Any instance of any previously derived theorem may be entered on any line of a derivation. As justification, write the name of the theorem. (E.g. ‘T13’.) An instance of a theorem is what you get by considering the theorem as a pattern, and filling in the pattern uniformly with sentences. For example, T1 is ∴P→P. This validates the pattern ‘□→□’. Anything got by filling in that pattern, putting the same sentence in for each occurrence of □, can be written on any line of a derivation. For example, you can write: (S→W) → (S→W)

21.

T1

More complicated theorems are more useful. For example, T4 is: (P→Q) → ((Q→R) → (P→R)) This gives us the pattern: (□→○) → ((○→△) → (□→△)) Putting ~R in for □, (U→V) for ○, and W for △ we have: (~R→(U→V)) → (((U→V)→W) → (~R→W)) Here is an example of a use of this theorem. Suppose you wish to do a derivation for this argument: R→~S ~S→~T ∴ R→~T A very short derivation could be given using theorem 4: 1. 2. 3. 4.

Show R→~T (R→~S) → ((~S→~T) → (R→~T)) (~S→~T) → (R→~T) R→~T

T4 2 pr1 mp 3 pr2 mp dd

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CHAPTER 1 SECTION 12

EXERCISES 1. Produce short derivations for these arguments using instances of the theorems listed above. a.

X → ~(Y → Z) ∴ (Y → Z) → ~X

b.

R → (~P→S) R → ~P ∴ R→S

c.

~(R → (S→T)) R→P P → (Q → (S→T)) ∴ ~Q

d.

Q→R R→S ∴ Q→S

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Answers to the Exercises -- CHAPTER 1

Answers to the Exercises -- Chapter 1 SECTION 1 1. a

Sentence, official notation ~~~P | ~~P | ~P | P

b

Sentence, informal notation ~Q→~R /\ ~Q ~R | | Q R

c d e

Not a sentence; it is impossible to construct "~→" Not a sentence; the rules of formation do not allow you to enclose a negated formula in parentheses, only conditionals. Sentence, informal notation (P→Q) → (R→~Q) /\ (P→Q) (R→~Q) /\ /\ P Q R ~Q | Q

f g

h

Not a sentence; conditionals contained inside other conditionals must be surrounded by parentheses. Not a sentence; this is the same as sentence (f) with extra parentheses put on the outside; If (g) were a sentence, then (f) would be a sentence in informal notation; since (f) is not a sentence, (g) can't be either. Sentence; informal notation (~S→R) → ((~R→S) → ~(~S→R)) /\ ~S→R (~R→S) → ~(~S→R) /\ /\ ~S R ~R→S ~(~S→R) | /\ | S ~R S ~S→R | /\ R ~S R | S

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i

Sentence; informal notation P → (Q→P) /\ P Q→P /\ Q P

SECTION 2 1. a b c d e 2. a

S→R, because "only if" immediately precedes the consequent R→S, because "provided that" is equivalent to "if", and "if" immediately precedes the antecedent ~S, because "won't" means the same as "will not", the sentence only contains one negation indicator S→R, because "only if" immediately precedes the consequent R→S, because "given that" is equivalent to "if" and "if" immediately precedes the antecedent Susan will be late only provided that it rains Susan will be late only if it rains S→R

b

Only on condition that it rains will Susan be late Only if it rains will Susan be late S→R

c

Susan will be late only in case it rains Susan will be late only if it rains S→R

d

Susan will be late only if it rains S→R

e

It is not the case that Susan will be late ~S

SECTION 3 1. a

If Veronica doesn’t leave William won’t either If Veronica doesn’t leave William won’t leave If Veronica doesn’t leave then William won’t leave ~V→~W

b

William will leave if Yolanda does, provided that Veronica doesn’t William will leave if Yolanda does, if Veronica doesn’t [leave] If Veronica doesn't [leave], then (William will leave if Yolanda does) If Veronica doesn't [leave], then (if Yolanda [leaves], then William will leave) ~V → (Y→W)

c

If Yolanda doesn’t leave, then Veronica will leave only if William doesn’t ~Y → (V → ~W)

d

If Yolanda doesn’t leave then Veronica will leave, given that William doesn’t If Yolanda doesn’t leave then Veronica will leave, if William doesn’t

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If William doesn’t [leave], then (if Yolanda doesn’t leave then Veronica will leave) ~W → (~Y→V) 2. a

William will leave if Veronica does If Veronica [leaves], then William will leave V→W

b

Veronica won't leave if William does If William [leaves], then Veronica won't leave W → ~V

c

If Veronica leaves, then if William doesn't leave, Yolanda will leave If Veronica leaves, then (~W → Y) V → (~W → Y)

d

If Veronica doesn't leave if William doesn't, then Yolanda won't If (if William doesn't [leave], then Veronica doesn't leave), then Yolanda won't [leave] (~W → ~V) → ~Y

e

William won’t leave provided that Veronica doesn’t leave William won’t leave if Veronica doesn’t leave If Veronica doesn't leave, then William won't leave ~V → ~W

f

If William leaves, then if Veronica leaves so will Yolanda If William leaves, then (if Veronica leaves, then Yolanda will leave) W → (V → Y)

g

William will leave only if if Veronica leaves then so will Yolanda William will leave only if (if Veronica leaves then so will Yolanda) William will leave → (if Veronica leaves then so will Yolanda) William will leave → (if Veronica leaves then Yolanda will leave) W → (V → Y)

h

William will leave only if Veronica leaves, only provided that Yolanda will leave (William will leave only if Veronica leaves) only if Yolanda will leave (W → V) → Y

SECTION 4 1. a

b c d e f g h i

None of the above; it might look like a modus tollens inference, but the second premise, Q, would have to first be changed to ~~Q by applying double negation; so while the argument is valid, it is not a one-step application of modus tollens. None of the above Double negation Modus ponens Modus tollens None of the above Modus ponens; the consequent of the conditional does not need to be an atomic sentence, it can be molecular as well. None of the above; it may look like a modus tollens inference, but the second premise is not actually the negation of the consequent of the first premise. Double negation and none of the above are both good answers; the conclusion can be inferred by double negation from the first premise, but since the second premise is not involved in that

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inference, the whole argument is not a double-negation inference 2. a b c d e f

g h i

In all cases we can validly infer any sentence which results from putting '~~' in front of either premise by the rule double negation. Such results are not enumerated below. ~X may be inferred by modus ponens. X may be inferred by double negation; ~~W may be inferred by modus tollens. Nothing additional (R→X) may be inferred by modus ponens. Nothing additional; Modus tollens cannot be applied because the second premise is not actually the negation of the consequent of the first premise. (W→X) may be inferred from the first premise by double negation; this must be done before you apply modus tollens with the second premise, so you can't apply modus tollens in one step to get ~W. Nothing additional; if you apply double negation to the second premise you can then apply modus tollens as a second step to get ~W. Nothing additional (W→X) follows by double negation.

SECTION 5 1. Only errors are listed. First derivation Line 6 -- line 6 is not available at line 6; derivation can be corrected by writing "5 dn". Second derivation Line 3 -- when justifying writing a premise, no line citation is given Line 4 -- the sentence on line 2 is not the negation of the consequent of the sentence on line 3; in this case we would need to first apply dn to line 2 as an intermediate step. Then we could apply mt with line 3, which would result in ~~P. P could then be inferred on the following line by dn. Line 5 -- two lines must be cited with mp; the sentence inferred does follow from line 4 together with the first premise, but the first premise must be cited somehow. Line 7 -- "5 6 mt" would result in ~R rather than ~~R. Line 9 -- "5 8 mp" is OK, but the derivation is not done; we set out to show ~R, but line 9 displays ~Q, so we cannot conclude the derivation at this point and so it is incorrect to write "dd" to mark the conclusion of the derivation of ~R. Third derivation Line 3 -- "~S" is not one of the premises. Line 7 -- neither line 5 nor line 6 is a conditional, so mt cannot possibly apply to that pair of lines. 2. In each case the derivation displayed does not represent the only possible derivation; many alternate, equally correct derivations can be given. P Q → ~P R→Q ∴ ~R

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1. 2. 3. 4. 5. 6. 7. 8.

Show ~R P ~~P Q → ~P ~Q R→Q ~R

pr 2 dn pr 3 4 mt pr 5 6 mt 7 dd

In this derivation and those below, the "dd" could occur at the end of the previous line.

W → ~(V→~Y) X → (V→~Y) V→Y (V→Y) → X ∴ ~W 1. Show ~W 2. V→Y 3. (V→Y) → X 4. X 5. X → (V→~Y) 6. V→~Y 7. ~~(V→~Y) 8. W → ~(V→~Y) 9. ~W 10.

pr pr 2 3 mp pr 4 5 mp 6 dn pr 7 8 mt 9 dd

(W→Z) → (Z→W) (Z→W) → ~X P→X ~~P ∴ ~(W→Z) 1. Show ~(W→Z) 2. ~~P 3. P 4. P→X 5. X 6. ~~X 7. (Z→W) → ~X 8. ~(Z→W) 9. (W→Z) → (Z→W) 10. ~(W→Z) 11.

pr 2 dn pr 3 4 mp 5 dn pr 6 7 mt pr 8 9 mt 10 dd

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SECTION 6 1. Only errors are listed. Derivation a All correct Derivation b Line 3 -- Line 1 is not available at line 3 because when line 3 is written, line 1 is still an un-cancelled show line. Line 7 -- No problem with the use of cd to box and cancel, but mt cannot be applied to lines 5 and 6 because line 6 does not contain the negation of the consequent of line 5; you would have to add a line and apply double negation to line 6 first. Derivation c Line 2 -- You can only assume the antecedent of the conditional to be shown. Line 3 -- While the sentence on line 3 does logically follow from line 2 and premise 1, you can't apply mp to line 2 alone. Line 9 -- The application of dn to line 8 is OK, but you can't end a conditional derivation on a line that does not contain the consequent of the conditional you set out to show. 2. In each case the derivation displayed does not represent the only possible derivation; many alternate, equally correct derivations can be given. a.

P → (Q → (R→S)) ~Q → ~R R ∴ P→S

1. Show P → S 2. P 3. P → (Q → (R→S)) 4. Q → (R→S) 5. R 6. ~~R 7. ~Q → ~R 8. ~~Q 9. Q 10. R→S 11. S 12. b.

ass cd pr 2 3 mp pr 5 dn pr 6 7 mt 8 dn 4 9 mp 5 10 mp 11 cd

Q → ~(R→S) P → (R→S) ~Q → R ∴ P→S

1. Show P → S 2. P 3. P → (R→S) 4. R→S 5. ~~(R→S) 6. Q → ~(R→S) 7. ~Q 8. ~Q → R 9. R 10. S 11.

ass cd pr 2 3 mp 4 dn pr 5 6 mt pr 7 8 mp 4 9 mp 10 cd

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Answers to the Exercises -- CHAPTER 1

c.

U → (U→V) ~R → ~(U→V) R → ~S ∴ U → ~S

1. Show U → ~S 2. U 3. U → (U→V) 4. U→V 5. ~~(U→V) 6. ~R → ~(U→V) 7. ~~R 8. R 9. R → ~S 10. ~S 11. 3. a

P→T ~X → ~T ~S → ~X ∴ P→S

1. Show P → S 2. P 3. P→T 4. T 5. ~~T 6. ~X → ~T 7. ~~X 8. ~S → ~X 9. ~~S 10. S 11. b

ass cd pr 2 3 mp 4 dn pr 5 6 mt pr 7 8 mt 9 dn 10 cd

T→S Y → (S→P) P → ~X Y ∴ T → ~X

1. Show T → ~X 2. T 3. T→S 4. S 5. Y → (S→P) 6. Y 7. S→P 8. P 9. P → ~X 10. ~X c

ass cd pr 2 3 mp 4 dn pr 5 6 mt 7 dn pr 8 9 mp 10 cd

ass cd pr 2 3 mp pr pr 5 6 mp 4 7 mp pr 8 9 mp cd

S → ~T ~S → ~R ~R → X Y→T ∴ ~X → ~Y

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Answers to the Exercises -- CHAPTER 1

1. Show ~X → ~Y 2. ~X 3. ~R → X 4. ~~R 5. ~S → ~R 6. ~~S 7. S 8. S → ~T 9. ~T 10. Y→T 11. ~Y

ass cd pr 2 3 mt pr 4 5 mt 6 dn pr 7 8 mp pr 9 10 mt cd

SECTION 7 1. Only errors are listed. Derivation a All correct Derivation b Line 3 -- The sentence on line 3 is not a premise. Line 4 -- Line 2 is not the negation of the consequent of line 3 so mt doesn't apply. You would first have to apply dn to line 2. Even in that case the result would be ~~S rather than ~S. Line 5 -- "ass id" may only appear on the line immediately following a show line. Line 7 -- The mt inference is OK, but 5 and 7 do not directly contradict so id is used incorrectly. The derivation could be concluded on line 7 with "4 id" since 4 and 7 contradict directly. Derivation c Line 2 -- The sentence on the line is not the negation (or the un-negation, for that matter) of the show line. Line 4 -- There is no way to apply mt with lines 2 and 3. Line 6 -- 2 is not the negation of the consequent of 5 (though it should have been). Line 7 -- There is no way that you can apply mt with lines 5 and 6; 2 and 7 don't contradict directly, so it is premature to conclude with id. 2. In each case the derivation displayed does not represent the only possible derivation; many alternate, equally correct derivations can be given. a.

~Q → R S → ~R ~S → Q ∴ Q

1. Show Q 2. ~Q 3. ~Q → R 4. R 5. ~~R 6. S → ~R 7. ~S 8. ~S → Q 9. Q 10.

ass id pr 2 3 mp 4 dn pr 5 6 mt pr 7 8 mp 2 9 id

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Answers to the Exercises -- CHAPTER 1

b.

(P→Q) → R S → (P→Q) ~S → R ∴ R

1. 2. 3. 4. 5. 6. 7. 8. 9.

Show R ~R (P→Q) → R ~(P→Q) S → (P→Q) ~S ~S → R R

c.

~P → (R→S) (R→S) → T ~T Q → (R→S) ∴ ~(P → Q)

1. Show ~(P → Q) 2. P→Q 3. ~T 4. (R→S) → T 5. ~(R→S) 6. Q → (R→S) 7. ~Q 8. ~P 9. ~P → (R→S) 10. R→S 11.

ass id pr 2 3 mt pr 4 5 mt pr 6 7 mp 2 8 id

ass id pr pr 3 4 mt pr 5 6 mt 2 8 mt pr 8 9 mp 5 10 id

SECTION 8 1. Only errors are listed. Derivation a All correct Derivation b Line 8 -- At this point in the derivation, line 5 is still an un-cancelled show line so line 4 can't be cited to conclude the sub-derivation; you could, however, use the rule r (repetition) to repeat line 4 within the sub-derivation and then use the repeated line to conclude the sub-derivation with id. Derivation c Line 6 -- Line 6 is not available on line 6. The problem would b resolved if we put 5 for 6. Line 13 -- Strictly speaking there is no error here, but it was un-necessary to repeat line 2 in order to apply id; we could have just cited "2 12 id" because line 2 is not separated from line 13 by any uncancelled show lines (only by cancelled ones). 2. In each case the derivation displayed does not represent the only possible derivation; many alternate, equally correct derivations can be given. a.

P → (Q→R)

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Answers to the Exercises -- CHAPTER 1

S→Q ∴ S → (P→R) 1. Show S → (P→R) 2. S 3. Show P→R 4. P 5. P → (Q→R) 6. Q→R 7. S→Q 8. Q 9. R 10. 11 b.

ass cd ass cd pr 4 5 mp pr 2 7 mp 6 8 mp 9 cd 3 cd

(P→Q) → Q P→R Q → ~Q ∴ ~(R → Q)

1. Show ~(R → Q) 2. R→Q 3. Show P → Q 4. P 5. P→R 6. R 7. Q 8. (P→Q) → Q 9. Q 10. Q → ~Q 11. ~Q c.

ass id ass cd pr 4 5 mp 2 6 mp cd pr 3 8 mp pr 9 10 mp 9 id

(U → V) → (W→X) U→Z ~V → ~Z X→Z ∴ W→Z

1. Show W → Z 2. W 3. Show U → V 4. U 5. U→Z 6. Z 7. ~~Z 8. ~V → ~Z 9. ~~V 10. V 11. (U → V) → (W→X) 12. W→X 13. X 14. X→Z 15. Z

ass cd ass cd pr 4 5 mp 6 dn pr 7 8 mt 9 dn cd pr 3 11 mp 2 12 mp pr 13 14 mp cd

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Answers to the Exercises -- CHAPTER 1

SECTION 9 1. a

P→R Q → ~R ~Q → Q ∴ P→Q

1. 2. 3. 4. 5. 6. 7. 8. 9.

Show P → Q P P→R R ~~R Q → ~R ~Q ~Q → Q Q

ass cd pr 2 3 mp 4 dn pr 5 6 mt pr 7 8 mp cd

The only change was to conclude with cd instead of id. b

1. 2. 3. 4. 5. 6. 7. 8. 9.

Q→U Q → ~U R→Q R ∴ P Show P ~P R R→Q Q Q→U U Q → ~U ~U

ass id pr pr 3 4 mp pr 5 6 mp pr 5 8 mp 7 id

The only change was to add an assumption for id. c

U → (V→W) X→U ~X → W ∴ V→W

1. Show V → W 2. ~(V→W) 3. U → (V→W) 4. ~U 5. X→U 6. ~X 7. ~X → W 8. W 9. Show V → W 10. V 11. W 12. Only added lines 9-12.

ass id pr 2 3 mt pr 4 5 mt pr 6 7 mp ass cd 8 r cd 2 9 id

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2. a

P→R Q → ~R ~Q → Q ∴ P→Q

1. 2. 3. 4. 5. 6.

Show P → Q P R ~~R ~Q Q b

1. 2. 3. 4. 5.

1. 2. 3. 4. 5.

Q→U Q → ~U R→Q R ∴ P Show P ~P Q U ~U

c

ass cd 2 pr1 mp 3 dn 4 pr2 mt 5 pr3 mp cd

ass id pr4 pr3 mp 3 pr1 mp 3 pr2 mp 4 id

U → (V→W) X→U ~X → W ∴ V→W Show V → W ~(V→W) ~U ~X W

ass id 2 pr1 mt 3 pr2 mt 4 pr3 mp cd

3 P Q ∴ P→Q 1. 2.

Show P → Q Q

pr2 cd

P ~Q ∴ ~(P→Q) 1. 2. 3. 4.

Show ~(P→Q) P→Q Q ~Q

ass id pr1 2 mp pr2 3 id

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~P Q ∴ P→Q 1. 2.

Show P → Q Q

pr2 cd

~P ~Q ∴ P→Q 1. 2. 3.

Show P → Q P ~P

ass cd pr1 2 id

SECTION 10 1.

S (R→S) → W ∴ W

1. 2. 3. 4.

Show W Show R → S S W

2.

P → (S→R) P → (W→S) W→P ∴ W→R

1. 2. 3. 4. 5. 6. 7.

Show W → R W P S→R W→S S R

3.

(P→Q) → S S→T ~T → Q ∴ T

1. 2. 3. 4. 5. 6. 7.

Show T ~T Q Show P→Q Q S T

SECTION 11

pr1 cd 2 pr2 mp dd

as cd 2 pr3 mp 3 pr1 mp 3 pr2 mp 2 5 mp 4 6 mp cd

ass id 2 pr3 mp 3 r cd 4 pr1 mp 6 pr2 mp 2 id Derivations for theorems are not given here.

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Answers to the Exercises -- CHAPTER 1

SECTION 12 1. a

X → ~(Y → Z) ∴ (Y → Z) → ~X

1. 2. 3.

Show (Y → Z) → ~X (X → ~(Y → Z)) → ((Y → Z) → ~X) (Y → Z) → ~X

b

R → (~P→S) R → ~P ∴ R→S Show R → S (R → (~P→S)) → ((R→~P) → (R→S)) (R→~P) → (R→S) R→S

1. 2. 3. 4.

c

1. 2. 3. 4. 5. 6. 7. 8. 9.

T6 2 pr1 mp 3 pr2 mp dd

~(R → (S→T)) R→P P → (Q → (S→T)) ∴ ~Q Show ~Q Q ~(R → (S→T)) → R R P Q → (S→T) S→T ~(R → (S→T)) → ~(S→T) ~(S→T)

d

1. 2. 3. 4.

T14 2 pr1 mp dd

ass id T21 3 pr1 mp 4 pr2 mp 5 pr3 mp 2 6 mp T22 8 pr1 mp 7 id

Q→R R→S ∴ Q→S Show Q → S (Q→R) → ((R→S) → (Q→S)) (R→S) → (Q→S) Q→S

T4 2 pr1 mp 3 pr2 mp dd

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CHAPTER 2 SECTION 1

Chapter Two Sentential Logic with 'and', 'or', if-and-only-if' 1 SYMBOLIC NOTATION In this chapter we expand our formal notation by adding three two-place connectives, corresponding roughly to the English words 'and', 'or' and 'if and only if': ∧ ∨ ↔

and or if and only if

Conjunction: The first of these, '∧', is the conjunction sign; it has the same logical import as 'and'. It goes between two sentences to form a complex sentence which is true if both of the parts (called 'conjuncts') are true, and is otherwise false: □ ○ (□∧○) T T T T F F F T F F F F Disjunction: The disjunction sign, '∨', makes a sentence that is true in every case except when its parts (its disjuncts) are both false. This corresponds to one use (the "inclusive" use) of 'or' in English: □ T T F F

○ T F T F

(□ ∨ ○) T T T F

Biconditional: The biconditional sign, '↔', states that both of the parts making it up (its constituents) are the same in truth value. It works like this: □ T T F F

○ T F T F

(□ ↔ ○) T F F T

Each of these new connectives behaves syntactically just like the conditional sign, '→': you make a bigger sentence out of two sentences plus a pair of parentheses: (□∧○) (□ ∨ ○) (□ ↔ ○) Our expanded definition of a sentence in official notation is now: Chapter Two SYMBOLIC SENTENCES • • •

Any capital letter between 'P' and 'Z' is a symbolic sentence. (Numerical subscripts may also be used, as 'P3', 'Q24'.) If □ is a symbolic sentence, so is ~□ If □ and ○ are symbolic sentences, so are (□→○), (□∧○), (□ ∨ ○), and (□ ↔ ○).

Nothing is a symbolic sentence for purposes of chapter 2 unless it can be generated by the clauses given above.

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CHAPTER 2 SECTION 1

As before, we allow ourselves informally to omit the outer parentheses when the sentence occurs alone on a line. It is also customary (and convenient) to omit parentheses around conjunctions or disjunctions when they are combined with a conditional or biconditional sign. The sentence: P∧Q → R is to be considered to be an informally worded conditional whose antecedent is a conjunction: (P∧Q) → R If we want to make a conjunction whose second conjunct is a conditional, we must use parentheses around the parts of the conditional: P ∧ (Q→R) Likewise, this sentence: P ↔ Q∨R is an informally written biconditional whose second constituent is a disjunction: P ↔ (Q∨R) If we wish to write a disjunction whose first disjunct is a biconditional, we need to use parentheses around the biconditional: (P↔Q) ∨ R. Finally, we may use two or more conjunction signs or disjunction signs (but not a mix of conjunctions with disjunctions) as abbreviations for what you get by restoring the parentheses by grouping the left parts together, so that: 'P ∧ Q ∧ R' is an abbreviation for '(P∧Q) ∧ R'.

Informal Conventions Outermost parentheses may be omitted. Conjunction signs or disjunction signs may be used with conditional signs or biconditional signs with the understanding that this is short for a conditional or biconditional which has a conjunction or disjunction as a part. For example: P∨Q → R is informal notation for (P∨Q) → R P ↔ Q∧R is informal notation for P ↔ (Q∧R) Repeated conjuncts or disjuncts without parentheses are short for the result of putting parentheses around the part to the left of the last conjunction or disjunction sign. For example: P∨Q∨R is informal notation for (P∨Q) ∨ R P∧Q∧R is informal notation for (P∧Q) ∧ R

Sentences with the new connectives may be parsed as we did in the previous chapter: P∧Q → R 2 P∧Q R 2 P Q

P ↔ Q∨R 2 P Q∨R 2 Q R

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CHAPTER 2 SECTION 1

~(P∧Q) → (R ↔ P∨Q) 2 ~(P∧Q) R ↔ P∨R | 2 P∧Q R P∨R 2 2 P Q P R

~~(R ↔ (P → ~Q)) | ~(R ↔ (P → ~Q)) | R ↔ (P → ~Q) 2 R P→~Q 2 P ~Q | Q

Determining Truth Values Using such parsings, there is a mechanical way to determine whether any given sentence is true or false if you know the truth values of the sentence letters making it up. First, make a parse tree as above by taking the sentences on any given line and writing their immediate parts below them. A parse tree for '(P∧Q) → (P∨R)' is: (P ∧ Q) → (P ∨ R) 2 (P ∧ Q) (P ∨ R) 2 2 P Q P R Then write the truth values of the sentence letters below them. For example, if P and Q are both true but R false, you would have: (P ∧ Q) → (P ∨ R) 2 (P ∧ Q) (P ∨ R) 2 2 P Q P R T T T F Then go up the parse tree, placing a truth value under the major connective of each sentence based on the truth values of its parts given below. For example, the truth value under '(P ∧ Q)' would be 'T' because it is a conjunction, and both of its parts are T:

(P ∧ Q) → (P ∨ R) 2 (P ∧ Q) (P ∨ R) T 2 2 P Q P R T T T F

Filling in the remaining parts gives you a truth value for the whole sentence at the top:

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CHAPTER 2 SECTION 1

( (P ∧ Q) → (P ∨ R) ) T 2 (P ∧ Q) (P ∨ R) T T 2 2 P Q P R T T T F Sometimes not all of the parse tree needs to be filled out; this happens when partial information below a sentence is sufficient to decide its truth value. In the example just given it is not necessary to figure out the truth value of '(P ∧ Q)', since the conditional on the top line is determined to be true based on the information that '(P ∨ R)' is true. So the following parse tree is sufficient to show that the main sentence is true if the sentence letters have the indicated truth values: (P ∧ Q) → (P ∨ R) T 2 (P ∧ Q) (P ∨ R) T 2 2 P Q P R T F EXERCISES 1. For each of the following state whether it is a sentence in official notation, or a sentence in informal notation, or not a sentence at all. If it is a sentence, parse it as indicated above. a. b. c. d. e. f. g. h. i.

P↔Q→R ~Q↔~R ~(Q↔R) P∧Q∨R (P→Q) ∨ (R→~Q) P ↔ (Q∧R) → Q P∧Q → (Q→R∨Q) P ↔ (P↔Q∧R) P ∨ (Q→P)

2. If 'P' and 'Q' are both true and 'R' is false, what are the truth values of the official or informal sentences in 1? (Use the parses that you give in 1 to guide the determination of truth values.)

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CHAPTER 2 SECTION 2

2 ENGLISH EQUIVALENTS OF THE CONNECTIVES Conjunctions: The word 'and' is equivalent to the symbol '∧'. There are other locutions of English that may also be equivalent to '∧', although they are sometimes used to communicate something additional. For example: The book is short, and it is interesting The book is short, but it is interesting The book is short, although it is interesting The book is short, even though it is interesting The book is short; it is interesting Some of these sentences suggest that if a book is short, you probably won't find it interesting. But all that they literally say is that it is both short and interesting. If it isn't short, what you have said is false, and if it isn't interesting then what you have said is false, but if it is both short and interesting, what you have said is true, even if possibly misleading. Conjunctions: □ ∧ ○ □ and ○ both □ and ○ □ but ○ □ although ○ although □, ○ □ even though ○ even though □, ○ □;○ In certain cases, use of a relative pronoun is logically equivalent to a use of '∧': the sentence 'Maria, who was late, greeted the vice-counsel' is equivalent to 'Maria was late ∧ Maria greeted the vice-counsel'. Disjunctions: The English word 'or' can be taken in two ways: inclusively or exclusively. If you are asked to contribute food or money, you will probably take this as saying that you may contribute either or both; the invitation is inclusive. But if a menu says that you may have soup or salad the normal interpretation is that you may have either, but not both; the offer is exclusive. The difference in logical import appears in the first row here: □ T T F F

○ T F T F

(□ inclusive-or ○) T T T F

(□ exclusive-or ○) F T T F

If the English 'or' can be read either inclusively or exclusively, we will need to have a convention for how to interpret it when it is used in exercises. Our convention will be that 'or' is always meant inclusively when it is used in problems and examples in this text. That is, it coincides in logical import with our disjunction sign '∨'. A common synonym of 'or' is 'unless'. The sentence 'Wilma will leave unless there is food' is false if there is no food but Wilma doesn’t leave; otherwise it is true, just like 'or' when read inclusively. Disjunctions: □ ∨ ○ □ or ○ either □ or ○ □ unless ○ Biconditionals: We will see below that a biconditional sign is equivalent to two conditionals made from its constituents. The sentence '(□ ↔ ○)' is equivalent to:

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CHAPTER 2 SECTION 2

(□ → ○) ∧ (○ → □) This can be read in English as '○ if □, and ○ only if □'; thus it is often pronounced 'if and only if'. The English phrase 'just in case' or 'exactly in case' are sometimes used to state the equivalence of two claims; the biconditional can be used to symbolize them: The game will be called off just in case it rains: The game will be played exactly in case it is sunny:

Q↔R P↔S

Biconditionals: □ ↔ ○ □ if and only if ○ □ exactly on condition that ○ □ just in case ○

EXERCISES 1. For each of the following sentences say which symbolic sentence it is equivalent to. a. It will rain, but the game will be played anyway. R∧P R→P R↔P b. Willa drove or got a ride W∨R W↔R c. Robert, who didn't get a ride, was tardy ~R → T ~R ∧ T d. It rained; the sell-a-thon was called off R↔S R∧S e. The quilting bee will be called off just in case it rains Q∧R Q↔R Q→R R→Q 2. Symbolize each of the following using this translation scheme: S V R Q T a. b. c. d. e.

Sally will walk Veronica will give Sally a ride It will rain Barbara will come with Quincy Barbara will come with Tom

Sally will walk or Veronica will give her a ride. Exactly on condition that it rains will Sally walk Although it will rain, Sally will walk Barbara will come with Quincy or Tom Barbara will come with Quincy; Sally will walk

3. What are the truth values of the sentences in 2 when all of the simple sentences are false?

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CHAPTER 2 SECTION 3

3 COMPLEX SENTENCES Complex sentences of English generally translate into complex sentences of the logical notation. As usual, it is important to be clear about the grouping of clauses in the English sentence. The following sentence is a simple conjunction: P∧Q

Polk and Quincy were presidents

The following sentence is also a conjunction, one of whose conjuncts is a negation: Polk, but not Quincy, was a president.

P ∧ ~Q

This is a negation of a conjunction: Not both Polk and Quincy were presidents.

~(P ∧ Q)

This is a simple disjunction: Either Polk or Quincy was president.

P∨Q

This is a complex sentence, with at least two different but equivalent symbolizations. Neither Polk nor Quincy was president. One symbolization is the negation of 'Either Polk or Quincy was president; in this symbolization 'neither' means 'not either': ~(P ∨ Q). An equivalent symbolization is a conjunction of negations; 'neither P nor Q' is equivalent to "not P and not Q": ~P ∧ ~Q The fundamental principles for our new connectives are:

and, or, if and only if When any of these expressions occurs between sentences, it gives rise to a conjunction, disjunction, or biconditional. The constituents of the conjunction, disjunction, or biconditional are symbolizations of sentences immediately to the left and to the right of 'and', 'or', or 'if and only if'. When 'either' occurs with 'or', the symbolization of the expression enclosed between 'either' and 'or' is a disjunct. Likewise, When 'both' occurs with 'and', the symbolization of the expression enclosed between 'both' and 'and' is a conjunct. 'neither □ nor ○' is equivalent to 'not (either □ or ○)'. As in chapter 1, these principles do not eliminate all ambiguity. The sentence 'Wilma will leave and Steve will stay or Tom will dance' is ambiguous between these two symbolizations: W & (S∨T) (W&S) ∨ T The use of 'either' will sometimes disambiguate; the only symbolization of 'Wilma will leave and either Steve will stay or Tom will dance' is: W & (S∨T) This is because 'either' and 'or' exactly enclose 'Steve will stay', and so 'S' must be a disjunct. But it is not a disjunct in '(W&S) ∨ T'. Commas play their usual role of grouping items on each side. The sentence 'Wilma will leave and Steve will stay, or Tom will dance' has only the symbolization:

Chapter 2 -- 7

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CHAPTER 2 SECTION 3

(W&S) ∨ T Conjunction and disjunction signs inside of sentences: Sometimes 'and' and 'or' occur within sentences, as in: Wilma sang and danced Tom or Sam left In such cases you need to fill in a missing part to get a sentence that we already know how to symbolize. Sometimes 'and' or 'or' occurs inside a simple sentence, where only the subject is conjoined or disjoined, and there is a single predicate, or only the predicate is conjoined or disjoined, and there is a single subject. If you fill in a copy of the shared part, you will get a synonymous sentence that we already know how to symbolize. These are some examples: Wilma sang and danced Tom or Sam left

Wilma sang and [Wilma] danced Tom [left] or Sam left

If there is a 'both' or an 'either, it ends up on the front: Both Tom and Sam left Either Tom or Sam left

Both Tom [left] and Sam left Either Tom [left] or Sam left

Wilma both sang and danced Wilma either sang or danced

Both Wilma sang and [Wilma] danced Either Wilma sang or [Wilma] danced.

There may also be a 'not' after the compound subject, or before a compound predicate. If the negation is after a compound subject, it forms part of the predicate, and it is filled in with that predicate: Wilma or Veronica didn't sing

Wilma [didn't sing] or Veronica didn't sing.

If the negation is before a compound predicate, it yields a negation sign that applies to the whole compound: Wilma didn't sing or dance Wilma didn't sing and dance

~ (Wilma sang or danced) ~ (Wilma sang and danced)

The parts inside the parentheses are then expanded as usual: Wilma didn't sing or dance Wilma didn't sing and dance

~ (Wilma sang or [Wilma] danced) ~ (Wilma sang and [Wilma] danced)

Compounds within simple sentences affect how sentences are grouped after symbolization: When connectives occur inside otherwise simple sentences, the symbolizations of the sentences form a unit. For example, the sentence 'Ruth tap-dances or sings and she plays the clarinet' must be grouped like this: (T ∨ S) & P This is because the disjunction with 'T' and 'S' must be a unit. In 'Ruth tap-dances or she sings and plays the clarinet' the opposite happens; you must have: T ∨ (S & P) because the conjunction with 'S' and 'P' must form a unit.

Chapter 2 -- 8

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CHAPTER 2 SECTION 3

Synonyms of 'and', 'or', and 'if and only it' are subject to the conditions described above. Here are some illustrations: If neither Wilma nor Sally attends, either Robert or Peter will be bored. If neither Wilma [attends] nor Sally attends, either Robert [will be bored] or Peter will be bored. If neither W nor S, either R or P ~(W∨S) → (R∨P) The 'neither' and the 'either' made units, and the comma was redundant. A slightly more complex case: If neither Wilma nor Sally attends, either Robert or Peter, but not Tom, will be bored. If neither Wilma [attends] nor Sally attends, either Robert [will be bored] or Peter [will be bored], but Tom will not be bored. If neither W nor S, either R or P, but not T ~(W∨S) → (R∨P) & ~T Here the 'either' made a unit, so the 'but ' could not take 'Peter would be bored' as its left conjunct. That is, the symbolization could not be this: ~(W∨S) → (R ∨ (P&~T)) Likewise, the original phrase 'either Robert or Peter, but not Tom, will be bored' consists of three simple sentences all sharing the 'will be bored', so it could not be pulled apart, as in: (~(W∨S) → (R∨P)) & ~T Some additional examples: Either Robert or Tom will attend, but not both Either Robert [will attend] or Tom will attend, but not both [will attend] Either R or T, but not R and T (R∨T) ∧~(R∧T) Robert will attend if Sally does, but she won't attend if neither Tom nor Wilma attend. Robert will attend if Sally does [attend], but she won't attend if neither Tom [attends] nor Wilma attends. R if S, but not S if neither T nor W (S→R) ∧(~(T∨W) → ~S) Neither Sally nor Robert will run, but if either Tom or Quincy run, Veronica will win. Neither S nor R, but if either T or Q, V ~(S∨R) ∧(T∨Q → V). Given that Sally and Robert won't both run, Tom will run exactly if Q does. Given that not both S and R, T exactly if Q. ~(S∧R) → (T↔Q) A variety of English expressions that we have not mentioned affect how a sentence is to be symbolized. Examples: Quincy will whistle if Reggie sings without Susan singing or Susan sings without Reggie, but he won't whistle if they both sing

Chapter 2 -- 9

Version of Aug 2013

CHAPTER 2 SECTION 3

Q if R and not S or S and not R, but not Q if S and R ((R∧~S) ∨ (S∧~R) → Q) ∧(S∧R → ~Q) Here 'Reggie sings without Susan singing' means that Reggie sings and Susan doesn't sing. If Sally runs, Rob will run, in which case Theodore will leave (S→R) ∧ (R→T) Here ' in which case' means "if Rob runs". If a symbolization of a sentence is a correct one, then it and the English sentence being symbolized must agree in truth value no matter what truth values the simple sentences have. If they agree for every assignment of truth values, then the symbolization is correct. If not, it is incorrect. (To tell whether an English sentence is true or false given a specification of truth values for its simple parts you must rely on your understanding of English. To tell whether a symbolic sentence is true or false given the truth values of its sentential letters, you parse it and figure out its truth value as in section 1.)

EXERCISES 1. If 'P' is true and both 'Q' and 'R' are false, what are the truth values of the following? (In answering, give a parse tree for the sentence.) a. b. c. d. e.

~(P ∨ (Q ∧ R)) ~P ∨ (Q ∧ R) ~(P ∨ R) ↔ ~P ∨ R ~Q ∧ (P ∨ (Q↔R)) P → (~Q ↔ (~R → Q))

For questions 2 and 3, use this translation scheme: V Veronica will leave W William will leave Y Yolanda will leave 2. For each of the following say which of the proposed translations is correct. a. Veronica won't leave if and only if William won’t leave ~(V ↔ ~W) ~V ↔ ~W V ↔ ~~W b. William and Veronica will both leave if Yolanda does, provided that Veronica doesn’t Y∧~V → W∧V (Y→W∧V) → ~V ~V → (Y→W∧V) c. Unless Yolanda leaves, Veronica or William will leave Y ∨ (W ∨ V) Y → (W ∨ V) Y↔W∧V d. Either Yolanda leaves and Veronica doesn't, or Veronica leaves and William doesn’t (Y ↔ ~V) ∨ (V ↔ ~W) (Y ∧ ~V) ∨ (V ∧ ~W) Y ∧ ~V ↔ V ∧ ~W

Chapter 2 -- 10

Version of Aug 2013

CHAPTER 2 SECTION 3

3. For each of the following produce a correct symbolization a. Only if Veronica doesn't leave will William leave, or Veronica and William and Yolanda will all leave b. If neither William nor Veronica leaves, Yolanda won't either c. If William will leave if Veronica leaves, then he will surely leave if Yolanda leaves d. Neither William nor Veronica nor Yolanda will leave 4. What are the truth values of 3a-d if Veronica leaves but neither William nor Yolanda leaves? For question 5 use this translation scheme: R W Q

Sally will run Sally will win Sally will quit

5. For each of the following produce a correct symbolization a. Sally will run and win unless she quits b. Sally will win exactly in case she runs without quitting c. Sally, who will run, will win if she doesn't quit d. Sally will run and quit, but she will win anyway

Chapter 2 -- 11

Version of Aug 2013

CHAPTER 2 SECTION 4

4 RULES Each new connective comes with two new rules. As earlier, it should be obvious from the truth-table descriptions of each connective that instances of these rules are formally valid arguments. Conjunction rules: Rule s (simplification) □∧○ ∴ □

or

□∧○ ∴○

□ ○ ∴ □∧○

or

Rule mtp (modus tollendo ponens) □ ∴○∨□

□∨○ ~○ ∴ □

Biconditional rules: Rule bc (biconditional-to-conditional) □↔○ ∴ □→○

or

or

□∨○ ~□ ∴○

Rule cb (conditionals-to-biconditional)

□↔○ ∴○→□

□→○ ○→□ ∴ □↔○

Simplification indicates that if you have a conjunction, you may infer either conjunct. For example, both of these valid arguments are instances of rule s: P∧W ∴ P by rule s ∴ W by rule s

Polk was a president and so was Whitney

∴ Polk was a president ∴ Whitney was a president

Adjunction indicates that if you have any two sentences, you may infer their conjunction, in either order. For example, these valid arguments are instances of rule adj: Polk was a president Whitney was a president ∴ Polk was a president and so was Whitney ∴Whitney was a president and so was Polk

P W ∴ P∧W ∴ W∧P

This derivation illustrates how the conjunction rules are used: P∧Q ∴ Q∧P 1. Show Q ∧ P 2. 3. 4.

Q P Q∧P

pr1 s pr1 s 2 3 adj dd

Addition indicates that from any sentence you may infer its disjunction with any other sentence. Polk was a president ∴ Polk was a president or Whitney was ∴ Whitney was a president or Polk was

P ∴ P∨W ∴ W∨P

Chapter 2 -- 12

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CHAPTER 2 SECTION 4

Rule add lets you add any disjunct, no matter how irrelevant. So from 'Cynthia left' you may infer 'Cynthia left ∨ Fido barked'. This is legitimate because '∨' is used inclusively, and all that you need for a disjunction to be true is that either disjunct be true. So if 'Cynthia left' is true, 'Cynthia left ∨ Fido barked' must be true too. Modus tollendo ponens indicates that from a disjunction and the negation of one of its disjuncts you may infer the other disjunct. Polk was a president or Whitney was Whitney wasn't a president ∴ Polk was a president

P∨W ~W ∴ P

by rule mtp

Polk was a president or Whitney was Polk wasn't a president ∴ Whitney was a president

P∨W ~P ∴ W

by rule mtp

Note that the following is not an instance of modus tollendo ponens: Whitney was a president or Truman was Truman was a president ∴ Whitney wasn't a president

W∨T T ∴ ~W

For mtp you need the negation of a disjunct. In the case given, if 'T' and 'W' were both true, then the argument would have true premises and a false conclusion. Here is a derivation illustrating the disjunction rules. It is a derivation for this argument: P R ∨ ~P ∴ R∨S 1. Show R ∨ S 2. 3. 4.

~~P R R∨S

pr1 dn 2 pr2 mtp 3 add dd

Biconditional-to-conditional indicates that from a biconditional you may infer either of the corresponding conditionals: Polk was a president if and only if Whitney was ∴ If Polk was a president, so was Whitney ∴ If Whitney was a president, so was Polk

P↔W ∴ P→W ∴ W→P

by rule bc by rule bc

Conditionals-to-biconditional indicates that from two conditionals where the antecedent of one is the consequent of the other, and vice versa, you may infer a bicondtional containing the parts of the conditionals: If Polk was a president, so was Whitney If Whitney was a president, so was Polk ∴ Polk was a president if and only if Whitney was

P→W W→P ∴ P↔W

by rule cb

Here are two more derivations using our new rules: S∧P P∨Q → ~R Q∨R ∴ Q

Chapter 2 -- 13

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CHAPTER 2 SECTION 4

1. Show Q 2. 3. 4. 5.

P P∨Q ~R Q

pr1 s 2 add 3 pr2 mp 4 pr3 mtp dd

R ↔ ~P ~Q ↔ R ∴ P↔Q 1. Show P ↔ Q 2.

Show P → Q

3. 4. 5. 6. 7. 8. 9.

P ~~P R → ~P ~R ~Q → R ~~Q Q

10.

Show Q → P

11. 12. 13. 14. 15. 16. 17.

Q ~~Q R → ~Q ~R ~P → R ~~P P

18.

P↔Q

ass cd 3 dn pr1 bc 4 5 mt pr2 bc 6 7 mt 8 dn cd ass cd 11 dn pr2 bc 12 13 mt pr1 bc 14 15 mt 16 dn cd 2 10 cb dd

EXERCISES 1. For each of the following arguments, say which rule it is an instance of (or say "none"). a.

P ∨ ~Q Q ∴ P

b.

~P ∧ Q ∴ ~P

c.

~~(P→Q) ∴ P→Q

d.

~P∨Q ~Q ∴ ~P

e.

~P → ~Q ~Q → ~P ∴ ~Q ↔ ~P

f.

P∨Q ~R ∴ P

g.

~~P ↔ R ∴ R → ~~P

h.

Q ∴ ~P ∨ Q

i.

P∨Q ∴ Q

2. Given the sentences below, say what can be inferred in one step by s, mtp, bc, cb using all of the premises. a.

~W → ~X ~X → ~W ∴ ?

b.

~W ∨ ~X ~~X ∴ ?

c.

W→X ~W ∴ ?

d.

~W ∧ ~X ∴ ?

e.

W ↔ ~X ∴ ?

f.

W∨X ∴ ?

Chapter 2 -- 14

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CHAPTER 2 SECTION 5

5 SOME DERIVATIONS USING RULES S, ADJ, CB Since there are new connectives it is useful to expand our strategy hints from Chapter 1: Additional Strategy Hints When trying to derive a conjunction, derive the conjuncts and then use adj. When trying to derive a biconditional, derive the corresponding conditionals and use cb. These strategy hints will be put to use below, as we extend our list of Theorems from Chapter 1. Theorem 24 is the commutative law for conjunction; it says that turning the conjuncts of a sentence around produces a logically equivalent sentence: T24

P∧Q ↔ Q∧P

"commutative law for conjunction"

This is easy to derive if you follow the last two strategy hints. You will be deriving a biconditional, so you will try to derive both conditionals: P∧Q → Q∧P and Q∧P → P∧Q and then combine them using rule cb. While deriving each conditional you will derive a conjunction by deriving its conjuncts and then using rule adj. Rule s is used whenever you want to get one of the conjuncts of an existing conjunction alone. 1. Show P∧Q ↔ Q∧P 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Show P∧Q → Q∧P P∧Q P Q Q∧P

ass cd 3s 3s 4 5 adj cd

Show Q∧P → P∧Q Q∧P Q P P∧Q P∧Q ↔ Q∧P

ass cd 8s 8s 9 10 adj cd 2 7 cb dd

The next theorem is the associative law for conjunction; it says that regrouping successive conjuncts produces a logically equivalent sentence. The strategy here is the same as that above: to derive the biconditional you derive the corresponding conditionals, and use rule cb. In deriving the conditionals you derive conjunctions using rule adj. Again, rule s is used to simplify conjunctions that you already have.

Chapter 2 -- 15

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CHAPTER 2 SECTION 5

T25 P∧ (Q∧R) ↔ (P∧Q)∧ R

"associative law for conjunction"

1. Show P∧ (Q∧R) ↔ (P∧Q)∧ R 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

Show P∧ (Q∧R) → (P∧Q)∧ R P∧ (Q∧R) P Q∧R Q R P∧Q (P∧Q) ∧R

ass cd 3s 3s 5s 5s 4 6 adj 7 8 adj cd

Show (P∧Q)∧ R → P∧ (Q∧R) (P∧Q)∧ R R P∧Q P Q Q∧R P∧ (Q∧R) P∧ (Q∧R) ↔ (P∧Q)∧ R

ass cd 3s 3s 5s 5s 4 6 adj 7 8 adj cd 2 10 cb dd

The next theorem is T26: T26

(P→Q) ∧(Q→R) → (P→ R)

"hypothetical syllogism"

Notice that T26 and T4 from the previous chapter are both called "hypothetical syllogism". T4 (P→Q) → ((Q→R) → (P→R))

"hypothetical syllogism"

These two theorems are closely related. They are related to one another as the following two patterns: □∧○→△ □ → (○ → △) where each theorem has 'P→Q' in place of □, 'Q→R' in place of ○, and 'P→R' in place of △. Our next theorem says that these two patterns are equivalent. T27

(P∧Q → R) ↔ (P → (Q→R))

"exportation"

The derivation of this theorem is also relatively straightforward: derive two conditionals and put them together by rule cb. Each conditional itself has conditionals as parts, so the derivation calls for two conditional subderivations (one of which itself contains another conditional subderivation).

Chapter 2 -- 16

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CHAPTER 2 SECTION 5

1. Show (P∧Q → R) ↔ (P→ (Q→R)) 2.

Show (P∧Q → R) → (P→ (Q→R))

3. 4.

P∧Q → R Show P→ (Q→R)

ass cd

5. 6.

P Show Q → R

ass cd

7. 8. 9.

Q P∧Q R

ass cd 5 7 adj 3 8 mp cd

10.

6 cd

11.

4 cd

12. Show (P→ (Q→R) → (P∧Q → R) 13. 14.

P → (Q→R) Show P∧Q → R

ass cd

P∧Q P Q→R Q R

15. 16. 17. 18. 19.

ass cd 15 s 13 16 mp 15 s 17 18 mp cd

20.

14 cd

21. (P∧Q → R) ↔ (P→ (Q→R))

2 12 cb dd

This derivation is complex. It may be useful to see how we might think up how to construct it. First, our main strategy is to derive a biconditional by deriving two conditionals. So our plan predicts that the derivation will have this overall structure: 1. Show (P∧Q → R) ↔ (P→ (Q→R)) 2.

(P∧Q → R) → (P→ (Q→R))

12. (P→ (Q→R) → (P∧Q → R) 21. (P∧Q → R) ↔ (P→ (Q→R))

2 12 cb dd

Line 2 will require a conditional derivation, and so will line 12. So the completed derivation will take this form: 1. Show (P∧Q → R) ↔ (P→ (Q→R)) 2. 3.

Show (P∧Q → R) → (P→ (Q→R)) P∧Q → R

ass cd

P → (Q→R)

xxx cd

12. Show (P→ (Q→R) → (P∧Q → R) 13.

P → (Q→R)

ass cd

P∧Q → R

xxx cd

21. (P∧Q → R) ↔ (P→ (Q→R))

2 12 cb dd

Chapter 2 -- 17

Version of Aug 2013

CHAPTER 2 SECTION 5

Lines 3-11 and 14-20 are taken up with completing the subderivations. Each of these itself uses a conditional subderivation, giving the following structure: 1. Show (P∧Q → R) ↔ (P→ (Q→R)) 2.

Show (P∧Q → R) → (P→ (Q→R))

3. 4.

P∧Q → R Show P→ (Q→R)

5.

ass cd

P

ass cd

10.

xxx cd

11.

4 cd

12. Show (P→ (Q→R) → (P∧Q → R) 13. 14.

P → (Q→R) Show P∧Q → R

15.

ass cd

P∧Q

ass cd

19.

xxx cd

20.

14 cd

21. (P∧Q → R) ↔ (P→ (Q→R))

2 12 cb dd

The rest of the work is filling in the remaining subderivations. It is often useful to develop a derivation as we did here by first sketching its overall structure, and then flesh it out with details afterwards.

EXERCISES 1. Produce derivations for theorems T28-T30, T33, T36-37, which are included among the theorems stated here: T28

(P∧Q → R) ↔ (P∧~R → ~Q)

T29

(P→ Q∧R) ↔ (P→Q) ∧(P→R)

T30

(P→Q) → (R∧P → R∧Q)

T31

(P→Q) → (P∧R → Q∧R)

T32

(P→R) ∧(Q→S) → (P∧Q → R∧S)

"Leibniz's praeclarum theorema"

T33

(P→Q) ∧(~P→Q) → Q

"separation of cases; constructive dilemma"

T34

(P→Q) ∧(P→~Q) → ~P

T35

(~P→R) ∧(Q→R) ↔ ((P→Q) → R)

T36

~(P ∧ ~P)

T37

(P→Q) ↔ ~(P ∧ ~Q)

"distribution of → over ∧"

Chapter 2 -- 18

Version of Aug 2013

CHAPTER 2 SECTION 6

6 ABBREVIATING DERIVATIONS It is useful in writing derivations to be able to combine two or more steps into one. For example, here is a derivation in which double negation is used twice: P ~Q → ~P Q→R ∴ R 1. Show R 2. ~~P 3. ~~Q 4. Q 5. R

pr1 dn 2 pr2 mt 3 dn 4 pr3 mp dd

One can shorten this derivation by two steps by combining the double negations with other rules, like this: 1. Show R 2. ~~Q 3. R

pr1 dn pr2 mt 2 dn pr3 mp dd

The meanings of the notations at the end of the lines are: 2. " pr1 dn pr2 mt "

take pr1 and double negate it; then combine the result with pr2 by mt to get ~~Q

3. "2 dn pr3 mp "

double (un)negate the sentence on line 2; then combine the result with pr3 using modus ponens

Here is a more highly abbreviated derivation. P∧Q R → ~Q S∨~R → T ∴ T∧P 1. Show T ∧ P 2.

~R

pr1 s dn pr2 mt

simplify pr1 to get Q, then double negate Q to get ~~Q; use mt on this and pr2 to get ~R

3.

T

apply add to the sentence on line 2 to get S ∨ ~R; then apply mp to that and pr3 to get T

4.

T∧P

simplify pr1 to get P and then adjoin this with the sentence on line 3 to get T∧P.

dd

Abbreviations of this sort may always be interpreted by the following "decoding procedure", starting at the left and moving right: A line number or premise number gives you a sentence -- the sentence on that line. Rule r also gives you a sentence -- the sentence on the line cited. A sentence followed by 'dn', 's', 'add' or 'bc' gives you the result of applying that rule to that sentence. (The old sentence is no longer available for further use.) Two sentences followed by 'mp', 'mt', 'adj', 'cb' give you the result of applying that rule to them.

Chapter 2 -- 19

Version of Aug 2013

CHAPTER 2 SECTION 6

If you can apply this decoding and end up with the sentence on the line which has the abbreviations at its end, the line is correct. If you can't, the line is not correct. (There is sometimes more than one way to apply a rule to a sentence, so there may be many ways to use the decoding process. If at least one way of using it ends you up with the sentence on the line, the abbreviation is correct; otherwise it is incorrect.) Applied to the abbreviations on lines 2, 3 and 4 above the decoding looks like this. We work from the left. First, the leftmost 'pr1' is replaced by the first premise: 2.

pr1

s

dn

pr2

mt

2.

P∧Q

s

dn

pr2

mt

dn

pr2

mt

Then rule s acts on this to produce 'Q': 2.

P∧Q

2.

s

Q

dn

pr2

mt

pr2

mt

pr2

mt

Then double negation turns 'Q' into '~~Q': 2.

Q

dn

2.

~~Q

Then pr2 is replaced by the second premise: 2.

~~Q

pr2

mt

2.

~~Q

R → ~Q

mt

Finally, rule mt acts on '~~Q' and 'R → ~Q' to give you '~R', which is the sentence that actually appears on line 2: 2.

~~Q

pr2

2.

mt

~R

Line 3 is decoded by the same process: 3.

2

pr3

mp

3.

~R

pr3

mp

3.

S ∨ ~R

pr3

mp

3.

S ∨ ~R

S∨~R → T

mp

3.

T

Chapter 2 -- 20

Version of Aug 2013

CHAPTER 2 SECTION 6

Likewise for line 4: 4.

pr1

s

3

4.

P∧Q

s

3

4.

P

3

4.

P

T

4.

T∧P

A long string of abbreviations can be difficult to decode, so we will confine ourselves to simple cases.

EXERCISES 1. Use the method of abbreviating derivations to produce shortened derivations for T38, T40-43. T38

P∧Q ↔ ~(P → ~Q)

T39

~(P∧Q) ↔ (P → ~Q)

T40

~(P → Q) ↔ P∧~Q

"negation of conditional"

T41

P ↔ P∧P

"idempotence for ∧"

T42

P∧~Q → ~(P→Q)

"negation of conditional"

T43

~P → ~(P∧Q)

T44

~Q → ~(P∧Q)

Chapter 2 -- 21

Version of Aug 2013

CHAPTER 2 SECTION 7

7 USING THEOREMS AS RULES In Chapter 1 we learned a way to introduce instances of previously derived theorems into a derivation. Theorems are even more useful when they are used to justify rules. The fundamental principle is that if a theorem has been derived that has the form of a conditional, it can be cited as a rule which allows you to infer one sentence from another whenever the conditional made from those sentences is an instance of the theorem. A theorem of conditional form: ∴ □→○ justifies a rule of this form: □ ∴ ○ For example, T13 ("transposition") is (P→Q) → (~Q→~P). This validates the rule: □→○ ∴ ~○ → ~□ We name such a rule by writing 'R' in front of the name of the theorem being used. An example of a use of a theorem as a rule is: 8. S → T 9. ~T → ~S

8 RT13

Here are two arguments, and derivations, that use some theorems from Chapter 1 as rules. ~(Q∧~R) → P P→Q R → ~P ∴ ~(Q → R) 1. Show ~(Q → R) 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Q→R (Q→R) → (P→R) P→R (R→~P) → (P→~P) P → ~P ~P ~~(Q∧~R) Q ∧ ~R Q R ~R

ass id pr2 RT4 2 3 mp 4 RT4 pr3 5 mp 6 RT20 pr1 7 mt 8 dn 9s 2 10 mp 9 s 11 id

T4 is (P→Q) → ((Q→R) → (P→R))

T20 is (P→~P) → ~P

S→T T → (Q → P) S→Q ∴ S→P 1. 2. 3. 4.

Show S → P S → (Q→P) (S→Q) → (S→P) S→P

pr1 pr2 RT4 2 RT6 pr3 3 mp dd

Chapter 2 -- 22

T4 is (P→Q) → ((Q→R) → (P→R)) T6 is (P→(Q→R)) → ((P→Q) → (P→R))

Version of Aug 2013

CHAPTER 2 SECTION 7

Theorems can be used to make rules in two more ways. One way applies when a theorem is a biconditional. Since a biconditional is logically equivalent to two conditionals, it makes sense to use the theorem as if it were two conditionals.

A theorem of biconditional form: ∴□↔○ validates both of these rules: □ ∴ ○

○ ∴ □

An example is T27: '(P∧Q→R) ↔ (P→(Q→R))' which validates both of these: □∧○ → △ ∴ □ → (○→△)

RT27

RT27

□ → (○→△) ∴ □∧○ → △

A final additional way to use theorems as rules is possible when a theorem is a conditional whose antecedent is a conjunction. This gives a rule which has multiple premises.

A theorem of this form: ∴ □∧○ → △ justifies a rule of this form: □ ○ ∴ △ An example is T26:

(P→Q) ∧(Q→R) → (P→R)

which validates: RT26

□→○ ○→△ ∴ □→△

These options combine, so that if one side of a biconditional is a conjunction, it validates a rule with multiple premises. For example, T38 (below) is 'P∧Q ↔ ~(P→~Q)', and one of the rules that it validates is: RT38 □ ○ ∴ ~(□→~○)

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EXERCISES 1. For each of the following derivations, determine which lines are correct and which incorrect. (In assessing a line, assume that previous lines are correct.) ∴ ((U→V) → S) → (~S → U) 1. Show ((U→V) → S) → (~S → U) 2. 3. 4. 5. 6. 7.

(U→V) → S Show ~S → U ~S ~S → ~(U→V) ~(U → V) U

8.

ass cd ass cd 2 RT13 4 5 mp 6 RT21 cd

T13 is (P→Q) → (~Q → ~P) T21 is ~(P→Q) → P

3 cd

∴ ~V ∧ (W → V∧U) → (~W → ~S) 1. Show ~V ∧ (W → V∧U) → (~W → ~S) 2. 3. 4. 5. 6. 7. 8.

~V ∧ (W → V∧U) ~V ~(V∧U) W → V∧U ~W ~(W∧S) ~W → ~S

ass cd 2s 3 RT43 2s 4 5 mt 6 RT43 7 RT39 cd

T43 is ~P → ~(P∧Q)

T39 is ~(P∧Q) ↔ (P→~Q)

Some more theorems: T45 P∨Q ↔ (~P→Q) T46 (P→Q) ↔ ~P ∨ Q

"definition of → in terms of ∨"

T47 P ↔ P∨P

"idempotence for ∨"

T48 (P∨Q) ∧(P→R) ∧(Q→S) → R∨S T49 (P∨Q) ∧(P→R) ∧(Q→R) → R

"separation of cases"

T50 (P→R) ∧(Q→R) ↔ (P∨Q→R) 2. Construct a derivation for T45, and then use RT45 to derive T46 3. Construct derivations for T47 and T48, and construct a derivation for T49 using RT47 and RT48 4. Use RT49 in constructing a derivation for T50. 5. Derive T53. Some additional theorems are given here for reference. T51

(P∨Q) ∧(P→R) ∧(~P∧Q→R) → R

T52

(P→R) ∧(~P∧Q→R) ↔ (P∨Q→R)

T53

P∨Q ↔ Q∨P

"commutative law for ∨"

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T54

P ∨ (Q∨R) ↔ (P∨Q) ∨ R

"associative law for ∨"

T55

(P → Q∨R) ↔ (P→Q) ∨ (P→R)

"distribution of → over ∨"

T56

(P→Q) → (R∨P → R∨Q)

T57

(P→Q) → (P∨R → Q∨R)

T58

(P→Q) ∨ (Q→R)

T59

P ∨ ~P

T60

(P→R) ∨ (Q→R) ↔ (P∧Q → R)

T61

P∧ (Q∨R) ↔ (P∧Q) ∨ (P∧R)

"distribution"

T62

P∨(Q∧R) ↔ (P∨Q) ∧(P∨R)

"distribution"

"excluded middle"

8 DERIVED RULES We now have over fifty theorems that can be used as rules, with more to come. There are too many of these to remember easily. It is customary to isolate a small number of rules based on the theorems and give them special names, and use these rules frequently in derivations. In this section we look at five of these. Rule nc (negation of conditional): 'Negation of conditional' is an easy-to-remember name for rule RT40 (which is "~(P→Q) ↔ P∧~Q"). It applies in either of these forms: Rule nc

□ ∧ ~○ ∴ ~(□→○)

~(□→○) ∴ □ ∧ ~○

This rule is often useful when you are trying to derive a conditional when a conditional derivation isn't working for you. Instead of assuming the antecedent of the conditional in order to use cd, assume the negation of the conditional for an indirect derivation. Then turn this negated conditional into a conjunction of the antecedent with the negation of its consequent. This gives you a lot to work with in continuing the derivation. As an example, suppose you are trying to validate this argument: R→Q R∨S S→R ∴ P→Q You begin the derivation: 1. Show P → Q You may now assume 'P' for purposes of doing a conditional derivation. But 'P' does not occur among the premises, and you may not see how to proceed. Instead of trying a conditional derivation, begin an indirect derivation: 1. Show P → Q 2.

~(P→Q)

ass id

Then apply the derived rule nc:

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1. Show P → Q 2. ~(P→Q) 3. P ∧ ~Q

ass id 2 nc

It is now fairly easy to simplify off '~Q', and use it to derive a contradiction: 1. Show P → Q 2. ~(P→Q) 3. P ∧ ~Q 4. ~Q 5. ~R 6. S 7. R

ass id 2 nc 3s 4 pr1 mt 5 pr2 mtp 6 pr3 mp

So you can finish the indirect derivation: 1. Show P → Q 2. 3. 4. 5. 6. 7.

~(P→Q) P ∧ ~Q ~Q ~R S R

ass id 2 nc 3s 4 pr1 mt 5 pr2 mtp 6 pr3 mp 5 id

Rule cdj (conditional as disjunction): This rule is constituted by T45 and T46, which together assert the equivalence of a conditional with a disjunction whose left disjunct is the negation (or unnegation) of the antecedent of the conditional and whose right disjunct is the consequent. Rule cdj has four cases: Rule cdj □→○ ∴ ~□ ∨ ○

~□ ∨ ○ ∴ □→○

~□ → ○ ∴ □∨○

□∨○ ∴ ~□ → ○

This rule can be useful when attempting to derive a disjunction. Instead of deriving the disjunction directly, derive the conditional whose antecedent is the (un)negation of the left disjunct and whose consequent is the other disjunct. This can usually be done using a conditional derivation. Then turn the result of the conditional derivation into the disjunction you are after using derived rule cdj. Here is a derivation of T54 using cdj together with T53, which was derived in the exercises for the last section. The overall structure of the derivation is to derive the biconditional by using two conditional derivations to get the corresponding conditionals, and then use rule cb.

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T54

P ∨ (Q∨R) ↔ (P∨Q) ∨ R 1. Show P ∨ (Q∨R) ↔ (P∨Q) ∨ R Show P ∨ (Q∨R) → (P∨Q) ∨ R

2.

P ∨ (Q∨R) Show ~R → (P∨Q)

3. 4. 5. 6.

~R Show ~P → Q

7. 8. 9.

~P Q∨R Q

10.

P∨Q

11. 12.

R ∨ (P∨Q) (P∨Q) ∨ R

ass cd ass cd ass cd 3 7 mtp 5 8 mtp cd 6 cdj cd

 cdj

4 cdj 11 RT53 cd

 cdj

13. Show (P∨Q) ∨ R → P ∨ (Q∨R) LIKE LINES 3-12 24.

P ∨ (Q∨R) ↔ (P∨Q) ∨ R

 cdj twice

2 13 cb

Rule sc (separation of cases): This is a combination of RT33 and RT49. It validates these inferences: Rule sc

□∨○ □→△ ○→△ ∴ △

□→△ ~□ → △ ∴△

The first form of rule sc (on the left) says that if you are given that at least one of two cases hold (the first premise), and if each of them imply something (the second and third premises), then you can conclude that thing. The second form of rule sc (on the right) applies when one of the two cases is the negation of the other. Then their disjunction (P ∨ ~P) is logically true, and needn't be stated as an additional premise. (See below for illustration.) Rule sc is especially useful when other attempts to produce a derivation have failed. For example, if you have a disjunction on an available line, then see if you can do two conditional derivations, each starting with one of the disjuncts, and each reasoning to the desired conclusion. If you can do this, the first form of sc applies. As an example, suppose you are given this argument: V∨W W → ~X ~U → X ∴ U∨V It may not be apparent how to proceed. So consider separation of cases. You have available a disjunction, 'V ∨ W', which is the first premise. If you can derive both V → U∨V and W → U∨V, the rule sc will give you the desired conclusion:

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1. Show U ∨ V 2. 3. 4. 5. 6. 7. 8. 9. 10.

Show V → U∨V V U∨V

Show W → U∨V W ~X U U∨V U∨V

ass cd 6 pr2 mp 7 pr3 mt dn 8 add cd pr1 2 5 sc dd

When you don't have a disjunction to work with, you may be able to use the second form of sc. Suppose you have this argument: R∧S → Q R→S ∴ R→Q In applying the second form of sc, you need to choose something which will serve as the antecedent for a conditional whose consequent is the desired conclusion, and whose negation will also serve as the antecedent for a conditional whose consequent is the desired conclusion. What should you choose? Often there is more than one choice that will work. In the case we are given, 'R' will work for this purpose. That is, you will indeed be able to derive both of these:

R → (R→Q) ~R → (R→Q) The second form of rule sc will then give you the desired conclusion: 1. Show R → Q 2. 3. 4. 5.

Show R → (R→Q) R S Q

 derive 'R → (R→Q)' ass cd 3 pr2 mp 3 4 adj pr1 mp cd

6.

Show ~R → (R → Q)

7. 8.

~R Show R → Q

9. 10.

R ~R

11. 12.

 derive '~R → (R→Q)'

ass cd ass cd 7 r id 8 cd

R→Q

2 6 sc

dd

 apply the second form of sc

Rule dm (DeMorgan's): This is a very useful rule. It lets you replace negations of conjunctions with modified disjunctions, and vice versa. It consists of any application of the rules based on theorems T63T66:

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T63

P ∧ Q ↔ ~(~P∨~Q)

T64

P ∨ Q ↔ ~(~P∧~Q)

T65

~(P∧Q) ↔ ~P ∨ ~Q

T66

~(P∨Q) ↔ ~P ∧ ~Q

So it allows any of the following inferences: Rule dm

□∧○ ∴ ~(~□∨~○) ∴

~(~□∨~○) □∧○

□∨○ ∴ ~(~□∧~○)

~(□∧○) ∴ ~□ ∨ ~○

~(□∨○) ∴ ~□ ∧ ~○

~(~□∧~○) □∨○

~□ ∨ ~○ ∴ ~(□∧○)

~□ ∧ ~○ ∴ ~(□∨○)

It may be easiest to remember these forms by remembering T63 and T64 in this form: A negation of a

conjunction disjunction

is equivalent to the

disjunction of the negations of its parts. conjunction

DeMorgan's rule can be handy when you are trying to derive a disjunction. To use it, you assume the negation of the disjunction for an indirect derivation. Rule dm lets you turn that negation into a conjunction, and then you have both conjuncts to use in deriving a contradiction. Example: P→U P∨Q Q→V ∴ U∨V 1. Show U ∨ V 2. 3. 4. 5. 6. 7.

~(U∨V) ~U ∧ ~V ~P Q V ~V

ass id 2 dm 3 s pr1 mt 4 pr2 mtp 5 pr3 mp 3 s 6 id

Rule nb (negation of biconditional): This rule is an application of T90: ~(P↔Q) ↔ (P↔~Q) The rule sanctions these inferences: Rule nb ~(□↔○) ∴ □ ↔ ~○

□ ↔ ~○ ∴ ~(□↔○)

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The first form is handy if you have the negation of a biconditional. The rule lets you infer a biconditional, which simplifies into two conditionals, which can be very useful. Here is an example: ~(P↔Q) ~Q ∴ P 1. Show P 2. 3. 4. 5.

~(P ↔ Q) P ↔ ~Q ~Q → P P

pr 2 nb 3 bc 4 pr2 mp

dd

The second form is handy if you want to derive the negation of a biconditional. Just derive the related biconditional, say by using conditional derivations to derive the associated conditionals. Example: P → (R↔Q) R → ~Q S→Q ~R → S ∴ ~P 1. Show ~P 2.

Show ~Q → R

3. 4. 5. 6.

~Q ~S ~~R R

7. 8. 9.

R ↔ ~Q ~(R↔Q) ~P

ass cd 3 pr3 mt 4 pr4 mt 5 dn cd pr2 2 cb 7 nb 8 pr1 mt

dd

EXERCISES 1. For each of the following derivations, determine which lines are correct and which incorrect. (In assessing a line, assume that previous lines are correct.) a. ∴

(U→S) → Q P∨R → S ~(T→Q) ~P

1. Show ~P 2. 3. 4. 5. 6. 7.

T ∧ ~Q ~(U→S) U ∧ ~S ~(P∨R) ~P ∧ ~R ~P

pr3 nc 2 s pr1 mt 3 nc 4 s pr2 mt 5 dm 6 s dd

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b.

~X ∨ W ~(V ↔ W) ~(W ↔X) ∨ V ∴ ~W 1. Show ~W 2. 3. 4.

W X

ass cd 3 pr1 mtp

5.

Show X → W

6. 7.

X X→W

8. 9. 10. 11.

c.

Show W → X

ass cd pr1 cdj dd

W↔X V V ↔ ~W ~W

2 5 bc 8 dn pr3 mtp pr2 nb 9 10 mp dd

(X →U) → (Y→Z) ~(Y ∨ ~Z) ∴ ~U 1. Show ~U 2. 3. 4. 5. 6.

~Y ∧ Z ~(Y → Z) ~(X → U) X ∧ ~U ~U

pr2 dm 2 nc 3 4 mt 4 nc 5 s dd

2. Construct correct derivations for each of the following arguments using derived rules when convenient. a.

U∧V → X ~V → Y X∨Y → Z ∴ ~Z → ~U

b.

(X→Y) → Z ~Z V→Y ∴ ~V

c.

P∨Q Q→S U ∨ ~S P∨S → R R→U ∴ U

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9 OFFICIAL CONDITIONS FOR DERIVATIONS Let us summarize here what we can now use in constructing an unabbreviated derivation. UNABBREVIATED DERIVATIONS A derivation from a set of sentences P consists of a sequence of lines that is built up in order, step by step, where each step is in accordance with these provisions: • • • •

• •

• •

Show line: A show line consists of the word "Show" followed by a symbolic sentence. A show line may be introduced at any step. Show lines are not given a justification. Premise: At any step, any symbolic sentence from the set P may be introduced, justified with the notation "pr". Theorem: At any step, an instance of a previously proved theorem may be entered with the name of the theorem given as justification. (e.g. "T32") Rule: At any step, a line may be introduced if it follows by a rule from sentences on previous available lines; it is justified by citing the numbers of those previous lines and the name of the rule. This includes the following basic rules: r mp mt dn s adj add mtp bc cb It also includes rules based on previously derived theorems, where the name of a rule based on a theorem is "R" followed by the name of the theorem; e.g. "RT32". If the appropriate enabling theorems have been derived, these rules are also available for use: nc cdj sc dm nb Direct derivation: When a line (which is not a show line) is introduced whose sentence is the same as the sentence on the closest previous uncancelled show line, one may, as the next step, write "dd" following the justification for that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. Assumption for conditional derivation: When a show line with a conditional sentence is introduced, as the next step one may introduce an immediately following line with the antecedent of the conditional on it; the justification is "ass cd". Conditional derivation: When a line (which is not a show line) is introduced whose sentence is the same as the consequent of the conditional sentence on the closest previous uncancelled show line, one may, as the next step, write "cd" at the end of that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. Assumption for indirect derivation: When a show line is introduced, as the next step one may introduce an immediately following line with the [un]negation of the sentence on the show line; the justification is "ass id". Indirect derivation: When a sentence is introduced on a line which is not a show line, if there is a previous available line containing the [un]negation of that sentence, and if there is no uncancelled show line between the two sentences, as the next step you may write the line number of the first sentence followed by "id" at the end of the line with the second sentence. Then you cancel the closest previous "show", and box all sentences below that show line, including the current line.

Except for steps that involve boxing and canceling, every step introduces a line. When writing out a derivation, every line that is introduced is written directly below previously introduced lines. Optional variant: When boxing and canceling with direct or conditional derivation, the "dd" or "cd" justification may be written on a later line which contains no sentence at all, and which is followed by the number of the line that completes the derivation. With indirect derivation, the "id" justification may be written on a later line which contains no sentence at all, and which is followed by the numbers of the two lines containing contradictory sentences. In all cases, the lines cited must be available from the later line.

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Some additional strategic hints Now that we have connectives in addition to the negation and conditional signs, we can give some general hints for doing derivations containing them. These have all been illustrated above, and they will simply be stated here for convenience. First are strategies that are often useful for deriving certain forms of sentences. If you want to derive a Conjunction □ ∧ ○ Derive each conjunct (perhaps by id) and adjoin them If you want to derive a Disjunction □ ∨ ○ Derive either disjunct and use add. Assume '~(□ ∨ ○)' for id, and use dm. Derive '~□ → ○', perhaps by cd, and use cdj If you want to derive a Biconditional □ ↔ ○ Derive each conditional and use cb. If you want to derive a Negation of a conjunction ~(□ ∧ ○) Use id. If you want to derive a Negation of a disjunction ~(□ ∨ ○) Derive '~□ ∧ ~○' and use dm. Perhaps assume '□ ∨ ○' for id, and try to derive both '□ → P∧~P' and '○→P∧~P'. Then use sc (applied to the assumed '□ ∨ ○' and the conditionals) to derive 'P∧~P'. If you want to derive a Negation of a biconditional ~(□ ↔ ○) Derive '□ ↔ ~○' and use nb. Then there are situations in which you have available a certain form of sentence, and want to know how to make use of it. If you have available a Conjunction □ ∧ ○ Simplify and use the conjuncts singly. If you have available a Disjunction □ ∨ ○ Try to derive the negation of one of the disjuncts, and use mtp Derive the conditionals '□ → △' and '○ → △', where '△' is something you want to derive. Then use sc with the disjunction and two conditionals. If you have available a Biconditional □ ↔ ○ Infer both conditionals and use them with mp, mt, and so on. If you have available a Negation of a conjunction ~(□ ∧ ○) Use dm to turn this into '~□ ∨ ~○', and then try to derive either '□' or '○' to use mtp. If you have available a Negation of a disjunction ~(□ ∨ ○) Use dm to turn this into '~□ ∧ ~○'; then simplify and use the conjuncts singly. If you have available a Negation of a biconditional ~(□ ↔ ○) Use nb to turn this into '□ ↔ ~○', and use bc to get the corresponding conditionals.

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EXERCISES 1. Construct correct derivations for each of the following arguments. a.

~(P ↔ Q) R∨P ~Q → R ∴ R

b.

W→U ~W → V ∴ U∨V

c.

P ∨ (Q∧S) R∨Q S ∨ ~P Q → ~S ∴ R

10 TRUTH TABLES AND TAUTOLOGIES The sentence 'P ∨ ~P' is logically true. It is true in all logically possible situations. This can be established by simple reasoning: Although there are an infinite number of logically possible situations, they fall into two classes. One class consists of situations in which 'P' is true, and the other consists of situations in which 'P' isn't true. In any situation in the first class, 'P ∨ ~P' is true because it is a disjunction whose first disjunct is true. In any situation in the second class, 'P' is not true in that situation, and so its negation, '~P' is true, and again 'P ∨ ~P' is true because it is a disjunction which has a true disjunct. So in either class of situations 'P ∨ ~P' is true. This pattern of reasoning can be summed up using a truth table. The table begins with listing the two options for the truth value of 'P' in a class of situations: situations in which 'P' is true  situations in which 'P' is false 

P T F

~P

The truth value of '~P' is determined in each class: situations in which 'P' is true  situations in which 'P' is false 

P T F

~P F T

and that information determines the truth value of 'P ∨ ~P' in each class: situations in which 'P' is true  situations in which 'P' is false 

P T F

~P F T

P ∨ ~P T T

This is an example in which no matter what truth value the simple parts of the sentence have, the sentence itself is true. Such a sentence is called a "tautology":

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Tautology: A sentence is a tautology if no matter what truth values are assigned to its simple parts, the definitions of the connectives used in the sentence determine that the sentence is true. The truth table just given shows that 'P ∨ ~P' is a tautology, because it shows that 'P ∨ ~P' is true no matter how truth values are assigned to its atomic parts. This use of truth tables can be applied to a sentence of any degree of complexity. Here is an example showing that 'P∧Q → Q' is a tautology. We begin by listing all of the possible combinations of truth values that 'P' and 'Q' might have. There are four of these: the sentences are both true, the first is true and the second false, the first is false and the second true, or they are both false: P T T F F

Q T F T F

P∧Q

P∧Q → Q

This assignment of truth values to 'P' and to 'Q' determines the truth values of 'P ∧ Q' and of 'P∧Q → Q'; for 'P∧Q → Q': P Q P∧Q P∧Q → Q T T T T T F F T F T F T F F F T In this table there are all T's under 'P∧Q → Q', showing that it is a tautology. To handle sentences of arbitrary numbers of sentence letters, we need to have a systematic way of representing all of the possible combinations of truth values that the sentence letters can receive. One way of doing this is to list all of the atomic parts of the sentence on the top of the table. Then, underneath the rightmost letter, write alternations of T and F: P

Q

R T F T F T F T F

Under the next letter to its left write alterations of TT and FF: P

Q T T F F T T F F

R T F T F T F T F

Under the next, write alterations of TTTT and FFFF:

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P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

Do this until the leftmost letter has gone through one whole set of alterations. If there is one sentence letter, only two rows are required. If there are two, the table will contain four rows. If three, then eight. n And so on. There are always 2 rows in the table when there are n sentence letters. Next, write the sentence to be tested, and underneath it write in each row the truth value that it has when its parts have the truth values appearing on that row. An example is: P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

P∧Q → P∧R T F T T T T T T

 not a tautology

Since there is a row (the second row) in which 'P∧Q → P∧R' does not have a T, that sentence is not a tautology. If there is a T in every row, it is a tautology, as in this case: P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

P∧Q → P∨R T T T T T T T T

This method is completely mechanical and it always yields an answer in a finite amount of time. If a sentence is a tautology, then you need to fill in every one of its rows to show that every one is T. But if a sentence is not a tautology, you need only find one row in which the sentence comes out F. For example, the following partial truth table shows that 'P∨Q ↔ R∨Q' is not a tautology:

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P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

P∨Q ↔ R∨Q

F

If a sentence is not a tautology and if you can identify what assignment of truth values to the simple parts will show this, you needn't set up a whole truth table. Just give a single row: P T

Q F

R F

P∨Q ↔ R∨Q F

and state that this assignment of truth values to sentence letters makes the sentence false. It turns out that every theorem of the first two chapters of this text is a tautology. This is because the rules and techniques in these chapters only allow derivations of tautologies when no premises are used. It is also a fact that any tautology can be derived by the rules we have. So we have two ways to show tautologies: theorems and truth tables, and we have one way to show non-tautologies: truth tables.

EXERCISES 1. Use truth tables or truth value assignments to determine whether each of these is a tautology. a. b. c. d. e. f.

(R↔S) ∨ (R↔~S) R ↔ (S↔R) R ∨ (S∧T) → R ∧ (S∨T) ~U → (U→~V) (~R↔R) → S (S∧T) ∨ (S∧~T) ∨ ~S

11 TAUTOLOGICAL VALIDITY It is easy to show by doing a derivation that this argument is valid: P∧P ∴ P It is also possible to show that the argument is valid using a technique like that of truth tables. Just show that there is no logically possible situation in which the premise is true and the conclusion false. This can be done as follows: All logically possible situations can be divided into two classes. In one class of situations, 'P' is true; no situation of this sort can be one in which the argument has true premises and a false conclusion, because in any of these situations the conclusion, 'P', is true. In all other situations, 'P' is false. But then so is 'P∧P'. So none of these are situations in which the argument has true premises and a false conclusion. So it is valid. Generalizing, we can say that an argument is valid whenever the premises "tautologically imply" the conclusion. This relation is called Tautological Validity. It is defined as follows:

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An argument is tautologically valid if and only if there is no assignment of truth values to its atomic parts which make the premises all true and the conclusion false.

There is a mechanical way to test an argument to see if it is tautologically valid. Just create a truth table in which all of the premises and conclusion appear at the top of some column. If there is no row in which all of the premises have T's under them and the conclusion has an F, then the argument is tautologically valid. If there is such a row, then the argument is not tautologically valid; instead, we say that it is tautologically invalid. Suppose that we are wondering whether this argument: P → ~Q R ↔ P∧Q Q∨R ∴ Q∧R is tautologically valid. Here is a truth table to test this: P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

P → ~Q F F T T T T T T

R ↔ P∧Q T F F T F T F T

Q∨R T T T F T T T F

Q∧R T F F F T F F F

There is in fact a row (the sixth) in which the premises are all true and the conclusion is false. So the argument is not tautologically valid. Using the same technique, we can show that this argument:

P → ~Q R ↔ P∧Q Q∨R P↔R

is tautologically valid. The truth table is: P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

P → ~Q F F T T T T T T

R ↔ P∧Q T F F T F T F T

Q∨R T T T F T T T F

P↔R T F T F F T F T

Here, there is no row in which the premises are all true and the conclusion 'P ↔ R' is false. So that argument is tautologically valid. Although we haven’t shown this here, if there is a derivation using the rules and techniques of Chapters 1 and 2 showing that an argument is valid, then the argument is indeed tautologically valid. And vice versa:

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CHAPTER 2 SECTION 11

if an argument is tautologically valid, then there is a derivation to show that the argument is valid using the techniques of chapters 1 and 2. So we have two ways to show that an argument is tautologically valid: it can be done with a truth table or with a derivation. We have only one way to show that some argument is not tautologically vaild: find a way to assign truth values to the sentential letters so that all the premises are true and the conclusion false.

EXERCISES For each of the following arguments, either show that it is tautologically valid, or show that it is tautologically invalid. a.

U∧V → X ~V → U X∨V → U ∴ V → ~U

b.

(X→Y) → Z ~Z ∴ ~Y

c.

~(P ↔ Q) R∨P ~Q → R ∴ R

d.

S∨T W∨S ~T ∨ ~S ∴ ~S

e.

W→U ~W → V ∴ U∨V

f.

P ↔ ~Q Q → R∨P R → ~Q ∨ ~P ∴ Q∨R

g.

P ∨ (Q∧S) S∨Q S ∨ ~P ∴ S

h.

P ∧ (Q∨S) S∨Q S∨P ∴ S

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CHAPTER 2 SUMMARIES

BASIC AND DERIVED RULES FROM CHAPTERS 1 AND 2 BASIC RULES OF CHAPTER 1 Repetition:

□ ∴ □

Modus Ponens:

□→○ □ ∴○

Modus Tollens:

□→○ ~○ ∴ ~□

Double negation:

□ ∴ ~~□

or

~~□ ∴□

BASIC RULES OF CHAPTER 2 Conjunction rules: Rule s (simplification) □∧○ ∴ □

or

□∧○ ∴○

or

Rule mtp (modus tollendo ponens) □ ∴○∨□

Biconditional rules: Rule bc (biconditional-to-conditional) □↔○ ∴ □→○

or

□ ○ ∴ □∧○

□↔○ ∴○→□

Chapter 2 -- 40

□∨○ ~○ ∴ □

or

□∨○ ~□ ∴○

Rule cb (conditionals-to-biconditional) □→○ ○→□ ∴ □↔○

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CHAPTER 2 SUMMARIES

THEOREMS USED AS RULES A theorem of conditional form: ∴ □∧○ → △ justifies a rule of this form: □ ○ ∴ △ A theorem of biconditional form: ∴□↔○ justifies whatever both conditionals justify.

DERIVED RULES (May be used if the theorems on which they are based have been derived.) Rule nc

□ ∧ ~○ ∴ ~(□→○)

~(□→○) ∴ □ ∧ ~○ Rule cdj

□→○ ∴ ~□ ∨ ○ Rule sc

~□ ∨ ○ ∴ □→○

~□ → ○ ∴ □∨○

□∨○ ∴ ~□ → ○

□∨○ □→△ ○→△ ∴ △

Rule dm

□∧○ ∴ ~(~□∨~○) ∴ Rule nb

~(~□∨~○) □∧○

□∨○ ∴ ~(~□∧~○)

~(□∧○) ∴ ~□ ∨ ~○

~(□∨○) ∴ ~□ ∧ ~○

~(~□∧~○) □∨○

~□ ∨ ~○ ∴ ~(□∧○)

~□ ∧ ~○ ∴ ~(□∨○)

~(□↔○) ∴ □ ↔ ~○

□→△ ~□ → △ ∴△

□ ↔ ~○ ∴ ~(□↔○)

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CHAPTER 2 SUMMARIES

UNABBREVIATED DERIVATIONS A derivation from a set of sentences P consists of a sequence of lines that is built up in order, step by step, where each step is in accordance with these provisions: • • • •

• •

• •

Show line: A show line consists of the word "Show" followed by a symbolic sentence. A show line may be introduced at any step. Show lines are not given a justification. Premise: At any step, any symbolic sentence from the set P may be introduced, justified with the notation "pr". Theorem: At any step, an instance of a previously proved theorem may be entered with the name of the theorem given as justification. (e.g. "T32") Rule: At any step, a line may be introduced if it follows by a rule from sentences on previous available lines; it is justified by citing the numbers of those previous lines and the name of the rule. This includes the following basic rules: r mp mt dn s adj add mtp bc cb It also includes rules based on previously derived theorems, where the name of a rule based on a theorem is "R" followed by the name of the theorem; e.g. "RT32". If the appropriate enabling theorems have been derived, these rules are also available for use: nc cdj sc dm nb Direct derivation: When a line (which is not a show line) is introduced whose sentence is the same as the sentence on the closest previous uncancelled show line, one may, as the next step, write "dd" following the justification for that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. Assumption for conditional derivation: When a show line with a conditional sentence is introduced, as the next step one may introduce an immediately following line with the antecedent of the conditional on it; the justification is "ass cd". Conditional derivation: When a line (which is not a show line) is introduced whose sentence is the same as the consequent of the conditional sentence on the closest previous uncancelled show line, one may, as the next step, write "cd" at the end of that line, draw a line through the word "Show", and draw a box around all the lines below the show line, including the current line. Assumption for indirect derivation: When a show line is introduced, as the next step one may introduce an immediately following line with the [un]negation of the sentence on the show line; the justification is "ass id". Indirect derivation: When a sentence is introduced on a line which is not a show line, if there is a previous available line containing the [un]negation of that sentence, and if there is no uncancelled show line between the two sentences, as the next step you may write the line number of the first sentence followed by "id" at the end of the line with the second sentence. Then you cancel the closest previous "show", and box all sentences below that show line, including the current line.

Except for steps that involve boxing and canceling, every step introduces a line. When writing out a derivation, every line that is introduced is written directly below previously introduced lines. Optional variant: When boxing and canceling with direct or conditional derivation, the "dd" or "cd" justification may be written on a later line which contains no sentence at all, and which is followed by the number of the line that satisfies the conditions for direct or conditional derivation. With indirect derivation, the "id" justification may be written on a later line which contains no sentence at all, and which is followed by the numbers of the two lines containing contradictory sentences. In all cases, the lines cited must be available from the later line.

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CHAPTER 2 SUMMARIES

STRATEGY HINTS Try to reason out the argument for yourself. Begin with a sketch of an outline of a derivation, and then fill in the details. Write down obvious consequences. When no other strategy is obvious, try indirect derivation.

To derive:

Try this:

Conjunction □∧○

Derive each conjunct, and adjoin them

Disjunction □∨○

Derive either disjunct and use add. (Often this is not possible.) Assume '~(□ ∨ ○)' for id and immediately use dm. Derive '~□ → ○' and use cdj

Conditional □→○

Use cd

Biconditional □↔○

Derive both conditionals and use cb.

Negation of conjunction ~(□ ∧ ○)

Use id.

Negation of disjunction ~(□ ∨ ○)

Derive '~□ ∧ ~○' and use dm. Perhaps assume '□ ∨ ○' for id and try to derive both '□ → P∧~P' and '○→P∧~P'. Then use sc (applied to the assumed '□ ∨ ○' and the conditionals) to derive 'P∧~P'.

Negation of conditional ~(□ → ○)

Use id.

Negation of biconditional ~(□ ↔ ○)

Derive '□ ↔ ~○' and use nb.

If you have this available:

Try this:

Conjunction □∧○

Simplify and use the conjuncts singly.

Disjunction □∨○

Try to derive the negation of one of the disjuncts, and use mtp.. Derive the conditionals '□ → △' and '○ → △', where '△' is something you want to derive. Then use sc with the disjunction and two conditionals.

Conditional □→○

Try to derive the antecedent to set up mp, or derive the negation of the consequent, to set up mt.

Biconditional □↔○

Infer both conditionals and use them with mp, mt, and so on.

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CHAPTER 2 SUMMARIES

Negation of conjunction ~(□ ∧ ○)

Use dm to turn this into '~□ ∨ ~○', and then try to derive either '□' or '○' to use mtp.

Negation of disjunction ~(□ ∨ ○)

Use dm to turn this into '~□ ∧ ~○'; then simplify and use the conjuncts singly.

Negation of conditional ~(□ → ○)

Use nc to derive '□ ∧ ~○', then simplify and use the conjuncts singly.

Negation of biconditional ~(□ ↔ ○)

Use nb to turn this into '□ ↔ ~○', and use bc to get the corresponding conditionals.

ALL THEOREMS FROM CHAPTERS 1 AND 2 T1

P→P

T2

Q → (P→Q)

T3

P → ((P→Q) → Q)

T4

(P→Q) → ((Q→R) → (P→R))

Syllogism

T5

(Q→R) → ((P→Q) → (P→R))

Syllogism

T6

(P → (Q→R)) → ((P→Q) → (P→R))

Distribution of → over →

T7

((P→Q) → (P→R)) → (P→(Q → R))

Distribution of → over →

T8

(P→ (Q→R)) → (Q → (P→R))

Commutation

T9

(P → (P→Q)) → (P→Q)

T10

((P→Q) → Q) → ((Q→P) → P)

T11

~~P → P

Double negation

T12

P → ~~P

Double negation

T13

(P→Q) → (~Q → ~P)

Transposition

T14

(P → ~Q) → (Q → ~P)

Transposition

T15

(~P → Q) → (~Q → P)

Transposition

T16

(~P → ~Q) → (Q → P)

Transposition

T17

P → (~P→Q)

T18

~P → (P→Q)

T19

(~P→P) → P

T20

(P→~P) → ~P

T21

~(P→Q) → P

T22

~(P→Q) → ~Q

T23

((P→Q) → P) → P

Peirce's law

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CHAPTER 2 SUMMARIES

T24

P∧Q ↔ Q∧P

"commutative law for conjunction"

T25

P∧ (Q∧R) ↔ (P∧Q)∧ R

"associative law for conjunction"

T26

(P→Q) ∧(Q→R) → (P→ R)

"hypothetical syllogism"

T27

(P∧Q → R) ↔ (P → (Q→R))

"exportation"

T28

(P∧Q → R) ↔ (P∧~R → ~Q)

T29

(P→ Q∧R) ↔ (P→Q) ∧(P→R)

T30

(P→Q) → (R∧P → R∧Q)

T31

(P→Q) → (P∧R → Q∧R)

T32

(P→R) ∧(Q→S) → (P∧Q → R∧S)

"Leibniz's praeclarum theorema"

T33

(P→Q) ∧(~P→Q) → Q

"separation of cases; constructive dilemma"

T34

(P→Q) ∧(P→~Q) → ~P

T35

(~P→R) ∧(Q→R) ↔ ((P→Q) → R)

T36

~(P ∧ ~P)

T37

(P→Q) ↔ ~(P ∧ ~Q)

T38

P∧Q ↔ ~(P → ~Q)

T39

~(P∧Q) ↔ (P → ~Q)

T40

~(P → Q) ↔ P∧~Q

"negation of conditional"

T41

P ↔ P∧P

"idempotence for ∧"

T42

P∧~Q → ~(P→Q)

"negation of conditional"

T43

~P → ~(P∧Q)

T44

~Q → ~(P∧Q)

T45

P∨Q ↔ (~P→Q)

T46

(P→Q) ↔ ~P ∨ Q

"definition of → in terms of ∨"

T47

P ↔ P∨P

"idempotence for ∨"

T48

(P∨Q) ∧(P→R) ∧(Q→S) → R∨S

T49

(P∨Q) ∧(P→R) ∧(Q→R) → R

T50

(P→R) ∧(Q→R) ↔ (P∨Q→R)

T51

(P∨Q) ∧(P→R) ∧(~P∧Q→R) → R

T52

(P→R) ∧(~P∧Q→R) ↔ (P∨Q→R)

T53

P∨Q ↔ Q∨P

"commutative law for ∨"

T54

P ∨ (Q∨R) ↔ (P∨QR) ∨ R

"associative law for ∨"

T55

(P → Q∨R) ↔ (P→Q) ∨ (P→R)

"distribution of → over ∨"

T56

(P→Q) → (R∨P → R∨Q)

T57

(P→Q) → (P∨R → Q∨R)

T58

(P→Q) ∨ (Q→R)

"distribution of → over ∧"

"separation of cases"

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CHAPTER 2 SUMMARIES

T59

P ∨ ~P

T60

(P→R) ∨ (Q→R) ↔ (P∧Q → R)

T61

P ∧ (Q∨R) ↔ (P∧Q) ∨ (P∧R)

"distribution"

T62

P ∨ (Q∧R) ↔ (P∨Q) ∧(P∨R)

"distribution"

T63

P ∧ Q ↔ ~(~P∨~Q)

"de morgans"

T64

P ∨ Q ↔ ~(~P∧~Q)

"de morgans"

T65

~(P∧Q) ↔ ~P ∨ ~Q

"de morgans"

T66

~(P∨Q) ↔ ~P ∧ ~Q

"de morgans"

T67

~P∧~Q → ~(P∨Q)

T68

P ↔ (P∧Q) ∨ (P∧~Q)

T69

P ↔ (P∨Q) ∧(P∨~Q)

T70

Q → (P∧Q ↔ P)

T71

~Q → (P∨Q ↔ P)

T72

(P→Q) ↔ (P∧Q ↔ P)

T73

(P→Q) ↔ (P∨Q ↔ Q)

T74

(P↔Q)∧ P → Q

T75

(P↔Q)∧ Q → P

T76

(P↔Q)∧ ~P → ~Q

T77

(P↔Q)∧ ~Q → ~P

T78

(P → (Q↔R)) ↔ ((P→Q) ↔ (P→R))

T79

(P → (Q↔R)) ↔ (P∧Q ↔ P∧R)

T80

(P↔Q) ∨ (P↔~Q)

T81

(P↔Q) ↔ (P→Q)∧ (Q→P)

T82

(P↔Q) ↔ ~((P→Q) → ~(Q→P))

T83

(P↔Q) ↔ (P∧Q) ∨ (~P∧~Q)

T84

P∧Q → (P↔Q)

T86

((P↔Q) → R) ↔ (P∧Q → R)∧ (~P∧~Q → R)

T87

~(P↔Q) ↔ (P∧~Q)∨(~P∧~Q)

T88

P∧~Q → ~(P↔Q)

T89

~P∧Q → ~(P↔Q)

T90

~(P↔Q) ↔ (P↔~Q)

T91

P↔P

T92

(P↔Q) ↔ (Q↔P)

T93

(P↔Q) ∧(Q↔R) → (P↔R)

T94

(P ↔ (Q↔R)) ↔ ((P↔Q) ↔ R)

"excluded middle"

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CHAPTER 2 SUMMARIES

T95

(P→Q) ↔ ((P↔R) ↔ (Q↔R))

T96

(P↔Q) ↔ (~P↔~Q)

T97

(P↔R)∧ (Q↔S) → ((P→Q) ↔ (R→S))

T98

(P↔R)∧ (Q↔S) → (P∧Q ↔ R∧S)

T99

(P↔R)∧ (Q↔S) → (P∨R ↔ Q∨S)

T100

(P↔R)∧ (Q↔S) → ((P↔Q) ↔ (R↔S))

T101

(Q↔S) → ((P→Q) ↔ (P→S)) ∧((Q→P) ↔ (S→P))

T102

(Q↔S) → (P∧Q ↔ P∧S)

T103

(Q↔S) → (P∨Q ↔ P∨S)

T104

(Q↔S) → ((P↔Q) ↔ (P↔S))

T105

P∧ (Q↔R) → (P∧Q ↔ R)

T106

(P → (Q→R)) ↔ ((P→Q) → (P→R))

T107

(P →→R)) ↔ (Q → (P→R))

T108

(P →(P→Q)) ↔(P→Q)

T109

((P→Q) → Q) ↔ ((Q→P) → P)

T110

P ↔ ~~P

T111

(P→Q) ↔ (~Q→~P)

T112

(P→~Q) ↔ (Q→~P)

T113

(~P→Q) ↔ (~Q→P)

T114

(~P→P) ↔ P

T115

(P→~P) ↔ ~P

T116

(P∧Q) ∨ (R∧S) ↔ (P∨R) ∧(P∨S) ∧(Q∨R) ∧(Q∨S)

T117

(P∨Q) ∧(R∨S) ↔ (P∧R) ∨ (P∧S) ∨ (Q∧R) ∨ (Q∧S)

T118

(P∨Q) ∧(R∨S) ↔ (~P∧~R) ∨ (~P∧S) ∨ (Q∧~R) ∨ (Q∧S)

T119

(P∨~P) ∧Q ↔ Q

T120

(P∧~P) ∨ Q ↔ Q

T121

P ∨ (~P∧Q) ↔ P∨Q

T122

P ∧ (~P∨Q) ↔ P∧Q

T123

P ↔ P ∨ (P∧Q)

T124

P ↔ P ∧ (P∨Q)

T125

(P→ Q∧R) → (P∧Q ↔ P∧R)

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CHAPTER 2 Answers to the Exercises

Answers to the Exercises -- Chapter 2 SECTION 1 Exercises 1 and 2 answered together: a. Not a sentence b. Informal notation ~Q↔~R F /\ ~Q ~R F T | | Q R T F c.

Official notation ~(Q↔R) T | Q↔R F /\ Q R T F

d. e.

Not a sentence Informal notation (P→Q) ∨ (R→~Q) T /\ P→Q R→~Q T /\ /\ P Q R ~Q F | Q

f. g.

Not a sentence Informal notation P∧Q → (Q→R∨Q) T /\ P∧Q Q→R∨Q /\ T P Q /\ Q R∨Q T /\ R Q T

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CHAPTER 2 Answers to the Exercises

h.

Informal notation P ↔ (P↔Q∧R) F /\ P P↔Q∧R T F /\ P Q∧R T F /\ Q R F

i.

Informal notation P ∨ (Q→P) T /\ P Q→P T /\ Q P

SECTION 2 1. a. b. c. d. e.

R∧P W∨R ~R ∧ T R∧S Q↔R

2 and 3 answered together a. S∨V false b. R↔S true c. R∧S false d. Q∨T false e. Q∧S false

SECTION 3 1. a.

b.

~(P ∨ (Q ∧ R)) F | P ∨ (Q ∧ R) T /\ P Q∧R T /\ Q R ~P ∨ (Q ∧ R) F /\ ~P Q∧R F F | /\

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CHAPTER 2 Answers to the Exercises

P Q R T F F c.

~(P ∨ R) ↔ ~P ∨ R T /\ ~(P ∨ R) ~P ∨ R F F | /\ P∨R ~P R T F F /\ | P R P T T

d.

~Q ∧ (P ∨ (Q↔R)) T /\ ~Q P ∨ (Q↔R) T T | /\ Q P Q↔R F T /\ Q R F F

e.

P → (~Q ↔ (~R → Q)) F /\ P ~Q ↔ (~R → Q) T F /\ ~Q ~R → Q T F | /\ Q ~R Q F T F | R F

2. a. b. c. d. 3. a.

~V ↔ ~W; "won't" is a negation with narrow scope. ~V → (Y→W∧V); "both" gives rise to a conjunction with narrow scope since it splits the names from the predicate. The comma prevents Y and V from occurring together. Y ∨ (W ∨ V); "unless" is a disjunction sign. (Y ∧ ~V) ∨ (V ∧ ~W) Only if Veronica doesn't leave will William leave, or Veronica and William and Yolanda will all leave. (Only if Veronica doesn't leave will William leave) ∨ (Veronica and William and Yolanda will leave) (William will leave → Veronica doesn't leave) ∨ (V ∧ W ∧ Y)

(W → ~V) ∨ (V ∧ W ∧ Y)

b.

If neither William nor Veronica leaves, Yolanda won't either If neither William [leaves] nor Veronica leaves, [then] Yolanda won't [leave] ~(W ∨ V) → ~Y

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c.

If William will leave if Veronica leaves, then he will surely leave if Yolanda leaves If (William will leave if Veronica leaves) then ([William] will leave if Yolanda leaves) (V → W) → (Y → W)

d.

Neither William nor Veronica nor Yolanda will leave ~(W ∨ V ∨ Y)

4. " Veronica leaves but neither William nor Yolanda leaves" corresponds to the truth-value assignment: V --- true; W --- false; Y --- false. We use parse trees to compute the truth values of the complex sentences. a.

(W → ~V) ∨ (V ∧ W ∧ Y) T /\ W → ~V V ∧ W ∧ Y T F /\ /\ W ~V V∧W Y F F F F | /\ V V W T T F

b.

~(W ∨ V) → ~Y T /\ ~(W ∨ V) ~Y F T | | W∨V Y T F /\ W V F T

c.

(V → W) → (Y → W) T /\ V→W Y→W F T /\ /\ V W Y W T F F F

d.

~(W ∨ V ∨ Y) F /\ W∨V∨Y T /\ W∨V Y T F /\ W V F T

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5. a.

Sally will run and win unless she quits (Sally will run and [Sally will] win) ∨ ([Sally] quits) (R ∧ W) v Q

b.

Sally will win exactly in case she runs without quitting Sally will win exactly in case (she runs [and doesn't] quit) W ↔ (R ∧ ~Q)

c.

Sally, who will run, will win if she doesn't quit Sally will run, and Sally will win if she doesn't quit R ∧ (~Q → W)

d.

Sally will run and quit, but she will win anyway Sally will run and quit, and she will win (R ∧ Q) ∧ W

SECTION 4 1. a. b. c. d. e. f. g. h. i.

None; if we had ~~Q instead of Q it would be an instance of MTP. Simplification Double Negation MTP CB None. BC Addition None

2. a. b. c. d. e. f.

~W ↔ ~X by CB; also ~X ↔ ~W by CB ~W by MTP Nothing ~W by S; also ~X by S W → ~X by BC; also ~X → W by BC Nothing

SECTION 5 Derivations of numbered theorems not given SECTION 6 Derivations of numbered theorems not given SECTION 7 1. a. b.

All fine In line 8, the sentence that can be inferred from 7 by RT39 is W → ~S.

2, 3, 4, 5: Derivations of numbered theorems not given

SECTION 8 1. a. b.

c.

All fine Line 4: MTP does not apply; Line 8: BC (biconditional to conditional) does not apply; we could use CB; Line 11: MP does not apply to biconditionals; you have to split the biconditional into conditionals first using BC. Line 2: the result of applying DM to pr2 is ~Y ∧ ~~Z rather than ~Y ∧ Z. Line 3: NC doesn't apply; the NC would generate line 3 if line 2 were Y ∧ ~Z. Line 4: Line 4 is not available at line 4; it may not be cited to justify itself. The sentence could be

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generated by applying MT to line 3 and pr1. 2. a.

U∧V → X ~V → Y X∨Y → Z ∴ ~Z → ~U

1. Show ~Z → ~U 2. ~Z 3. ~(X ∨ Y) 4. ~X ∧ ~Y 5. ~X 6. ~Y 7. ~(U ∧ V) 8. ~U ∨ ~V 9. ~~V 10. ~U

b.

1. 2. 3. 4. 5.

(X→Y) → Z ~Z V→Y ∴ ~V Show ~V ~(X→Y) X ∧ ~Y ~Y ~V

c.

ass cd pr3 2 mt 3 dm 4s 4s pr1 5 mt 7 dm 6 pr2 mt 8 9 mtp cd

pr1 pr2 mt 2 nc 3s 4 pr3 mt dd

P∨Q Q→S U ∨ ~S P∨S → R R→U ∴ U

1. Show U 2. Show P → U 3. P 4. P∨S 5. R 6. U 7. Show Q → U 8. Q 9. S 10. ~~S 11. U 12. U 13.

ass cd 3 add 4 pr4 mp 5 pr5 mp cd ass cd 8 pr2 mp 9 dn 10 pr3 mtp cd pr1 2 7 sc 12 dd

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CHAPTER 2 Answers to the Exercises

SECTION 9 1. a.

~(P ↔ Q) R∨P ~Q → R ∴ R

1. 2. 3. 4. 5. 6. 7. 8.

Show R ~R P P ↔ ~Q P → ~Q ~Q R

b.

W→U ~W → V ∴ U∨V Show U ∨ V Show ~U → V ~U ~W V U∨V

1. 2. 3. 4. 5. 6.

c.

ass id 2 pr2 mtp pr1 nb 4 bc 3 5 mp 6 pr3 mp 2 7 id

ass cd 3 pr1 mt 4 pr2 mp cd 2 cdj dd

P ∨ (Q∧S) R∨Q S ∨ ~P Q → ~S ∴ R

1. Show R 2. ~R 3. Q 4. ~S 5. ~Q ∨ ~S 6. ~(Q∧S) 7. P 8. ~~P 9. S 10.

ass id 2 pr2 mtp 3 pr4 mp 4 add 5 dm 6 pr1 mtp 7 dn 8 pr3 mtp 4 9 id

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CHAPTER 2 Answers to the Exercises

SECTION 10 (R↔S) ∨ (R↔~S); tautology

1. a.

(R↔S) ∨ (R↔~S) T T T T

R T T F F

S

b.

R ↔ (S↔R); not a tautology

T F T F

R T

R ↔ (S↔R) F

S F

R ∨ (S∧T) → R ∧ (S∨T); not a tautology

c. R F

S T

T T

R ∨ (S∧T) → R ∧ (S∨T) F

~U → (U→~V); tautology

d. U T T F F

~U → (U→~V) T T T T

V T F T F

e.

(~R↔R) → S; tautology

R T T F F

S

f.

(S∧T) ∨ (S∧~T) ∨ ~S; tautology

(~R↔R) → S T T T T

T F T F

T T T F F

(S∧T) ∨ (S∧~T) ∨ ~S T T T T

S T F T F

SECTION 11 a.

U T

U∧V → X ~V → U X∨V → U ∴ V → ~U V T

X T

NO

U∧V → X T

~V → U T

X∨V → U T

V → ~U F

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b.

X T T T T F F F F

Y T T F F T T F F c.

P T T T T F F F F

S T

U T T T T F F F F

R T F T F T F T F

~(P ↔ Q) F F T T T T F F

W T

W T F T F T F T F

R F

~Y F F T T F F T T

R∨P T T T T T F T F

~Q → R T T T F T T T F

R T F T F T F T F

~T ∨ ~S T

~S F

NO

S∨T T

W∨S T YES

W→U T T T T F T F T

P ↔ ~Q Q → R∨P R → ~Q ∨ ~P ∴ Q∨R Q F

~Z F T F T F T F T

YES

W→U ~W → V ∴ U∨V V T T F F T T F F

f.

(X→Y) → Z T F T T T F T F

S∨T W∨S ~T ∨ ~S ∴ ~S T F

e.

Z T F T F T F T F

YES

~(P ↔ Q) R∨P ~Q → R ∴ R Q T T F F T T F F

d.

P T

(X→Y) → Z ~Z ∴ ~Y

P ↔ ~Q T

~W → V T T T F T T T F

U∨V T T T T T T F F

NO

Q → R∨P T

R → ~Q ∨ ~P T

Chapter 2 -- 56

Q∨R F

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CHAPTER 2 Answers to the Exercises

g.

P T T T T F F F F

Q T T F F T T F F h.

P T

P ∨ (Q∧S) S∨Q S ∨ ~P ∴ S S T F T F T F T F

P ∨ (Q∧S) T T T T T F F F

P ∧ (Q∨S) S∨Q S∨P ∴ S Q T

S F

P ∧ (Q∨S) T

YES

S∨Q T T T F T T T F

S ∨ ~P T F T F T T T T

S T F T F T F T F

S∨P T

S F

NO

S∨Q T

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CHAPTER 3 SECTION 1

Chapter Three Name letters, Predicates, Variables and Quantifiers 1 NAME LETTERS AND PREDICATES In chapters 1 and 2 we studied logical relations that depend only on the sentential connectives: '~', '→', '∧', '∨', '↔'. The atomic sentences -- those that contain no connectives -- were symbolized by sentential letters, and we paid no attention to any internal structure that they might have. It is now time to study that structure. The Predicate Calculus is a system of logic that studies the ways in which sentences are constructed out of name letters, predicates, variables, and quantifiers, as well as connectives. We have already studied connectives; in this section we introduce name letters, predicates, variables, and quantifiers. In our logical symbolism, name letters are written as the small letters: a, b, c, d, e, f, g, h (and with subscripts, such as ‘c3’). Any small letter between 'a' and 'h' can be used as a name letter. Name letters in the logical symbolism correspond to names of English: Carlos, Agatha, Dr. Samuelson, Ms. Bernstein, Madame Curie, David Rockefeller, San Diego, Germany, UCLA, General Electric, Microsoft, Google, Macy's, The Los Angeles Times, I-405, Memorial Day, the FBI, ... Any one of these may be symbolized by means of a name letter: h c g

Henry California General Electric

The simplest way to make a sentence containing a name letter is to combine it with a one-place predicate. One-place predicates appear in our logical symbolism as the capital letters from A to O (and with subscripts, such as ‘G2’). One-place predicates correspond roughly to grammatical predicates in English; in the following examples, the underlined phrases would be symbolized as one-place predicates: Agatha is clever. Henry is a giraffe. Ferdy dances well. Georgia is a state Ann will run for re-election. (The parts that are not underlined are symbolized with name letters.) Whereas English proper names are usually capitalized, the logical name letters that represent them are not, and whereas English predicates are typically not capitalized, the logical predicates that represent them are capitalized. There is nothing "logical" about this reverse convention; it is an historical accident, but it has now become part of the tradition of symbolic logic Further, in the usual formulations of the predicate calculus the predicate comes before the name letter, instead of after it as in English. This, too, is an historical accident. So the sentences given above can be symbolized as follows: Agatha is clever. Henry is a giraffe. Ferdy dances well. Georgia is a state. Ann will run for re-election.

Ca Gh Df Ag Ea

A one-place ("monadic") predicate is any capital letter between 'A' and 'O' (optionally with a numerical subscript). A name letter is any small letter from 'a' to 'h' (optionally with a numerical subscript). An atomic sentence may be formed by writing a one-place predicate followed by a name letter.

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CHAPTER 3 SECTION 1

EXERCISES 1. Symbolize each of the following sentences: a. b. c. d. e. f.

Fred is an orangutan. Gertrude is an orangutan but Fred isn't. Tony Blair will speak first. Gary lost weight recently; he is happy. Felix cleaned and polished. Darlene or Abe will bat clean-up.

We assume that a one-place predicate is true of certain things, and that a name letter stands for a unique thing. A sentence consisting of a one-place predicate together with a name letter is true if and only if the predicate is true of the thing that the name letter stands for. Thus, taking the examples listed above, we assume that 'C' is true of all and only clever things, that 'a' stands for Agatha (presumably a person or animal), and then: Ca is true if and only if Agatha is one of the clever things that the predicate is true of. Similarly, if `G' is true of giraffes, then `Gh' is true if Henry is one of the giraffes. If `E' is true of the things that will run for re-election, and if 'a' stands for Ann, then `Ea' is true if and only if Ann will run for reelection. Predicates are generally true of several specific things, but a predicate might be true of only one thing ('is a moon of the earth') or might not be true of anything at all. If there are in fact no dragons, the sentence: Df

Fred is a dragon

contains a predicate 'D' that is true of nothing at all. This means that the sentence `Df' will be false, no matter who or what `Fred' stands for. In this chapter we assume that each name letter in our logical symbolism stands for a unique thing. This assumption is an idealization, for it is not true that the words of English that we are representing by name letters always succeed in naming something. If there is no such person as Paul Bunyan, then `Paul Bunyan' is a "name" that names nothing at all. In some systems of logic it is possible to use name letters which do not stand for anything; these systems of logic are called "free logics". (They are called "free" because they are "free of" the assumption that the name letters they contain actually stand for things.) Free logics are a bit more complicated than standard logic. (Studies of free logic assume that the reader is already acquainted with the standard logic taught here.) In this text we assume that any name letter that we use stands for something. EXERCISES 2. Symbolize each of the following, assuming: `D' is true of doctors 'L' is true of people who are in love 'h' stands for Hans 'a' stands for Amanda a. b. c. d. f.

Hans is a doctor but Amanda isn't. Hans, who is a doctor, is in love Hans is in love but Amanda isn't Neither Hans nor Amanda is in love Hans and Amanda are both doctors.

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CHAPTER 3 SECTION 1

3. Symbolize each of the following, using: 'L' for things that live in Brea 'D' for things that drive to school a. b. c. d. e.

Eileen and Cosi both live in Brea. Eileen drives to school, and so does Hank. If Hank lives in Brea then he drives to school; otherwise he doesn't drive to school. If David and Hank both live in Brea then David drives to school but Hank doesn't. Neither Hank nor Eileen live in Brea, yet each of them drives to school.

2 QUANTIFIERS, VARIABLES, AND FORMULAS So far, we have no means at all in our symbolism to express generalities. We can say that Pedro is a doctor, and we can say that Pedro is wealthy, but we cannot say that everyone is a doctor, or that every doctor is wealthy. Nor can we deny that everyone is a doctor, or say that some doctor isn't wealthy. We cannot even express these claims. In order to express generalities we will introduce quantifiers and variables. Variables: Any small letter from 'i' to 'z' is a variable; also small letters between 'i' and 'z' with numerical subscripts. The universal quantifier sign is '∀'. The existential quantifier sign is '∃'. A quantifier is either quantifier sign followed by a variable: ∀x ∀z ∀s ∃x ∃z ∃s Here is how we use quantifiers. Suppose that we wish to say -- as some philosophers have said -- that everything in the universe is either mental or physical. Suppose that `M' is the one-place predicate `is mental', and `H' is the one-place predicate `is physical'. Then we symbolize the claim that everything is either mental or physical as follows: ∀x(Mx ∨ Hx). The initial `∀x' is a universal quantifier phrase. This is followed by something, `(Mx ∨ Hx)', which we will call a symbolic formula. A formula is just like a symbolic sentence except that instead of a name letter following each predicate we may have a variable, such as `x' above. The displayed formula says that everything satisfies a certain condition. The universal quantifier is responsible for the "everything" part, and the combination of variables and predicates tells us what the condition is. In the case in point, the condition is that it is either mental or physical: ∀x (Mx ∨ Hx)

Everything is such that

it is either mental or physical

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CHAPTER 3 SECTION 2

An existential quantifier can appear in a formula in the same place that a universal quantifier may appear: ∃x (Mx ∨ Hx)

Something is such that

it is either mental or physical

In order to construct sentences in our new extended notation, we begin by defining what a symbolic formula is. Intuitively, a symbolic formula is like a sentence, except that it may contain variables in places where name letters otherwise would appear. We use the word 'term' to cover both name letters and variables. Terms: Any name letter or variable is a term. So 'a' and 'x' are both terms. A formula is built up in steps, as follows: Sentence letters: Any sentence letter is an atomic formula.

Atomic formulas: A one-place predicate followed by a term is an atomic formula. Thus, if F is a one-place predicate and 'a' is a name letter, then 'Fa' is an atomic formula; If 'F' is a one-place predicate and 'x' is a variable then 'Fx' is an atomic formula. Both 'Henry is a giraffe' and 'x is a giraffe' are symbolized as atomic formulas: Gh

Gx Molecular formulas: If □ and ○ are formulas, then the following are molecular formulas: ~□

(□∧○)

(□∨○)

(□→○)

(□↔○)

Here are some molecular formulas: ~Gh

~Gx

(Gx ∧ Fa)

(Gx ∨ Jc)

(Gh → Jy)

(~Fa ↔ Ga) → Hx

We can also make formulas out of other formulas by "generalizing" them with quantifiers: Quantified formulas: If □ is a formula, and 'x' is a variable, then these are quantified formulas: ∀x□ ∃x□ Examples of quantified formulas are: ∀xGx

∃xFx

∀y(Gy→Fy)

∃w~(Gw ∧ ~Fb)

∀v(~Jx ↔ Fv)

Once a quantified formula is constructed, it may be used as input to any of these provisions. So, given that the examples above are formulas, we can make new formulas by combining them with connectives: (∀xGx ∧ ∃xFx)

(∃xFx ∨ ∀y(Gy→Fy))

∀y(Gy→Fy)

Chapter Three -- 4

(P ∧ ∃xFx)

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CHAPTER 3 SECTION 2

We may informally omit parentheses exactly as we did in the last chapter, to produce informal notation: ∀xGx ∧ ∃xFx

∃xFx ∨ ∀y(Gy→Fy)

(Note that '∀yGy→Fy', is a conditional; it is not equivalent to '∀y(Gy→Fy)', which is a universal generalization of a conditional.) Likewise, we can add a quantifier to a formula that already has one or several quantifiers within it: ∀x(Gx → ∃yFy)

∀x~∃y(Gx ∨ ~Fy)

∀x∀y∀z(Gx → Fz)

A formula is anything that can be constructed by means of the above provisions for atomic formulas, molecular formulas, and quantified formulas. Nothing else is a formula. Every formula is either atomic, or it has a main connective or a quantifier with scope over the whole formula. The main connective or quantifier in a formula is the last connective or quantifier that was added in constructing the formula. Formulas may be parsed as in chapters 1 or 2. Some examples are: ∀x(Gx → ∃yFy) | (Gx → ∃yFy) 2 Gx ∃yFy | Fy

∀x~∃y(Gx ∨ ~Fy) | ~∃y(Gx ∨ ~Fy) | ∃y(Gx ∨ ~Fy) | (Gx ∨ ~Fy) 2 Gx ~Fy | Fy

EXERCISES 1. For each of the following, say whether it is a formula in official notation, or in informal notation, or not a formula at all. If it is a formula, parse it. a. ~∀x(Fx → (Gx ∧ Hx)) b. ∃x~~Gx → Hx ∨ ∃yGy c. ~(Gx ↔ ~Hx) d. ∀xGx ∧ ∃Hx e. Fa → (Gb ↔ Hc) f. ∀x(Gx ↔ x ∨ Ha) g. ∀x(Gx ↔ Hx) → Ha ∧ ∃zKz

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CHAPTER 3 SECTION 3

3 SCOPE AND BINDING In the following we will need to distinguish a symbol from an occurrence of that symbol. For example, the formula: ∀xFx contains one variable, the variable 'x', which occurs twice in the formula. It has one occurrence as part of the quantifier, and one occurrence immediately following the predicate 'F'. It will be important to be able to say when an occurrence of a quantifier binds an occurrence of a variable. This can be given a precise explanation in terms of the scope of an occurrence of a quantifier. The scope of an occurrence of a quantifier includes itself along with the formula to which it was prefixed when constructing the whole formula. Here are some occurrences of quantifiers and their scopes, indicated by underlining. (The line immediately under a quantifier occurrence indicates its scope.) ∀x Fx ∀x(Fx → Gx) ∃xFx ∧ ∃y(Gy ∧ Hy) ∃x(Fx ∧ ∀yGy) ∃x(Fx ∧ ∃y(∃zGz ∧ Hy)) Using the notion of the scope of a quantifier, we can say when a quantifier occurrence binds an occurrence of a variable in a formula: A quantifier occurrence binds an occurrence of a variable if the variable occurrence is within the scope of the quantifier occurrence the variable occurrence is the same letter as the one in the quantifier occurrence the variable occurrence is not already bound by another quantifier occurrence within the scope of the first quantifier occurrence (Notice that a variable occurrence that is part of a quantifier is automatically bound by that quantifier.) The arrows here indicate which variables are bound by the quantifier: ∀x(Fx → Gx) The initial quantifier binds both occurrences of 'x' because (1) they are within its scope, (2) they are the same letter as the one in the quantifier itself, and (3) they are not already bound by another quantifier in the formula. These examples are similar: ∃xFx ∧ ∃y(Gy ∧ Hy) ∃x(Fx ∧ ∀yGy) ∃x(Fx ∧ ∃y ( ∃zGz ∧ Hy))

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CHAPTER 3 SECTION 3

∃x(Fx ∧ ∃y ( ∃zGz ∧ Hy ∧ Hx))

The following example illustrates a case in which an occurrence of 'x' (the last one) is not bound by the initial quantifier '∃x′, even though it is within its scope. This is because there is another quantifier inside that already binds that occurrence of 'x': ∃x(Fx ∧ ∃x ( ∃zGz ∧ Hx))

Using the notion of a quantifier binding an occurrence of a variable, we can define what a sentence is: A sentence is any formula in which every occurrence of a variable in the formula is bound by an occurrence of a quantifier in the formula. A variable occurrence that is not bound is called "free". So a sentence can also be defined as a formula that contains no free occurrences of variables. All of the examples given above are sentences. The following formulas are not sentences because certain occurrences of variables in them are not bound any of their quantifiers: ∀x(Fy → Gx)

no quantifier contains 'y'

∃xFx ∧ ∃y(Gx ∧ Hy)

the scope of the initial quantifier does not include the second 'x'

∃x(Fx ∧ ∃y (∃zGz ∧ Hz))

the scope of the quantifier with 'z' does not extend far enough

∃x(Fx ∧ ∃x(∃zGz ∧ Hy))

no quantifier contains 'y'

EXERCISES 1. For each of the following, say whether it is a sentence, a formula that is not a sentence, or not a formula at all. (Include sentences and formulas in informal notation as sentences and formulas.) If it is a sentence or formula, indicate which quantifiers bind which variables. a. b. c. d. e. f. g. h. i. j.

∃x(Fx ∧ ∀y(Gy ∨ Hx)) ∃y(Hy ∧ ∃~zHz) ∃z~(Hz ∧ Gx ∧ ∃xIx) ~(~Gx → ∀y(Jx ∧ Ky ↔ Lx)) ∃xGx ↔ ∃y(Gy ∧ Hx) ∀x(Gx → ∀y(Hy → ∀z(Iz → Hx ∧ Gz))) ∀x∃y(Hx ↔ ~Gy) ∀xy(Gx ∧ Hy → Kx) ∀x(Gx ∧ ∃y → Hx ∧ Jy) ∀x∃y∀z(Gx ↔ ∃w(Hw ∧ ~Hx ∧ Gy))

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CHAPTER 3 SECTION 4

4 MEANINGS OF THE QUANTIFIERS What do quantifiers mean? This can be answered indirectly by giving a way to read symbolic formulas in English. We already know how to read the parts of formulas without quantifiers or variables; we have: Gh Ea Gh ∧ Ea Gh → Ea

Henry is a giraffe Ann with run for reelection Henry is a giraffe and Ann will run for reelection. If Henry is a giraffe then Ann will run for reelection.

We can read a quantified formula by adding these provisions: Read any universal quantifier as "everything is such that", while reading any variable that it binds as a pronoun which has the 'everything' as its antecedent. Read any existential quantifier as "something is such that" while reading any variable that it binds as a pronoun which has the 'something' as its antecedent. Here are some examples: ∀xGx ∃x(Gx ∧ Ex) ∀x(Gx → Ex)

everything is such that it is a giraffe something is such that it is a giraffe and it will run for reelection everything is such that if it is a giraffe then it will run for reelection

These readings are stilted, and sometimes cumbersome. But they are accurate paraphrases of the symbolic notation. Often there are more natural ways to word an English sentence. For example, these are all equivalent: ∃x(Gx ∧ Ex) something is such that it is a giraffe and it will run for reelection something is a giraffe which will run for reelection some giraffe will run for reelection Likewise, these are all equivalent: ∀x(Gx → Ex) everything is such that if it is a giraffe then it will run for reelection everything, if it is a giraffe, will run for reelection every giraffe will run for reelection As in the case of connectives, we need to distinguish carefully between the official definition of the quantifiers and the question of how best to read them in English. The official definition of the quantifiers has to do with the truth-values of the sentences that are produced using them: Definitions of the quantifiers To tell whether or not a sentence of the form ∀x(...x...x...) is true: Remove the initial universal quantifier. Pretend that the variable it was binding is a name letter. If you now have a sentence that is true no matter what the pretend name stands for, then the original sentence is true; otherwise it is false. To tell whether or not a sentence of the form ∃x(...x...x...) is true: Remove the initial existential quantifier. Pretend that the variable it was binding is a name letter. If there is something that the pretend name could stand for such that the sentence you now have is true, then the original sentence is true; otherwise it is false.

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CHAPTER 3 SECTION 4

To apply this to the example 'Everything is either mental or physical': Begin with the sentence: ∀x(Mx ∨ Hx). Erase the initial quantifier, yielding: Mx ∨ Hx. Now pretend that `x' is a name letter, and ask ourselves: Is `Mx ∨ Hx' true no matter what `x' stands for? If the answer is yes, then the original sentence `∀x(Mx ∨ Hx)' is true; otherwise `∀x(Mx ∨ Hx)' is false. This test explains why we read `∀x(Mx ∨ Hx)' in English as `Everything is either mental or physical'. It is because the test for the truth of `∀x(Mx ∨ Hx)' succeeds if everything is indeed either mental or physical, and it fails if not everything is either mental or physical. To see that this is so, compare the meaning of the English sentence with the official statement of the conditions under which the symbolized version is true: Suppose that certain philosophers are right, and everything is either mental or physical. Then if we treat `x' as a name letter, the phrase `Mx ∨ Hx' must be true no matter what `x' stands for. Because it can only stand for something that is mental or physical (that's all there is), and if it stands for something mental the first disjunct is satisfied, and if it stands for something physical then the second disjunct is satisfied. Suppose on the other hand that not everything is either mental or physical. (Suppose, as some philosophers have argued, that the number 4 is neither a mental thing nor a physical thing.) Then if we treat `x' as a name letter, we will not find that the phrase `Mx ∨ Hx' is true no matter what `x' stands for. For if `x' stands for the number 4, neither disjunct will be satisfied. These considerations do not settle the question of whether everything is either mental or physical. Instead they show that there is an equivalence between the truth-value, in English, of the sentence `Everything is either mental or physical', and the truth-value, according to our official account, of the predicate calculus sentence `∀x(Mx ∨ Hx)'. EXERCISES 1. Suppose that `A' stands for `is a sofa', `B' stands for `is well-built' and `C' stands for `is comfortable'. For each of the following sentences, produce an accurate but "cumbersome" reading in English as well as a natural idiomatic reading if possible. a. b. c. d.

∃x(Ax ∧ Bx) ∀x(Ax → Bx) ∀x(Ax ∨ Bx) ∃x~Ax

e. f. g. h.

∀y~Ay ∀z(Az ∧ Bz → Cz) ∃xCx ∧ ∀yBy ∃x(Cx → ∀yBy)

2. Assume that all giraffes are friendly, and that some giraffes are clever and some aren't. What are the truth-values of these sentences? a. ∀x(Gx → Fx) b. ∀x(Gx → Cx) c. ∃x(~Fx ∧ Gx)

d. ∃y(Fy ∧ Cy) e. ∃z(Gz ∧ Cz) f. ∀x(Gx → ~Gx)

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CHAPTER 3 SECTION 5

5 SYMBOLIZING SENTENCES WITH QUANTIFIERS 5A CATEGORICAL SENTENCES The ancient Greek philosopher Aristotle is generally credited with the invention of formal logic. He devised a fairly complete and accurate study of the logical relations among sentences of a certain special sort. These are called "categorical" sentences, and they include any sentence which has one of the following forms (with Aristotle's titles): Universal affirmative: Particular affirmative: Universal negative: Particular negative:

Every A is B Some A is B No A is B Some A is not B

These categorical sentences are only a few of the forms that can be represented in modern predicate logic, but they are simple and basic, and their treatment provides a nice introduction to the symbolism. A universal affirmative sentence of the form: Every A is B is represented in the predicate calculus as: ∀x(Ax → Bx). You can judge the adequacy of this for yourself by comparing the reading of the symbolic version with the English form; that is, compare: Everything is such that if it is an A then it is B with: Every A is B. The question to ask for logical purposes is: Is there any possible situation in which these two sentences differ in truth-value? If they agree in all logically possible situations, then the proposed symbolization is a good one; otherwise not. Here is some reasoning that suggests the symbolization is a good one: Suppose that in some possible situation every A is B. Then, in that situation everything will be such that if it's an A then it is B. Suppose on the other hand that not every A is B. Then there will be something that is an A but is not B. So it won't be true that everything is such that if it's an A it is B. Traditionally, the main reservation expressed about this symbolization concerns a possible situation in which there are no A's at all. Suppose that a naturalist is uncertain about whether or not there are any friendly elephants, but is willing to assert: Every friendly elephant is an herbivore. Suppose that there are in fact no friendly elephants. Then is what the naturalist said true or false? If we accept the proposed symbolization above, we will represent the naturalist as having said something true. Let us see why this is so. The proposed symbolization is: ∀x(x is a friendly elephant → x is an herbivore), that is: ∀x(Fx ∧ Ex → Hx). If there are no friendly elephants, this sentence will be true, because, treating `x' as a name letter, the following is true no matter what `x' stands for: Fx ∧ Ex → Hx. It is true because no matter what `x' stands for, the antecedent is false (because there are no friendly elephants). Chapter Three -- 10

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CHAPTER 3 SECTION 5

Is that a proper treatment of the English sentence that was asserted? The consensus on this matter seems to be "sometimes yes, sometimes no." That is, sometimes when we say "Every A is B" we presuppose or imply that there are some A's, and sometimes we are neutral on this. In this text we will always take the weaker interpretation, supposing that "Every A is B" does not commit you to there being any A's. It is true, not false, if there are no A's. This is just a convention (a widely adopted one) for our convenience. (If you want a version of 'Every A is B' that does commit you to there being A's, you can write instead: '∃xAx ∧ ∀x(Ax → Bx)'.) The particular affirmative form -- "Some A is B" -- is easy to symbolize; it gets represented as: ∃x(Ax ∧ Bx), that is, "Something is such that it is both A and B." Plural forms of categorical sentences are symbolized just like the singular forms: All A's are B Some A's are B

Every A is B Some A is B

∀x(Ax → Bx) ∃x(Ax ∧ Bx)

This might seem wrong if you think that the use of the plural in English commits you to the view that there is more than one A which is B. (The symbolized version has no such commitment.) The answer seems to be that we sometimes use the plural to convey the thought that there is more than one A, but sometimes we are neutral about this. In this text we will adopt the weaker interpretation, which makes "Some A's are B" true whenever there is at least one A that is B. The universal negative form is: No A is B There are two equally natural ways to symbolize this. One way depends on noticing that "No A is B" is equivalent to saying "Every A is not B," which can be symbolized as: ∀x(Ax → ~Bx). The other way is to notice that "No A is B" is equivalent to denying that "At least one A is B," and symbolizing the sentence as: ~∃x(Ax ∧ Bx). Soon we will be able to prove that these two forms are logically equivalent. There are two traps to beware of when symbolizing categorical sentences. They both involve trying to make the symbolizations of "universal" and "particular" sentences look alike. Suppose that we want to symbolize: Some dogs are brown. It will not be correct to symbolize this as: ∃x(Dx → Bx), that is: Something is such that if it's a dog then it's brown. This would be wrong because in some possible situations the symbolized version would differ in truth-value from the English version. Consider a possible situation which is just like the actual one except that all dogs are black, white, or grey. The English sentence 'Some dogs are brown' would be false in that situation. But the symbolized version would be true in that situation. It would be true for the totally irrelevant reason that not everything is a dog!!! Remember the official account of the existential quantifier; the sentence: Chapter Three -- 11

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∃x(Dx → Bx) is true if there is something to let `x' stand for which makes this true: Dx → Bx. But that's easy; just let `x' stand for some thing that is not a dog -- and then we have a conditional whose antecedent is false. And such a conditional is true. The symbolized version is automatically true if there is anything that isn't a dog, whereas the English sentence is not automatically true in such a situation. So the symbolization is not a good one to use for that English sentence. The other trap is to try to symbolize: Every A is B as: ∀x(Ax ∧ Bx). For example, you might try to symbolize: Every dog is a mammal as: ∀x(Dx ∧ Mx). It is easy to see that this cannot be a correct symbolization, for the English sentence is true, whereas the symbolized version is false. The symbolized version says: Everything is such that it is a dog and it is a mammal, that is: Everything is both a dog and a mammal. But you are not both a dog and a mammal, so the symbolic sentence is false. So the symbolic sentence is not a correct way to represent the English sentence we are trying to symbolize, `All dogs are mammals', since the English sentence is true. The right way to translate the English sentence is the way discussed above: ∀x(Dx → Mx). EXERCISES 1. Symbolize these sentences. a. Every handsome elephant is friendly. b. No handsome elephant is friendly. c. Some elephants are not handsome. d. Some handsome elephants are friendly. e. Each friendly elephant is handsome. f. A handsome elephant is not friendly. g. No friendly elephant is handsome.

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5B COMPLEX CATEGORICAL FORMS Many sentences are constructed out of categorical forms. An example is: Every brown dog is happy and well-fed To symbolize this sentence, notice that the sentence in fact is a universal affirmative sentence; it just happens to have a complex antecedent and a complex consequent. So begin by using the pattern for universal affirmatives: ∀x(x is a brown dog → x is happy and well-fed) Then complete the symbolization by filling in the details in the antecedent and consequent: ∀x(Bx∧Dx → Hx∧Fx) (The combination Adjective + Noun, such as 'brown dog', gets symbolized as a conjunction. For the cases under consideration in this text, that is always the way to symbolize a combination consisting of an adjective modifying a noun.) This example is similar: Some brown dog isn't either happy or lively. Its overall form is that of a particular affirmative: ∃x(x is a brown dog ∧ x isn't either happy or lively) Its symbolization is then got by filling in the details in the conjuncts: ∃x(Bx ∧ Dx ∧ ~(Hx ∨Lx)) Some other examples like this are: No dog is happy unless every dog is well-fed ∀x(x is a dog → ~x is happy) unless ∀x(x is a dog → x is well-fed) ∀x(Dx → ~Hx) ∨ ∀x(Dx → Fx) Each dog is happy unless it isn't well-fed ∀x(x is a dog → x is happy unless x is not well-fed) ∀x(Dx → Hx ∨ ~Fx) As we have seen, categorical sentences can themselves be combined with connectives. Another example is: If every dog is well-fed, and every dog is an animal, and every animal is happy, then every dog is both well-fed and happy. This is a complex of categorical sentences: If ∀x(Dx → Fx) and ∀y(Dy → Ay) and ∀z(Az → Hz) then ∀z(Dz → Fz ∧ Hz) that is: ∀x(Dx → Fx) ∧∀y(Dy → Ay) ∧∀z(Az → Hz) → ∀z(Dz → Fz ∧ Hz) Sometimes a sentence is apparently ambiguous, but variable binding resolves the ambiguity. This happens in the example Each dog is happy unless it isn't well-fed We decided above to include the 'unless' as part of the consequent of the quantified conditional. We might try instead to make 'unless' be the major connective: Chapter Three -- 13

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∀x(x is a dog → x is happy) unless x isn't well-fed ∀x(Dx → Hx) ∨ ~Fx However, this leaves the 'x' unbound by the quantifier. You have a formula that is not a sentence, and there is no way to interpret the unbound occurrence of 'x'. Whenever a symbolization of an ordinary meaningful English sentence ends up with a variable that is not bound by any quantifier, the symbolization will not be correct.

EXERCISES 2. Suppose that `A' stands for `is a U.S. state', `C' for `is a city', `L' for `is a capital', and `E' for `is in the Eastern time zone'. What are the truth values of these sentences? a. ∀x(Cx → Lx) b. ∃x(Cx ∧ Lx) c. ∃x(Cx ∧ Lx ↔ Ex) d. ∀x(Cx ∧ Ex → Ax) e. ~∃x(Ax ∧ Ex) f. ∃x(Cx ∧ Ex) ∧ ∃x(Cx ∧ ~Ex) g. ∃x(Cx ∧ Ex ∧ Ax) h. ~∃x(Cx ∧ ~Cx) 3. Symbolize the following sentences: a. b. c. d. e. f. g. h. i. j. k. l. m. n. o.

All giraffes are spotted. All clever giraffes are spotted. No clever giraffes are spotted. Every giraffe is either spotted or drab. Some giraffes are clever. Some spotted giraffes are clever. Some giraffes are clever and some aren't. Some spotted giraffes aren't clever. No spotted giraffe is clever but every unspotted one is. Every clever spotted giraffe is either wise or foolhardy. Either all spotted giraffes are clever, or all clever giraffes are spotted. Every clever giraffe is foolhardy. If some giraffes are wise then not all giraffes are foolhardy. All giraffes are spotted if and only if no giraffes aren't spotted. Nothing is both wise and foolhardy.

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5C "ONLY" In chapter 1 we looked at how 'only' affects the symbolization of conditionals. The same word occurs in connection with quantification. Consider the sentence: Only dogs are happy Reflection on what this says indicates that it could be symbolized the same as: Any non-dog isn't happy and thus as: ∀x(~Dx → ~Hx) But intuitively the sentence is also equivalent to: Anything that's happy is a dog ∀x(Hx → Dx) Fortunately, we will be able to prove later that these two forms are equivalent. Recall that the effect of 'only' on 'if' is to reverse antecedent and consequent. Something like that occurs here too; compare the sentences: All dogs are happy Only dogs are happy

∀x(Dx → Hx) ∀x(Hx → Dx)

They look pretty much the same except that the antecedent and consequent of the quantified conditional are switched. Here are some examples of symbolizations of sentences using 'only': Dogs can run, but only birds can fly. ∀x(Dx → Cx) ∧∀x(Fx → Bx) Only birds can fly, but not all of them can. ∀x(Fx → Bx) ∧~∀x(Bx → Fx) Dogs are happy and frisky; giraffes are happy, but only the well-fed ones are frisky. ∀x(Dx → Hx ∧ Fx) ∧∀x(Gx → Hx) ∧∀x(Gx ∧ Fx → Ex) (Using 'E' for 'is well-fed'.) Notice that the last conjunct is not symbolized as: ∀x(Fx → Gx ∧ Ex) This would say that everything that is frisky is a well-fed giraffe, which is not what is intended. The point is that among giraffes only the well-fed ones are frisky. The last conjunct could also be symbolized as: ∀x(Gx → (Fx → Ex)) The word 'only' can create ambiguity. Consider the sentence: Only brown dogs are happy This could be read as saying that everything that is happy is a brown dog: ∀x(Hx → Bx ∧ Dx) or it could be read as saying that among dogs, every happy one is brown: Chapter Three -- 15

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∀x(Dx → (Hx → Bx)) Usually we don't notice such ambiguity since it is usually clear from context which is meant. Emphasis also helps; saying "Only brown dogs are happy" indicates that among dogs, only the brown ones are happy. Out of context, the sentence is simply ambiguous.

EXERCISES 4. Symbolize these sentences. If a sentence is ambiguous, give all pertinent symbolizations. a. Only friendly elephants are handsome b. If only elephants are friendly, no giraffes are friendly c. Only the brave are fair. d. If only elephants are friendly then every elephant is friendly e. All and only elephants are friendly. f. If every elephant is friendly, only friendly animals are elephants g. If any elephants are friendly, all and only giraffes are nasty h. Among spotted animals, only giraffes are handsome. i. Among spotted animals, all and only giraffes are handsome j. Only giraffes frolic if annoyed.

5D RELATIVE CLAUSES Relative clauses modify nouns, as adjectives do, although relative clauses are typically more complex. There are two sorts of relative clause: restrictive and non-restrictive, illustrated by: Non-restrictive Restrictive

Dogs, which are frisky, are cute Dogs which are frisky are cute

Non-restrictive relative clauses do not affect the noun they follow; instead they are used to insert a comment in addition to what the main sentence says. The main sentence of the non-restrictive example is that dogs are cute, and the additional comment is that they are frisky. The entire sentence is used to make both of these claims. If we want to capture the whole content of a sentence with a non-restrictive relative clause the best we can do is to conjoin the two claims: Dogs are frisky ∧ Dogs are cute

∀x(Dx → Fx) ∧∀x(Dx → Cx)

A restrictive relative clause restricts the content of the noun to which it is adjoined. In the restrictive example above, it is frisky dogs that are said to be cute, not dogs in general. The symbolization is: Dogs which are frisky are cute

∀x(Dx ∧ Fx → Cx)

You can usually tell a non-restrictive relative clause, for it is set off from its surroundings by commas before and after it. When there are no commas, we assume in this text that the reading is restrictive. Restrictive relative clauses are like adjectives, in that in logical form they are conjoined with the noun that they modify. In the above example 'dogs which are frisky' becomes the conjunction 'Dx ∧ Fx'. When the relative clause is more complex, it gives you something complex to conjoin to the part originating with the noun that is modified. This is seen in: Every dog which is neither cute nor frisky is not happy. ∀x(Dx ∧ ~(Cx ∨ Fx) → ~Hx)

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EXERCISES 5. Symbolize these sentences. a. Every giraffe which frolics is happy b. Only giraffes which frolic are happy c. Only giraffes are animals which are long-necked. d. If only giraffes frolic, every animal which is not a giraffe doesn't frolic. e. Some giraffe which frolics is long-necked or happy. f. No giraffe which is not happy frolics and is long-necked. g. Some giraffe is not both long-necked and happy.

5E IMPLICIT UNIVERSAL QUANTIFIERS In the symbolizations we have considered so far, symbolic universal quantifiers have originated naturally from "universal" quantifier words of English. For example, the universal quantifier is often used in symbolizing a sentence with one of the words 'each', 'every', 'all' in it, and the position of the English quantifier word often corresponds to the position of the symbolic quantifier. In 'Every A is B' the English sentence begins with 'every' and its symbolization begins with '∀x'. Sometimes a universal quantification originates with an English indefinite article 'a' or 'an'. This happens in: A dog that is well-fed is happy. This sentence is most naturally treated as conveying a universal claim, that any dog that is well-fed is happy: ∀x(Dx ∧ Fx → Hx) This is in spite of the fact that the indefinite article often conveys an existential claim, as in: A girl left early ∃x(Gx ∧ Lx) A good test for this is whether the indefinite article can be paraphrased by 'each'; this is natural in the first example, but not in the second. A more interesting case is when an indefinite article occurs inside a sentence, indicating a universal quantification with scope over the whole sentence. This happens in: If a dog is well-fed, it is happy This appears to be a conditional of the form: a dog is well-fed → it is happy But that won't do, since there is nothing to bind the variable that comes from the 'it' in the consequent. Instead, the indefinite article indicates a universal quantification of dog, with the rest of the sentence within its scope. That is, it has the form: ∀x(x is a dog → (x is well-fed → x is happy)) ∀x(Dx →( Fx → Hx)) This happens in the following two examples as well. In the first: A giraffe is wise if and only if it's not foolhardy.

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This has a logical form something like: Every giraffe is such that it is wise if and only if it is not foolhardy ∀x(Gx → (Wx ↔ ~Fx)) This sentence is similar: A brown dog is frisky only if it is happy Every brown dog is such that it is frisky only if it is happy ∀x(Bx ∧ Dx → (Fx → Hx)) The idea that indefinite phrases sometimes correspond to universal quantifiers with wide scope applies also to plural indefinites -- to plural nouns or noun phrases which have no article or quantifier word before them. An example is: If dogs are well-fed, then they are happy ∀x(x is a dog → (x is well-fed → x is happy)) ∀x(Dx → (Ex → Hx))

EXERCISES 6. Symbolize the following sentences. a. If a giraffe is happy then it frolics unless it is lame. b. A monkey frolics unless it is not happy. c. Among giraffes, only happy ones frolic. d. All and only giraffes are happy if they are not lame. e. A giraffe frolics only if it is happy. f. Only giraffes frolic if happy. g. All monkeys are happy if some giraffe is. h. Cute monkeys frolic. i. Giraffes run and frolic if and only if they are blissful and exultant. j. If those who are healthy are not lame, then if they are exultant, they will frolic. k. Only giraffes and monkeys are blissful and exultant. l. The brave are happy. m. If a giraffe frolics, then no monkey is blissful unless it is. n. Giraffes and monkeys frolic if happy.

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6 DERIVATIONS WITH QUANTIFIERS Our first step in including quantificational sentences in derivations is to extend all of the rules from chapters 1 and 2 to include formulas which have free variables. Although we continue to use derivations for arguments consisting entirely of sentences, it will be essential to also allow formulas inside of the derivations. In this section we introduce three rules for quantifiers. Rule ui (universal instantiation): The first rule is simple; it says that if everything satisfies a certain condition, any particular thing satisfies that condition. That is, from any universally quantified formula one may infer the result of removing the initial quantifier, and replacing every occurrence of the variable that it was binding by a name letter or by a variable: Rule ui: (universal instantiation): ∀x ...x...x... ∴ …b…b…

∀x ...x...x... ∴ …y…y…

Every occurrence of 'x' that '∀x' was binding must be replaced with the same name or variable. An example of this rule is to validate the argument from 'everything is either mental or physical' to 'Disneyland is either mental or physical': ∀x(Mx ∨ Px) ∴ Ma ∨ Pa

by rule ui

A more typical application would be to use rule ui to validate an inference like this: Every giraffe is happy Fido is a giraffe ∴ Fido is happy ∀x(Gx → Hx) Gf ∴ Hf A derivation using rule ui to validate this argument could go like this: 1. Show Hf 2. Gf → Hf 3. Hf

pr1 ui pr2 2 mp dd

The universal instantiation step takes us from "everything is such that if it is a giraffe then it is happy" to "if Fido is a giraffe then Fido is happy". Modus ponens does the rest. In using rule ui the quantifier must be on the front of the formula and it must have scope over the whole formula. If it has a narrower scope, then it is fallacious to apply the rule. For example, this inference is not permitted: ∀xFx → Fg ∴ Fb → Fg

If everything is happy, Gertrude is happy If Betty is happy, Gertrude is happy

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Rule eg (existential generalization): The second rule is the reverse of the first, using the existential quantifier instead of the universal. It says that if a particular thing satisfies a certain condition, then something satisfies it. That is, from any formula one may infer the result of replacing some occurrences of a name letter or a variable in it by a new variable, putting an existential quantifier on the front using that variable. Rule eg (existential generalization): ...b...b... ∴ ∃x…x…b…

...y...y... ∴ ∃x…x…y…

(You need not replace every occurrence of 'b' or of 'y' by 'x'.) For example, if Fido is a brown dog, then something is a brown dog: Bf ∧ Df ∴ ∃x(Bx ∧ Dx) The existential quantifier that is put on the front must have scope over the whole formula. If the formula you start with is in informal notation, you may need to restore the dropped parentheses before applying the rule, as we did here. Here is a little derivation that uses both of these rules. It validates the argument: Every dog is happy Fido is a dog ∴ Something is happy ∀x(Dx → Hx) Df ∴ ∃xHx 1. Show ∃xHx 2. 3. 4.

Df → Hf Hf ∃xHx

pr1 ui 2 pr2 mp 3 eg dd

There is a difference between Rules ui and eg. When using rule ui, you must replace every occurrence of the variable that the initial quantifier binds with a name or variable. For example, you cannot do this: ∀x(Dx → Hx) ∴ Dx → Hb That is: Everything is such that if it is a dog then it is happy.

∴ If it is a dog then Bob is happy

Rule eg is different. When using rule eg you needn't replace all of the occurrences. For example, from: Bob is happy or Bob is sad you may infer Something is such that Bob is happy or it is sad. This conclusion looks odd, but it should be clear that it follows logically.

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Here is another example of a derivation using both of our new rules: Fido is a dog Every dog is happy ∴ Some dog is happy 1. Show ∃x(Dx ∧ Hx) 2. 3. 4. 5.

Df → Hf Hf Df ∧ Hf ∃x(Dx ∧ Hx)

pr2 ui 2 pr1 mp pr1 3 adj 4 eg dd

There is a constraint on both of these rules: there must be no "capturing". If a new variable appears in the conclusion of either rule that was not there previously, it must not be "captured" by a quantifier in the formula. Specifically, if a new variable appears, none of its new occurrences may be bound by a quantifier already in the formula. For example, this use of rule eg is not permitted: Df ∧ ∀x(Hf → Gx) ∴ ∃x(Dx ∧ ∀x(Hx → Gx))  the universal quantifier captures the variable 'x' that replaces the second 'f' No capturing: When using rule ui or rule eg a new variable must not be introduced if some of its new occurrences are bound by a quantifier in the original formula. You will not often encounter cases of capturing; they usually happen by accident. The possibility of capturing can be avoided by always choosing a variable that does not already occur in the formula.

EXERCISES 1. Symbolize these arguments and produce derivations for them. a.

The sky is blue Everything that is blue is pretty ∴ Something is pretty

b.

Every hyena is grey. Every hyena is an animal Jenny is a hyena ∴ Some animal is grey

c.

If some hyena is grey, every hyena is grey Every scavenger is grey Jenny is a hyena and a scavenger Kathy is a hyena ∴ Kathy is grey

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Rule ei (existential instantiation): Our third rule is rule ei (existential instantiation). It works just like universal instantiation, except that (1) it applies to an existential quantifier, (2) you must instantiate to a variable, not to a name letter, and (3) you must use a variable that has not already occurred in the derivation or in any of the premises that have been cited in the derivation. This rule is meant to capture the following kind of reasoning. Suppose that you are given the information: Every dog is happy Something is a dog and you wish to infer that something is happy. You are not told that any particular named thing is a dog; you just know that there are some. You might reason as follows: By the second premise, there are some dogs. Call one of them "z". Then z is happy (by the first premise), so something is happy. What you did was to choose a label, 'z', for some dog, without specifying which dog it is. Then you made inferences using that label, ending up with a conclusion that does not contain the label. The label was just a device to reason with. It was important that you chose a label that was not already assigned to something. If you used an already existing name for the label, that could lead to fallacies. For example, consider this bad argument: Every dog is happy Something is a dog Fluffy is a cat ∴ Some cat is happy It would be wrong to reason like this: By the second premise, there are some dogs. Call one of them "Fluffy". Then Fluffy is happy (by the first premise). Also, Fluffy is a cat (third premise). So some cat is happy. By using the name 'Fluffy' for one of the dogs you were implicitly assuming that Fluffy was a dog. That assumption is not justified. Formally we get around such an unjustified assumption by using only variables for labels, and by requiring that these variables are not already used for something else. We accomplish this by requiring that the new variable not have occurred already in the derivation: Rule ei: (existential instantiation): ∴

∃x ...x...x... …y…y…

You must replace every occurrence of 'x' that '∃x' was binding. The variable 'y' must not occur in the existentially quantified formula itself, or in previous lines in the derivation, or in a premise that has been cited on a previous line. Here now is a derivation using all of our new rules: ∀x(Bx ∧ Dx → Ex) ∃x(Dx ∧ Fx) ∀y(Fy → By) ∴ ∃z(Dz ∧ Ez)

Every brown dog is well-fed. Some dog is frisky Everything frisky is brown ∴ Some dog is well-fed

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1. Show ∃z(Dz ∧ Ez) 2. 3. 4. 5. 6. 7. 8. 9.

Du ∧ Fu Fu → Bu Bu Bu ∧ Du Bu ∧ Du → Eu Eu Du ∧ Eu ∃z(Dz ∧ Ez)

pr2 ei pr3 ui 2 s 3 mp 2 s 4 adj pr1 ui 5 6 mp 2 s 7 adj 8 eg dd

('u' has not already occurred in the derivation)

The reader should check to see that each of the new rules is properly used. This derivation illustrates an important strategy rule. Often you will have an opportunity to apply ei to introduce a variable, and then use ui to instantiate to that variable. In the derivation just given, ei introduces 'u' on line 2 and ui is used twice to instantiate to 'u', on lines 3 and 6. The strategy rule is that when this is a possibility, you should always apply rule ei before you apply rule ui. Strategy hint: When using both ei and ui to instantiate to the same variable, apply rule ei before rule ui. This is because if you try using ui first, you will not then be able to use ei to instantiate to the same variable, because the variable will not then be new. For example, suppose that you started the above derivation with: 1. Show ∃z(Dz ∧ Ez) 2. Fu → Bu 3. Du ∧ Fu

pr3 ui pr2 ei

Line 3 is fallacious because you have instantiated to 'u', but 'u' has already occurred in the derivation, which violates the constraint that the variable used in ei must be new. Here is a straightforward illustration of our three rules: Every crook who steals a lot and doesn't get caught is affluent.. No crook who gets caught is affluent. Some lucky crooks steal a lot. Some crooks who aren't lucky don't steal a lot. Every crook who isn't lucky gets caught. Every crook who is lucky doesn't get caught. ∴ Some crooks are affluent and some aren't. ∀x(Cx ∧ Ex ∧ ~Gx → Ax) ∀x(Cx ∧ Gx → ~Ax) ∃x(Lx ∧ Cx ∧ Ex) ∃x(Cx ∧ ~Lx ∧ ~Ex) ∀x(Cx ∧ ~Lx → Gx) ∀x(Cx ∧ Lx → ~Gx) ∴ ∃x(Cx ∧ Ax) ∧∃x(Cx ∧ ~Ax) (In doing this derivation recall that 'P ∧ Q ∧ R' is informal notation for '((P ∧ Q) ∧R)'.)

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1. Show ∃x(Cx ∧ Ax) ∧ ∃x(Cx ∧ ~Ax) Lz ∧ Cz ∧ Sz Cz ∧ Lz → ~Gz Sz Lz Cz ~Gz Cz ∧ Sz ∧ ~Gz → Az Az Cz ∧ Az ∃x(Cx ∧ Ax) Cu ∧ ~Lu ∧ ~Eu Cu ∧ ~Lu Cu ∧ ~Lu → Gu Gu Cu ∧ Gu → ~Au ~Au Cu ∧ ~Au ∃x(Cx ∧ ~Ax) ∃x(Cx ∧ Ax) ∧∃x(Cx ∧ ~Ax)

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

pr3 ei pr6 ui 2s 2ss 2ss 5 6 adj 3 mp pr1 ui 6 4 adj 7 adj 8 mp 6 9 adj 10 eg pr4 ei 12 s pr5 ui 13 14 mp pr2 ui 13 s 15 adj 16 mp 13 s 17 adj 18 eg 11 19 adj dd

Notice that the ei step in line 2 precedes the ui steps in lines 3 and 8, and that the ei step in line 12 precedes the ui steps in lines 14 and 16.

EXERCISES 2. Here is a fallacious derivation to validate this argument: ∃x(Nx ∧ Ex) ∃x(Nx ∧ Ox) ∴ ∃x(Nx ∧ Ox ∧ Ex)

some number is even some number is odd some number is both odd and even

Identify the error in the derivation. 1. Show ∃x(Nx ∧ Ox ∧ Ex) 2. 3. 4. 5.

Nz ∧ Ez Nz ∧ Oz Nz ∧ Oz ∧ Ez ∃x(Nx ∧ Ox ∧ Ex)

pr1 ei pr2 ei 2 s 3 adj dd

3. Produce derivations for each of the following (be careful to obey the strategy rule just given): a. theorem T202: ∴ ∀x(Fx → Gx) → (∃xFx → ∃xGx) b. half of T203: ∴ ∃x~Fx → ~∀xFx c. half of T204: ∴ ∀x~Fx → ~∃xFx T201 T202 T203 T204

∀x(Fx → Gx) → (∀xFx → ∀xGx) ∀x(Fx → Gx) → (∃xFx → ∃xGx) ~∀xFx ↔ ∃x~Fx ~∃xFx ↔ ∀x~Fx

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7 UNIVERSAL DERIVATIONS We have two instantiation rules, one for each quantifier, and we have a generalization rule for the existential quantifier. It is customary and useful to have some kind of universal generalization rule as well. For example, someone might want to reason as follows: Every dog is a mammal Every mammal is an animal ∴ Every dog is an animal A natural approach might be like this. Let x be anything whatsoever. Instantiating the first premise tells us that if x is a dog, it is a mammal; and instantiating the second premise tells us that if x is a mammal, it is an animal. So using techniques from chapter 1, we may infer that if x is a dog, x is an animal. Now since 'x' was chosen to represent anything whatever, we can infer that everything is such that if it is a dog it is an animal. That is, every dog is an animal. What we want to capture is the idea that if you can show something for any arbitrarily chosen thing, it holds for everything. Something like: Dx → Ax ∴ ∀x(Dx → Ax)

because x can be anything at all

For technical reasons, this principle will be formulated not as a rule, but as a special kind of derivation. It will take the form that if you want to show a universal claim, and you succeed in showing that it holds for a variable, x, then if x is completely arbitrary, you may box and cancel the show line for the universal claim. So the above reasoning will take this form: If you have a derivation of this form: Show ∀x(Dx → Ax) ::::: ::::: Dx → Ax

where x is completely arbitrary

Then you can box and cancel Show ∀x(Dx → Ax) ::::: ::::: Dx → Ax

ud

where x is completely arbitrary

The requirement that x be completely arbitrary is realized by the technical requirement that 'x' shall not have occurred free anywhere in the derivation available from the show line, or in a premise cited in such a line.

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The "ud" notation is the name of our new form of derivation: Universal Derivation (UD): If you have a derivation of the following form: Show ∀x . . . x . . . x . . . ::::: ::::: ...x...x... Then if there are no uncancelled show lines in between the first and last lines displayed, and if 'x' does not occur free on any line in the derivation available from the show line, or in any premise cited in such a line, you may box and cancel, using the notation 'ud'. The reasoning suggested above may now be incorporated into a derivation like this: ∀x(Dx → Mx) ∀y(My → Ay) ∴ ∀z(Dz → Az)

Every dog is a mammal Every mammal is an animal ∴ Every dog is an animal

1. Show ∀z(Dz → Az) 2. 3. 4. 5. 6. 7.

Dz → Mz Mz → Az Show Dz → Az Dz Mz Az

8.

pr1 ui pr2 ui ass cd 2 5 mp 3 6 mp

cd

4 ud

The reader should check that this derivation meets the conditions necessary for a ud derivation. In a previous exercise we proved half of theorem 203. The other half of T203 is more difficult, but we can do it using a universal derivation. ∴ ~∀xFx → ∃x~Fx It is easy to begin the derivation, setting up a conditional derivation: 1. Show ~∀xFx → ∃x~Fx 2. 3.

~∀xFx ?????

ass cd

With no other guide, our strategy rules say to try id: 1. Show ~∀xFx → ∃x~Fx 2. 3.

~∀xFx Show ∃x~Fx

ass cd

4. ~∃x~Fx ass id 5. ??? Again there is no clear way to proceed. Since we are trying to derive any contradiction, it is natural to try to derive the unnegation of line 2:

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1. Show ~∀xFx → ∃x~Fx 2. 3. 4. 5. 6.

~∀xFx Show ∃x~Fx

ass cd

~∃x~Fx Show ∀xFx

ass id

?????

We can in fact show this universally quantified formula, by means of a universal derivation. We only need to show the part folowing the quantifier: Fx. 1. Show ~∀xFx → ∃x~Fx 2. ~∀xFx 3. Show ∃x~Fx 4. 5.

~∃x~Fx Show ∀xFx

6. 7.

Show Fx ???

ass cd ass id

The rest is easy by means of another indirect derivation: 1. Show ~∀xFx → ∃x~Fx 2. 3. 4. 5. 6. 7. 8. 9.

~∀xFx Show ∃x~Fx ~∃x~Fx Show ∀xFx Show Fx ~Fx ∃x~Fx ~∃x~Fx

ass cd ass id

ass id 7 eg 4r

We can now complete the universal derivation: 1. Show ~∀xFx → ∃x~Fx 2. ~∀xFx 3. Show ∃x~Fx

ass cd

4. 5.

~∃x~Fx Show ∀xFx

ass id

6.

Show Fx

7. 8. 9.

~Fx ∃x~Fx ~∃x~Fx

10. 11. 12.

ass id 7 eg 4 r 8 id 6 ud

~∀xFx

2 r 6 id 3 cd

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The rest is straightforward: 1. Show ~∀xFx → ∃x~Fx 2. ~∀xFx 3. Show ∃x~Fx

ass cd

4. 5.

~∃x~Fx Show ∀xFx

ass id

6.

Show Fx

7. 8. 9.

~Fx ∃x~Fx ~∃x~Fx

10. 11. 12.

ass id 7 eg 4 r 8 id 6 ud

~∀xFx

2 r 6 id 3 cd

EXERCISES 1. Produce derivations for each of the following (be careful to obey the strategy rule just given): a. theorem T201: ∴ ∀x(Fx → Gx) → (∀xFx → ∀xGx) b. half of T204: ∴ ~∃xFx → ∀x~Fx (similar to the derivation of half of T203) c. half of theorem T205: ∴ ∀zFx → ~∃x~Fx

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8 SOME DERIVATIONS Many derivations take a common form. You begin with quantified sentences, and you remove quantifiers. Then you manipulate formulas using the techniques from chapters 1 and 2. Finally, you restore the quantifiers. In some cases this is straightforward: Every bear is friendly Some bear is dangerous ∴ Something dangerous is friendly ∀x(Px → Qx) ∃y(Py ∧ Ry) ∴ ∃z(Rz ∧ Qz) First we remove quantifiers using instantiation rules, being careful to apply ei before ui when that is possible: 1. Show ∃z(Rz ∧ Qz) 2. Pu ∧ Ru 3. Pu → Qu

pr2 ei pr1 ui

We choose to use 'u' in the universal instantiation step because it gives us something useful. Choosing other variables or names would be correct, but not useful. Now we use sentential rules to get a formula that we can existentially quantify: 4. 5.

Qu Ru ∧ Qu

2 s 3 mp 2 s 4 adj

Now we are in a position to existentially quantify line 5 to get the desired conclusion: 6.

∃z(Rz ∧ Qz)

5 eg

We can then box and cancel: 1. Show ∃z(Rz ∧ Qz) 2. 3. 4. 5. 6.

Pu ∧ Ru Pu → Qu Qu Ru ∧ Qu ∃z(Rz ∧ Qz)

pr2 ei pr1 ui 2 s 3 mp 2 s 4 adj 5 eg dd

Strategy hint: When a line is available that begins with a universal or existential quantifier, apply an instantiation rule, ei or ui, to derive an instance. When the conclusion is a universally quantified formula, it will very likely be derived by using a universal derivation. When a universal derivation is used, it is usually best to set up that derivation as early as possible. Consider this example: Every jaguar is a fast cat Every cat is an animal ∴ Every jaguar is a fast animal. ∀x(Jx → Fx ∧ Cx) ∀x(Cx → Ax) ∴ ∀x(Jx → Fx ∧ Ax)

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Our initial show line is a universally quantified sentence: 1. Show ∀x(Jx → Fx ∧ Ax) We can derive line 1 if we can show the formula that you get by removing its initial quantifier. So set that up as a show line: 2. Show Jx → Fx ∧ Ax This is a conditional, so try conditional derivation: 3.

Jx

ass cd

The rest of the conditional derivation is relatively straightforward: 4. 5. 6. 7. 8.

Jx → Fx ∧ Cx Fx ∧ Cx Cx → Ax Ax Fx ∧ Ax

pr1 ui 3 4 mp pr2 ui 5 s 6 mp 5 s 7 adj

We have derived the consequent of the conditional to be shown; after boxing and canceling we have: 1. Show ∀x(Jx → Fx ∧ Ax) 2. Show Jx → Fx ∧ Ax 3. 4. 5. 6. 7. 8.

Jx Jx → Fx ∧ Cx Fx ∧ Cx Cx → Ax Ax Fx ∧ Ax

ass cd pr1 ui 3 4 mp pr2 ui 5 s 6 mp 5 s 7 adj cd

Since line 2 has been shown, we may infer line 1 by universal derivation: 1. Show ∀x(Jx → Fx ∧ Ax) 2. 3. 4. 5. 6. 7. 8.

Show Jx → Fx ∧ Ax Jx Jx → Fx ∧ Cx Fx ∧ Cx Cx → Ax Ax Fx ∧ Ax

9.

ass cd pr1 ui 3 4 mp pr2 ui 5 s 6 mp 5 s 7 adj cd 2 ud

When the conclusion has both universal and existential quantifiers, the strategy is essentially to combine those above, applying whichever strategy is relevant at the time. Consider this argument: For every giraffe, there is a leopard which is happy if and only if it (the giraffe) is. For every leopard, there is a monkey that is happy if and only if it (the leopard) is. ∴ For every giraffe, there is a monkey which is happy if and only if it (the giraffe) is. ∀x(Gx → ∃y(Ly ∧ (Hy ↔ Hx))) ∀x(Lx → ∃y(My ∧ (Hy ↔ Hx))) ∴ ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx))) The conclusion to be shown is universally quantified, so set up a universal derivation. In fact, this should generally be done as early as possible.

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Strategy hint: If a universal derivation is to be used to show a universally quantified formula, ∀x□, set it up as early as possible, by inserting a Show line with ∀x□, and then proceed to derive the part following the quantifier, namely □. It is often convenient to immediately follow the show line containing ∀x□ by another containing □. This is done in line 2 here: 1. Show ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx))) 2. Show Gx → ∃y(My ∧ (Hy ↔ Hx)) Line 2 is a conditional, so try conditional derivation: 3. Gx ass cd Universally instantiating the first premise and using modus ponens is a natural thing to try: 4. Gx → ∃y(Ly ∧ (Hy ↔ Hx)) pr1 ui 5. ∃y(Ly ∧ (Hy ↔ Hx)) 3 4 mp We now have derived an existentially quantified formula, and there are some universally quantified ones in the premises. Generally, when both rules ei and ui are possible, as we stated above, you should use rule ei first. This is because rule ei introduces a variable which must be brand new in the derivation. If you do ei first, then you can do ui using the variable introduced by ei. But if you do ui first, you cannot do ei using that variable. In our derivation, the "ei before ui" strategy is relevant. Apply ei to line 5 using a variable that does not already occur in the derivation: 6.

Lz ∧ (Hz ↔ Hx)

We can now make use of our second premise to get: 7.

Lz → ∃y(My ∧ (Hy ↔ Hz))

pr2 ui

We can obviously use line 6 to get the consequent of line 7. That consequent is also existentially quantified, so we apply ei: 8. 9.

∃y(My ∧ (Hy ↔ Hz) Mu ∧ (Hu ↔ Hz)

6 s 7 mp 8 ei

Now look over what we have and what we want. We are in a conditional derivation, and we need to show '∃y(My ∧ (Hy ↔ Hx))' to complete that derivation. This formula is existentially quantified, and so we will probably derive it by existentially generalizing something. That is, we will existentially generalize something of the form: M_ ∧ (H_ ↔ Hx) We already have something very close to that, on line 9; we could get what we want by deriving a formula just like line 9 but with 'x" instead of 'z'. So suppose we try to derive ' Mu ∧ (Hu ↔ Hx)'. We already have the left conjunct, so the job is to derive the right conjunct 'Hu ↔ Hx'. This is a biconditional, so we need to derive two conditionals, probably by conditional derivation, and then put them together by cb. That in fact is easy to do: 10. 11. 12. 13. 14.

Show Hu → Hx Hu Hz Hx

ass cd 9 s bc 11 mp 6 s bc 12 mp cd

Show Hx → Hu

15. 16. 17.

Hx Hz Hu

18.

Hu ↔ Hx

ass cd 6 s bc 15 mp 9 s bc 16 mp cd 10 14 cb Chapter Three -- 31

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To finish, we only need to put line 18 together with the first conjunct on line 9, and existentially generalize: 19. 20.

Mu ∧ (Hu ↔ Hx) ∃y(My ∧ (Hy ↔ Hx))

9 s 18 adj 19 eg

This completes our conditional derivation, so we now have: 1. Show ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx))) 2.

Show Gx → ∃y(My ∧ (Hy ↔ Hx))

3. 4. 5. 6. 7. 8. 9. 10.

Gx Gx → ∃y(Ly ∧ (Hy ↔ Hx)) ∃y(Ly ∧ (Hy ↔ Hx)) Lz ∧ (Hz ↔ Hx) Lz → ∃y(My ∧ (Hy ↔ Hz)) ∃y(My ∧ (Hy ↔ Hz) Mu ∧ (Hu ↔ Hz) Show Hu → Hx

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

ass cd pr1 ui 3 4 mp 5 ei pr2 ui 6 s 7 mp 8 ei

Hu Hz Hx

ass cd 9 s bc 11 mp 6 s bc 12 mp cd

Show Hx → Hu Hx Hz Hu

ass cd 6 s bc 15 mp 9 s bc 16 mp cd

Hu ↔ Hx Mu ∧ (Hu ↔ Hx) ∃y(My ∧ (Hy ↔ Hx))

10 14 cb 9 s 18 adj 19 eg cd

Line 2 has now been shown by the conditional derivation. Now we only need to add line 21, and box and cancel, finishing the universal derivation. 1. Show ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx))) 2.

Show Gx → ∃y(My ∧ (Hy ↔ Hx)) [[DETAILS ABOVE]]

21.

2 ud

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EXERCISES 1. Symbolize these arguments and provide derivations to validate them. Give an explicit scheme of abbreviation for each. a.

If history is right, then if anyone was strong, Hercules was strong. Only those who work out are strong, and only those with self-discipline work out. ∴ If Hercules does not have self-discipline, then either history is not right or nobody is strong.

b.

If some giraffes are not happy, then all giraffes are morose. Some giraffes ponder the mysteries of life. ∴ If some giraffes are not morose, then some who ponder the mysteries of life are happy.

c.

There is not a single critic who either likes art or can paint. Some level-headed people are critics. Anyone who can't paint is uneducated. ∴ Some level-headed people are uneducated.

d.

No astronaut is a good dancer. Every singer is warm-blooded. If something is warm-blooded and is not a good dancer, then nothing that is either a singer or an astronaut is exultant. ∴ If some astronaut is a singer, then no singer is exultant.

e.

All students who have a sense of humor or are brilliant seek fame. Anyone who seeks fame and is brilliant is insecure. Whoever is a mogul is brilliant. ∴ Every student who is a mogul is insecure.

f.

There is a monkey that is happy if and only if some giraffe is happy. There is a monkey that is happy if and only if some giraffe is not happy. All monkeys are happy. ∴ It is not the case that either every giraffe is happy or none are.

g.

For every astronaut that writes poetry, there is one that doesn't. For every astronaut that doesn't write poetry, there is one that does. ∴ If there are any astronauts, some write poetry and some don't.

2. Derive theorems 203-208, 231-232. T203 T204 T205 T206 T207 T208 T209 T210

~∀xFx ↔ ∃x~Fx ~∃xFx ↔ ∀x~Fx ∀xFx ↔ ~∃x~Fx ∃xFx ↔ ~∀x~Fx ∃x(Fx ∨ Gx) ↔ ∃xFx ∨ ∃xGx ∀x(Fx ∧ Gx) ↔ ∀xFx ∧ ∀xGx ∃x(Fx ∧ Gx) → ∃xFx ∧ ∃xGx ∀xFx ∨ ∀xGx → ∀x(Fx ∨ Gx)

:::::::: T231 T232

∀xFx ↔ ∀yFy ∃xFx ↔ ∃yFy

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9 DERIVED RULES We have looked at formulas that have quantifiers on their front, or quantifiers that end up on front after a step such as modus ponens. Things are different if those quantifiers are preceded by a negation sign. Consider the following simple derivation: Every A is B. Nothing is both B and C. So every A isn't C. ∀x(Ax → Bx) ~∃x(Bx ∧ Cx) ∴ ∀x(Ax → ~Cx) This is intuitively valid, but deriving it requires slightly indirect reasoning. Our conclusion is universally quantified, so we set up a universal derivation right away: 1. Show ∀x(Ax → ~Cx) 2. Show Ax → ~Cx This is a conditional, so we try conditional derivation: 3.

Ax

ass cd

We can spell out some obvious consequences of what we have by instantiating the first premise and doing modus ponens: 4. 5.

Ax → Bx Bx

pr1 ui 3 4 mp

The second premise in fact is not anything we can make use of by applying any of our quantifier rules. Some other approach is needed. At this point it is useful to fall back on a technique from chapter 1; we are trying to derive '~Cx", so try to derive it by indirect derivation: 6. 7.

Show ~Cx Cx

ass id

We are not in a position to use the second premise directly, but we can use it indirectly by deriving something that contradicts it. This is simple in two lines: 8. 9.

Bx ∧ Cx ∃x(Bx ∧ Cx)

Now we complete our indirect derivation with: 10.

~∃x(Bx ∧ Cx)

pr2 9 id

boxing and canceling to get: 1. Show ∀x(Ax → ~Cx) 2. Show Ax → ~Cx 3. Ax 4. Ax → Bx 5. Bx 6. Show ~Cx 7. 8. 9. 10.

Cx Bx ∧ Cx ∃x(Bx ∧ Cx) ~∃x(Bx ∧ Cx)

ass cd pr1 ui 3 4 mp ass id 5 7 adj 8 eg pr2 9 id

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This essentially completes the derivation. For line 6 has completed the conditional derivation that starts on line 2, and once the 'show' on line 2 is cancelled, line 1 follows by universal derivation: 1. Show ∀x(Ax → ~Cx) 2. 3. 4. 5. 6. 7. 8. 9. 10.

Show Ax → ~Cx Ax Ax → Bx Bx Show ~Cx Cx Bx ∧ Cx ∃x(Bx ∧ Cx) ~∃x(Bx ∧ Cx)

ass cd pr1 ui 3 4 mp ass id 5 7 adj 8 eg pr2 9 id

11.

6 cd

12.

2 ud

This kind of indirect strategy is typical of how to handle derivations with sentences that begin with negated quantifiers when we use only our basic rules for quantifiers. However, it is usually more useful to use some derived rules that let us replace initial negated quantifiers by unnegated ones of the opposite sort, which may be used directly. The rule called quantifier negation does this. It lets you replace a negated initial quantifier by the opposite quantifier followed by a negation. If we lump in all applications of double negation, we get eight cases: Rule qn (Quantifier negation) ~∀xFx ∴ ∃x~Fx

~∃xFx ∴ ∀x~Fx

∀xFx ∴ ~∃x~Fx

∃xFx ∴ ~∀x~Fx

~∀x~Fx ∴ ∃xFx

~∃x~Fx ∴ ∀xFx

∀x~Fx ∴ ~∃xFx

∃x~Fx ∴ ~∀xFx

These derived rules are based on T203-206, which are given in the last set of exercises. Here is how we can use rule qn to shorten the derivation above. We begin as before: ∀x(Ax → Bx) ~∃x(Bx ∧ Cx) ∴ ∀x(Ax → ~Cx) 1. Show ∀x(Ax → ~Cx) 2. Show Ax → Cx 3. Ax 4. Ax → Bx 5. Bx

ass cd pr1 ui 3 4 mp

Now instead of introducing a subderivation to make indirect use of the second premise, we apply rule qn to that premise and then make direct use of the result; this lets us proceed quickly to get the desired '~Cx':

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1. Show ∀x(Ax → ~Cx) 2.

Show Ax → ~Cx

3. 4. 5. 6. 7. 8. 9.

Ax Ax → Bx Bx ∀x~(Bx ∧ Cx) ~(Bx ∧ Cx) ~Bx ∨ ~Cx ~Cx

ass cd pr1 ui 3 4 mp pr2 qn  8 ui 9 dm 5 dn 8 mtp cd

10.

2 ud

The advantage is not just that the derivation is two lines shorter, but the reasoning is simpler, and it is easier to think up. For that reason we have this strategy hint: Strategy hint: If an available formula begins with a negation sign immediately followed by a quantifier which has scope over the rest of the formula, convert it to a more useful formula by applying rule qn to it. Here is another example of the use of rule qn. We are given this argument to validate: ~∃x(Ax ∧ Bx) ∀y(Ay ↔ ~Cy) ∀y(Dy → By) ~∀xCx ∴ ∃x~Dx Neither the first nor the fourth premise may be used as an input to one of the basic quantifier rules. However, rule qn turns them into useful forms. 1. Show ∃x~Dx 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

∃x~Cx ~Ck Ak ↔ ~Ck Ak ∀x~(Ax ∧ Bx) ~(Ak ∧ Bk) ~Ak ∨ ~Bk ~Bk Dk → Bk ~Dk ∃x~Dx

pr4 qn  2 ei pr2 ui 4 bc 3 mp pr1 qn  6 ui 7 dm 5 dn 8 mtp pr3 ui 9 10 mt 11 eg dd

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Rule av: There is another useful derived rule, though one not so often used. Given our explanation of quantifiers, our choice of bound variables is irrelevant; one is as good as another. This is made explicit in derived rule av ("alphabetic variance"). The rule says that alphabetically varying the choice of a bound variable used with an initial quantifier yields an equivalent formula. In particular: Rule av (alphabetic variance) From a formula of the form '∀x . . . x . . x . . .', where the initial quantifier has scope over the whole formula, you may infer '∀y . . . y . . y . . .', which is the result of changing the variable 'x' in the quantifier to another variable, 'y', and changing all variables inside the first formula that are bound by the initial quantifier to 'y'. Likewise if the initial quantifier is '∃' instead of '∀'. Constraint: No capturing is allowed. That is, this inference is not permitted if a new occurrence becomes bound by a quantifier inside of the original formula. As an example, from ∀z(Dz ∧ Ez → ∃u(Du ∨ Fz)) you may infer ∀w(Dw ∧ Ew → ∃u(Du ∨ Fw)). But you may not infer ∀u(Du ∧ Eu → ∃u(Du ∨ Fu)) because that violates the no capturing rule. Rule av is based on theorems T231 and T232, proved in the exercises. Here is a situation in which rule av is useful. Suppose you are given the argument: ∀z(Dz ∧ Ez → ∃u(Du ∨ Fz)) ∀x(Dx → ~Fx) ∴ ∀u(Du → ~Eu) A natural derivation might go like this. The conclusion is universally quantified, so set up a universal derivation: 1. Show ∀u(Du → ~Eu) 2. Show Du → ~Eu This is a conditional, so set up a conditional derivation: 3.

Du

ass cd

You now need to show ~Eu, and it is natural to set up an indirect derivation to show this: 4. 5.

Show ~Eu Eu

ass id

Now universally instantiate the first premise: 6.

Du ∧ Eu → ∃u(Du ∧ Fu)

pr1 ui

Oops, you can't do that! The 'u' following the 'F' gets captured by the quantifier in the consequent of the conditional. So what can we do? Different ideas might be tried, but it is easy if we use rule av. Don't start out to derive the conclusion, because it uses a variable that gets you in trouble. Instead, derive a Chapter Three -- 37

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sentence that is exactly like the conclusion, but one that uses a different variable. Then use rule av to change this into the desired conclusion. Here is a derivation which reaches a sentence just like the conclusion except for using a different variable: 1. Show ∀u(Du → ~Eu) 2.

Show ∀w(Dw → ~Ew)

3.

Show Dw → ~Ew

4. 5. 6. 7. 8. 9. 10. 11. 12.

Dw ass cd Show ~Ew Ew ass id Dw ∧ Ew → ∃u(Du ∧ Fw) pr1 ui ∃u(Du ∧ Fw) 4 6 adj 7 mp Ds ∧ Fw 8 ei Fw 9s Dw → ~Fw pr2 ui ~Fw 4 11 mp 10 id

13.

 no capturing occurs here

3 ud

Because we used 'x' instead of 'u', we did not encounter any capturing problems in applying rule ui. Now we merely apply rule av to line 2, and we are done: 1. Show ∀u(Du → ~Eu) 2.

Show ∀w(Dw → ~Ew)

3.

Show Dw → ~Ew

4. 5. 6. 7. 8. 9. 10. 11. 12.

Dw ass cd Show ~Ew Ew ass id Dw ∧ Ew → ∃u(Du ∧ Fw) pr1 ui ∃u(Du ∧ Fw) 4 6 adj 7 mp Ds ∧ Fw 8 ei Fw 9s Dw → ~Fw pr2 ui ~Fw 4 11 mp 10 id

13.

3 ud

14. ∀u(Du → ~Eu)

2 av dd

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EXERCISES 1. Provide derivations for these arguments. a.

~∃x(Ax ∨ Bx) ∀x∀y(Gx ∧ Hy → By) ∃xGx ∴ ∀x~Hx

b.

∃x(Hx ∧ ~∃y(Gy ∧ Hx)) ∴ ∀y~Gy

c.

∀x(Ax → ∀y(Bx ↔ By)) ∃zBz ∴ ∀y(Ay → By)

d.

~∀x(Dx ∨ Ex) ∃x(Fx ↔ ~Ex) → ∀zDz ∴ ∃x~Fx

e.

Jc ∧ ~Jd ∀xKx ∨ ∀x~Kx ∃x(Jx ∧ Kx) → ∀x(Kx → Jx) ∴ ~Kc

2. Provide derivations for these theorems: T229 ∃x(∃xFx → Fx) T230 ∃x(Fx → ∃xFx) T234 ∀x((Fx → Gx) ∧ (Gx → Hx) → (Fx → Hx)) T235 ∀x(Fx → Gx) ∧ ∀x(Gx → Hx) → ∀x(Fx → Hx) T236 ∀x(Fx ↔ Gx) ∧ ∀x(Gx ↔ Hx) → ∀x(Fx ↔ Hx) T237 ∀x(Fx → Gx) ∧ ∀x(Fx → Hx) → ∀x(Fx → Gx ∧ Hx) T238 ∀xFx → ∃xFx T242 ~∀x(Fx → Gx) ↔ ∃x(Fx ∧ ~Gx) T243 ~∃x(Fx ∧ Gx) ↔ ∀x(Fx → ~Gx) T248 ∃xFx ∧ ∃x~Fx ↔ ∀x∃y(Fx ↔ ~Fy)

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10 INVALIDITIES In chapters 1 and 2 we studied tautological validity, which is formal validity that is due to how sentences are built up out of sentential letters and connectives. An argument which is tautologically valid is definitely valid. However, an argument may be valid even if it is not tautologically valid if its validity is due to something in addition to how it is built up with connectives. We have seen examples of such arguments in this chapter, arguments such as: ∀xFx ∴ Fa In this chapter we have studied the kind of formal validity which is due to how formulas are built up out of names, monadic (one-place) predicates, variables, connectives and quantifiers. We call such validity "MPC validity" ("monadic predicate calculus validity"). Derivations using the methods of chapters 1-3 show that the arguments they validate are MPC valid. An argument which is MPC valid is definitely valid. (An argument may be valid even if it is not MPC valid if its validity is due to something in addition to how it is built up from names, variables, monadic predicates, quantifiers, and connectives. Some examples of this are: Some boy fed every cat

∴ Every cat was fed by a boy There are infinitely many prime numbers ∴ There is at least one prime number.

Dr. Jekyll is tall Dr. Jekyll is Mr. Hyde ∴ Mr. Hyde is tall

Even though MPC validity is not the whole story, it remains an important kind of validity.) So far in this chapter we have learned how to show that arguments are MPC valid by means of giving derivations which validate the arguments. We have not yet focused on how to show that an argument is not MPC valid. To do that we may describe a logically possible situation in which the argument has true premises and a false conclusion. It is convenient in doing this to consider very "small" situations -- that is, situations in which only a small number of things exist. To illustrate this, suppose we are given this argument: There are some fibers Every fiber is green Something isn't green ∴ Everything green is a fiber Its MPC form is: ∃xFx ∀x(Fx → Gx) ∃x~Gx ∴ ∀x(Gx → Fx) Now consider the following "small" situation: There are three things: The first is a fiber; the others are not. The first and the second are green; the third is not. In this situation the first premise, '∃xFx', is true because the first thing is a fiber. The second premise, '∀x(Fx → Gx)', is true because there is only one fiber, and it is green. The third premise is true because something isn't green (the third thing). The conclusion is false because not everything that is green is a Chapter Three -- 40

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fiber -- the second thing is green but not a fiber. So this is a situation in which the argument has true premises and a false conclusion, and so it is not MPC valid. If we reflect on this technique, we see that all that we need to show that an argument is not MPC valid is that its logical structure permits this kind of situation. It is sufficient to interpret the argument as applied to a situation in which there are three things, where we interpret ‘F’ as holding of the first (and no other), and we interpret ‘G’ as holding of the first and second (but not the third). If you can interprert an argument so that its premises come out true when so interpreted, and its conclusion false, this is enough to show that an argument of the given form isn't MPC valid. We call such an interpretation a counter-example for the argument. We can describe such a counter-example by using a format like this: Universe:

First thing Second thing Third thing

F: {the first thing} G: {the first thing, the second thing} This indicates how many things there are in the situation, and it gives the "extensions" of 'F' and 'G'. The extension of a predicate is just the set of things it is true of in that interpretation. So the information above tells us that 'F' is true of the first thing and of nothing else, and it tells us that 'G' is true of the first and second things, and not of the third. It doesn’t matter at all what these things are. For specificity, it is often convenient to use integers: Universe:

0 1 2

F: {0} G: {0, 1} This information describes a counter-example for the original argument, because it describes, in minimal terms, the structure of a situation in which the premises of the argument are true and the conclusion false. Here are some more arguments that are not MPC valid, and counter-examples for them. Counter-example #2: ∃x(Fx ∧ ~Gx) ∀x(Hx → ~Gx) ∃x(Hx ∧ Fx) ∴ ∀x(Fx → ~Gx) Universe:

0 1 2

F: {0, 1} G: {0, 2} H: {1} The first premise is true in this interpretation because 'F' is true of 1 and G isn't. The second premise is true because everything that 'H' is true of, namely 1, 'G' is not true of, and the third premise is true because both 'F' and 'H' are true of 1. But the conclusion is not true, because not everything that 'F' is true of is something that 'G' is not true of; 0 is an example. (Removing 2 from the universe will also yield a counter-example.)

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If the argument contains name letters, we indicate what they stand for in the given universe: Counter-example #3: ∀x(Ax → (Bx ↔ Cx)) Bk ∧ ~Ck ∴ ∀x~Ax Universe: A: B: C: k:

0 1 {1} {0, 1} {1} 0

The first premise is true because whatever 'A' is true of, namely 1, is such that 'B' and 'C' are both true of it, so their biconditional comes out true. The second premise is true because 'B' is true of what 'k' stands for, namely 0, and 'C' isn't. The conclusion is false because 'A' is not false of everything; it is true of 1. Counter-example #4: ∀x∃y(Ax ↔ By) ∃xBx ∧ ∃x~Bx ∀x(Ax → ~Cx) ∴ ~∀xCx Universe:

0 1

A: { } B: {0} C: {0, 1}

The first premise is true because everything is such that something is such that 'A' is true of the first if and only if 'B' is true of the second. In fact, 'A' is true of nothing at all. And no matter what there is, there is something that 'B' is not true of, namely 1. So there is always something that makes the biconditional true. The second premise is clearly true since 'B' is true of something, namely 0, and 'B' is also false of something, namely 1. The third premise is true because 'A' is true of nothing, so that every instance is a conditional with a false antecedent. The conclusion is false because 'C' is indeed true of everything. Thinking up counter-examples: If you believe that an argument is not MPC valid, how do you think up a counter-example? There is a mechanical way to do this (described below), but it is too complex to be useful in many cases. So we will usually have to be creative. Some general observations may be useful in guiding our creativity. One approach that is often used is to build up the counter-example a piece at a time, guided by what is needed to make the premises true and conclusion false. Suppose we are given this argument: ∀x~(Fx ↔ Hx) ∃x(Hx ∧ Gx) ∃x(Hx ∧ ~Gx) ∴ ∀x(Fx → Gx) So far, we don't know what will be in the universe. Begin by asking what is needed to make the conclusion false. In this case, what is needed is that there be something that 'F' is true of and 'G' is not. So write this: F: {0} G: { }

The notation "" at the right is not part of the counter-example; it is merely a reminder to yourself that when constructing the counter-example you should not add 0 to the list of things that 'G' is true of, because that could make the conclusion true. Chapter Three -- 42

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CHAPTER 3 SECTION 10

Now consider the first premise; this says that whatever there is in the universe, 'F' and 'H' must disagree about it. This must be kept in mind as a constraint on what can be in the counter-example. So far, in fact, it tells us that since 'F' is true of 0, 'H' must not be: F: {0} G: { } H: { }

Next, consider the second premise: 'G' and 'H' are both true of something. It can't be 0, so fill in 1: F: {0} G: {1} H: {1}

Next, the third premise; this says that there is something that 'H' is true of which 'G' is not true of. It can't be 0 because 'H' cannot be true of 0. It can't be 1 because 'G' is true of 1. So there must be a third thing: F: {0} G: {1}

H: {1, 2}

At this point we have all of the information we need. This is our proposed counter-example: Universe:

0 1 2

F: {0} G: {1} H: {1, 2} If you check through the parts of the argument, you will see that the premises are all true and the conclusion false. Sometimes if you start with no predicate being true of anything, a counter-example falls into your lap. Here is such a case. The argument is: ∀x(Jx → Kx ∨ Hx) ~∀x(~Kx → Jx) ~∃x(Kx ∧ ~Hx) Hc → ∃xJx ∴ ~∃x(Hx ∨ ~Jx) Begin with this minimal proposed counter-example: Universe:

0

H: { } J: { } K: { } c: 0 Let us see what we need to add to what the predicates are true of to make this a counter-example. The first premise is already true because it is a quantified conditional with an antecedent that is false for each thing in the universe. The second premise is true because its unnegation '∀x(~Kx → Jx)' is false. This is false because the part following the quantifier: '~Kx → Jx' is not true for every way of treating 'x' like a name; it is false when 'x' stands for 0. The third is true because there is nothing that is K. The fourth is true because it is a conditional with a false antecedent. And the conclusion is false because there is indeed something that is either H or not J; 0 is not J, so it is either H or not J. In short, the counterexample works as stated. (Usually, of course, more work will be needed.)

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You may sometimes wonder how many things to put into the universe in order to produce a counterexample. There is no best way to determine this; usually you just put more things in when that seems to be required by the premises being true and the conclusion false. There is, however, an upper limit on what you need. If there is only one predicate letter in the argument, then you will need no more than two things in the universe. If there are two predicate letters, you will need no more than four. If there are three predicate letters, you will need no more than eight things. And so on. There is a formula for this: if there n are n predicate letters, if there is a counter-example, there is one using no more than 2 things. Name letters have no effect on the number of things needed. If there are only two predicate letters, and thirteen name letters, then if there is a counter-example at all, there is one with four or fewer things. (Of course, if there are four things and thirteen name letters, several different names will have to stand for the same things. But that's OK.) So here is a mechanical way to come up with a counter-example. Decide, by the formula above, the maximum number of things needed in the universe for a counter-example. For example, suppose that 2 there are two monadic predicates in the argument. A universe of size 2 , that is, 4, will do. Now just consider what choices there may be for the extension of predicate 'F'. There are 16 options: { }, {0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1,3}, {2,3}, {0, 1, 2}, {0, 1, 3}, {0, 2, 3}, {1, 2, 3}, {0, 1, 2, 3} There are also 16 options for 'G'. So there are 16×16 = 256 options for possible counter-examples. If you just check these out, one at a time, you are sure to find one if one exists. If there are three monadic predicates there are 4,096 options. And so on. (Exercise for the reader: In the above calculation we have supposed that if there is a counter-example, we can find one using a maximum size universe. We have ignored the possibility that there is, say, a counter-example using a universe of size 3 but none using a universe of size 4. Why are we justified in making that assumption?) EXERCISES 1. Give counter-examples for each of the following arguments. a.

∀x(Ax → ∃y(By ∧ ~Ay)) ~∀xBx ~∃x(Bx ∧ Cx) ∴ ∃x(Ax ∧ Cx)

b.

∃x(Dx ∧ Ex ∧ ~Fx) ∃x(~Dx ∧ ~Ex) ∀x(Ex → Dx ∨ Fx) ∴ ∀x(Dx ∧ Ex → ~Fx)

c.

∃x(Fx ∧ Gx) ∃x(Fx ∧ ~Gx) ∃x(~Fx ∧ Gx) ∴ ∀x(~Fx → Gx)

d.

∀x∃y(Fx ↔ (Gy ∨ Fx)) ∴ ~∃xFx → ~∃xGx

e.

Ha ∧ ~Hb ∀x(Kx → Hx ∧ Jx) ∃x(Jx ∧ ~Kx) ∴ ∃x(Hx ∧ ~Jx)

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CHAPTER 3 SECTION 11

11 EXPANSIONS In constructing counter-examples it is sometimes difficult to assess the truth value of a sentence in the counter-example, especially when it contains overlapping quantifiers. For example, ask yourself whether the following is a legitimate counter-example to this argument: ∀x∃y(Ax ↔ ~Ay) ∃x(Ax ∧ Bx) ∴ ∀xAx Universe:

0

1

2

A: {0} B: {0} It is clear that this makes the conclusion false, and the second premise true. What about the first premise? It makes that true too. The first premise says that every thing in the universe is such that, there is a thing in the universe such that it isn't A if the first thing is A, and it is A if the first thing isn't. This is in fact true in the counter-example. But this may not be obvious to you. If not, there is a mechanical way to answer such a question. It resembles truth tables in that it will automatically give you a yes or no answer, but it may involve complexity. The technique is based on the idea that if there are a small number of things in the universe, then a universally quantified claim is equivalent to a conjunction of unquantified claims got by removing the quantifier and applying each resulting claim to a thing in the universe. And an existentially quantified claim, in turn, is equivalent to a disjunction of such claims that are applied to each thing in the universe. Let us introduce a convention for naming things in a universe. When there are three things the names will be 'i0', 'i1', and 'i2', where: 'i0' stands for 0 'i1' stands for 1 'i2' stands for 2. (If there are fewer things, leave out 'i2', or both 'i1' and 'i2'. If there are more things add 'i4', 'i5', and so on.) Now consider the sentence '∀xAx'. This says that everything in the universe is A. This is equivalent to saying that the first thing is A and the second thing is A and the third thing is A. That is, it is equivalent to the conjunction: ∀xAx

is equivalent to

Ai0 ∧ Ai1 ∧ Ai2

It is easy to check that this conjunction is false, because not all conjuncts are true. The second premise is '∃x(Ax ∧ Bx)'. This is equivalent to saying that either the first thing is both A and B, or the second thing is, or the third. That is, the quantified sentence is equivalent to this disjunction: ∃x(Ax ∧ Bx)

is equivalent to

(Ai0 ∧ Bi0) ∨ (Ai1 ∧ Bi1) ∨ (Ai2 ∧ Bi2)

It is easy to check that this disjunction is true, because at least one disjunct is true; the first disjunct is true. The first premise, '∀x∃y(Ax ↔ ~Ay)', is more interesting. It is universally quantified, so it is equivalent to the following conjunction: ∃y(Ai0 ↔ ~Ay) ∧ ∃y(Ai1 ↔ ~Ay) ∧ ∃y(Ai2 ↔ ~Ay) It may be easy to determine that this is true. The first conjunct is true because there is something which is not A if and only if 0 is A. We know that 0 is A, and there is indeed at least one thing which is not A; for example, 1 is not A. The second conjunct is true because there is something which is not A if and only if 1 is A. We know that 1 is not A, and there is indeed at least one thing which is A; for example, 0 is A. Chapter Three -- 45

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CHAPTER 3 SECTION 11

The third conjunct is just like the second; it is true because there is something which is not A if and only if 2 is A. We know that 2 is not A, and there is indeed at least one thing which is A; for example, 0 is A. Even this, however, is a bit subtle. There is a way to make it even more mechanical. Namely, each existentially quantified biconditional is equivalent to a disjunction. So this three-part conjunction: ∃y(Ai0 ↔ ~Ay) ∧ ∃y(Ai1 ↔ ~Ay) ∧ ∃y(Ai2 ↔ ~Ay) is equivalent to this: ( (Ai0 ↔ ~Ai0) ∨ (Ai0 ↔ ~Ai1) ∨ (Ai0 ↔ ~Ai2) ) ∧ ( (Ai1 ↔ ~Ai0) ∨ (Ai1 ↔ ~Ai1) ∨ (Ai1 ↔ ~Ai2) ) ∧ ( (Ai2 ↔ ~Ai0) ∨ (Ai2 ↔ ~Ai1) ∨ (Ai2 ↔ ~Ai2) ) And this is easy to evaluate. There are only three atomic sentences in this complex sentence: 'Ai0', 'Ai1', and 'Ai2'. The first of these is true, and the others are false. It is thus easy to evaluate the biconditionals: ( (Ai0 ↔ ~Ai0) ∨ (Ai0 ↔ ~Ai1) ∨ (Ai0 ↔ ~Ai2) ) ∧ false true true ( (Ai1 ↔ ~Ai0) ∨ (Ai1 ↔ ~Ai1) ∨ (Ai1 ↔ ~Ai2) ) ∧ true false false ( (Ai2 ↔ ~Ai0) ∨ (Ai2 ↔ ~Ai1) ∨ (Ai2 ↔ ~Ai2) ) true false false Each disjunction has a true disjunct, so each is true. So the conjunction of the disjunctions is also true. That is, the whole sentence, which is equivalent to '∀x∃y(Ax ↔ ~Ay)', is true. This process is tedious, but completely mechanical. If the counter-example has a universe of only one thing, then this device is very easy to apply. Consider this argument and the accompanying counter-example: ∀x∀y(Jx ↔ ∃z(Kz ↔ Jy)) ∴ ∀xJx Universe:

0

J: { } K: {0} It is clear that the conclusion is false, because 'J' is not true of 0. The premise is universally quantified, so it is equivalent to a conjunction of all of its instances using names of things in the universe. Since there is only one thing in the universe, this conjunction has only one conjunct. So: ∀x∀y(Jx ↔ ∃z(Kz ↔ Jy))

is equivalent to

∀y(Ji0 ↔ ∃z(Kz ↔ Jy))

But that in turn has a simpler equivalent: ∀y(Ji0 ↔ ∃z(Kz ↔ Jy))

is equivalent to

Ji0 ↔ ∃z(Kz ↔ Ji0)

In 'Ji0 ↔ ∃z(Kz ↔ Ji0)' the existentially quantified formula on the right is equivalent to a disjunction with only one disjunct, so we finally have: Ji0 ↔ (Ki0 ↔ Ji0) The truth values of the parts of this sentence are:

Chapter Three -- 46

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CHAPTER 3 SECTION 11

'Ji0 ↔ (Ki0 ↔ Ji0)' false true false and the whole sentence is true, as desired. One more example. Consider the argument, and proposed counter-example: ∀x∃y(Fx ∨ Gy) ~∀xFx ~∀xGx ∴ ~∃xGx Universe:

0

1

F: {0} G: {1} It is pretty clear that this proposed counter-example makes the conclusion false, since something is G, namely, 1. The third premise is true since not everything is G; 0 isn't G. Likewise, the second premise is true since not everything is F; 1 is not F. What about the first? If you are not certain, you can expand it. In this proposed counter-example, the sentence '∀x∃y(Fx ∨ Gy)', which starts with a universal quantifier, is equivalent to this conjunction: ∃y(Fi0 ∨ Gy) ∧ ∃y(Fi1 ∨ Gy) Each of the existentially quantified sentences is equivalent to a disjunction, so we have: ((Fi0 ∨ Gi0) ∨ (Fi0 ∨ Gi1)) ∧ ((Fi1 ∨ Gi0) ∨ (Fi1 ∨ Gi1)) evaluating the parts we have: ((Fi0 ∨ Gi0) ∨ (Fi0 ∨ Gi1)) ∧ ((Fi1 ∨ Gi0) ∨ (Fi1 ∨ Gi1)) true false true true false false false true true

true

Each conjunct is true, so the sentence is itself true.

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CHAPTER 3 SECTION 11

EXERCISES 0. For each of the following arguments use the method of expansions to determine whether the following is a counterexample for it or not.

Universe:

0

1

2

F: {0} G: {0, 2} H: {2} a.

∀x(Hx → ∃y(Fy ∧ ~Hy)) ~∀xFx ~∃x(Fx ∧ Gx) ∴ ∃x(Hx ∧ Gx)

b.

∃x(Gx ∧ Hx ∧ ~Fx) ∃x(~Gx ∧ ~Hx) ∀x(Hx → Gx ∨ Fx) ∴ ∀x(Gx ∧ Hx → ~Fx)

c.

∃x(Fx ∧ Gx) ∃x(Fx ∧ ~Gx) ∃x(~Fx ∧ Gx) ∴ ∀x(~Fx → Gx)

d.

∀x∃y(Fx ↔ (Gy ∨ Fx)) ∴ ~∃xFx → ~∃xGx

e.

Ha ∧ ~Hb ∀x(Fx → Hx ∧ Gx) ∃x(Gx ∧ ~Fx) ∴ ∃x(Hx ∧ ~Gx)

Chapter Three -- 48

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CHAPTER 3 RULES

BASIC RULES AND DERIVATION TECHNIQUES FOR CHAPTER 3

Rule ui: (universal instantiation): ∴

∀x ...x...x... …b…b…

∀x ...x...x... …y…y…

Every occurrence of 'x' that '∀x' was binding must be replaced with the same name or variable. A new variable must not be introduced if some of its new occurrences are bound by a quantifier in the original formula.

Rule eg (existential generalization): ...b...b... ∴ ∃x…x…b…

...y...y... ∴ ∃x…x…b…

(You need not replace every occurrence of 'b' or of 'y' by 'x'.) A new variable must not be introduced if some of its new occurrences are bound by a quantifier in the original formula.

Rule ei: (existential instantiation): ∴

∃x ...x...x... …y…y…

You must replace every occurrence of 'x' that '∃x' was binding. The variable 'y' must not already occur in the derivation or in a premise cited in the derivation..

Universal derivation: If you have a derivation of the following form: Show ∀x . . . x . . . x . . . ::::: ::::: ...x...x... Then if there are no uncancelled show lines in between the first and last lines displayed, and if 'x' does not occur free anywhere in the derivation that is available from the show line (or in a premise that has been cited on an available line), you may box and cancel, using the notation 'ud'.

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CHAPTER 3 RULES

DERIVED RULES

Rule qn (Quantifier negation) ~∀xFx ∴ ∃x~Fx

~∃xFx ∴ ∀x~Fx

∀xFx ∴ ~∃x~Fx

∃xFx ∴ ~∀x~Fx

~∀x~Fx ∴ ∃xFx

~∃x~Fx ∴ ∀xFx

∀x~Fx ∴ ~∃xFx

∃x~Fx ∴ ~∀xFx

Rule av (alphabetic variance) From a formula of the form '∀x . . . x . . x . . .', where the initial quantifier has scope over the whole formula, you may infer '∀y . . . y . . y . . .', which is the result of changing the variable 'x' in the quantifier to another variable, 'y', and changing all variables inside the first formula that are bound by the initial quantifier to 'y'. Likewise if the initial quantifier is '∃' instead of '∀'. Constraint: No capturing is allowed. That is, this inference is not permitted if the new variable becomes bound by a quantifier inside of the original formula.

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CHAPTER 3 STRATEGIES

STRATEGY HINTS All of the strategy hints from chapters 1 and 2 still apply.

These are new:

To derive:

Try this:

Universal Quantification ∀x□

Set up a universal derivation. Write a show line containing ∀x□, and then immediately follow this with a show line containing □. When the second show is cancelled, use rule ud to cancel the first. Or write a show line with '∀x□', and then assume '~∀x□' for an indirect derivation. Turn this into '∃x~□', and proceed from there.

Existential Quantification ∃x□ Negation of a Universal Quantification ~∀x□ Negation of an Existential Quantification ~∃x□

Derive an instance and then use rule eg. Or write a show line with '∃x□', and then assume '~∃x□' for an indirect derivation. Turn this into '∀x~□', and proceed from there. State a show line with '~∀x□', and then assume '∀x□' for an indirect derivation. Or derive '∃x~□' and apply derived rule qn. State a show line with '~∃x□', and then assume '∃x□' for an indirect derivation. Or derive '∀x~□' and apply derived rule qn.

If you have this available:

Try this:

Universal Quantification ∀x□

Use rule ui to derive an instance. (But use rule ei first if that is an option.)

Existential Quantification ∃x□

Use rule ei to derive an instance.

Negation of a Universal Quantification ~∀x□

Use derived rule qn to turn this into an existential quantification.

Negation of an Existential Quantification ~∃x□

Use derived rule qn to turn this into a universal quantification.

Use rule av if necessary: If you are having difficulty with capturing when you use rule ui or ei, change what you are trying to derive to an alphabetic variant. Complete the derivation, and then use derived rule av to convert this into a derivation of what you are after.

Chapter Three -- 51

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CHAPTER 3 THEOREMS

CHAPTER 3 THEOREMS

LAWS OF DISTRIBUTION: T201 T202 T207 T208 T209 T210 T211 T212 T213 T214

∀x(Fx → Gx) → (∀xFx → ∀xGx) ∀x(Fx → Gx) → (∃xFx → ∃xGx) ∃x(Fx ∨ Gx) ↔ ∃xFx ∨ ∃xGx ∀x(Fx ∧ Gx) ↔ ∀xFx ∧ ∀xGx ∃x(Fx ∧ Gx) → ∃xFx ∧ ∃xGx ∀xFx ∨ ∀xGx → ∀x(Fx ∨ Gx) (∃xFx → ∃xGx) → ∃x(Fx → Gx) (∀xFx → ∀xGx) → ∃x(Fx → Gx) ∀x(Fx ↔ Gx) → (∀xFx ↔ ∀xGx) ∀x(Fx ↔ Gx) → (∃xFx ↔ ∃xGx)

LAWS OF QUANTIFIER NEGATION T203 T204 T205 T206

~∀xFx ↔ ∃x~Fx ~∃xFx ↔ ∀x~Fx ∀xFx ↔ ~∃x~Fx ∃xFx ↔ ~∀x~Fx

LAWS OF CONFINEMENT T215 T216 T217 T218 T219 T220 T221 T222 T223 T224 T225 T226

∀x(P∧Fx) ↔ P∧∀xFx ∃x(P∧Fx) ↔ P∧∃xFx ∀x(P∨Fx) ↔ P∨∀xFx ∃x(P∨Fx) ↔ P∨∃xFx ∀x(P→Fx) ↔ (P→∀xFx) ∃x(P→Fx) ↔ (P→∃xFx) ∀x(Fx→P) ↔ (∃xFx→P) ∃x(Fx→P) ↔ (∀xFx→P) ∀x(Fx↔P) → (∀xFx→P) ∀x(Fx↔P) → (∃xFx→P) (∃xFx↔P) → ∃x(Fx↔P) (∀xFx↔P) → ∃x(Fx↔P)

LAWS OF VACUOUS QUANTIFICATION T227 T228 T229 T230

∀xP ↔ P ∃xP ↔ P ∃x(∃xFx → Fx) ∃x(Fx → ∀xFx)

LAWS OF ALPHABETIC VARIANCE T231 T232

∀xFx ↔ ∀yFy ∃xFx ↔ ∃yFy

Chapter Three -- 52

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CHAPTER 3 THEOREMS

OTHER T233 T234 T235 T236 T237 T238 T239 T240 T241 T242 T243 T244 T245 T246 T247 T248

(Fx→Gx) ∧ (Gx→Hx) → (Fx→Hx) ∀x((Fx → Gx) ∧ (Gx → Hx) → (Fx → Hx)) ∀x(Fx → Gx) ∧ ∀x(Gx → Hx) → ∀x(Fx → Hx) ∀x(Fx ↔ Gx) ∧ ∀x(Gx ↔ Hx) → ∀x(Fx ↔ Hx) ∀x(Fx → Gx) ∧ ∀x(Fx → Hx) → ∀x(Fx → Gx ∧ Hx) ∀xFx → ∃xFx ∀xFx ∧ ∃xGx → ∃x(Fx∧Gx) ∀x(Fx→Gx) ∧ ∃x(Fx∧Hx) → ∃x(Gx∧Hx) ∀x(Fx→Gx∨Hx) → ∀x(Fx →Gx) ∨ ∃x(Fx∧Hx) ~∀x(Fx → Gx) ↔ ∃x(Fx ∧ ~Gx) ~∃x(Fx ∧ Gx) ↔ ∀x(Fx → ~Gx) ~∃xFx → ∀x(Fx→Gx) ~∃xFx ↔ ∀x(Fx→Gx) ∧ ∀x(Fx→~Gx) ~∃xFx ∧ ~∃xGx → ∀x(Fx↔Gx) ∃x(Fx→Gx) ↔ ∃x~Fx ∨ ∃xGx ∃xFx ∧ ∃x~Fx ↔ ∀x∃y(Fx ↔ ~Fy)

Chapter Three -- 53

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Answers to Exercises -- Chapter 3

Answers to the Exercises -- Chapter 3 SECTION 1 1. a. Fred is an orangutan. Of b. Gertrude is an orangutan but Fred isn't. Gertrude is an orangutan [and] Fred is not [an orangutan]. Og ∧ ~Of c. Tony Blair will speak first. Fb d. Gary lost weight recently; he is happy. Gary lost weight recently [and] [Gary] is happy. Lg ∧ Hg e. Felix cleaned and polished. Felix cleaned and [Felix] polished. Cf ∧ Of f. Darlene or Abe will bat clean-up. Darlene [will bat clean-up] or Abe will bat clean-up. Bd ∨ Ba 2. 'D' 'L' 'h' 'a'

is true of doctors is true of people who are in love stands for Hans stands for Amanda

a.

Hans is a doctor but Amanda isn't. Hans is a doctor [and] Amanda is not [a doctor] Dh ∧ ~Da

b.

Hans, who is a doctor, is in love Hans is in love [and Hans] is a doctor Lh ∧ Dh

c.

Hans is in love but Amanda isn't Hans is in love [and] Amanda is [not in love] Lh ∧ ~La

d.

Neither Hans nor Amanda is in love [It is not the case that] (Hans [is in love] or Amanda is in love) ~(Lh ∨ La)

f.

Hans and Amanda are both doctors. Hans is a doctor [and] Amanda is a doctor. Dh ∧ Da

3. 'L' for things that live in Brea 'D' for things that drive to school

Chapter Three -- 54 -- Answers to the Exercises

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Answers to Exercises -- Chapter 3

a.

Eileen and Cosi both live in Brea. Eileen lives in Brea and Cosi loves in Brea Le ∧ Lc

b.

Eileen drives to school, and so does Hank. Eileen drives to school and hank drives to school De ∧ Dh

c.

If Hank lives in Brea then he drives to school; otherwise he doesn't drive to school. (If Hank lives in Brea then he drives to school) [and] (otherwise he doesn't drive to school) (If Hank lives in Brea then he drives to school) [and] ([if Hank doesn't live in Brea then] he doesn't drive to school)

(Lh → Dh) ∧ (~Lh → ~Dh) d.

If David and Hank both live in Brea then David drives to school but Hank doesn't. If (David and Hank both live in Brea) then (David drives to school [and] Hank doesn't [drive to school])

(Ld ∧ Lh) → (Dd ∧ ~Dh) e.

Neither Hank nor Eileen live in Brea, yet each of them drives to school. Neither Hank nor Eileen live in Brea, [and] [Hank and Eileen] drive to school. ~(Lh ∨ Le) ∧ (Dh ∧ De)

SECTION 2 1. For each of the following, say whether it is a formula in official notation, or in informal notation, or not a formula at all. If it is a formula, parse it. a. Official notation ~∀x(Fx → (Gx ∧ Hx)) | ∀x(Fx → (Gx ∧ Hx)) | (Fx → (Gx ∧ Hx)) /\ Fx (Gx ∧ Hx) /\ Gx Hx b. Informal notation ∃x~~Gx → Hx ∨ ∃yGy /\ ∃x~~Gx Hx ∨ ∃yGy | /\ ~~Gx Hx ∃yGy | | ~Gx Gy | Gx c. Official notation ~(Gx ↔ ~Hx) | (Gx ↔ ~Hx)

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/\ Gx ~Hx | Hx d. Not a formula; a quantifier cannot occur outside a quantifier phrase. e. Informal notation Fa → (Gb ↔ Hc) /\ Fa (Gb ↔ Hc) /\ Gb Hc f. Not a formula; a variable can only occur in an atomic formula or a quantifier phrase, and never by itself. g. Informal notation ∀x(Gx ↔ Hx) → Ha ∧ ∃zKz /\ ∀x(Gx ↔ Hx) Ha ∧ ∃zKz | /\ Gx ↔ Hx Ha ∃zKz /\ | Gx Hx Kz

SECTION 3 1. a.

Sentence

∃x(Fx ∧ ∀y(Gy ∨ Hx))

b. c.

Not a formula; there is no way to form "∃~z" in our grammar. Formula ∃z~(Hz ∧ Gx ∧ ∃xIx)

d.

Formula

~(~Gx → ∀y(Jx ∧ Ky ↔ Lx))

e.

Formula

∃xGx ↔ ∃y(Gy ∧ Hx)

f.

Sentence

∀x(Gx → ∀y(Hy → ∀z ( Iz → Hx ∧ Gz)))

g.

Sentence

∀x ∃y( Hx ↔ ~Gy)

h. i. j.

Not a formula; there is no way to form "∀xy" in our grammar. Not a formula; "∃y" cannot stand on its own as a subformula. Sentence ∀x ∃y∀z(Gx ↔ ∃w(Hw ∧ ~Hx ∧ Gy))

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Answers to Exercises -- Chapter 3

SECTION 4 1. a. b. c. d. e. f.

Something is a sofa and is well built. There is a well-built sofa.. Everything is such that if it is a sofa then it is well-built. All sofas are well-built. Everything is either a sofa or is well-built. Everything is a sofa, unless it's well-built. Something is such that it is not a sofa. Something isn't a sofa. Everything is such that it is not a sofa. There are no sofas. Everything is such that if it is both bell-built and a sofa, then it is comfortable. Every well-built sofa is comfortable. g. Something is comfortable and everything is well-built. h. Something is such that if it is comfortable, then everything is well-built.

2. Assume that all giraffes are friendly, and that some giraffes are clever and some aren't. a. b. c. d. e. f.

∀x(Gx → Fx) ∀x(Gx → Cx) ∃x(~Fx ∧ Gx) ∃y(Fy ∧ Cy) ∃z(Gz ∧ Cz) ∀x(Gx → ~Gx)

True, since all giraffes are friendly. False, since not every giraffe is clever. False, since every giraffe is friendly. True, since giraffes are friendly, and some of them are clever. True, since some giraffes are clever. False, since not every giraffe isn't a giraffe. (In fact, no giraffe isn't a giraffe, but it only takes one to falsify the symbolic sentence.)

SECTION 5a 1. a. b. c. d. e. f. g.

Every Handsome Elephant is Friendly. ∀x((Hx ∧ Ex) → Fx) No handsome elephant is friendly. ~∃x((Hx ∧ Ex) ∧ Fx) Some elephants are not handsome. ∃x(Ex ∧ ~Hx) Some handsome elephants are friendly. ∃x((Hx ∧ Ex) ∧ Fx) Each friendly elephant is handsome. ∀x((Fx ∧ Ex) → Hx) A handsome elephant is not friendly. ∃x((Hx ∧ Ex) ∧ ~Fx) No friendly elephant is handsome. ~∃x((Fx ∧ Ex) ∧ Hx)

SECTION 5b 1. Suppose that `A' stands for `is a U.S. state', `C' for `is a city', `L' for `is a capital', and `E' for `is in the Eastern time zone'. What are the truth values of these sentences? a. b. c. d. e. f. g. h.

∀x(Cx → Lx) --- False; Los Angeles is a city but not a capital. ∃x(Cx ∧ Lx) --- True; Sacramento is a city and a capital. ∃x(Cx ∧ Lx ↔ Ex) --- True, because something makes the biconditional true, by making both sides false. For example, Los Angeles is not a capital, and it is not in the Eastern time zone. ∀x(Cx ∧ Ex → Ax) --- False; Philadelphia is not a state. ~∃x(Ax ∧ Ex) --- False; Delaware is a state in the Eastern time zone. ∃x(Cx ∧ Ex) ∧ ∃x(Cx ∧ ~Ex) --- True; Philadelphia is a city in the Eastern time zone and LA is a city outside the eastern time zone. ∃x(Cx ∧ Ex ∧ Ax) --- False; no city is also a state. ~∃x(Cx ∧ ~Cx) --- True. There is no city which isn't a city.

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Answers to Exercises -- Chapter 3

2. a. b. c. d. e. f. g.

h. i.

j. k. l. m. n. o.

All Giraffes are spOtted. ∀x(Gx → Ox) All Clever giraffes are spotted. ∀x(Gx ∧ Cx → Ox) No clever giraffes are spotted. ~∃x(Gx ∧ Cx ∧ Ox) Every giraffe is either spotted or Drab. ∀x(Gx → (Ox ∨ Dx)) Some giraffes are clever. ∃x(Gx ∧ Cx) Some spotted giraffes are clever. ∃x(Ox ∧ Gx ∧ Cx) Some giraffes are clever and some aren't. Some giraffes are clever and some [giraffes are not clever]. ∃x(Gx ∧ Cx) ∧ ∃x(Gx ∧ ~Cx) Some spotted giraffes aren't clever. ∃x(Ox ∧ Gx ∧ ~Cx) No spotted giraffe is clever but every unspotted one is. No spotted giraffe is clever [and] every un-spotted [giraffe] is [clever]. ~∃x(Ox ∧ Gx ∧ Cx) ∧ ∀x(~Ox ∧ Gx → Cx) Every clever spotted giraffe is either wIse or Foolhardy. ∀x(((Cx ∧ Ox) ∧ Gx) → (Ix ∨ Fx)) Either all spotted giraffes are clever, or all clever giraffes are spotted. ∀x(Ox ∧ Gx → Cx) ∨ ∀x(Cx∧Gx → Ox) Every clever giraffe is foolhardy. ∀x(Cx ∧ Gx → Fx) If some giraffes are wise then not all giraffes are foolhardy. ∃x(Gx ∧ Ix) → ~∀x(Gx → Fx) All giraffes are spotted if and only if no giraffes aren't spotted. ∀x(Gx → Ox) ↔ ~∃x(Gx ∧ ~Ox) Nothing is both wise and foolhardy. ~∃x(Ix ∧ Fx)

SECTION 5c 1. a.

Only Friendly Elephants are Handsome (ambiguous) i. ∀x(Hx → (Fx ∧ Ex)) ii. ∀x((Ex ∧ Hx) → Fx) b. If only elephants are friendly, no Giraffes are friendly ∀x(Fx → Ex) → ~∃x(Gx ∧ Fx) c. Only the Brave are fAir. ∀x(Ax → Bx) d. If only elephants are friendly then every elephant is friendly ∀x(Fx → Ex) → ∀x(Ex → Fx) e. All and only elephants are friendly. All elephants are friendly [and] Only elephants are friendly. ∀x(Ex → Fx) ∧ ∀x(Fx → Ex) f. If every elephant is friendly, only friendly Animals are elephants (ambiguous) i. ∀x(Ex → Fx) → ∀x(Ex → (Fx ∧ Ax)) ii. ∀x(Ex → Fx) → ∀x((Ex ∧ Ax) → Fx)

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Answers to Exercises -- Chapter 3

g.

h. i. j.

If any elephants are friendly, all and only giraffes are nasty If some elephants are friendly, (all giraffes are Nasty and only giraffes are nasty) ∃x(Ex ∧ Fx) → (∀x(Gx → Nx) ∧ ∀x(Nx → Gx)) Among spOtted animals, only giraffes are handsome. ∀x(Ox → (Hx → Gx)) Among spotted animals, all and only giraffes are handsome ∀x(Ox → ((Gx → Hx) ∧ (Hx → Gx)) Only giraffes frolic if annoyed. If a thing froLics if aNnoyed, it is a giraffe. ∀x((Nx → Lx) → Gx)

SECTION 5d 1. Symbolize these sentences. a.

Every Giraffe which Frolics is Happy ∀x(Fx ∧ Gx → Hx) b. Only giraffes which frolic are happy (ambiguous) i. ∀x(Gx ∧ Hx → Fx) ii. ∀x(Hx → Gx ∧ Fx) c. Only giraffes are Animals which are Long-necked. ∀x(Ax ∧ Lx → Gx) d. If only giraffes frolic, every animal which is not a giraffe doesn't frolic. ∀x(Fx → Gx) → ∀x(Ax ∧ ~Gx → ~Fx) e. Some giraffe which frolics is long-necked or happy. ∃x((Fx ∧ Gx) ∧ (Lx ∨ Hx)) f. No giraffe which is not happy frolics and is long-necked. ~∃x((~Hx ∧ Gx) ∧ (Fx ∧ Lx)) g. Some giraffe is not both long-necked and happy. ∃x(Gx ∧ ~(Lx ∧ Hx))

SECTION 5e 1. a. b. c. d. e. f. g. h. i.

If a Giraffe is Happy then it Frolics unless it is Lame. ∀x(Gx ∧ Hx → Fx ∨ Lx) A Monkey frolics unless it is not happy. ∀x(Mx → Fx ∨ ~Hx) Among giraffes, only happy ones frolic. ∀x(Gx → (Fx → Hx)) All and only giraffes are happy if they are not lame. ∀x(Gx ↔ (~Lx → Hx)) A giraffe frolics only if it is happy. ∀x(Gx ∧ Fx → Hx) or ∀x(Gx → (Fx→Hx)) Only giraffes frolic if happy. ∀x((Hx → Fx) → Gx) All monkeys are happy if some giraffe is. ∃x(Gx ∧ Hx) → ∀x(Mx → Hx) Cute monkeys frolic. ∀x(Cx ∧ Mx → Fx) Giraffes ruN and frolic if and only if they are Blissful and Exultant. ∀x(Gx → (Nx ∧ Fx ↔ Bx ∧ Ex))

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Answers to Exercises -- Chapter 3

j. k. l. m. n.

If those who are heAlthy are not lame, then if they are exultant, they will frolic. ∀x((Ax → ~Lx) → (Ex → Fx)) Only giraffes and monkeys are blissful and exultant. ∀x(Bx ∧ Ex → Gx ∨ Mx) The brave(I) are happy. ∀x(Ix → Hx) If a giraffe frolics, then no monkey is blissful unless it is. ∀x((Gx ∧ Fx) → (Bx ∨ ~∃y(My ∧ By))) Giraffes and monkeys frolic if happy. ∀x(Gx ∨ Mx → (Hx → Fx))

SECTION 6 1. a.

The sky is Blue Everything that is blue is prEtty ∴ Something is pretty Be ∀x(Bx → Ex) ∴ ∃xEx

1 2 3 4

Show ∃xEx Be → Ee Ee ∃xEx

b.

Every Hyena is Grey. Every hyena is an Animal Jenny is a hyena ∴ Some animal is grey

pr2 ui 2 pr1 mp 3 eg dd

∀x(Hx → Gx) ∀x(Hx → Ax) He ∴ ∃x(Ax ∧ Gx)

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Answers to Exercises -- Chapter 3

1 2 3 4 5 6 7

Show ∃x(Ax ∧ Gx) He → Ge He → Ae Ge Ae Ae ∧ Ge ∃x(Ax ∧ Gx)

c.

If some Hyena is Grey, every hyena is grey Every sCavenger is grey Jenny is a hyena and a scavenger Kathy is a hyena ∴ Kathy is grey

pr1 ui pr2 ui pr3 2 mp pr3 3 mp 4 5 adj 6 eg dd

∃x(Hx ∧ Gx) → ∀x(Hx → Gx) ∀x(Cx → Gx) He ∧ Ce Ha ∴ Ga 1 2 3 4 5 6 7 8 9 10 11

Show Ga Show ∃x(Hx ∧ Gx) He Ce Ce → Ge Ge He ∧ Ge ∃x(Hx ∧ Gx) ∀x(Hx → Gx) Ha → Ga Ga

pr3 s pr3 s pr2 ui 4 5 mp 3 6 adj 7 eg dd 2 pr1 mp 9 ui 10 pr4 mp dd

2. The error is at line 3. It is not permissible to use EI to get an instance of pr2 in the variable z because z occurs already on line 2; this would violate the restriction on EI. 3. No derivations are given for named theorems.

SECTION 8 1. Symbolize these arguments and provide derivations to validate them. Give an explicit scheme of abbreviation for each. a.

If history is right (P), then if anyone was strOng, hercules was strong. Only those who work out (M) are strong, and only those with self-Discipline work out. ∴ If Hercules does not have self-discipline, then either history is not right or nobody is strong.

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Answers to Exercises -- Chapter 3

P → (∃xOx → Oh) ∀x(Ox → Mx) ∧ ∀x(Mx → Dx) ∴ ~Dh → (~P ∨ ~∃xOx) 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Show ~Dh → (~P ∨ ~∃xOx) ~Dh Mh → Dh ~Mh Oh → Mh ~Oh Show ~P ∨ ~∃xOx Show ~~P → ~∃xOx ~~P P ∃xOx → Oh ~∃xOx ~P ∨ ~∃xOx

b.

If some Giraffes are not Happy, then all giraffes are Morose. Some giraffes pOnder the mysteries of life. ∴ If some giraffes are not morose, then some who ponder the mysteries of life are happy.

ass cd pr2 s ui 2 3 mt pr2 s ui 4 5 mt

ass cd 9 dn p1 10 mp 6 11 mt cd 8 cdj dd 7 cd

∃x(Gx ∧ ~Hx) → ∀x(Gx → Mx) ∃x(Gx ∧ Ox) ∴ ∃x(Gx ∧ ~Mx) → ∃x(Ox ∧ Hx) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Show ∃x(Gx ∧ ~Mx) → ∃x(Ox ∧ Hx) ∃x(Gx ∧ ~Mx) Gi ∧ ~Mi Show ~∀x(Gx → Mx) ∀x(Gx → Mx) Gi → Mi Mi ~Mi ~∃x(Gx ∧ ~Hx) Gj ∧ Oj Show Hj ~Hj Gj ∧ ~Hj ∃x(Gx ∧ ~Hx) ~∃x(Gx ∧ ~Hx) Oj ∧ Hj ∃x(Ox ∧ Hx)

ass cd 2 ei ass id 5 ui 3 s 6 mp 3 s 7 id 4 pr1 mt pr2 ei ass id 10 s 12 adj 13 eg 9 r id 10 s 11 adj 16 eg cd

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Answers to Exercises -- Chapter 3

c.

There is not a single Critic who either Likes art or can pAint. Some level-Headed peOple are critics. Anyone who can't paint is unEducated. ∴ Some level-headed people are uneducated. ∀x(Cx → ~(Lx ∨ Ax)) ∃x((Hx ∧ Ox) ∧ Cx) ∀x(Ox → (~Ax → ~Ex)) ∴ ∃x((Hx ∧ Ox) ∧ ~Ex)

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Show ∃x((Hx ∧ Ox) ∧ ~Ex) (Hi ∧ Oi) ∧ Ci Ci Hi Oi Ci → ~(Li ∨ Ai) ~(Li ∨ Ai) ~Li ∧ ~Ai ~Ai Oi → (~Ai → ~Ei) ~Ai → ~Ei ~Ei (Hi ∧ Oi) ∧ ~Ei ∃x((Hx ∧ Ox) ∧ ~Ex)

d.

No Astronaut is a good Dancer. Every sInger is warm-Blooded. If something is warm-blooded and is not a good dancer, then nothing that is either a singer or an astronaut is Exultant. ∴ If some astronaut is a singer, then no singer is exultant.

pr2 ei 2s 2ss 2ss pr1 ui 3 6 mp 7 dm 8s pr3 ui 5 10 mp 9 11 mp 2 s 12 adj 13 eg dd

∀x(Ax → ~Dx) ∀x(Ix → Bx) ∃x(Bx ∧ ~Dx) → ∀x((Ix ∨ Ax) → ~Ex) ∴ ∃x(Ax ∧ Ix) → ∀x(Ix → ~Ex)

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Answers to Exercises -- Chapter 3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Show ∃x(Ax ∧ Ix) → ∀x(Ix → ~Ex) ∃x(Ax ∧ Ix) Show ∀x(Ix → ~Ex) Show Ix → ~Ex Ix Ai ∧ Ii Ai → ~Di Ii → Bi ~Di Bi ∧ ~Di ∃x(Bx ∧ ~Dx) ∀x((Ix ∨ Ax) → ~Ex) (Ix ∨ Ax) → ~Ex Ix ∨ Ax ~Ex

e.

All stuDents who have a sense of Humor or are Brilliant seek Fame. Anyone who seeks fame and is brilliant is Insecure. Whoever is a Mathematician is brilliant. ∴ Every student who is a mathematician is insecure.

ass cd

ass cd 2 ei pr1 ui pr2 ui 6 s 7 mp 6s 8 mp 9 adj 10 eg 11 pr3 mp 12 ui 5 add 13 14 mp cd 4 ud 3 cd

∀x((Dx ∧ (Hx ∨ Bx)) → Fx) ∀x(Fx ∧ Bx → Ix) ∀x(Mx → Bx) ∴ ∀x((Dx ∧ Mx) → Ix) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Show ∀x((Dx ∧ Mx) → Ix) Show (Dx ∧ Mx) → Ix Dx ∧ Mx Dx Mx Mx → Bx Bx Hx ∨ Bx Dx ∧ (Hx ∨ Bx) (Dx ∧ (Hx ∨ Bx)) → Fx Fx Fx ∧ Bx Fx ∧ Bx → Ix Ix

ass cd 3s 3s pr3 ui 5 6 mp 7 add 4 8 adj pr1 ui 9 10 mp 7 11 adj pr2 ui 12 13 mp cd 2 ud

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Answers to Exercises -- Chapter 3

f.

There is a Monkey that is Happy if and only if some Giraffe is happy. There is a monkey that is happy if and only if some giraffe is not happy. All monkeys are happy. ∴ It is not the case that either every giraffe is happy or none are. ∃x(Mx ∧ (Hx ↔ ∃x(Gx ∧ Hx)) ∃x(Mx ∧ (Hx ↔ ∃x(Gx ∧ ~Hx)) ∀x(Mx → Hx) ∴ ~(∀x(Gx → Hx) ∨ ∀x(Gx → ~Hx))

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Show ~(∀x(Gx → Hx) ∨ ∀x(Gx → ~Hx)) ∀x(Gx → Hx) ∨ ∀x(Gx → ~Hx) Mi ∧ (Hi ↔ ∃x(Gx ∧ Hx) Mj ∧ (Hj ↔ ∃x(Gx ∧ ~Hx) Mi → Hi Mj → Hj Hi Hj ∃x(Gx ∧ Hx) ∃x(Gx ∧ ~Hx) Gk ∧ Hk Gm ∧ ~Hm Show ~∀x(Gx → Hx) ∀x(Gx → Hx) Gm → Hm Hm ~Hm ∀x(Gx → ~Hx) Gk → ~Hk ~Hk Hk

g.

For every Astronaut that writes pOetry, there is one that doesn't. For every astronaut that doesn't write poetry, there is one that does. ∴ If there are any astronauts, some write poetry and some don't.

ass id pr2 ei pr3 ei pr4 ui pr4 ui 3 s 5 mp 3 s 6 mp 3 s bc 7 mp 4 s bc 7 mp 9 ei 10 ei ass id 14 ui 12 s 15 mp 12 s id 2 13 mtp 18 ui 11 s 19 mp 11 s id

∀x((Ax ∧ Ox) → ∃x(Ax ∧ ~Ox)) ∀x((Ax ∧ ~Ox) → ∃x(Ax ∧ Ox)) ∴ ∃xAx → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)

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Answers to Exercises -- Chapter 3

1 Show ∃xAx → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox) 2 ass cd ∃xAx 3 Ai 2 ei 4 T59 Oi ∨ ~Oi 5 Show Oi → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox) 6 Oi ass cd 7 3 6 adj Ai ∧ Oi 8 7 eg ∃x(Ax ∧ Ox) 9 Pr1 ui Ai ∧ Oi → ∃x(Ax ∧ ~Ox) 10 7 9 mp ∃x(Ax ∧ ~Ox) 11 8 10 adj cd (∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)) 12 Show ~Oi → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox) 13 ~Oi ass cd 14 13 3 adj Ai ∧ ~Oi 15 14 eg ∃x(Ax ∧ ~Ox) 16 Pr2 ui Ai ∧ ~Oi → ∃x(Ax ∧ Ox) 17 14 16 mp ∃x(Ax ∧ Ox) 18 15 17 adj cd (∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)) 19 4 5 12 sc ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox) 20 19 cd

SECTION 9 1. a.

~∃x(Ax ∨ Bx) ∀x∀y(Gx ∧ Hy → By) ∃xGx ∴ ∀x~Hx

1 2 3 4 5 6 7 8 9 10 11 12

Show ∀x~Hx ~∀x~Hx ∃xHx Hi Gj Gj ∧ Hi Gj ∧ Hi → Bi Bi ∀x~(Ax ∨ Bx) ~(Ai ∨ Bi) ~Ai ∧ ~Bi ~Bi

ass id 2 qn 3 ei pr3 ei 4 5 adj pr2 ui ui 6 7 mp pr1 qn 9 ui 10 dm 11 s 8 id

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Answers to Exercises -- Chapter 3

b.

∃x(Hx ∧ ~∃y(Gy ∧ Hx)) ∴ ∀y~Gy

1 2 3 4 5 6 7 8 9

Show ∀y~Gy ~∀y~Gy ∃yGy Hi ∧ ~∃y(Gy ∧ Hi) Hi ~∃y(Gy ∧ Hi) Gj Gj ∧ Hi ∃y(Gy ∧ Hi)

c.

∀x(Ax → ∀y(Bx ↔ By)) ∃zBz ∴ ∀y(Ay → By)

1 2 3 4 5 6 7 8 9 10 11

Show ∀y(Ay → By) Show ∀i(Ai → Bi) Show Ai → Bi Ai Ai → ∀y(Bi ↔ By) ∀y(Bi ↔ By) Bj Bi ↔ Bj Bi

d.

~∀x(Dx ∨ Ex) ∃x(Fx ↔ ~Ex) → ∀zDz ∴ ∃x~Fx

ass id 2 qn pr1 ei 4s 4s 3 ei 5 7 adj 8 eg 6 id

∀y(Ay → By)

ass cd pr1 ui 4 5 mp pr2 ei 6 ui 8 bc 7 mp cd 3 ud 2 av dd

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Answers to Exercises -- Chapter 3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Show ∃x~Fx ~∃x~Fx ∀xFx ∃x~(Dx ∨ Ex) ~(Di ∨ Ei) ~Di ∧ ~Ei ~Di ~Ei Show Fi → ~Ei ~Ei Show ~Ei → Fi Fi Fi ↔ ~Ei ∃x(Fx ↔ ~Ex) ∀zDz Di

e.

Jc ∧ ~Jd ∀xKx ∨ ∀x~Kx ∃x(Jx ∧ Kx) → ∀x(Kx → Jx) ∴ ~Kc

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Show ~Kc Kc Show ~∀x~Kx ∀x~Kx ~Kc Kc ∀xKx Jc ∧ Kc ∃x(Jx ∧ Kx) ∀x(Kx → Jx) Kd → Jd Kd Jd ~Jd

ass id 2 qn pr1 qn 4 ei 5 dm 6s 6s 8 r cd 3 ui cd 9 11 cb 13 eg 14 pr2 mp 15 ui 7 id

ass id ass id 4 ui 2 r 5 id 3 pr2 mtp pr1 s 2 adj 8 eg 9 pr3 mp 10 ui 7 ui 11 12 mp pr1 s 13 id

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Answers to Exercises -- Chapter 3

SECTION 10 1. a.

∀x(Ax → ∃y(By ∧ ~Ay)) ~∀xBx ~∃x(Bx ∧ Cx) ∴ ∃x(Ax ∧ Cx)

Universe: {1, 2, 3} A: {1} B: {2} C: {3} b.

∃x(Dx ∧ Ex ∧ ~Fx) ∃x(~Dx ∧ ~Ex) ∀x(Ex → Dx ∨ Fx) ∴ ∀x(Dx ∧ Ex → ~Fx)

Universe: {1, 2, 3} D: {1, 2} E: {1, 2} F: {1} c.

∃x(Fx ∧ Gx) ∃x(Fx ∧ ~Gx) ∃x(~Fx ∧ Gx) ∴ ∀x(~Fx → Gx)

Universe: {1, 2, 3, 4} F: {2, 3} G: {1, 2} d.

∀x∃y(Fx ↔ (Gy ∨ Fx)) ∴ ~∃xFx → ~∃xGx

Universe: {1, 2} F: { } G: {1} e.

Ha ∧ ~Hb ∀x(Kx → Hx ∧ Jx) ∃x(Jx ∧ ~Kx) ∴ ∃x(Hx ∧ ~Jx)

Universe: {1, 2} H: {1} J: {1,2} K: { } a --- 1 b --- 2

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Answers to Exercises -- Chapter 3

SECTION 11 1. For each of the following arguments use the method of expansions to determine whether the following is a counterexample for it or not. Universe: F: G: H: a: b: a.

0

1

2

{0} {0, 2} {2} 2 0

∀x(Hx → ∃y(Fy ∧ ~Hy)) ~∀xFx ~∃x(Fx ∧ Gx) ∴ ∃x(Hx ∧ Gx)

The conclusion expands to: (Ha1 ∧ Ga1) ∨ (Ha2 ∧ Ga2) ∨ (Ha3 ∧ Ga3) which is true because Ha3 and Ga3 are true. Since we have a true conclusion, we don't have a counterexample. b.

∃x(Gx ∧ Hx ∧ ~Fx) ∃x(~Gx ∧ ~Hx) ∀x(Hx → Gx ∨ Fx) ∴ ∀x(Gx ∧ Hx → ~Fx)

The conclusion expands to: (Ga1 ∧ Ha1 → ~Fa1) ∧ (Ga2 ∧ Ha2 → ~Fa2) ∧ (Ga3 ∧ Ha3 → ~Fa3) which is true because the first conjunct has a false antecedent, the second conjunct has a false antecedent, and the third conjunct has a true consequent. Since we have a true conclusion, we don't have a counterexample. c.

∃x(Fx ∧ Gx) ∃x(Fx ∧ ~Gx) ∃x(~Fx ∧ Gx) ∴ ∀x(~Fx → Gx)

The second premise expands to: (Fa1 ∧ ~Ga1) ∨ (Fa2 ∧ ~Ga2) ∨ (Fa3 ∧ ~Ga3) which is false because each disjunct is false. Since we have a false premise we don't have a counterexample.

Chapter Three -- 70 -- Answers to the Exercises

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Answers to Exercises -- Chapter 3

d.

∀x∃y(Fx ↔ (Gy ∨ Fx)) ∴ ~∃xFx → ~∃xGx

The conclusion expands to: ~(Fa1 ∨ Fa2 ∨ Fa3) → ~(Ga1 ∨ Ga2 ∨ Ga3) which is true because the antecedent is false because its leftmost disjunct is true. Since we have a true conclusion we don't have a counterexample. e.

Ha ∧ ~Hb ∀x(Fx → Hx ∧ Gx) ∃x(Gx ∧ ~Fx) ∴ ∃x(Hx ∧ ~Gx)

The second premise expands to: (Fa1 → Ha1 ∧ Ga1) ∧ (Fa2 → Ha2 ∧ Ga2) ∧ (Fa3 → Ha3 ∧ Ga3) which is false because the first conjunct has a true antecedent and a false consequent. Since we have a false premise, we don't have a counterexample.

Chapter Three -- 71 -- Answers to the Exercises

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CHAPTER 4 SECTION 1

Chapter Four Many-Place Predicates In chapter 3 we studied the concept of formal validity in the Monadic Predicate Calculus, validity that is due to the logical forms that can be expressed using names, variables, connectives, quantifiers, and oneplace predicates. In this chapter the restriction to monadic (one-place) predicates is lifted. We are now focusing simply on Predicate Calculus formal validity: validity that is due to the logical forms that can be expressed using logical signs plus predicates of any number of places.

1 MANY-PLACE PREDICATES In earlier chapters we used predicate letters that combine with one name to make a sentence: Antarctica is peaceful Fido is a giraffe Cynthia ran

Ea Gf Ac

There are also expressions that combine with two names to form sentences: Andria is taller than Bill Cynthia is a friend of David Fred sees Bella or with three names, or more: Cary gave Fido to Andy Egbert sent Beatrice to Compton Fred drove Anna to Chicago with David To accommodate these expressions we use predicate letters that are followed by two or more names or variables enclosed in parentheses. Some examples with names are: Andria is taller than Bill Cynthia is a friend of David Fred sees Bella Cary gave Fido to Andy Egbert sent Beatrice to Compton

T(ab) F(cd) S(fb) G(cfa) S(ebc)

These are atomic sentences, on a par with atomic sentences consisting of a single sentence letter or of a predicate letter followed by a name. They also occur with variables to form atomic formulas: Andria is taller than x x is a friend of y z sees Bella Cary gave x to y z sent u to v

T(ax) F(xy) S(zb) G(cxy) S(zuv)

We can use any capital letter for a many-place predicate (adding subscripts if desired). You can tell whether a predicate letter is being used as a one-place predicate or a many-place predicate by seeing what follows it. If it is followed by a single name or variable, it is being used as a one-place predicate; if it is followed by a pair of parentheses containing names or variables, it is being used as a many-place predicate. An atomic formula is now either a sentence letter alone, or a one-place predicate letter followed by a name or variable, or any predicate letter followed by a pair of parentheses containing any number of names or variables. These new atomic formulas combine with connectives and quantifiers as in the previous chapter, yielding formulas such as:

Chapter 4 -- 1

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CHAPTER 4 SECTION 1

∃xA(bx) ∀yB(yy) ∀x(Ax → B(xd)) etc We are now using parentheses for two different purposes: to surround the terms following a many-place predicate symbol, and to surround molecular formulas. It is common to use either parentheses or square brackets for the latter purpose, and using square brackets instead of parentheses sometimes increases readability. So we will often write complex formulas as follows: ∀x[Ax → B(xd)] ∀x∃y[[A(xy)→B(yx)] ↔ A(yx)∧B(yx)] ∃xF(ax) ↔ ∃y[G(ay) ∧ G(yx)] Our official account of formulas is now: A sentence letter is any capital letter between 'P' and 'Z' (perhaps with a subscript). A one-place predicate is any capital letter between 'A' and 'O' (perhaps with a subscript). A many-place predicate is any capital letter between 'A' and 'Z' (perhaps with a subscript). An atomic formula is: a sentence letter alone, a one-place predicate letter followed by one name or one variable, or any predicate letter followed by a pair of parentheses containing any number of names or variables. If '□' and '△' are formulas, so are: ~□ (□∧△) (□∨△) (□→△) (□↔△) ∃x□ ∀x□

EXERCISES 1. Which of the following are formulas in official notation? Which are formulas in informal notation? Which are not formulas at all? a. b. c. d. e. f. g. h. i.

~~F(xa) [∀xG(bx) → ~∃yG(yx)] ∀xG(bx) → ~∃yG(yx) ~Fa ∧ ~G(aa) ∧~Fb ∧ Gxb ~F(a) ∨ ~G(ab) ~Fa ∨ ~Gab ~∃x[~Fx → ∀yG(yy)] ∃x∀y~Fxy ∃x∃yF[xy]

Chapter 4 -- 2

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CHAPTER 4 SECTION 2

2 SYMBOLIZING SENTENCES USING MANY-PLACE PREDICATES It should be clear how to symbolize atomic formulas, given the practice from chapter 3 and the explanations above. The only additional complexity has to do with the order of the terms. When symbolizing a transitive verb, such as ‘likes’ it is convenient to use the same order in the symbolization as in the English, so that ‘Ann likes Bill’ becomes ‘L(ab)’. But when there are more places we need to be careful about the order. For example both of these English sentences say the same thing, but they use a different order: Ann sent Bill her dog Ann sent her dog to Bill It is convenient to be explicit about the ordering, by specifying e.g. S():  sent  to  With this understanding, both of the English sentences above would be symbolized: S(adb) Symbolizing complex sentences with many-place predicates mostly involves the same techniques as those we used earlier. For example, when there is one quantificational expression and a name, the symbolizations follow the same patterns as before. Some examples: Every giraffe is happy Every giraffe sees Fido

∀x[Gx → Hx] ∀x[Gx → S(xf)]

Some dog is spotted Some dog loves Bobby

∃x[Dx ∧ Sx] ∃x[Dx ∧ L(xb)]

The pattern is similar when the name is the subject of the sentence: Fido sees every dog Bobby loves some dog

∀x[Dx → S(fx)] ∃x[Dx ∧ L(bx)]

When there are two quantificational expressions, the translations may often be produced in stages: Some dog likes every cat

Partial translation: ∃x[Dx ∧ x likes every cat]

Then 'x likes every cat' is handled just as if 'x' were a name: x likes every cat

∀y[Cy → L(xy)]

The whole sentence then has the form: ∃x[Dx ∧ x likes every cat ]

Some dog likes every cat

∃x[Dx ∧ ∀y[Cy → L(xy)] ]

Some additional examples are: Some dog chased a cat Some dog chased no cat Every dog chased a cat Every dog chased no cat

∃x[Dx ∧ x chased a cat] ∃x[Dx ∧ x chased no cat] ∀x[Dx → x chased a cat] ∀x[Dx → x chased no cat]

∃x[Dx ∧ ∃y[Cy ∧ H(xy)]] ∃x[Dx ∧ ~∃y[Cy ∧ H(xy)]] ∀x[Dx → ∃y[Cy ∧ H(xy)]] ∀x[Dx → ~∃y[Cy ∧ H(xy)]]

Some examples with three-place predicates, using ‘G()’ for ‘ gave  to ’: Some nurse gave a doll to a child ∃x[Nx ∧ x gave a doll to a child] ∃x[Nx ∧ ∃y[Dy ∧ x gave y to a child]] ∃x[Nx ∧ ∃y[Dy ∧ ∃z[Cz ∧ x gave y to z]]] ∃x[Nx ∧ ∃y[Dy ∧ ∃z[Cz ∧ G(xyz)]]]

Chapter 4 -- 3

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CHAPTER 4 SECTION 2

No child gave a doll to a nurse ~∃x[Cx ∧ x gave a doll to a nurse] ~∃x[Cx ∧ ∃y[Dy ∧ x gave y to a nurse]] ~∃x[Cx ∧ ∃y[Dy ∧ ∃z[Nz ∧ x gave y to z]]] ~∃x[Cx ∧ ∃y[Dy ∧ ∃z[Nz ∧ G(xyz)]]] Every child gave some doll to every nurse ∀x[Cx → x gave some doll to every nurse] ∀x[Cx → ∃y[Dy ∧ x gave y to every nurse]] ∀x[Cx → ∃y[Dy ∧ ∀y[Ny → x gave y to z]]] ∀x[Cx → ∃y[Dy ∧ ∀y[Ny → G(xyz)]]] Sometimes in English the wording is unclear regarding which quantificational expression has wider scope; that is, which quantificational expression has the other within its scope. For example, the sentence: Some freshman dated every sophomore can be read in two ways. One is the "super-dater" reading, which says that there is a certain freshman who dated every sophomore. Its symbolization is: ∃x[Fx ∧ x dated every sophomore]

∃x[Fx ∧ ∀y[Oy → D(xy)]]

Here, the quantifier '∃x' which originates from the 'some freshman' has widest scope, and the '∀y' which originates from the 'every sophomore' is within the scope of '∃x'. The other reading expresses the more natural situation, which merely says that for every sophomore, some freshman dated him/her: ∀y[Oy → some freshman dated y]

∀y[Oy → ∃x[Fx ∧ D(xy)]]

In this symbolization the '∀y' which originates from the 'every sophomore' has widest scope, and the quantifier '∃x' which originates from the 'some freshman' is within the scope of '∀y'. A slightly more complicated example of this is: Every ambulance went to a location in a mall Using ‘I()’ for ‘ is in ’ and ‘W()’for ‘ went to ’, this could mean that all the ambulances went to the same location: ∃y[My ∧ ∃x[Lx ∧ I(xy) ∧ ∀z[Az → W(zx)]]] "there is a mall, and a location in it, and every ambulance went there" This gives the '∃y' widest scope, and within its scope the '∃x' has wider scope than the '∀z'. Or it could mean that they were sent to locations in the same mall, though not necessarily to the same location: ∃y[My ∧ ∀z[Az → ∃x[Lx ∧ I(xy) ∧ W(zx)]]] "there is a mall and every ambulance was sent to some location in it" In this symbolization the '∃y' still has widest scope, but the '∃x' and '∀z' are interchanged. Or it could merely mean that each ambulance was sent to some location in some mall: ∀z[Az → ∃y[My ∧ ∃x[Lx ∧ I(xy) ∧ W(zx)]]] "every ambulance is such that there is a mall and a location in it and the ambulance went there" In this symbolization the '∀z' now has widest scope, and the '∃x' is within the scope of the '∃y'. All three symbolizations have the same ingredients; they differ with respect to how those ingredients are arranged. Certain words have as their main function indicating that the quantifier they occur with has a wide scope. An example is 'certain' in: Every reporter admired a certain car.

Chapter 4 -- 4

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CHAPTER 4 SECTION 2

The 'certain' gives the existential quantifier with 'car' wide scope: ∃x[Cx ∧ ∀y[Ey → A(yx)]] The phrase 'the same' can also work in this way, as in: Every reporter admired the same car. ∃x[Cx ∧ ∀y[Ey → A(yx)]] At every hoe-down the same fiddler played the same song ∃x[fiddler x ∧ ∃y[song y ∧ at every hoe-down x played y]] ∃x[fiddler x ∧ ∃y[song y ∧ ∀z[hoe-down z → x played y at z]]] ∃x[Fx ∧ ∃y[Gy ∧ ∀z[Hz → P(xyz)]]] As in earlier chapters, some linguistic constructions have no obvious rationale in terms of their parts. An example is the medieval example 'No man lectures in Paris unless he is a fool'. Here are different but equivalent symbolizations (using 'L()' for ' lectures in ' and 'a' for 'Paris'): ∀x[Mx → ~L(xa)∨Fx] ~∃x[Mx∧L(xa)∧~Fx]

EXERCISES 1. Symbolize each of the following: a. b. c. d. f. g. h. i. j. k. l. m. n. o. p. q.

Hans sees every doctor but Amanda doesn't see any doctor. Hans, who owns a dog, doesn't own a cat. Hans loves Amanda but she doesn't love him. Neither Hans nor Amanda has a cat. Some hyena and some giraffe like each other. Some giraffe likes every baboon. Some giraffe that likes every baboon likes no hyena. Some giraffe likes every baboon that likes no hyena Some giraffe likes every baboon that likes it Eileen resides in a big city.

Eileen and Betty both reside in the same city. If Hank resides in Brea then he attends UCLA; otherwise he doesn't attend UCLA. If David and Hank both live in Brea then David attends a private school and Hank attends a public school. Nobody who comes from Germany attends a California school. No giraffe likes Fido unless it is crazy Nobody gives a book to a freshman unless it is inexpensive

Chapter 4 -- 5

Version of Aug 2013

CHAPTER 4 SECTION 3

3 DERIVATIONS Adding many-place predicates to the notation has no effect on the rules of inference; they are already adequate as they stand. Here are two examples, using familiar techniques. Any giraffe that is taller than Harriet is taller than every zebra. Some giraffes aren't taller than some zebras. So there is a giraffe that is not taller than Harriet. ∀x[Gx ∧ T(xh) → ∀y[Ey → T(xy)]] ∃x[Gx ∧ ∃y[Ey ∧ ~T(xy)]] ∴ ∃x[Gx ∧ ~T(xh)] 1. Show ∃x[Gx ∧ ~T(xh)] 2. 3. 4. 5. 6. 7.

Gu ∧ ∃y[Ey ∧ ~T(uy)] Gu ∃y[Ey ∧ ~T(uy)] Ev ∧ ~T(uv) Gu ∧ T(uh) → ∀y[Ey → T(uy)] Show ~T(uh)

8. 9. 10. 11. 12. 13.

T(uh) Gu ∧ T(uh) ∀y[Ey → T(uy)] Ev → T(uv) T(uv) ~T(uv)

pr2 ei 2s 2s 4 ei pr1 ui ass id 3 8 adj 6 9 mp 10 ui 5 s 11 mp 5 s 12 id

14. Gu ∧ ~T(uh) 15. ∃x[Gx ∧ ~T(xh)]

dd

Betty scolded every dog that chased a cat. Betty is a jeweler. Some dog that chased Cleo was grey. Cleo is a cat. So a jeweler scolded some grey dog. ∀x[Dx ∧ ∃y[Cy ∧ H(xy)] → S(bx)] Jb ∃x[Dx ∧ H(xc) ∧ Gx] Cc ∴ ∃x[Jx ∧ ∃y[Dy ∧ Gy ∧ S(xy)]] 1. Show ∃x[Jx ∧ ∃y[Dy ∧ Gy ∧ S(xy)]] 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

Du ∧ H(uc) ∧ Gu Du ∧ ∃y[Cy ∧ H(uy)] → S(bu) H(uc) Cc ∧ H(uc) ∃y[Cy ∧ H(uy)] Du Du ∧ ∃y[Cy ∧ H(uy)] S(bu) Du ∧ Gu Du ∧ Gu ∧ S(bu) ∃y[Dy ∧ Gy ∧ S(by)] Jb ∧ ∃y[Dy ∧ Gy ∧ S(by)] ∃x[Jx ∧ ∃y[Dy ∧ Gy ∧ S(xy)]

pr3 ei pr1 ui 2ss pr4 4 adj 5 eg 2ss 7 6 adj 8 3 mp 2 s 7 adj 9 10 adj 11 eg pr2 12 adj 13 eg dd

Chapter 4 -- 6

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CHAPTER 4 SECTION 3

When existential quantifiers combine with universal ones, in general, a formula beginning with an existential followed by a universal is stronger than a universal followed by an existential. This simple derivation illustrates this: ∃x∀yF(xy) ∴ ∀y∃xF(xy)

something forces everything everything is forced by something

1. Show ∀y∃xF(xy) 2.

Show ∃xF(xy)

3. 4. 5.

∀yF(uy) F(uy) ∃xF(xy)

6.

pr1 ei 3 ui 4 eg dd 2 ud

If we try to prove the reverse: ∀y∃xF(xy) ∴ ∃x∀yF(xy) we won't be able to. The obvious strategy would be to set up a universal derivation, trying to derive 'F(uy)' in order to show '∀yF(uy)'. If we could do this, we could existentially generalize, and the derivation would be done. So we set things up for that: 1. Show ∃x∀yF(xy) 2. Show ∀yF(xy) 3. ∃xF(xy) 4. F(uy)

pr1 ui 3 ei ud?????

But this is not in the right form for a universal derivation; there is an 'F(uy)' when a 'F(xy)' is needed. And there is no way to get 'F(xy)' from '∃xF(xy)' by ei, since ei requires that the variable be new. One might naturally then try to derive the conclusion indirectly, by an indirect derivation: 1. Show ∃x∀yF(xy) 2. ~∃x∀yF(xy) 3. ∀x~∀yF(xy) 4. ~∀yF(xy) 5. ∃y~F(xy) 6. ∃xF(xy) 7. ??????????

ass id 2 qn 3 ui 4 qn pr1 ui

The problem at this point is that when one uses ei on lines 5 and 6, different variables need to be used, and there seems to be no way to derive a contradiction. In fact, the argument is invalid, and so no derivation will work. This will be shown in section 9.

Chapter 4 -- 7

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CHAPTER 4 SECTION 3

STRATEGY HINTS: The strategy hints given at the end of chapters 2 and 3 remain unchanged. They are repeated here for convenience. Try to reason out the argument for yourself. Begin with a sketch of an outline of a derivation, and then fill in the details. Write down obvious consequences. When no other strategy is obvious, try indirect derivation.

To derive:

Try this:

Conjunction

Derive each conjunct, and adjoin them

□∧○ Disjunction

Derive either disjunct and use add. (Often this is not possible.)

□∨○ Conditional □→○ Biconditional

Assume '~(□ ∨ ○)' for id and immediately use dm. Derive '~□ → ○' and use cdj Use cd Derive both conditionals and use cb.

□↔○ Negation of conjunction

Use id.

~(□ ∧ ○) Negation of disjunction

Derive '~□ ∧ ~○' and use dm.

~(□ ∨ ○) Negation of conditional ~(□ → ○) Negation of biconditional

Perhaps assume '□ ∨ ○' for id and try to derive both '□ → P∧~P' and '○→P∧~P'.

Then use sc (applied to the assumed '□ ∨ ○' and the conditionals) to derive 'P∧~P'. Use id. Derive '□ ↔ ~○' and use nb.

~(□ ↔ ○)

If you have this available:

Try this:

Conjunction

Simplify and use the conjuncts singly.

□∧○ Disjunction

Try to derive the negation of one of the disjuncts, and use mtp..

□∨○

Derive the conditionals '□ → △' and '○ → △', where '△' is something you want to derive. Then use sc with the disjunction and two conditionals.

Conditional

Try to derive the antecedent to set up mp, or derive the negation of the consequent, to set up mt.

□→○ Biconditional

Infer both conditionals and use them with mp, mt, and so on.

□↔○ Negation of conjunction

Use dm to turn this into '~□ ∨ ~○', and then try to derive either '□' or '○' to use mtp.

~(□ ∧ ○) Negation of disjunction

Use dm to turn this into '~□ ∧ ~○'; then simplify and use the conjuncts singly.

Chapter 4 -- 8

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CHAPTER 4 SECTION 3

Negation of conditional

Use nc to derive '□ ∧ ~○', then simplify and use the conjuncts singly.

~(□ → ○) Negation of biconditional

Use nb to turn this into '□ ↔ ~○', and use bc to get the corresponding conditionals.

~(□ ↔ ○)

To derive:

Try this:

Universal Quantification

Set up a universal derivation. Write a show line containing ∀x□, and

∀x□

then immediately follow this with a show line containing □. When the second show is cancelled, use rule ud to cancel the first. Or write a show line with '∀x□', and then assume '~∀x□' for an indirect derivation. Turn this into '∃x~□', and proceed from there.

Existential Quantification ∃x□

Derive an instance and then use rule eg. Or write a show line with '∃x□', and then assume '~∃x□' for an indirect derivation. Turn this into '∀x~□', and proceed from there.

Negation of a Universal Quantification

State a show line with '~∀x□', and then assume '∀x□' for an indirect derivation.

~∀x□ Or derive '∃x~□' and apply derived rule qn. Negation of an Existential Quantification

State a show line with '~∃x□', and then assume '∃x□' for an indirect derivation.

~∃x□ Or derive '∀x~□' and apply derived rule qn.

If you have this available:

Try this:

Universal Quantification

Use rule ui to derive an instance. (But use rule ei first if that is an option.)

∀x□ Existential Quantification

Use rule ei to derive an instance.

∃x□ Negation of a Universal Quantification

Use derived rule qn to turn this into an existential quantification.

~∀x□ Negation of an Existential Quantification

Use derived rule qn to turn this into a universal quantification.

~∃x□

Use rule av if necessary: If you are having difficulty with capturing when you use rule ui or ei, change what you are trying to derive to an alphabetic variant. Complete the derivation, and then use derived rule av to convert this into a derivation of what you are after.

Chapter 4 -- 9

Version of Aug 2013

CHAPTER 4 SECTION 3

EXERCISES Show each of the following arguments to be valid. 1.

∀x∀y∀z[S(xy) ∧ S(yz) → S(xz)] S(bc) ∧ S(ab) ∴ S(ac)

2.

∀x∀y[Ax ∧ By → [S(xy) ↔ ~S(yx)]] ∴ ∀x∀y[Ax ∧ By → [S(xy) ∨ S(yx)]]

3.

∀x∃yS(xy) ∀x∀y[Cx ∧ S(xy) → Dy] ∀x∀y[Dx ∧ S(yx) → Dy] ∴ ~∃x[Cx ∧ ~Dx]

4.

∃xEx ∧ ∃x~Ex ∀x∀y[Ex ∧ S(xy) → Ey] ∴ ∃x∃y~S(xy)

5.

∀x∀y[S(xy) ↔ S(yx)] ∃x∃y[Ax ∧ By ∧ S(xy)] ∴ ∃x∃y[By ∧ Ax ∧ S(yx)]]

6.

∃x[Ax ∧ ∀y[By → S(xy)]] ∀x∀y[Bx ↔ Cy] ∴ ∀x[Cx → ∃yS(yx)]

7. Prove the following biconditional theorems.

T251 ∀x∀yF(xy) ↔ ∀y∀xF(xy)

T252 ∃x∃yF(xy) ↔ ∃y∃xF(xy)

T255 ∀x∀yF(xy) ↔ ∀y∀xF(yx)

T261 ∀x[∃yF(xy)→∃yG(xy)] ↔ ∀x∀y∃z[F(xy)→G(xz)]

Chapter 4 -- 10

Version of Aug 2013

CHAPTER 4 SECTION 4

4 THE RULE "INTERCHANGE OF EQUIVALENTS" Although no new rules are needed when many-place predicates are added, some new rules are convenient. One of these is called "Interchange of Equivalents". This rule allows us to change any part of a formula to a known equivalent part. For example, it allows us to change: ∀x[Gx ∨ ~~H(xx)] into: ∀x[Gx ∨ H(xx)] by changing the part '~~H(xx)' to the equivalent 'H(xx)'. We know that '~~H(xx)' is equivalent to 'H(xx)' because rule dn says so. This new rule is given by: Rule ie ("interchange of equivalents") If we have a rule stating that a certain formula '□' is equivalent to another formula '○', then from any available line whose formula contains '□' we may infer a line with the same formula but with '□' changed to '○'. The justification consists of writing 'ie' followed by '/' and the name of the rule giving the equivalence. A rule establishes that one formula is equivalent to another if the rule can be applied to either to infer the other. These rules all establish equivalents: dn nc cdj dm nb qn av

All theories that treat definite descriptions as terms agree on how to treat proper definite descriptions. They obey the rule that if the description is proper, the descriptive part is true when its variables (the occurrences bound by '℩') are replaced by the definite description itself. For example, if 'the book that Betty wrote' is proper, then the book that Betty wrote is indeed a book, and Betty did write it: ∃z∀x[Bx∧W(bx) ↔ x=z]

properness of 'book that Betty wrote'

∴ B℩x[Bx∧W(bx)] ∧ W(b ℩x[Bx∧W(bx)])

the book that Betty wrote is a book ∧ Betty wrote the book that Betty wrote

The pattern is always the same: if there is a unique such-and-such, then the such-and-such is such-andsuch. When 'such-and-such' is complex, so is the application of this rule. Rule for Proper Descriptions (prd) If there is an available line or premise stating that the definite description '℩x○' is proper: ∃z∀x[○ ↔ x=z] then one may infer the formula that you get by taking '○' alone, replacing every free occurrence of 'x' in it by '℩x○'. As justification cite the earlier line number plus 'prd'.

For example, above we saw that the definite description ℩x[Bx∧W(bx)] is proper, because this is true: ∃z∀x[Bx∧W(bx) ↔ x=z]. Applying the rule, we infer the result of replacing every free occurrence of 'x' in 'Bx∧W(bx)' by '℩x[Bx∧W(bx)]'; that is, we infer: B℩x[Bx∧W(bx)] ∧ W(b ℩x[Bx∧W(bx)]).

Chapter 6 -- 5

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CHAPTER 6 SECTION 3

EXERCISES 1.

What can be inferred from the statements that say that these definite descriptions are proper? The spy who loved me The tallest giraffe to fly to the moon The number whose square root is the same as its cube root The boy such that he and the girl who saw him both sang The largest gift given to UCLA The big blue tuba

Chapter 6 -- 6

Version of Aug 2013

CHAPTER 6 SECTION 4

4 SYMBOLIZING ORDINARY LANGUAGE Often when we use a definite description in speech the description is proper only when limited to what is under discussion when it is used. For example, if Maria owns a dog, we may, in speaking to her, say 'The dog is hungry'. Here we do not intend to be speaking about all animals on earth, or even all animals in town. If we were, the description, 'the dog', namely, '℩xDx', would not be proper -- for there exist many dogs, not just one. However, if it is clear in the context in which we say 'the dog' that we are only speaking about things of immediate concern to us, then among those things there may indeed be exactly one thing that is a dog, and this would make our definite description proper. When Frege introduced notation for definite descriptions it was within a project of providing a logical foundation for mathematics. Typically, before introducing a definite description he would prove it to be proper. This approach worked well for his enterprise. When it comes to using logical notation to symbolize statements made in ordinary language, things are different. For example, suppose that somebody says: The dog that Betty owns chased the cat that Fred owns. One would usually infer from such an utterance that Betty owns a dog, that Fred owns a cat, that a dog chased a cat, and many other things. This is because using definite descriptions in such a sentence usually presupposes that the definite descriptions are proper. But if you symbolize the sentence above in the most straightforward way in our current logical notation, there is no such assumption. That is, if you symbolize the sentence above as: H(the dog that Betty owns, the cat that Fred owns) i.e.

H(℩x[Dx∧O(bx)] ℩x[Cx∧O(fx)])

there is nothing in the symbolization to indicate that the definite descriptions are proper. As a result, you cannot infer, for example, that some dog chased some cat: ∃x∃y[Dx∧Cy∧H(xy)] So if you want to symbolize everything that is communicated by a use of the English sentence above, you will need to add to your symbolization the assumption that the description is proper, something like: H(℩x[Dx∧O(bx)] ℩x[Cx∧O(fx)]) ∧ '℩x[Dx∧O(bx)]' is proper ∧ '℩x[Cx∧O(fx)]' is proper which in the case under discussion will be: H(℩x[Dx∧O(bx)] ℩x[Cx∧O(fx)]) ∧ ∃z∀x[Dx∧O(bx) ↔ x=z] ∧ ∃z∀x[Cx∧O(fx) ↔ x=z] On the other hand, not every use of a definite description presupposes that the description is proper. Consider the negation of the above statement: The dog that Betty owns didn't chase the cat that Fred owns. One could reasonably follow up this assertion with "in fact, Betty doesn't even own a dog!". Some people think the displayed sentence can be used either so as to presuppose that the descriptions are proper, or used so as not to presuppose this. There is no real consensus on this issue. EXERCISES Symbolize each of the following arguments (i) so as to include the claim that the definite descriptions are proper, and then (ii) so as not to include that claim. Assess each symbolized argument; if it is valid, produce a derivation to show this; if not, produce a counterexample. [You may want to read section 6 below before producing the counter-example.] 1. 2. 3.

The hyena that a lion chased fled.

∴ A hyena fled.

The winner will congratulate the loser.

∴ There will be a winner.

The cat that Maria owns chased a mouse that ate the fig.

∴ A cat that Maria owns chased a mouse that ate a fig.

Chapter 6 -- 7

Version of Aug 2013

CHAPTER 6 SECTION 5

5 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: IMPROPER DESCRIPTIONS Any complete theory of definite descriptions has to say how to handle improper definite descriptions -descriptions whose descriptive part is not satisfied by anything at all, such as: the planet between Mercury and Venus or descriptions that are satisfied by more than one thing, such as: the planet between Earth and the sun. One natural response is that these are terms that don't refer to anything at all. Some systems of logic work in this way -- they take improper definite descriptions to be terms that fail to refer. This is certainly a feasible approach, but it is a complicated one if we want a way to treat improper definite descriptions that fits in with the rules that we already have in this text. From chapter 3 on, we have made the idealization that our closed terms (simple names and complex terms containing operation symbols) each refer to a single thing. We will continue that idealization here, in the understanding that in the case of definite descriptions this is clearly artificial. Our artificial technique for handling improper definite descriptions was suggested over a century ago by the logician Gottlob Frege. The technique is to arbitrarily choose something for all improper descriptions to stand for. We then assume that any improper definite description stands for this thing. This thing can be anything -- the number zero, your pet dog, the tallest giraffe in the San Diego Zoo, the left front burner of the stove on which I cooked oatmeal today. Since the thing is arbitrarily chosen, we will not identify it in any further way, say by assigning a simple name to it, or applying a predicate to it. Any name that we are using might actually name the artificially chosen thing, but nothing in our logic tells us so. In spite of not knowing what it is, we can easily refer to this arbitrarily chosen thing with an appropriate complex term. We just need to use a definite description that we know to be improper. A natural example is to use the definite description: the thing that is not identical to itself ℩x x≠x Since nothing can fail to be identical to itself, this definite description has to be improper. This is a logical truth, for the statement that the description is improper is: ∴ ~∃z∀x[x≠x ↔ x=z] and we can easily produce a derivation to show that this is a theorem of logic: 1.

Show ~∃z∀x[x≠x ↔ x=z]

2. 3. 4. 5.

∃z∀x[x≠x ↔ x=z] ℩x x≠x ≠ ℩x x≠x ℩x x≠x = ℩x x≠x

ass id 2 prd sid 3 4 id

 putting '℩x x≠x' in for both occurrences of 'x' in 'x≠x'.

Our single rule for improper definite descriptions says that if a definite description is improper, it refers to whatever '℩x x≠x' refers to: Rule for Improper Descriptions (imd) If there is a statement on an available line or premise stating that the definite description '℩x○' is improper: ~∃z∀x[○ ↔ x=z] Then one may infer: ℩x○ = ℩x x≠x Justification: cite the earlier line number plus 'imd'

Chapter 6 -- 8

Version of Aug 2013

CHAPTER 6 SECTION 5

A description must either be proper or improper, although based on information given to us we may not know which. We do know this much, however: if the definite description does not refer to the chosen object, it must be proper: ℩x○ ≠ ℩x x≠x ∴ ∃z∀x[○ ↔ x=z] This is a trivial consequence of the rule for improper descriptions: 1. Show ∃z∀x[○ ↔ x=z] 2. 3. 4. 5.

~∃z∀x[○ ↔ x=z] ℩x○ = ℩x x≠x ℩x○ ≠ ℩x x≠x

ass id 2 imd pr1 3 4 id

One must be careful not to make a similar but invalid inference. Given that ℩x○ = ℩x x≠x one may not infer from this that '℩x○' is improper. Since the chosen object may be anything at all, it might be the dog that Cynthia bought. In this case the definite description 'the dog that Cynthia bought' refers properly to the chosen object. That is, we have: ℩x[Dx∧B(cx)] = ℩x x≠x where the definite description '℩x[Dx∧B(cx)]' is proper: ∃z∀x[Dx∧B(cx) ↔ x=z] So the chosen object can be referred to by proper descriptions, in addition to improper ones. Some applications of the rule for improper descriptions are relatively straightforward. An example is: ∃x∃y[x≠y∧Fx∧Fy] ~∃xGx ∴ ℩xFx = ℩xGx 1. Show ℩xFx = ℩xGx 2.

Show ~∃z∀x[Fx ↔ x=z] ∃z∀x[Fx ↔ x=z] ∀x[Fx ↔ x=i] u≠v∧Fu∧Fv Fu ↔ u=i u=i Fv ↔ v=i v=i u=v u≠v

3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Show ~∃z∀x[Gx ↔ x=z] ∃z∀x[Gx ↔ x=z] ∀x[Gx ↔ x=j] Gj ↔ j=j Gj ∃xGx ~∃xGx

13. 14. 15. 16. 17. 18. 19. 20. 21.

ass id 3 ei pr1 ei ei 4 ui 5 s s 6 bp 4 ui 5 s 8 bp 7 9 LL 5 s s 10 id

℩xFx = ℩x x≠x ℩xGx = ℩x x≠x ℩xFx = ℩xGx

ass id 13 ei 14 ui sid 15 bp 16 eg pr2 17 id 2 imd 12 imd 19 20 LL dd

Chapter 6 -- 9

 rule imd  rule imd

Version of Aug 2013

CHAPTER 6 SECTION 5

Often you will be given an argument whose premises contain definite descriptions that may be either proper or improper. If you can prove that a definite description is proper, you can often use that to prove other desired things. Likewise, if you can prove that a definite description is improper, you can often use that to prove other desired things. But sometimes you cannot prove either of these things, because not enough information is given to decide. You may still be able to use both strategies just described: you must both (i) infer what you want to infer using the assumption that the definite description is proper, and also (ii) infer what you want to infer using the assumption that the definite description is improper. If you can do this, you can use the rule for separation of cases to get the desired conclusion. ∀x[Hx → Gx] F℩xFx → G℩xFx ~H℩xFx → ℩xFx ≠ ℩xx≠x ∴ ∃xGx 1. Show ∃xGx 2.

Show ∃z∀x[Fx ↔ x=z] → ∃xGx ∃z∀x[Fx ↔ x=z] F℩xFx G℩xFx ∃xGx

3. 4. 5. 6. 7.

Show ~∃z∀x[Fx ↔ x=z] → ∃xGx ~∃z∀x[Fx ↔ x=z] ℩xFx = ℩xx≠x H℩xFx G℩xFx ∃xGx

8. 9. 10. 11. 12. 13.

ass cd 3 prd pr2 4 mp 5 eg cd

∃xGx

ass cd 8 imd 9 dn pr3 mt pr1 ui 10 mp 11 eg cd 2 7 sc dd

 rule sc

[You may be tempted to use separation of cases with the cases being identity with the chosen object ("℩xFx=℩xx≠x") and non-identity with the chosen object ("℩xFx≠℩xx≠x"). But this is not often useful, because if a definite description is identical to the chosen object, you still don't know whether it is proper or not, and so you can't use either rule prd or rule imd. It is usually better to take the cases to be the statements that the definite description is proper, and that it is improper.]

EXERCISES Produce derivations for the following arguments. 1.

∃x[Fx∧Gx∧Hx] ∃x[Fx∧Gx∧~Hx] ∴ ℩xFx = ℩xGx

2.

Fa ∧ Fb ∧ a≠b ∀x[Fx ∧ x≠a ↔ Gx] ℩xGx≠℩xFx ∴ ∀x[Gx → x=b]

3.

∀x∃y[R(xy) ∧ ∀z[z≠y → ~R(xz)]] ∃x∃y[x≠y∧ ∀z[z=x∨z=y]] ∴ ℩xR(bx)≠ ℩x~R(bx)

Chapter 6 -- 10

Version of Aug 2013

CHAPTER 6 SECTION 7

6 INVALIDITIES WITH DEFINITE DESCRIPTIONS The technique of giving counter-examples to show invalidity is unchanged from previous chapters, with one exception. Since what '℩x x≠x' stands for is not determined by its wording, when a counter-example is given the reference of '℩x x≠x' must be independently specified. Since rearranging things in the universe has no effect on counter-examples, it is simplest to always choose 0 to be the chosen object. We will abide by that custom here. Here are some counter-examples for arguments containing definite descriptions. The reader should review each to see how the counter-example works. ∃z∀x[Fx∧Gx ↔ x=z] ∃x[Fx∧~Gx] ∴ ∃y[Gy ∧ y≠℩x[Fx∧Gx]] Universe: {0, 1, 2} F: {1,2} G: {1} ℩x x≠x: 0 The first premise is true because the conjunction 'Fx∧Gx' is true of exactly one thing. The second premise is true because there is something, namely 2, that is F but not G. The conclusion is false because there isn't anything that is G and unequal to 1, which is what the proper description, '℩x[Fx∧Gx]', stands for. ∀y∃z∀x[R(xy) ↔ x=z] ∴ ∃y y=℩xR(xy) Universe {0,1} R: {, } ℩x x≠x: 0 The premise is true because whatever you pick for y, there is something -- namely, whatever y isn't -- such that everything bears R to y iff it is that thing that y isn't. The conclusion is false because nothing is the unique thing that is related to it by R: 0 isn't such a thing, and 1 isn’t such a thing. ~∀x[Fx ↔ Gx] ∴ ℩xFx ≠ ℩xGx Universe: {0, 1, 2} F: {} G: {1,2} ℩x x≠x: 0 The premise is true because F and G don't agree everywhere. (In fact, they agree nowhere.) The conclusion is false because both Fx and Gx are improper -- the first is true of nothing, and the second is true of two things -- so both definite descriptions stand for the chosen object, 0. EXERCISES Give counter-examples to show that these arguments are invalid: 1.

A℩xAx ∴ ∀x∀y(Ax∧Ay→x=y)

2.

∀x∃yR(xy) ∃x∀y[R(xy)→y≠a] ∴ ℩xR(xx) = ℩x x≠x

3.

℩xAx≠℩xBx ℩xBx≠℩xCx d(a)= ℩xBx ∀xd(d(x))=d(x) ∴ ℩xAx≠℩xCx

Chapter 6 -- 11

Version of Aug 2013

CHAPTER 6 SECTION 7

7 COUNTER-EXAMPLES WITH INFINITE UNIVERSES There are invalid arguments containing definite descriptions whose counter-examples require infinite universes. No new techniques are involved in giving such counter-examples. Here is an example of an invalid argument that can't be given a counter-example with a finite universe:

∀y∃z∀x[R(yx) ↔ x=z] ∀x∀y[x≠y → ℩zR(xz)≠℩zR(yz)] ∀x∃yR(yx)

A counter-example that shows it to be invalid is: Universe: {0, 1, 2, . . . } R(): =+1 The first premise is true here because for whatever you pick for y, there is something -- namely, y's successor -- such that everything is one greater than y iff it is y's successor. The second premise is true because whenever there are two different things, the unique thing that is one greater than the first is something other than the unique thing that is one greater than the second. And the conclusion is false because 0 is such that there is nothing such that it plus 1 is 0. EXERCISES Give counter-examples with infinite universes to show that these arguments are invalid: 1.

∀x∀y[x≠y → d(x)≠d(y)] ∴ ℩x~∃y x=d(y) = ℩x x≠x

2.

∀xR(x c(x)) ∀x∀y∀z[R(xy)∧R(yz)→R(xz)] ∴ ∀x∃yx=℩zR(yz)

3.

∀x∀y[b(x)=b(y)→x=y] ∃y∀x~y=℩z[z=b(x)] ∴ ∃x∃y[x≠y∧∀z[b(z)≠x∧b(z)≠y]]

Chapter 6 -- 12

Version of Aug 2013

Chapter Six -- Answers to the exercises 1 DEFINITE DESCRIPTIONS 1. Which of the following are well-formed formulas? B~℩xFx no negation cannot directly precede a term ~B℩xFx yes H(℩xFx ℩xFx) yes ∀x∀y(x=℩z∃u(G(zu)∧Au) ∧ G(y℩vCv) → x=y) yes ℩z(Bz∧∃yR(zy))=℩uDu yes ℩~Ax=~℩Ax no no variable following first ℩ ℩x~~Ax=~℩yBy no negation cannot directly precede a term R(℩xR(x℩yFy)y) yes

2 SYMBOLIZING SENTENCES WITH DEFINITE DESCRIPTIONS 1. Symbolize each of the following, 2. Read each symbolized sentence in stilted English using the recipe given above. Anna dated the tallest spy. D(a ℩x[Fx∧∀y[Fy∧y≠x→T(xy)]]) F: spy Anna dated the thing such that it is a spy and everything is such that if it is a spy and different from it then it is taller than it The person who put a bug in my drink will pay. I℩x[Ex∧∃y[By∧P(xy℩z[Dz∧H(az)])]] E: person P(xyz): x put y in z a: me H: have I: will pay The thing such that it is a person and something is such that it is a bug and it put it in the thing such that it is a drink and I have it will pay Beatrice likes the man who bought her a ring. L(b ℩x[Mx∧∃y[Iy∧B(xby)]]) Beatrice likes the thing such that it is a man and something is such that it is a ring and he bought her it Every giraffe loves the keeper who feeds it. ∀x[Gx→L(x ℩y[Ky∧F(yx)])] Everything is such that if it is a giraffe then it likes the thing such that it is a keeper and it feeds it Every giraffe loves the tallest keeper who feeds it. ∀x[Gx→L(x ℩y[Ky∧F(yx)∧∀z[z≠y∧Kz∧F(zx)→T(yz)]])] Everything is such that if it is a giraffe then it likes the thing such that it is a keeper and feeds it and everything is such that if it is another keeper who feeds it then it is taller than it The woman who studied did better than the woman who didn't study. B(℩x[Ax∧Ex] ℩x[Ax∧~Ex]) B: did better than A: woman E: studied The thing such that it is a woman and it studied did better than the thing such that it is a woman and it didn't study. Everybody honors the woman who gave birth to her/him. ∀x[Ex→H(x ℩y[Ay∧B(yx)])] Everything is such that if it is a person then it honors the thing such that it is a woman and it gave birth to it. The prize will be awarded to the person who spells the word correctly. A(℩xIx ℩y[Ey∧S(y ℩zFx)]) I: prize A: will be awarded to S: spells correctly F: word The thing such that it is a prize will be awarded to the thing such that it is a person and it spells correctly the thing such that it is a word. Every woman parked her own car. ∀x[Ax → P(x ℩y[Cy∧H(xy)])] Everything is such that if it is a woman then it parked the thing such that it is a car and she has it. Copyrighted material

Chapter 6 -- 13

Version of Aug 2013

3 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: PROPER DESCRIPTIONS 1.

What can be inferred from the statements that say that these definite descriptions are proper? The spy who loved me There is one and only one spy who loved me. The tallest giraffe to fly to the moon There is one and only one giraffe which flew to the moon and is taller than every other giraffe which flew to the moon. The number whose square root is the same as its cube root There is one and only one number such that its square root is its cube root.

The boy such that he and the girl who saw him both sang There is one and only one thing such that it is a boy and there is one and only one girl who saw it and they both sang. The largest gift given to UCLA There is one and only one gift given to UCLA which is larger than any other gift given to UCLA. The big blue tuba There is one and only one thing which is big and blue and is a tuba.

4 SYMBOLIZING ORDINARY LANGUAGE Symbolize each of the following arguments (i) so as to include the claim that the definite descriptions are proper, and then (ii) so as not to include that claim. Assess each symbolized argument; if it is valid, produce a derivation to show this; if not, produce a counterexample. [You may want to read section 6 below before producing the counter-example.] 1.

The hyena that a lion chased fled.

∴ A hyena fled.

(i) F℩x[Hx∧∃y[Iy∧C(yx)]] ∃z∀x[Hx∧∃y[Iy∧C(yx)]↔x=z] ∴ ∃x[Hx∧Fx]

1. Show ∃x[Hx∧Fx] 2. 3. 4.

H℩x[Hx∧∃y[Iy∧C(yx)]]∧∃y[Iy∧C(y℩x[Hx∧∃y[Iy∧C(yx)]])] H℩x[Hx∧∃y[Iy∧C(yx)]] ∧ F℩x[Hx∧∃y[Iy∧C(yx)]] ∃x[Hx∧Fx]

pr2 prd 2 s pr1 adj 3 eg dd

(ii) F℩x[Hx∧∃y[Iy∧C(yx)]] ∴ ∃x[Hx∧Fx] INVALID Universe: {0,1} F: {0} H: {} I: {} C: {} chosen object: 0 The premise is true because the description is improper, so it refers to 0, which 'F' is true of. The conclusion is false because nothing is both H and F.

Chapter 6 -- 14

Version of Aug 2013

2.

The winner will congratulate the loser.

∴ There will be a winner. (i)

C(℩xIx ℩xEx) ∃z∀x[Ix↔x=z] ∃z∀x[Ex↔x=z] ∴ ∃xIx 1.

Show ∃xIx

2. 3. (ii)

3.

I℩xIx ∃xIx

pr2 prd 2 eg dd

C(℩xIx ℩xEx) ∴ ∃xIx INVALID Universe: {0,1} I: {} E: {} C: {} chosen object: 0 The cat that Maria owns chased a mouse that ate the fig.

∴ A cat that Maria owns chased a mouse that ate a fig. (i)

∃x[Ex∧A(x ℩yFy) ∧ H(℩u[Cu∧O(au)]x)] E: mouse A: ate H: chased a: Maria ∃z∀x[Fx↔x=z]

∃z∀x[Cx∧O(ax) ↔ x=z]

∴ ∃v[Cv∧O(av)∧∃w[Ew∧∃r[Fr∧A(wr)]∧H(vw)]] 1. Show ∃v[Cv∧O(av)∧∃w[Ew∧∃r[Fr∧A(wr)]∧H(vw)]] C℩u[Cu∧O(au)] ∧ O(a ℩u[Cu∧O(au)]) Ei∧A(i ℩yFy) ∧ H(℩u[Cu∧O(au)]i) F℩yFy A(i ℩yFy) F℩yFy ∧ A(i ℩yFy) ∃r[Fr∧A(i r)] Ei ∧ ∃r[Fr∧A(i r)] Ei ∧ ∃r[Fr∧A(i r)] ∧ H(℩u[Cu∧O(au)]i) ∃w[Ew∧∃r[Fr∧A(wr)]∧H(℩u[Cu∧O(au)]w)] C℩u[Cu∧O(au)] ∧ O(a ℩u[Cu∧O(au)]) ∧ ∃w[Ew∧∃r[Fr∧A(wr)]∧H(℩u[Cu∧O(au)]w)] 12. ∃v[Cv∧O(av)∧∃w[Ew∧∃r[Fr∧A(wr)]∧H(vw)]] 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

(ii)

pr3 prd pr1 ei pr2 prd 3ss 4 5 adj 6 eg 3 s s 7 adj 3 s 8 adj 9 eg 2 10 adj 11 eg dd

∃x[Ex∧A(x ℩yFy) ∧ H(℩u[Cu∧O(au)]x)] ∴ ∃v[Cv∧O(av)∧∃w[Ew∧∃r[Fr∧A(wr)]∧H(vw)]] INVALID Universe: {0, 1, 2} F: {1} E: {2} A: {} C: {} O: {} H: {} chosen object: 0

Chapter 6 -- 15

Version of Aug 2013

ANSWERS for CHAPTER 6 SECTION 5

5 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: IMPROPER DESCRIPTIONS Produce derivations for the following arguments. 1.

∃x[Fx∧Gx∧Hx] ∃x[Fx∧Gx∧~Hx] ∴ ℩xFx = ℩xGx 1. Show ℩xFx = ℩xGx 2. 3. 4.

Fi∧Gi∧Hi Fj∧Gj∧~Hj Show ~∃z∀x[Fx↔x=z]

5. 6. 7. 8. 9. 10. 11. 12.

∃z∀x[Fx↔x=z] ∀x[Fx↔x=u] Fi↔i=u i=u Fj↔j=u j=u Hu ~Hu

pr1 ei pr2 ei ass id 5 ei 6 ui 2 s s 7 bp 6 ui 3 s s 9 bp 2 s 10 LL 3 s 10 LL 11 id

13. Show ~∃z∀x[Gx↔x=z] 14. 15. 16. 17. 18. 19. 20. 21.

∃z∀x[Gx↔x=z] ∀x[Gx↔x=v] Fi↔i=v i=v Fj↔j=v j=v Hv ~Hv

22. ℩xFx=℩xx≠x 23. ℩xGx=℩xx≠x 24. ℩xFx = ℩xGx

ass id 14 ei 15 ui 2 s s 16 bp 6 ui 3 s s 18 bp 2 s 19 LL 3 s 19 LL 20 id 4 imd 13 imd 22 23 LL dd

Chapter 6 -- 16

Version of Aug 2013

ANSWERS for CHAPTER 6 SECTION 5

2.

Fa ∧ Fb ∧ a≠b ∀x[Fx ∧ x≠a ↔ Gx] ℩xGx≠℩xFx ∴ ∀x[Gx → x=b] 1. Show ∀x[Gx → x=b] 2.

Show Gx → x=b

3. 4. 5. 6. 7. 8. 9 10. 11. 12.

Gx Show ~∃z∀x[Fx↔x=z] ∃z∀x[Fx↔x=z] ∀x[Fx↔x=i] Fa↔a=i a=i Fb↔b=i b=i a=b a≠b

ass cd as id 5 ei 6 ui pr1 s s 7 bp 6 ui pr1 s s 9 bp 8 10 LL pr1 s 11 id

13. 14. 15.

℩xFx=℩xx≠x ℩xGx≠℩xx≠x Show ∃z∀x[Gx↔x=z]

4 imd 13 pr3 LL

16. 17. 18.

~∃z∀x[Gx↔x=z] ℩xGx=℩xx≠x ℩xGx≠℩xx≠x

ass id 16 imd 14 r 17 id

19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

∀x[Gx↔x=j] Fb ∧ b≠a ↔ Gb b≠a Fb∧b≠a Gb Gb ↔ b=j b=j Gx↔x=j x=j x=b

15 ei pr2 ui pr1 s sm pr1 s s 21 adj 20 22 bp 19 ui 23 24 bp 19 ui 3 26 bp 25 27 LL cd 2 ud

Chapter 6 -- 17

Version of Aug 2013

ANSWERS for CHAPTER 6 SECTION 5

3.

∀x∃y[R(xy) ∧ ∀z[z≠y → ~R(xz)]] ∃x∃y[x≠y∧ ∀z[z=x∨z=y]] ∴ ℩xR(bx)≠ ℩x~R(bx) 1. Show ℩xR(bx)≠ ℩x~R(bx) 2. 3. 4.

i≠j ∧ ∀z[z=i ∨ z=j] R(bk) ∧ ∀z[z≠k→~R(bz)] Show ∃z∀x[R(bx)↔x=z]

5.

Show ∀x[R(bx)↔x=k]

6.

Show R(bx)↔x=k

7.

pr2 ei ei pr1 ui ei show 'R(bx)' is proper

Show R(bx)→ x=k

8. 9. 10. 11.

R(bx) x≠k→~R(bx) ~x≠k x=k

12.

ass cd 3 s ui 8 dn 9 mt 10 dn cd

Show x=k→R(bx)

13. 14.

x=k R(bx)

15.

R(bx)↔x=k

16. 17.

∃z∀x[R(bx)↔x=z]

ass cd 3 s 13 LL cd 7 12 cb dd 6 5

ud eg

dd

18. Show k=i → ∃z∀x[~R(bx)↔x=z] 19. 20.

k=i Show ∀x[~R(bx)↔x=j]

21.

show that if k=i then ass cd

Show ~R(bx)↔x=j

'~R(bx)' is proper

2 s 19 LL

Show ~R(bx) → x=j

22. 23. 24. 25. 26.

~R(bx) x≠k x≠i x=j

ass cd 3 s 23 LL 19 24 LL 2 s ui 25 bp cd

Show x=j → ~R(bx)

27. 28. 29. 30. 31.

x=j x≠i x≠k ~R(bx)

32.

~R(bx)↔x=j

ass cd 2 s 28 LL 19 29 LL 3 s ui 30 mp cd 21 27 cb

33.

dd

32 ud ∃z∀x[~R(bx)↔x=z]

34.

20 eg

cd

35. Show k=j → ∃z∀x[~R(bx)↔x=z]

show that if k=j then

LINES 36-51 FOLLOW THE PATTERN OF 19-34 52. 53. 54. 55. 56.

k=i∨k=j ∃z∀x[~R(bx)↔x=z] R(b ℩xR(bx)) ~R(b ℩x~R(bx)) ℩xR(bx)≠ ℩x~R(bx)

2 us 18 35 52 sc 4 prd 53 prd 54 55 LL dd

Chapter 6 -- 18

'~R(bx)' is proper conclude '~R(bx)' is proper  prd  prd

Version of Aug 2013

ANSWERS for CHAPTER 6 SECTION 6

6 INVALIDITIES WITH DEFINITE DESCRIPTIONS Give counter-examples to show that these arguments are invalid: 1.

A℩xAx ∴ ∀x∀y(Ax∧Ay→x=y) Universe: {0,1} A: {0,1} chosen object: 0

Since 'A' is true of two things, 'Ax' is improper, and so '℩xAx' stands for the chosen object, 0. Since 'A' is true of 0, the premise is true. The conclusion is false because 0 and 1 are both A. 2.

∀x∃yR(xy) ∃x∀y[R(xy)→y≠a] ∴ ℩xR(xx) = ℩xx≠x Universe: {0, 1, 2} R: {,,} a: 1 chosen object: 0

The description 'R(xx)' is proper, since 'R' holds of exactly one pair of identical things: . The definite description, '℩xR(xx)', then stands for 1, which is not the chosen object, 0, so the conclusion is false. The first premise is clearly true, and the second is true when 'x' is taken to be 2. 3.

℩xAx≠℩xBx ℩xBx≠℩xCx d〈a〉= ℩xBx ∀xd〈d〈x〉〉=d〈x〉 ∴ ℩xAx≠℩xCx Universe: {0,1} A: {1} C: {1} B: {0} a: 0 d〈0〉=0 d〈1〉=1 (chosen object: 0, though this is not important since all descriptions are proper)

Chapter 6 -- 19

Version of Aug 2013

ANSWERS for CHAPTER 6 SECTION 7

7 COUNTER-EXAMPLES WITH INFINITE UNIVERSES Give counter-examples with infinite universes to show that these arguments are invalid: 1.

∀x∀y[x≠y → d〈x〉≠d〈y〉] ∴ ℩x~∃yx=d〈y〉 = ℩xx≠x Universe: {-1, 0, 1, 2, . . . } d(): +1 chosen object: 0

The description '~∃yx=d〈y〉' in this universe is uniquely true of -1, so '℩x~∃yx=d〈y〉' refers to -1, which is distinct from 0 (the chosen object), so the conclusion is false. The premise is true since whenever x and y are different, so are x+1 and y+1. 2.

∀xR(xc(x)) ∀x∀y∀z[R(xy)∧R(yz)→R(xz)] ∴ ∀x∃yx=℩zR(yz) Universe: {0, 1, 2, . . . } R(): 