Logic Primer (1st edition)

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Acknowledgements We were unfortunately remiss in the first edition in failing to thank Harry Stanton for his encouragement to write this text. We gratefully acknowledge the comments we have received from colleagues who have taught from the first edition , particularly Jon K vanvig and Chris Menzel . A number of typo graphical errors were identified by an extremely meticulous self - study reader from the Midwest whose identity has become lost to us and whom we encourage to contact us again (and to accept our apologies) . Chris Menzel also receives credit for his extensive (and voluntary ) work on the web software. Finally , Amy Kind deserves special thanks for her help with the website and for her most useful comments on the manuscript for the second edition .

This excerpt is provided, in screen-viewable form, for personal use only by members of MIT CogNet. Unauthorized use or dissemination of this information is expressly forbidden. If you have any questions about this material, please contact [email protected]

This excerpt is provided, in screen-viewable form, for personal use only by members of MIT CogNet. Unauthorized use or dissemination of this information is expressly forbidden. If you have any questions about this material, please contact [email protected]

Preface To the Student The most important thing for you to know about this book is that it is designed to be used with a teacher. You should not expect to learn logic from this book alone (although it will be possible if you have had experience with formal systems or can make use of the web site at http :// mitpress.mit .edu/ LogicPrimer /) . We have deliberately reduced to a minimum the amount of explanatory material , relying upon your instructor to expand on the ideas. Our goal has been to produce a text in which all of the material is important , thus saving you the expense of a yellow marker pen. Consequently, you should never turn a page of this book until you understand it thoroughly . The text consists of Definitions , Examples, Comments, and Exercises. (Exercises marked with asterisks are answered at the back of the book .) The comments are of two sorts. Those set in full - size type contain material we deem essential to the text. Those set in smaller type are relatively incidental - the ideas they contain are not essential to the flow of the book , but they provide perspective on the two logical systems you will learn .

In this age of large classesand diminished personalcontact between studentsand their teachers , we hope this book promotesa rewarding learningexperience. To the Teacher We wrote this book because we were dissatisfied with the logic texts now available . The authors of those texts talk too much. Students neither need nor want page after page of explanation that require them to turn back and forth among statements of rules , examples, and discussion. They prefer having their teachers explain things to them- after all , students take notes. Consequently, one of our goals has been to produce a text of minimal chattiness, leaving to the

Preface

in Sb"uctor the task of providing explanations . Only an insb"uctor in a given classroom can be expected to know how best to explain the material to the students in that class, and we choose not to force upon the insb"uctor any particular mode of explanation .

Another reason our for dissatisfaction was that most texts contain material that we are not interestedin teachingin an introductory logic class. Some logic texts, and indeed some very popular ones, contain chapterson informal fallacies, theoriesof definition, or inductive logic, and some contain more than one deductive apparatus . Consequently , we found ourselvesordering texts for a single-semestercourse and covering no more than half of the material in them. This book is intended for a one-semestercourse in which propositional logic and predicatelogic are introduced, but no metatheory. (Any studentwho has masteredthe material in this book will be well preparedto take a second course on metatheory, using Lemmon' s classic, Beginning ' Logic, or evenTennants Natural Logic.) We prefer systemsof natural deductionto other ways of representing ' , and we have adopted Lemmon s technique of explicitly arguments tracking assumptionson each line of a proof. We find that this techniqueilluminates the relation betweenconclusionsand premises better than other devices for managingassumptions . Besidesthat, it allows for shorter, more elegant proofs. A given assumptioncan be dischargedmore than once, so that it need not be assumedagain in order to be dischargedagain. Thus, the following is possible, and there is no needto assumeP twice:

1 2 1,2

(1) (2) (3)

P~ (Q & R) P Q& R

assume assume from 1,2

Preface

1,2 1,2 1 1 1

(4) (5) (6) (7) (8)

Q R P-+Q P-+R (P - + Q) & (P - + R)

from 3 from 3 from 4, discharge2 from 5, discharge2 from 6,7

---

Clearly, the notion of subderivation has no application in such a system. The alternative approachinvolving subderivationsallows a given assumptionto be dischargedonly once, so the following is needed:

P-+ (Q & R) P Q& Q p~ Q P Q& R

(8) (9)

R P-+R

(10) (P~ Q) & (P~ R)

assume assume from 1,2 from 3 from 4, discharge2 assume from 1,6 (sameinferenceas at 3!) from 7 from 8, discharge6 from 5,9

The redundancyof this proof is obvious. Nonetheless , an instructor who preferssubderivationstyle proofs can useour systemby changing the rules concerningassumptionsetsas follows: (i ) Every line has the assumptionset of the immediately preceding line, except when an . (ii ) The only assumption available for assumption is discharged at a line is the highest-numberedassumptionin the discharge given , that line assumptionset. (iii ) After an assumptionhasbeendischarged number can never again appearin a later assumptionset. (In other

Preface

words , the assumption - set device becomes a stack or a flfSt - in - last - out memory device .)

There are a number of other differences between our system and Lemmon' s, including a different set of primitive rules of proof. What follows is a listing of the more significant differences betweenour ' systemand Lemmon s, togetherwith reasonswe prefer our system. .

Lemmon disallows vacuousdischargeof assumptions . We allow it . Thus it is correct in our systemto dischargean assumptionby reductio ad absurdum when the contradictiondoesnot dependon that assumption. Whenevervacuousdischargeoccurs, one can obtain a Lemmon-acceptable deduction by means of trivial additionsto the proof. We prefer to avoid theseadditions. (Note that Lemmon' s preclusion of vacuous discharge means that accomplishingthe same effect requires redundantsteps of & introduction and & -elimination. For instance, Lemmon requires (a) to prove P ~ Q ~ P, while we allow (b) . (a) 1 2 1,2 1,2 1

( 1) (2) (3) (4) (5)

P Q Q& P P Q~ P

assume assume from 1,2 from 3 from 4, discharge2

(b) 1 2 1

(1) (2) (3)

P Q Q~ P

assume assume

from 1, discharge2

xiii

Preface

Lemmon ' s characterization of proof entails that an argument has been established as valid only when a proof has been given in which the conclusion depends on all of the argument' s premises. This is needlessly restrictive , since in some valid arguments the conclusion is in fact provable from a proper subset of the premises. We remove this restriction , allowing a proof for a given argument to rest its conclusion on some but not all of the ' argument s premises. We have replaced Lemmon ' s primitive v -Elimination rule by what is normally known as Disjunctive Syllogism (DS ) . We realize that Lemmon ' s rule is philosophically preferable, as it is a pure rule ; however , DS is so much easier to learn that pedagogical considerations outweigh philosophical ones in this case. Despite the preceding point , we have kept the 3 -elimination rule used by Lemmon . Although slightly more complicated than the more common rule of 3 - Instantiation , this rule frees the student from having to remember to instantiate existential quantifications before instantiating universal quantifications . It also frees the student from having to examine the not- yet-reached conclusion of the argument, to determine which instantial names are unavailable for a given application of 3 - Instantiation . Furthermore , at any point in a proof using 3 -elimination , some argument has been proven . If the proof has reached a line of the form

m, ...,n

(k)

z

tilen tile sentencez has been establishedas provable from the premise set { m, . . .,n } . (Here the right-hand ellipsis indicates

xiv

Preface

which role was applied to yield z, and which earlier sentencesit was applied to.) This is quite useful in helping the student understand what is going on in a proof . In a system using 3 instantiation , however , this feature is absent: there are correct proofs some of whose lines do not follow from previous lines , since the role of 3 -instantiation is not a valid role . For instance, the following is the beginning of a proof using 3 -instantiation . 1 1

( 1) ( 2)

3xFx Fa

assumption 1 3 - instantiation

Line 2 does not follow from line 1. This difference between 3 elimination and 3 -instantiation can be put as follows : in an 3 elimination proof , you can stop at any time and still have a correct proof of some argument or other , but in an 3 -instantiation proof , you cannot stop whenever you like . It seems to us that

these implications of 3-instantiations invalidity outweigh the additional complexity of 3-elimination. In an 3-elimination system, not only is the systemsoundas a whole, but every rule is individually valid; this is not true for an 3-instantiationsystem. .

Whereas Lemmon requires that existentialization (existential generalization) replace all tokens of the generalizedname by tokensof the bound variable, we allow existentializationto pick up only someof the tokensof the generalizedname. We have abandoned Lemmon' s distinction between proper namesand arbitrary names, which is not essentialin a natural deductionsystem. The conditionson quantifier rules ensurethat the instantial name is arbitrary in the appropriatesense. ( We commenton this motivation for the conditionsin the text.)

Preface

In many cases, we have deliberately not used quotation marks to indicate that an expression of the formal language is being mentioned. In general, we use single quotes to indicate mention only when confusion might result. ( We hope no one is antagonized by this flaunting of convention . Trained philosophers may at first find the absence of quotes disconcerting , but we believe that we are making things easier without leading the student astray significantly .)

We have tried to presentthe material in a way that revealsclearly the systematicorganizationof the text. This mannerof presentationmakes it especiallyeasy for studentsto review the material when studying, and to look up particular points when the need arises. Consequently , thereis little discursiveprosein the text, and what seemedunavoidable . We hope to have produced a has been relegatedto the Comments small text that is truly studentoriented but that still allows the instructora maximumof flexibility in presentingthe material. TheSecondEdition With one exception, the changesto the second edition have been minimal. We haveaddeda ueatmentof identity to chapters3 and 4. In chapter3 this requiredmerely a slight modification to the definition of wff , somecommentson uanslation, and the inclusion of inuoduction and elimination rules for identity. The changesmadeto chapter4 are more extensive. In the first edition we avoided overt referenceto the object language/ metalanguage distinction and had no need to introduce into the specification of interpretations the extensions (denotations, referents) of names, but the inclusion of identity in the language necessitatesthem. To keep matters simple , when giving interpretations for sentences that involve identity we use italicized names in the metalanguage, and we recommend that no member of the universe of an interpretation be given more than one metalinguistic name. This makes it easy to specify whether or not two names of the object

Preface

language have the same extension in an interpretation , for the same metalinguistic name will be used for names denoting the same object . Expansions now involve the use of italicized names, so that strictly speaking they are not wffs of the object language. This does not affect their use in determining truth values of quantified wffs in an interpretation , and facilitates their use in determining truth values of wffs involving identity . ( We realize that italicization is not available for hand- written exercises, so we recommend that instructors adopt a convention such as underlining for blackboard presentations.) The addition of the material on identity is supplemented with new exercises in chapters 3 and 4. We have tried to organize the new material in such a way that an instructor who wishes to omit it can do so easily .

In chapter 1, a set of exerciseshas beeninsertedwhoseproofs do not require ~ I and RAA . That is, these proofs do not involve the . These exercises are intended to allow discharge of assumptions studentsto become comfortable with the remaining rules of proof before they are forced to learn the more complicatedmechanicsof ~ I andRAA . In chapter 3 we have waited until after the section on translations to introduce the notions of a wff s universalization , existentialization , and instance. This change reduces the chance of the student' s confusing the rules for constructing universally quantified wffs , where at least one occurrence of a name must be replaced by a variable , and univers alization , where all occurrences of the name must be replaced.

Preface

xvii

Web Support A variety of interactive exercises and an automated proof checker for the proof systems introduced in this book can be accessed at http :// mitpress.mit .edu/ Logic Primer/ . Use of the software requires nothing more than a basic web browser running on any kind of computer .

This excerpt is provided, in screen-viewable form, for personal use only by members of MIT CogNet. Unauthorized use or dissemination of this information is expressly forbidden. If you have any questions about this material, please contact [email protected]

This excerpt is provided, in screen-viewable form, for personal use only by members of MIT CogNet. Unauthorized use or dissemination of this information is expressly forbidden. If you have any questions about this material, please contact [email protected]

Chapterl Sentential Logic 1.1

notion. Basic logical

argument , premises , conclusion

Definition. An ARGUMENT is a pair of things: . a setof sentences , the PREMISES . a sentence , the CONCLUSION . Comment. All argumentshave conclusions, but not all argumentshave premises: the set of premisescan be the empty set! Later we shall examine this idea in somedetail. Comment . If the sentencesinvolved belongto English (or any other natural language), we need to specify that the premisesand the conclusion are sentencesthat can be true or false. That is, the premisesand the conclusionmust all be declarative(or indicative) sentencessuch as ' The cat is on the mat' or ' I am here' , and not sentencessuchas ' Is the cat on the mat?' (interrogative) or ' Come here!' (imperative) . We are going to construct some formal languagesin which every sentenceis either true or false. Thus this qualification is not presentin the definition above.

validity

Definition . An argument is VA Lm if and only if it is necessary that if all its premises are true, its conclusion is true.

Comment . The intuitive idea captured by this definition is this: If it is possiblefor the conclusionof an argument to be false when its premisesare all b1Je , then the argumentis not reliable (that is, it is invalid).

1 Chapter

If true premisesguaranteea true conclusion then the argumentis valid. Alternateformulation of the definition. An argumentis V A Lm if and only if it is impossible for all the premisesto be true while the conclusionis false. entaUment

Definition. When an argumentis valid we say that its premisesENTAll.., its conclusion.

soundness

Definition . An argument is SOUND I if valid and all its premises are b"ue.

andonly if it is

. It follows that all sound argumentshave Comment hue conclusions. Comment. An argument may be unsound in either of two ways : it is invalid , or it has one or more false premises. with validity rather Comment . The restof this bookis concerned than soundness.

Exercise1.1

Indicate whether each of the following sentencesis True or False.

i* ii * iii * iv* v*

Every premise of a valid argument is hUe. Every invalid argument has a false conclusion . Every valid argument has exactly two premises. Some valid arguments have false conclusions. Some valid arguments have a false conclusion despite having premises that are all hUe.

Chapterl

vi * vii * ... Vlll * ix * x*

A sound argument cannot have a false conclusion . Some sound arguments are invalid . Some unsound arguments have true premises. Premises of sound arguments entail their conclusions. If an argument has true premises and a true conclusion then it is sound.

1.2

A Formal Language

fonnal

Comment . To representsimilarities among arguments of a natural language, logicians introduce formal . The first formal languagewe will introduce languages is the language of sentential logic (also known as propositionallogic) . In chapter3 we introduce a more sophisticatedlanguage: that of predicatelogic.

language

vocabulary

for Sentential

Logic

. TheVOCABULARY OF SENTENTIAL Definition -

LOGICconsists of . SENTENCE LETTERS , . CONNECTIVES and , . PARENTHESES . sentence

Definition. A SENTENCE LEIf ER is anysymbol

letter

from the following list:

A, ... , Z, Ao'... , Zo'AI' ... ,Zit....

sentence variable

Comment . By the useof subscriptswe make available an infinite numberof sentenceletters. Thesesentence letters are also sometimescalled SENTENCE VARI ABLES , becausewe use them to stand for sentences of naturallanguages .

Chapter1

Definition. The SENTENTIAL CONNECTIVES (often just called CONNECTIVES ) are the members of the following list: - , & , v , - +, H .

connectives

Comment. The sentential connectives correspond to various words in natural languages that serve to connect declarative sentences.

tilde

-

The TILDE correspondsto the English ' It is not the casethat' . (In this casethe useof the term ' connective' is odd, since only one declarativesentenceis negated at a time.)

' ampersand & The AMPERSAND correspondsto the English Both ... and ... ' . The WEDGE correspondsto the English ' Either ... or . . . ' in its inclusivesense.

wedge

v

an - ow

~ The ARROW corresponds to the English

then

,

... .

double arrow

The DOUBLE - ARROW correspondsto the English ' ' if and only if .

Chapterl

Comment. Natural languagestypically provide more than one way to expressa given connectionbetweensentences . For instance, the sentence' John is dancing but Mary is sitting down' expresses the same logical relationship as ' John is dancing and Mary is sitting down' . The issue of translation from English to the formal languageis takenup in section 1.3.

) and ( . expression

The right and left parentheses are used as punctuation marks for the language .

Definition. An EXPRESSION of sentential logic is any sequenceof sentenceletters, sentential rives, or left andright parentheses . Examples. (P ~ Q) is an expressionof sententiallogic. )PQ~ - is alsoan expressionof sententiallogic. (3 ~ 4) is not an expressionof sententiallogic.

metavariable Definition. Greek letters such as , and 'I' are used as METAVARIABLES . They are not themselvesparts of the language of sentential logic, but they stand for expressionsof the language. Comment. (, ~ '1') is not an expressionof sentential : an expressionof logic, but it may be usedto represent sententiallogic.

ChapterI

Definition. A WELL - FORMED FORMULA ( WFF) of sententiallogic is any expressionthat accordswith the following sevenrules:

weD - formed

f OrDlu1a

( 1) A sentenceletter standing alone is a wff .

[Definition.The sentence lettersaretheATOMIC of thelanguage: of sententiallogic.] SENTENCES

atomic sentence

(2) If , is a wff , then the expression denoted by - , is also awIf . [Definition. A wff of this form is known as a NEGA TION , and - ~ is known asthe NEGATION OF ~.]

negation

(3) If ~ and 'I' are both wffs, then the expression denoted by (~ & '1') is a wff .

[Definition. A wff of this form is knownas a CONJUNCTION. ct>and'If areknownasthe left andright .] CONJUNCTS,respectively

conjunction

(4) If cj) and 'If are both wffs, then the expression denoted by (~ v '1/) is a wff .

disjunction

[Definition. A wff of this form is known as a DIS JUNCTION . ~ and '1/ are the left and right DISJUNCTS, respectively.]

(5) If (j) and 'I' are both wffs, then the expression denoted by (j) - + '1') is a wff .

Chapter1

conditional , antecedent , consequent

[Definition. A wffofthis Connis known as a O : NXnO NAL. Thewff , is knownas theANTECEDENT of the conditional . The wff 'If is known as the .] CONSEQUENTof theconditional

(6) If , and 'If are both wiTs, then the expression denoted by (cj) H '1') is a wff .

biconditional [Definition. A wff of this form is known as a BI CONDm ONAL. It is alsosometimes knownasan . EQUIVALENCE] (7) Nothingelseis awIf . binary and unary connectives

Definition. & , v , ~ , and H are BINARY CONNECTIVES , since they connect two wffs together. - is a UNARY CONNECTIVE , sinceit attachesto a single wff .

sentence

Definition. A SENTENCE of the fonnallanguage is a wff that is not part of a largerwff .

denial

Definition . The DENIAL of a wff (j) that is not a negation is - (j). A negation, - (j), has two DENIALS : (j) and - - (j).

Example. - (P - + Q) hasone negation: - - (P - + Q) It hastwo denials: (P - + Q) and - - (P - + Q) . (P - + Q) hasjust one denial: its negation, - (P - + Q).

Chapterl

Comment. The reason for introducing the ideas of a sentence and a denial will be apparent when the rules of proof are introduced in section 1.4.

Exercise 1.2.1 Which of die following expressions are wffs ? If an expression is a wff , say whedier it is an atomic sentence, a conditional , a conjunction , a disjunction , a negation, or a biconditional . For die binary connectives , identify die component wffs (antecedent, consequent , conjuncts , disjuncts , etc.) .

0 1* 00 n* 000 U1* iv * v* vi * vii * viii * ix * x* xi * xii * xiii * xiv * xv *

A (A (A ) (A - + B)

(A - + ( (A - + ( B - + C P & Q) - + R)

A & B) v (C -+ (0 H G ) - (A -+ B) - (P -+ Q) v - (Q & R) - (A) (- A) -+ B (- (P& P) & (PH (Q v - Q ) (- B v P) & C) H 0 v - G) -+ H (- (Q v - (B v (EH (0 V X )

1 Chapter parenthesisdropping conventions

Comment. For easeof reading , it is often convenient to drop parentheses from wffs , so long as no ambiguity results. If a sentence is surrounded by parenthesesthen these may be dropped. Example . p ~ Q will be read as shorthand for (P ~ Q) . Comment. Where parentheses are embedded within sentences we must be careful if we are to omit any parentheses. For example, the expression P & Q ~ R is potentially ambiguous between P & Q) ~ R) and (P & (Q ~ R . To resolve such ambiguities , we adopt the following convention: - binds more strongly than all the other connectives; & and v bind component expressions more strongly than ~ , which in turn binds its components more strongly than H .

. Examples - P & Q ~ R is readas - P & Q) ~ R). P ~ Q H R is readas P ~ Q) H R). P v Q & R is not allowed,asit is ambiguous between v P ( (Q & R and P v Q) & R). P ~ Q ~ R is notallowed,asit is ambiguous between (P ~ (Q ~ R and P ~ Q) ~ R). Comment. The expressions admitted by these parenthesis -dropping conventions are not themselves well fonned fonnulas of sentential logic .

1 Chapter

Exercise1.2.2 Rewrite all the sentencesin exercise1.2.1 above, using -dropping conventions. Omit any parentheses the parenthesis you can without introducingambiguity. Exercise 1.2.3 State whether each of the following is ambiguous or unambiguous, given the parenthesis-dropping conventions . In the unambiguous cases, write out the sentences and reinstate all omitted parentheses. 0 1* 00 n* 000 U1* iv *

vi * vii * viii * ix *

PH - QvR PvQ ~ R & S PVQ ~ RHS PvQ & R ~ - S P~ R & S~ T P~ Q ~ R ~ S P& QH - RvS - P& QVR ~ SHT P~ Q & - R H - Sv T ~ U P~ Q & - R ~ - Sv T H U

1.3

Translation of English to Sentential Wffs

translation

Definition . A TRANSLATION SCHEME for the lan guage of sentential logic is a pairing of sentence letters with sentencesof a natural language. The sentencesin a translation scheme should be logically simple . That

scheme

is , they should not contain any of the words corresponding to the sentential connectives.

ChapterI

logical form

Definition. The LOGICAL FORM of a sentenceof a natural language relative to a translation scheme is given by its translation into a wff of sententiallogic accordingto that translationscheme. Example. Under the translationscheme P: Johndoeswell at logic Q: Bill is happy The sentence If Johndoeswell at logic, then Bill is happy hasthe logical form (P ~ Q) . Comment. English provides many different ways of stating negations, conditionals, conjunctions, disjunctions , andbiconditionals. Thus, manydifferent sentences of English may havethe samelogical form.

sty Ustic varianu

Definition. If two sentencesof a naturallanguagehave the same logical form relative to a single translation scheme, they are said to be STYLISTIC VARIANTS of eachother. Comment. There are far too many stylistic variants of negations, conjunctions, disjunctions, conjunctions, and biconditionals to list here. The follow is a partial list of stylistic variantsin eachcategory.

1 Chapter

negations

conditionals

Let P b' anslate the sentence ' John is conscious.' Here are a few of the ways of expressing - P: John is not conscious. John is unconscious. It is not the case that John is conscious. It is false that John is conscious.

Stylistic variants whose logical fonn is <. - + '1'), where . is the antecedentand 'I' is the consequent include the following : If . , '1'. . only if '1'. . is a sufficient condition for '1'. . is sufficient for '1'. Providedthat . , '1'. 'I' providedthat . . 'I' on the condition that . . 'I' is a necessarycondition for , . 'I' is necessaryfor . . Whenever. , '1'. 'I' if . . Given that . , '1'. In case. , '1'. . only on the condition that '1'.

Chapterl

conjunctions

Variants widi logical form (. & '1') include die following : (j) and'll . Both (j) and'll .

cj), but'll. cj), although'll. cj)aswell as'II. Thoughcj), 'II. cj), also'll. disjunctions Variantswith logical fonD (. v 'II>include these: . or 'l' . Either . or '1'. . unless'1'. ' Comment. , unless '1" is also commonly translated as ( - 'I' - + , ) . The proof techniques introduced in section 1.4 can be used to show that this is equivalent to (, v '1') . biconditionals

Variants having the logical form (, H '1') include the

following: . . , , neither. nor ...

if andonly if '1/. is equivalentto '1/. . is necessaryand sufficient for '11 . just in case'11

of theforill ' Neither~ nor'1/' have Englishsentences thelogicalforill - (~ v '1/), or, equivalently , (- ~ & - '1/).

Chapter1

tenses

Comment. In English , the sentences ' Mary is dancing ' and ' Mary will dance' have different meanings because of the tenses of their respective verbs. In some cases, when one is analyzing arguments it is important to preserve the distinction between tenses. In other cases, the distinction can be ignored . In general, a judgment call is required to decide whether or not tense can be safely ignored .

Example. Considerthe following two arguments: A If Mary is dancing, Johnwill dance. Mary is dancing. Therefore, Johnis dancing. B

If Mary dances, John will dance. If John dances, Bill will dance. Therefore , if Mary dances, Bill will dance.

In A , if the difference between ' John will dance' and ' John is dancing ' is ignored , then the argument will look valid in translation . But this seems unreasonable

on inspectionof the English. In B , ignoring the difference between ' John will dance' and ' John dances' also makes the argument valid in translation . In this case , however , this seems reason -

able.

Chapter1

In the translation exercisesthat follow, assumethat tensedistinctionsmay be ignored. Exercise 1.3

Translatethe following sentencesinto the languageof sententiallogic.

for 1- 20

1* 2* 3* 4* 5* 6*

10*

12* 13*

Johnis dancingbut Mary is not dancing. If Johndoesnot dance, then Mary will not be happy. John' s dancingis sufficient to makeMary happy. John' s dancingis necessaryto makeMary happy. John will not danceunlessMary is happy. If John' s dancing is necessaryfor Mary to be happy, Bill will be unhappy. If Mary dancesalthough John is not happy, Bill will dance. If neitherJohnnor Bill is dancing, Mary is not happy. Mary is not happy unless either John or Bill is dancing. Mary will be happyif both Johnand Bill dance. Although neither John nor Bill is dancing, Mary is happy. If Bill dances, then if Mary dancesJohnwill too. Mary will be happyonly if Bill is happy.

I Chapter 14* 15

*

16* 17

*

18* 19* 20*

Neither Johnnor Bill will danceif Mary is not happy. If Mary dancesonly if Bill dancesand John dances only if Mary dances, then John dancesonly if Bill dances. Mary will danceif Johnor Bill but not both dance. If John dancesand so does Mary, but Bill does not, then Mary will not be happybut Johnand Bill will . Mary will be happyif and only if Johnis happy. Provided that Bill is unhappy, John will not dance unlessMary is dancing. If John danceson the condition that if he dancesMary dances, then he dances. ira P: Q: R: S: T:

21*

22* 23

*

24*

DSlation schemefor 21- 25 A purposeof punishmentis deterrence . Capital punishmentis an effectivedeterrent. Capital punishmentshouldbe continued. Capital punishmentis usedin the United States. A purposeof punishmentis rebibution.

If a purpose of punishment is deterrence and capital punishment is an effective deterrent , then capital punishment should be continued . Capital punishment is not an effective deterrental though it is used in the United States . Capital punishment should not be continued if it is not an effective deterrent , unless deterrence is not a purpose of punishment . If retribution is a purpose of punishment but deterrence is not , then capital punishment should not be continued .

Chapter1

25

*

Capital punishment should be continued even though capital punishment is not an effective deterrent provided that a purpose of punishment is retribution in addition to deterrence.

Primitive Rules of Proof turnstile

Definition. The TU RN STa Eis the symbol ~ .

sequent

Definition. A SEQUENT consists of a number of sentencesseparatedby commas(correspondingto the premises of an argument), followed by a turnstile, followed by another sentence(correspondingto the conclusionof the argument). Example. (P & Q) - + R, - R & P ~ - Q more than a Comment. Sequents are nothing convenient way of displaying arguments in the formal notation . The turnstile symbol may be read as ' therefore ' .

proof

Definition. A PROOF is a sequence of lines containing . Eachsentenceis either an assumptionor sentences the result of applying a rule of proof to earlier . The primitive rules of proof sentencesin the sequence are statedbelow.

Chapter1

Comment. The purpose of presenting proofs is to demonsuate unequivocally that a given set of premises entails a particular conclusion . Thus, when presenting a proof we associate three things with each sentence in the proof sequence:

annotation

On the right of the sentencewe provide an ANNO TADON specifying which rule of proof was applied to which earlier sentencesto yield the given sentence .

assumptionset On the far left we associatewith eachsentencean ASSUMPTIONSET containingthe assumptions on which the given sentencedepends. line number

Also on the left , we write the current LINE NUM BER of the proof.

Hneof proof

Definition. A sentenceof a proof, togetherwith its annotation setandthe line number , its assumption , is calleda LINE OF TIlE PROOF. . Example 1,2 (7) P - + Q & R i Linenumber i set Sentence Assumption

proof for a given argument

6 ~ I (3) Annotation

Definition. A PROOF FOR A GIVEN ARGUMENT is a proof whose last sentence is the argument's conclusion depending on nothing other than the ' arguments premises.

Chapter1

primitive

rules

assumption

Definition. The ten PRI Mm V E RULES OF -inttoPROOF are the rules assumption , ampersand -elimination, wedge-inttoduction, duction, ampersand wedge-elimination, arrow-inttoduction, arrow-elimination , reductio ad absurdum, double-arrow-inttoduction, anddouble-arrow-elimination, as describedbelow. Assume any sentence .

Annotation : Assumption set: Comment:

A The currentline number. Anything may be assumed at any time . However , some assumptions are useful and some are not !

Example.

(1) ampersand . intro

Pv Q

Given two sentences ( at lines m and n ) , conclude conjunction of them .

Annotation : Assumption set: Comment:

Also known as:

a

m, n & 1 The union of the assumption sets at lines m and n. The order of lines m and n in the proof is irrelevant . The lines referred to by m and n may also be the same. Conjunction (CONI ) .

I Chapter

. Examples 1 ( 1) 2 (2) 1,2 (3) 1,2 (4) 1 (5) ampersand -

elim

p Q P& Q Q& P P& P

Given a sentence that is a conjunction clude either conjunct . Annotation : Assumption set : Also known as:

wedge-intro

A A

1,2 &1 1,2 &1 1,1&1 ( at line m ), con -

m &E The sameasat line m. Simplification (S).

. Examples (a) 1 (1) P& Q 1 (2) Q" 1 (3) P

l &E l &E

(b) 1 1

l &E

(1) (2)

P& (Q ~ R) Q~ R

Given a sentence (at line m), conclude any disjunction having it as a disjunct . Annotation : Assumption set: Also known as:

m vI The same as at line m. Addition (ADD ) .

Chapterl

Examples.

wedge-elim

(a) 1 1 1

(1) (2) (3)

P Pv Q (RH - T) v P

A 1vI 1vI

(b) 1 1

(1) (2)

Q -+ R (Q -+ R) v (P & - g)

A 1 vI

Given a sentence(at line m) that is a disjunction and anothersentence(at line n) that is a denial of one of its disjunctsconcludethe other disjunct. Annotation : Assumption set: Comment : Also known as :

m, n v E The union of the assumptionsetsat lines m andn. The order of m and n in the proof is irrelevant. Modus Tollendo Ponens (MTP), Disjunctive Syllogism (OS) .

Examples.

(a) 1 2 1,2

(1) (2) (3)

Pv Q -P Q

A A

(b) 1 2 1,2

(1) (2) (3)

P v (Q -+ R) - (Q -+ R) P

A A

1,2 vB

1,2 vE

Chapter1

(c) 1 2 1,2 arrow - in tro

(1) (2) (3)

Pv - R R P

A A

1,2 v E

Given a sentence (at line n), conclude a conditional having it as the consequent and whose antecedent appears in the proof as an assumption (at line m) . Annotation :

n - +1 (m )

Assumption set :

Everything in the assumption set at line n excepting m, the line number where the antecedent was assumed. The antecedent must be present in the proof as an assumption . We this assumption speak of DISCHARGING when applying this rule . the number m in parentheses Placing indicates it is the discharged assumption . The lines m and n may be the same. Conditional Proof (CP) .

Comment :

Also known as:

. Examples (a) 1 (1) 2 (2) 1,2 (3) 1 (4)

- PvQ P Q P~ Q

A A

1,2 vE 3 ~ I (2)

ChapterI

(b) 1 2 2 (c) 1

armw - elim

(1) (2)

R P

A A

(3)

P~ R

1-+1(2)

( 1) (2)

P P-+P

A 1 - +1 ( 1)

Given a conditional sentence(at line m) and another sentencethat is its antecedent(at line n), concludethe consequentof the conditional. Annotation : Assumption set:

m, n - +E The union of the assumption sets at lines m and n-

Comment: Also known as:

. Example 1 ( 1) 2 (2) 1,2 (3)

The order of m and n in the proof is irrelevant. Modus Ponendo Ponens (MPP), Modus Ponens (MP), Detachment , Affirming the Antecedent.

P~ Q P Q

A A

1,2 -.+E

Chapterl

reductio ad absurdum

Alsoknownas:

IndirectProof(IP), - Intra! - Elim.

Examples.

(a) 1 2 3 1,3 1,2

(1) (2) (3) (4) (5)

---

(b) 1 2 3 2,3 1,2,3 1,3

P-+ Q -Q P Q -P

A A A 1,3 ~ E

PvQ -P - P~ - Q -Q P P

A A A

2,4 RAA(3)

2,3 -+E 1,4 vE 2,5 RAA(2)

I Chapter

(c) 1 2 3 2,3

(1) (2) (3) (4)

p Q -Q -p

A A A

2,3 RAA(1)

double - armw . Given two conditional sentences having the forms intro cI> - + 'I' and 'I' - + cI> (at lines m and n ), conclude a

biconditional with cI>on one side and 'I' on the other. Annotation :

m , n HI set: The union of the assumption sets at Assumption lines m and n. Comment: The order of m and n in the proof is irrelevant .

. Examples 1 (1) 2 (2) 1,2 (3) 1,2 (4)

P~ Q Q~ P PHQ QH P

A A

1,2 HI 1,2 HI

double-arrow - Given a biconditionalsentence, H ' I' ( at line m ), con eUm clude either . ~ 'If or 'If ~ . . Annotation : Assumption set : Also known as :

mHE the sameas at m. Sometimesthe rules HI and HE aresubsumedasDefinition of Biconditional (df.H ) .

Chapter1

. Examples 1 (1) PHQ 1 (2) P- + Q 1 (3) Q - +P

A tHE tHE

Comment . These ten rules of proof are truth preserving . Given b' Ue premises, they will always yield b' Ueconclusions. This entails that if a proof can be consb' Uctedfor a given argument, then the argument is valid . aid in the discoveryof proofs, . A numberof strategies Comment . Wedonotprovideanyprooffor practice butthereis no substitute ' s . We in this book- that is the instructor job discoverystrategies lack of should be no so there do provideplenty of exercises , . opportunityto practice Exercise 1.4.1 Fill in the blanks in the following proofs .

P, - Q ~ P& - Q 1 (1) P (2) - Q (3) P& - Q PvQ , - QvR,

(1) Pv Q (2) - Qv R (3) (4) Q (5)

1,3 vE 2,4

Nil .>~ ~,

~ > l 0. t -~t ~.

Qt 0 . ~ > l at to -Qt ~-~( t ~

N

N~

~

. ~ t ~

--- I Q: 0

.*

3,5 RAA

--- Qt 0

~ t oi ~ > 0 . 0

: *

I -

1 Chapter

Exercise 1.4.2 Give proofs for the following sequents . All of these proofs may be completedwithout using the rules - +1 or RAA.

81* 82 83* 84* 85 86 87* 88 89* 810

P v - R, - R - + S, - P ~ S P v - R, - R - + S,- P ~ S & - R P - + - Q, - Q v R - + - S, P & T ~ - S P & (Q & R), P & R - + - S, S v T ~ T P - + Q, P - + R, P ~ Q & R P, Q v R, - R v S, - Q ~ P & S - P, Rv - PHPVQ ~ Q (P H Q) - + RiP- + Q,Q - + P ~ R - P- + Q & R, - Pv S - + - TU & - P ~ (U & R) & - T (Q v R) & - S - + T, Q & U, - S v - U ~ T & U

Sequents and Derived Rules double turnstile

Comment. If a sequent has just one sentence on each side of a turnstile , a reversed turnstile may be inserted ( -i ) to represent the argument from the sentence on the right to the sentenceon the left .

Example. P ~ ~ P v P Comment. This examplecorrespondsto two sequents : p ~ p v P and P v P ~ P. You may read the example as saying ' P thereforeP or P, and P or P thereforeP' . When proving cI ~~ 'If, one must give two proofs: one for cI ~ 'If and one for 'If ~ cI.

ChapterI

. Example ProveP -i .- P v P. (a) ProveP .- P v P. 1 ( 1) P 1 (2) P v P

A 1v I

(b) ProveP v 1 ( 1) 2 (2) 1,2 (3) 1 (4)

A A 1,2 v E 2,3 RAA (2)

P.- P. Pv P -P P P

Exercise1.5.1 Give proofs for the following sequents , using the primitiverulesof proof.

S11* S12* S13 S14* S15 S16* S17* S18* S19 S20 S21* S22 S23 S24 S25

P -i .- - - P P - + Q, - Q .- - P P - + - Q, Q .- - P - P --+Q, - Q I- P - P --+- Q, Q I- P P --+Q, Q --+RIP --+R PI - Q --+P - PIP --+Q PIP - -+Q P --+Q, P --+- Q I- - P - P v Q ~I- P -:0+Q P v Q ~I- - P --+Q P v Q ~I- - Q --+P P v - Q ~I- Q --+P P v Q, P --+R, Q --+R I- R

DoubleNegation ModusTollendoTollens MTf MTf MTf HypotheticalSyllogism TrueConsequent FalseAntecedent FA Antecedent Impossible WedgeArrow (v~ ) v - + v - + v - +

Simple DUemma

Chapterl

S26* S27 S28* S29 S30 S31 S32* S33 S34 S35 S36 S37* S38* S39 S40 S41* S42* S43 S44 S45 S46 S47 S48 S49 S50 S51 S52

P v Q, P - + R, Q - + S ... R v S Complex Dilemma P - + Q, P - + Q ... Q SpecialDilemma - (P v Q) ~ ... - P & - Q DeMorgan' s Law - (P & Q) ~... - P v - Q OM P & Q ~ ' " ( P v Q) OM P v Q ~ ' " ( P & Q) OM - (P - + Q) ~ ... P & - Q NegatedArrow (Neg- +) - (P - + - Q) ~... P & Q Neg- + P - + Q ~ ... (P & Q) Neg- + P - + Q ~... (P & Q) Neg- + P & Q ~... Q & P & Commutativity P v Q ~ ... Q v P v Commutativity P H Q ~ ... Q H P H Commutativity P - + Q ~ ... Q - + P trausposition P & (Q & R) ~ ... (P & Q) & R & Associativity P v (Q v R) ~ ... (P v Q) v R v Associativity P & (Q v R) ~ ... (P & Q) v (P & R) &Iv Distribution P v (Q & R) ~... (P v Q) & (P v R) v/& Distribution P - + (Q - + R) ~ ... P & Q - + R Imp/ Exportation P H Q, P ... Q Biconditional Ponens P H Q, Q ... P BP P H Q, P ... Q Biconditional Tollens P H Q, - Q ... - P BT P H Q ~ ... - Q H - P Bi' Ira DSposition P H - Q ~... - P H Q Bi Trans - ( PH Q) ~ ... P H - Q NegatedH - ( P H Q) ~... - P H Q NegH

Chapter!

Exercise 1.5.2 Give proofs for the following sequents using the primitive rules of proof.

S53* S54 S55* S56* S57 S58 S59 S60 S61

S62 S63 substitution instance

PH Q -It- (P & Q) V (- P& - Q) P - + Q & R, R V - Q - + S & T, T HUt - P - + U (- P V Q) & R, Q - + St- P - + (R - + S) Q & R, Q - + P V S, - (S & R) t- P P - + R & Q, S - + - R V - Q t- S & P - + T R & P, R - + (S V Q), - (Q & P) t- S P & Q, R & - S, Q - + (P - + T) , T- +(R- +SvW) t- w R - + - P, Q, Q - + (P V - g) t- S - + - R P - + Q, P - + R, P - + S, T - + ( U- + (- V - + - S , Q - + T, R - + ( W- + V), V - + - W, W t- - P P H - Q & S, P & (- T - + - g) t- - Q & T PVQHP & Qt - PHQ Definition . A SUBSTITUTION INSTANCE of a sequent is die result of uniformly replacing its sentence letters with wffs .

Comment. This definition states that each occurrence of a given sentence letter must be replaced with the

samewff throughoutthe sequent. Example. The sequent P v Q J- - P - + Q hasas a substitutioninstancethe sequent (R & S) v Q J- - (R & S) - + Q accordingto the substitutionpattern P/ (R & g); Q/Q.

Chapter}

Comment. The given substitution pattern shows that the sentence letter P was replaced throughout the original sequent by the wff (R & S) , and the sentence letter Q was replaced throughout by itself .

Exercise1.5.3 Identify each of the following with a sequentin . exercise1.5.1 andidentifythesubstitution pattern i* ii * iii * iv* v* vi * vii * viii * ix* x* derived

R - + S -1.- - S - + - R - P - + Q v R, Q v R - + S.- - P - + S (P & Q) v R -1.- R v (P & Q) (PvQ) & (- Rv - S) -I'- PvQ) & - R) v PvQ) & - S) R v S -1.- - - (R v S) (P v R) & S -1.- - (P v R - + - S) P v (Q v R) -1.- - P - + Q v R - (P & Q) .- R - + - (P & Q) - P & Q) v (R & S -1.- - (P & Q) & - (R & S) P v (R v S), P - + Q & R, R v S - + Q & R .- Q & R rule

thatonehasprovedusingonly . Anysequent Comment be usedas a the primitiverulesmay subsequently DERIVEDRULEof proofif in the proof are the (i ) somesentences appearing premisesof the sequent,or (ii ) some sentencesappearing in the proof are the premisesof a substitutioninstanceof the sequent. In case (i ) the conclusion of the sequent may be assertedon the current line; in case(ii ) the conclusion . of the substitutioninstancemay be asserted

Chapterl

Annotation :

Assumptionset:

The line numbers of the premises followed by Sf , where S# is the number from the book, or the name of the sequent (see comment below ) . The union of the assumption sets of the premises.

Comment. All of the sequentsin exercise1.5.1 (SII 552) are used so frequently as rulesof proof thatthey have the names we have indicated. (Indeed, in some systems of logic some of our derived rules are given as

primitive rules.)

. Examples (a) ProveR v 5 - + T, - T ~ - R. 1 ( 1) R v 5 - + T 2 (2) - T 1,2 (3) - (R v 5) 1,2 (4) - R & - 5 1,2 (5) - R

A A

1,2 M1T 30M 4 &E

(b) ProveP v R ~ ST ~ - S .- T ~ - (P v R). 1 A ( 1) P v R ~ S 2 A (2) T ~ - S 1 S ~ (P v R) 1Trans (3) 1,2 2,3 HS (4) T ~ (P v R)

Chapter1

Comment . Requiring that the sequentto be used as a derived rule hasbeenproved using only primitive rules is unnecessarilyresbictive. If the sequentsare provedin a sbict order and no later sequent in the series is used in the proof of an earlier sequent , then no logical errors can result. We suggestthe strongerresbiction only becauseit is good practice to construct proofs using only the primitive rules. Exercise 1.5.4

Prove the following using either primitive or derived from the previous exercises . If you like a challenge , prove them again using primitive rules only . rules

S64 S65 S66* S67* S68 S69 S70 S71 S72 S73* S74 S75 S76* S77 S78* S79* S80* S81*

- p - + PiP PH Q -i .- - P - + Q) - + - (Q - + P P H Q -i .- P v Q - + P & Q PH Q -i .- - (P v Q) v - (- P v - Q) P H Q -i .- - (P & Q) - + - (P v Q) PH Q -i .- - (- (P & Q) & - (- P & - Q P v Q - + R & - P , Q v R, - R .- C - PHQ , P- + R, - R .- - QHR - P H - Q) H R), S - + P & (Q & T), R v (P & S) .- S & K - + R & Q (P & Q) v (R v S) .- P & Q) v R) v S P & (~ & - R), P- +(- S - + T), - S - + (T H R v Q) .- S P & - Q - +-R, (- S - + - P) H - R .- R H Q & (P & - S) P v Q, (Q - + R) & (- P v S), Q & R - + T .- T v S P - + QVR, (~ & S) v (T- +-P), - (- R - + - P) '- - T & Q P v Q, P - + (R - + - S), (- R H T) - + - P .- S & T - + Q (PH ~ - + - R, (- P& S) v (Q& 1) , SvT - + R .- Q - + P - S v (S & R), (S - + R) - + P .- P Pv ( RVQ ) , (R- +S) & (Q- +1) , S v T - + P v Q, - P .- Q

ChapterI

S82* S83* S84* S85* S86

(P - + Q) - + R, S - + (- Q - + T) t- R v - T - + (S - + R) P & Q - + R v St - (P - + R) v (Q - + S) (P- +Q) & (R - + f ), (PvR) & - (Q&R) t- (P& Q) & - R P & Q - + (R v S) & - (R & S), R & Q - + S, S - + R & Q) v (- R & - Q v - P t- P - + - Q - (P& --Q)v- (- R& -s), - S& --Q, T- +(- S- +- R& p) t- - T

1.6

Theorem.

theorem

Definition. A THEOREM is a sentence that can be proved from the empty set of premises. Comment. We can assert that a given sentence is a theorem by presenting it as the conclusion of a sequent with nQthing to the left of the turnstile .

. Example Provet- P& Q -+ Q & P. 1 (1) P& Q 1 (2) Q 1 (3) P 1 (4) Q & P (5) P& Q -+ Q & P

A 1 &E 1 &E 2,3 &1 4 - +1( 1)

Comment . Note that in step 5 we discharge assumption 1. Hence, the final conclusion rests on no . assumptions

ChapterI

Exercise 1.6.1 Provethe following theorems, (i ) using primitive rules only and (ii ) using primitive rules together with derivedrules establishedin a previousexercise.

Tl * T2* T3 T4* T5* T6 17 T8* T9* TIO* Til * T12

*

T13* T14* T15 T16 T17* TIS T19* T20 T21* T22 T23 T24 T25 T26 T27*

t- P ~ P Identity ExcludedMiddle t- P v - P Non-Contradiction t- - (P & - P) t- P ~ (Q ~ P) Weakening t- (P ~ Q) v (Q ~ P) Paradoxof Material Implication Double Negation t- P H - - P t- (PHQ ) H (QHP ) t- - (PHQ ) H (- PHQ ) Peirce's Law t- P ~ Q) ~ P) ~ P t- (P ~ Q) v (Q ~ R) t- (PHQ ) H (- PH - Q) ~ (- P ~ Q) & (R ~ Q) H (P ~ R) ~ Q ~ ~ ~ ~ ~ ~ ~

& Idempotence PH P& P v Idempotence V PH P P (PHQ ) & (RHS ) ~ P ~ R) H (Q ~ S (PHQ ) & (RHS ) ~ (P & RHQ & S) (PHQ ) & (RHS ) ~ (Pv R H Q V S) (PHQ ) & (RH S) ~ PHR ) H (QH S (PHQ) ~ (R~ P) H (R~ Q & P~ R) H (Q ~ R

.- (PHQ ) ~ (R & PHR & Q) .- (PHQ ) ~ (R V P H Rv Q) .- (PHQ ) ~ RHP ) H (RHQ .- P & (Q H R) ~ (P & Q H R) .- P~ (Q ~ R) H P~ Q) ~ (P~ R .- P ~ (Q ~ R) H Q ~ (P ~ R) '- P~ (P~ Q) HP ~ Q .- (P~ Q) ~ QH (Q ~ P) ~ P

1 Chapter

T28 T29 T30* T31* T32

*

T33 T34 T35 T36 T37 T38 T39

theorems as derived rules

~ P~ - QHQ ~ - P ~ - P~ PHP ~ (P& Q) v (R & S) H P v R) & (P v S & Q v R) & (Q v S ~ (P v Q) & (R v S) H P & R) v (P & S v Q & R) v (Q & S ~ (P~ Q) & (R ~ S) H - P & - R) v (- P & S v Q & - R) v (Q & S ~ (PV - P) & QHQ ~ (P& - P) VQHQ ~ P v (- P & Q) H P v Q ~ P& (- PVQ) HP & Q ~ PHPV (P& Q) ~ PHP & (PVQ ) ~ (P ~ Q & R) ~ (P & Q H P & R)

Comment. We now consider a special case of the use of sequents as derived rules. Since it is the conclusion of a sequent without premises, a theorem or a substitution instance of a theorem can be written as a line of a proof with an empty assumption set. For a theorem to be used this way , it must have been proved already by means of primitive rules alone. The annotation should be the name of the theorem or T# (the theorem' s number) .

Chapterl

. Example ProveP ~ Q, - p ~ Q .- Q. 1 ( 1) P ~ Q 2 (2) - p ~ Q (3) P v - p 1,2 (4) Q

A A

T2 1,2,3 SimDil

. In the precedingexample, the annotationfor Comment line 3 gives the number of the theorem introduced. Since this theorem has a name, the annotation ' ExcludedMiddle' would alsohavebeen . acceptable Comment. As with sequentintroductions. requiring that dleorems first

be proved using only

primitive

rules is unnecessarily

restrictive .

Exercise1.6.2 Using theoremsas derivedrules, attemptto construct alternativeproofs of sequentsappearingin exercise 1.5.4.

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Chapter2 Truth Tables 2.1 truth value

Definition. Truth and Falsity (abbreviatedT and F) are TRUm VALUES.

truth table

Comment . When an argumentis valid, its conclusion cannotbe false when its premisesare all true. One way to discoverwhetheran argumentis valid is to consider explicitly all the possible combinationsof truth values amongthe premisesand the conclusion. In this chapter we show how to do this. The idea is to assign truth valuesvariously to the sentencelettersof the argument and seehow the premisesand the conclusionturn out. The following rules, codified in TRUTH TABLES (17s ), enableus to do this. Comment. For this method to work , it has to be the case that the truth values of compound sentences are determined by the truth values of the sentence letters that appear in them.

truth functional connectives

Comment. All the sententialconnectives introducedin chapter I have the property described in the previous comment. Sincethe truth values of compoundsentencescontaining theseconnectives arefunctions of the truth values of the componentwffs, they are known as TRUTH -FUNCTIONAL CONNE Cfl VES. ( Not all English connectivesaretruth-functional.)

2 Chapter TT for negation

In order for a negation - , to be true , , must be false .

>

I

>

<

I

<

F

T T

F

Table2.1 Truthfunction for negation.

TT for conjunction

TT for disjunction

Tr for conditional

In order for a conjunction(cj) & '1/) to be true, both conjuncts cj) and'1/ mustbe true. In order for a disjunction(cj) v '1/) to be false, both disjuncts cIand 'II mustbe false. In order for a conditional (, - + 'If) to be false, the antecedent , must be true while the consequent 'If is false.

Tf for In order for a biconditional (, H 'If) to be true, , and biconditional ' II must have the same truth value .

Table2.2 Truthfunction.~.for the binary connectives

.

Chapter2

Comment. Observe that if a conditional ' s antecedent is false, then the conditional is troe no matter what the troth value of its consequent. Also , if its consequent is troe, then it is troe, regardless of the troth value of its antecedent. These are the troth table analogues of the derived rules False Antecedent and True Consequent.

Tfs for sentences

By means of these rules we can construct Tfs for compound wffs , exhibiting how their truth values are determined by the truth values of their sentence letters.

Example.

Table2.3 1Tfor the sentence(P ~ Q) v (- Q & R).

Chapter2

. By referring to the columnsfor P and Q, we Comment construct column (a), for (P - + Q), using the Tf for conditionals(seetable 2.2). Next, we constructcolumn (b), for - Q, (seetable 2.1) . Column (c), for (- Q & R) is constructed by referring to the columns for its conjuncts, - Q and R and using the Tf for conjunction (see table 2.2) . Finally, we construct columnd ), for (P - + Q) v (- Q & R), by referring to those for its disjuncts, (P - + Q) and (- Q & R) (seetable 2.2) . Comment.The column for a given componentof a sentence (other than the sentenceletters) is placed under that component's connective. For example, the column for (P ~ Q) in table 2.3 falls underits arrow. Exercise2.1

. Construct17s for the following sentences

i* ii * iii * iv * v* vi * vii * ... Vl11* ix * x*

Pv (- PvQ ) - (P & Q) v P - (P ~ Q) ~ P (P v Q) v (- P & Q) PvQ ~ Rv - P RH - Pv (R & Q) (P & Q H Q) ~ (Q ~ P) (P H - Q) H (- P H - Q) (PHQ ) H (PV R ~ (- Q ~ R (P & Q) v (R & S) ~ (P & R) v (Q & S)

For additionalpractice, construct 17s for wffs in chapter1.

2 Chapter

2.2

validity with TTs

Truth Tables for Sequents

' To determine a sequent s validity or invalidity , we construct a single Tf for the whole sequent. If there is a line in the Tf where all the premises are true and the conclusion is false, then the sequent is invalid . If there is no such line , it is valid .

valid example Table2.4 This sequentis valid sincethere is no line on which - p and Q - + (P & Q) are both true but - Q is false.

Chapter2

invalid example Table 2.5 This sequent is invalid since there is at least one line where - P - + Q and (R & P ) - + Q are both true but the conclusion is false - the fourth line .

invalidating assignment

Definition . An I NV A Lm A TIN GASSIGNMENT for a sequent is an assignment of truth and falsity to its sentence letters that makes the premises true and the conclusion false. Comment. From the Tf for an invalid sequent, you can read off an invalidating assignment. Find a row of the 17 where the premises are all true and the conclusion is false. The invalidating assignment is given at the left side of that row .

Example. An invalidatingassignmentfor the sequentin Table2.5 assignstruth to P and falsity to Q and R.

Chapter2 number of fines

Comment. When the sequent in question involves only two sentence letters, the Tf has exactly four lines ; three sentence letters requires eight lines. In general, when n sentence letters are present, the number of lines in the Tf is 2n.

incompatible premises

. Considerthis specialcase:if you constructa Comment 17 for the sequentP - + Q, Q - + R, P & - R ~ S you find that there is no line on which all the premisesare true. Consequently , there is no line on which the conclusionis false while all the premisesare true. Thus the sequentis valid.

Exercise2.2

Use Tfs to detennine whether each of the following sequents is valid . For each invalid one, find an invalidating assignment. For each valid one, give a proof .

i* ii * iii * iv * v* vi * vii *

P& - Q ~ - (PHQ ) P & (Q v R) ~ Q & (P v R) P& Q - + R ~ P- + R PvQ - +R ~ P- + R P- + QvR ~ P- + R (P - + - P) - + (- P - + P) ~ P Q - + R ~ (P - + Q) & (Q - + R) P v Q, P - + R, - S - + - Q ~ - P P - + Q, P - + R, - (- R - + Q) ~ P P H - Q, Q H - R, R H - S ~ PHS P v Q ~ (- P - + R) v (- Q - + R) P H (R - + P v - Q), - (R - + P v Q) ~ - Q - (R & - P - + Q v R) ~ - (Q H R) P - + (Q & R - + g), P, - S ~ - (Q & R)

viii * ix * x* xi * xii * xiii * xiv *

2 Chapter xv * xvi * xvii *

xviii *

- R ~ - Q, (- P& R) ~ - Q ~ - (PH - R & Q) S~ (fHP), Q~ (- SH - 1), - (P& RHT ~ S) ~ R & - Q Q ~ (P ~ R & - Q), - Q ~ - ( T V V), U & SH P ~ (S ~ - V) v - T Q V R ~ U & T, (P H Q), - (S V W) ~ P ~ QvV ~ (S & U) V (T & W)

Tautologies no premises

Comment . Another special case is a valid sequent without premises. In this case, validity requires that there be no lines of the Tf on which the conclusionis false, sinceno premisesarepresentto be considered.

a i ~ " i Q -

Table 2.6 A valid sequent without premises . 0 Q

2 Chapter

Table 2.7 An invalid sequent without premises.

tautology

Definition . A sentence cI> is a TAUTOLOGY (or, is TAUTOLOGOUS ) when the sequent that has no premises and has cI>as its conclusion is valid . Comment. When a sentence is a tautology , it cannot be false: its 17 has only Ts in the column for the sentence. Some sentences have only Fs appearing in their column of a 17 ; others have both Ts and Fs. The sentence appearing in table 2.6 is a tautology .

inconsistent and contingent

Definition . A sentencethat has only Fs in its column of a 17 is INCONSISTENT . A sentence that is neither tautologous nor inconsistent is CONTINGENT .

Chapter2

Comment. The sentenceappearingin table 2.7 is contingent .

Table 2.8 P is contingent , P & - P is inconsistent , and P v - P is tautologous .

Table2.9 ((P - + Q) - + P) - + P is tautologous.

2 Chapter

Tab~ 2.10 (P & Q) H (- P V - Q) is inconsistent . Exercise 2.3

Use Trs to establishthat all the theoremsconsidered in chapter aretautologies.

Indirect TTs indirecj TT

. 178 providea way Comment

to search systematicalry

. A shorterway of doing assignments indirect b "Uth table this is the (1Tf ). for invalidating

In an I1T , one attemptsto build invalidating assignments . When the sequentis valid, it is impossible to build an invalidating assignment(as in the first example below) .

Chapter2

In cases of invalid arguments, an invalidating assignment can be discovered (as in the second example below ) . Sometimes one must examine more than one assignment (as in the third example below ) .

easyvalid case Example. Considerthe sequent p ~ Q, - R ~ - Q ~ There false false

P T

is

only

- R

:

and

- +

must P

Q ,

for

way

must

- R TF

be

the

true

be

- +

-R ~ conclusion

and

- P

false

true

, as

shown

-

' -

- R

Q

-p (

- R

. That below

- +

- + is . R

- F ) to

be

must

be

.

- P

TFFFf

Having established these truth assignments, we now see if there is any way of making the premises all true that is compatible with this assignment. In other words , we need a value of Q to complete the following : -p p ~ Q, - R ~ - Q ' - - R ~ TT TFT TFF Ff The assignment indicated requires Q to be true, in order for the first premise to be true , but also requires - Q to be true (hence Q to be false), in order for the second premise to be true. This is the only way to make both premises true and the conclusion false, and it is impossible to achieve. Thus , there are no invalidating assignments, and the argument is valid .

Chapter2

easy invalid case

Example . The sequent below has a conditional conclusion . Thus , if the con~lusion is to be false, its antecedent must be hue and its consequent false.

P & - Q, T TTF

Q ~ R F TF

P ~ R TFF

The invalidating assignmentassignsT to P and F to Q

andR. harder case

Example. In the sequentbelow there are three ways to makethe conclusionfalse. Here is one of them: - P - + Q, - P - + - Q

~

P& Q FFT

On this assignment , the secondpremiseis false. Thus, . So we have failed to find an invalidating assignment we try a different way of making P & Q false: - P ~ Q, Ff

-P~ -Q Ff

r

P& Q TFF

arehue, sincetheybothhavefalse Here, bothpremises an . Thus, antecedents invalidatingassignment assigns T to P andF to Q.

Chapter2

Exercise2.4.1 Use I Trs to determinewhether exercise2.2 are valid or invalid.

sequents

given in

Exercise 2.4.2 Use I Trs to determine whether the following sequents are valid . For each invalid one, give an invalidating

. Foreachvalidone, constructa proof. assignment . 1* ..* 11 ...* 111 iv * v* vi * vii * ...* V111 ix * x* . Xl* xii * ... XUl* xiv * XV* xvi * xvii * xviii * xix * xx *

P - + Q, Q J- P P v Q, P J- Q P - + Q, - Q - + R J- P - + R P v - Q, - Q & R J- P & R P H Q v R, - Q J- - P P - + Q, (R - + S) - + - P J- Q v R P - + Q v R, Q - + S & T, - S J- - P P & - Q - + R, P H - R J- (Q & R) v P P - + Q & - R, - P v Q H S J- S - + - P v T - (P H Q), P - + R, Q - + S J- - R v S S - + Q, - S - + Q v T, T - + P J- P - + Q v R - Q - + S, S - + Q v - T, - T - + P J- Q - + P v R p ~ (- Q ~ - R & - S), - ( RH S), - Q J- - P P v Q, - (R ~ P) J- Q H (- T ~ - R v S) P & S ~ R, R v T, T ~ Q & P, - Q v U J- P ~ S v U - (P H Q), P ~ R, Q ~ S J- - R v S - (P~ - Q & R), - RH - PJ- P& Q (P ~ Q) & (- Q ~ P & R) ~ (S v T ~ - Q) J- Q ~ - (- S ~ T) .- (Pv - Q ~ - P & - Q) H - P QH - Q ' - PH - P

Chapter2

English Counterexamples counter -

example

Definition. An English COUNTEREXAMPLE for an invalid argumentor sequentis an argumentthat hasthe samelogical form as the original, but whosepremises are all obviously true and whose conclusion is obviouslyfalse.

. Example A counterexamplefor P -+ Q, Q J- Pis If Los Angeles is in Canada, then Los Angeles is in North America. Los Angelesis in North America. . Therefore, Los Angelesis in Canada Comment. The relationships of Los Angeles , Canada, and North America to one another are public knowledge . The premises are both obviously true, and the conclusion is obviously false.

. In consbuctinga counterexample Comment , it is not generallyuseful to consbuctthe premisesand the conclusion using either unspecific pronouns or . For example , given the invalid personalinformation above,onemightpresent sequent If it is raining then therearecloudsin the sky. Therearecloudsin the sky. Therefore, it is raining.

2 Chapter

Although one can see in a hypothetical situation that the premises might be true at the same time as the conclusion is false, the trouble with this argument as a counterexample is that the second premise is not obviously true (you may not be in a position to determine whether there are clouds in the sky) and likewise the conclusion is not obviously false. Similarly , the following is not useful: If my cousin is intelligent , she will pass logic . My cousin will pass logic . Therefore , my cousin is intelligent . Since it is not general knowledge who your cousin is and whether or not she is intelligent or will pass logic , this does not provide a clear counterexample to the given sequent.

Exercise2.5.1 Constructcounterexamplesfor the invalid sequentsin

chapter2. Exercise2.5.2 Give proofs, invalidating assignments , or counterexamples to establishthe validity or the invalidity of the following sequents :

I. ii ... III

p ~ Q, - Q v R, RIP - P v Q, - Q v R, - RIP PHQ , QH - RI - - PH - R

2 Chapter , IV V , VI " VII ", VIII , IX X XI ,, XII XIII , XIV XV XVI .. XVll XVl11 . XIX XX XXI .. XXll xxiii XXIV XXV

(Q - + P) - + R, - Q v S, - S ~ - R - + T P & (Q - + R), Q v - P, R v S - + T ~ T v U P H - Q, R v - Q, R H S ~ S v P PHQ , QH - R, R - + P~ - PH - R PHQ ~ (RHP ) H (PHQ ) - R H - Q, P v - Q, PHS ~ S v - R R H - Q, P v - Q, PHS ~ S & R (P - + Q) v (R - + S) ~ (P - + S) v (R - + Q) (P - + Q) & (R - + S) ~ (P - + S) & (R - + Q) P & Q, Q- +(R- +P), R - + (~ - +- Tv - W) , - S & T ~ W P& Q, Q - +(p- +R), R - + (~ - +- Tv - W) , - S & T ~ - W PvQ - +RvS, - (TvR ) - +S, (T - +P) & (R- +Q),- S ~ R - (Pv - Q), - P - + R v S, - S v - Q, R v - T - +W & (Y - + - Q) ~ - ( W- +Y) P v (QvR), S& T, - (~ v1) - + - P, (R- +W) & - W ~ Q P v (Q v R), S & - T, (R - + W) & - W ~ Q (PHQ ) H (- PH - R) ~ P- + (QHR ) P H Q, - (- R & P), R V S - + - (T & Q) ~ T - + ~ VQ) P H Q, R V - P, T & Q - + - R ~ - S & T - + - (P V Q) P & Q - + (R H S), - P - + - T, - (- R V S) ~ Q - + - T P & Q - + R, P & - R H Q V - S, T & (- Q & - R - + P), (T - + S) V (T - + R) ~ S & R Rv (p- +s)' T& - ~ (- TvW) - +-R, (S - + Q) & - Q ~ - P Rv (pvS), T& - ~ - (- TvW) - + - R, (S - + Q) & - Q ~ P

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Chapter 3 Predicate Lo Rit 3.1

A Formal Language for Predicate Logic Comment. Sentential logic allows us to study the logical relations among sentencesthat hold because of their sb"ucture, insofar as that sb"ucture is detennined by the presence of connectives. But sentential logic ' ' cannot handle the similarity between Kareem is tall ' ' and ' Akeem is tall , not to mention Someone is tall ' - these would be represented as P, Q , and R , as if they had nothing in common . We now introduce a new language that accommodates this further sb"ucture.

vocabulary

. TheVOCABULARYOF PREDICATE Definition of LOGICconsists . SENTENCE LEn ERS, . CONNECTIVES , . NAMES, . VARIABLES , . PREDICATE LEn ERS, . themENTIT YSYMBOL, . QUA N' CD' I KRS, and . PARENTHESES . Sentence letters, connectives, and parentheses are adopted from the language of sentential logic .

Chapter3

names

Definition. A NAME is a symbol from the following list: a, b , c , d,

variables

. ... ai ' bl ' CI ' dl ' ~ ' b2 '

Definition. A VARIABLE is a symbol from the following list: 0 , V, W , X , y , z ,

1'

V ' I WI ' Xl ' YI ' ZI '

2'

. . ..

Comment. Names and variables are used to refer to objects in much the same way as names and certain kinds of pronouns in English . Section 3.2 deals with translation between English and the language defined in this section. Comment. Where there is no possibility of confusion we shall sometimesuse lowercaseletters other than those listed above as names.

I -place

. A I -PLACE PREDICATE LETTERis Definition

predicate letter

any symbolfrom the following list: AI , ... , ZI , AJ , ... , ZJ, ... .

2-place

A 2-PLACE PREDICATE LETr ER is any symbol

n - place

from the list : following 2 A,...,Z2 ,A5 ,...,z5,....

In general, an n -PLACE PREDICATE LETTER is any symbolfrom the list

An, ... , Zn,~ , ... , ZO...

3 Chapter many-place

. Predicateletters with more than one place Comment are referred to as MANY -PLACE PREDICATE L E' nE R S. Predicate letters will sometimes be referredto as ' predicates'for short. Comment. In practice the superscripts can and will be omitted . Any of the capital letters may appear as sentence letters or predicate letters. It is usually possible to tell how a letter is being used in a wff by looking at the number of names or variables immediately following it . A capital letter with no names or variables is a sentence letter , one followed by one name or variable is a I -place predicate, and so on. ' Also , the letters ' R ' and ' 5 are sometimes reserved for 2-place predicates.

identity symbol

'=' isthemEN'rI TYSYM .The symbol Definition . BOL Comment. The identity symbol is usedto representthe relationship of numerical identity, such as, for example, that Mark Twain is identical to (i.e., one and the sameas) SamuelLanghorneClemens.

. The Greek letters a , P, 1, etc. are used as metavariables Comment METAVARIABLES for the names and variables of predicatelogic.

Chapter3

universal

" ' Definition. A UNIVERSAL QUANrD IER is any

quantifier

symbol of the form Va wherea is a variable. Comment . Universal quantifiers correspondto the ' ' English word every .

existential

Definition. An EXISTENTIAL any symbol of the form 3a wherea is a variable.

quantifIer

'' ~ QUAN I DIER is

Existentialquantifiers correspond to the Englishword ' some'. Comment .

expression

Definition. An EXPRESSION OF PREDICATE LOGIC is anysequence of itemsfrom the vocabulary of predicatelogic.

wffs

Definition.A WELL -FORMEDFORMULA of predicate with the logic is any expressionin accordance followingsevenrules:

(1) Sentenceletters are wtTs. (2) An n -place predicate letter followedI by n namesis a wff .

Chapter3

(3) Expressions of the forln a=Ji where a and Ji are names are wffs .

Comment. Although the placement of the identity symbol superficially resemblesthat of a connective, it is in fact a special two- place predicate. For historical reasonsalone it is placedbetweena. and Ii rather than in front of them. [Definition . Wffs of the form specifiedin rules 1- 3 are the ATOMIC SENTENCES of predicate logic. Thoseconforming to rule 3 are also known as mENTITY STATEMENTS.]

atomic sentence

. We adopt the practice of omitting superscripts Comment from predicateletters.

(4) Negations , conjunctions, disjunctions, conditionals, and biconditionals of wft"s are wft"s. Comment . The formation rules of chapter 1 are subsumedby this clause.

(5) If (j) is a wff , then the result of replacing at least one occurrence of a name in (j) by a new variable a (i.e., a not in (j) and prefixing 'v'a is a wff . universal wff

[Definition. Such wffs are called UNIVERSALLY '~ ~ QUAN I D lED wffs, or UNIVERSAL wffs.]

Chapter3

If cj) is a wff , then the result of replacing at least one occurrence of a name in cj) by a new variable a (i.e., a not in cj) and prefixing 3a is a wff . existential wff [Definition. Suchwffs are calledEXISTENTIALLY

~ QUANTDIED wffs, or EXISTENTIAL wffs.]

(7) Nothingelseis awIf . Examples. Wffs of this languageinclude the following : Fa v Fb) ~ Gab) 3yFy Vx ( Fx ~ Gx) VxVy (Rxy ~ Ryx) (3xFx H VxGx) - 3x(Fx H - VyGy) (3xFx ~ P) Vx3y Fyxb - a=b Vx x=x VxVy (x=y ~ y=x)

3 Chapter

Exercise3.1.1 Which of the following expressionsare wffs? If an expression is a wff , say whetherit is an atomic sentence , a negation, a conditional, a conjunction, a disjunction, a biconditional, a universal, or an existential. ( Note: Any wff must fall into exactly one of thesecategories.) .. 11* ...* 111 iv * v* vi * vii * viii * ix * x* xi * xii * xiii * xiv * xv * xvi * xvii * xviii * xix * xx * xxi * xxii * xxiii * xxiv * xxv *

Fz VxGac VxGcax

3x' Vy(Gxy & Gyx) ' Vx(Gxy H 3yHy) 3x(Ax ~ ' VxFxx ) 'v'x'v'y(Fxy ~ 'v'z(Hxyz & Jz 'v'xFxx H 'v'x'v'yFxy - 'v'x- 3z(Hz v Jx) Ga ~ 'v'x- (Ha v Fxx) p ~ Gab - (P & - 3xFx) 'v'x( Fx) & P 3y (Fyyy & P) 'v'xyz(Fzx H Hxyz) b=b (a=a) P=c Fa=Fa Vz(Fz ~ a=b) Vx (x=x) 3x(Fx=Gx) - Vx (Fx & 3y x=y) (- a=b H - Vx (Fxa & Pbx Vx3y (- x=y ~ y=- x)

Chapter3

quantifier convention

Comment. When a wff contains an uninterrupted sequence of quantifiers of the same type , existential or universal , it is often convenient to omit repetitions of 3 or 'v' .

Examples. The expression ' Vxyz(Fxy & Gyz H Hzx) will be readas shorthandfor ' Vx' Vy' Vz ( Fxy & Gyz H Hzx). The expression 3xyVzw( Fxyz & Gwx - + - Hzx) is to be readas 3x3y VzVw ( Fxyz & Gwx - + - Hzx) .

non-identity

We introduce the special symbol ~ that may be used to abbreviate statements of the fonD - a = (3 thus: a ~ (3. It will be useful to bear in mind that sentences of this fonD are negations, not atomic . Comment. As with the parenthesis-dropping conventions introduced in chapter 1, the formulas allowed by the conventions here are not strictly well -formed . They are merely acceptable abbreviations for wffs .

3 Chapter

openfonnula Definition . An OPEN FORMULA

is the result of replacing at least one occurrence of a name in a wff by a new variable (one not already occurring in the wft ) . Comment. Open formulas are not wffs and hence never appear as sentences in proofs . The notion of an open

formula is used to present the rules of proof for predicate logic .

Examples. Fx is an openformula. It occursaspart of the wff ' VxFx. Fxy is an openformula. It occursas part of the open formula 3yFxy, which in turn is part of the wff Vx3y Fxy.

scope

Definition. The SCOPE of a quantifierin a given formulais theshortestopenformulato theright of the quantifier. Examples. In the wff

(VxFx & 3y(Fy - + Gy , thescopeof Vx is theexpression Fx andthescopeof 3y is theexpression ( Fy- + Gy).

3 Chapter

In thewff 3y(Fy & Vz(Gz v - Rzy , thescopeof 3y is (Fy & Vz(Gz v - Rzy , but the scopeof "V'z is

(Gz v - Rzy).

wider and narrower scope

bound variable

Definition. A quantifier whose scopecontains another other quantifier is said to have WIDER SCOPE than the second. The secondis said to have NARROWER SCOPE than the first.

Definition. A variable, a , that is in the scope of a quantifier for that variable (i.eVa or 3a ) is called a BOUND VARIABLE . A variablethat is not boundby a quantifier is saidto be UNBOUND or FREE.

Exercise 3.1.2* Identify all the open fonnulas appearingin exercise 3.1. If an openfonnula appearsin an expressionthat is not well-fonned, give an exampleof a wff in which it might appear.

Exercise3.1. 3 In the following sentences , determinethe scopesof all quantifiers.

. 1 .. 11

' Vx(Px - + ' VzRxz) - ' VxPxH ' Vx' VzRxz

Chapter3

... m . IV V . VI .. VII ... VIII . IX X

VxPx- + Vz- VxRxz Vz(Px - + VxRxz) Vx3yFyxb 3y(Fy & Vz(Gz v - Rzy VxVy(Fxy - + Vz(Hxyz& Jz VxVy(Rxy - + Ryx) 3z3x(Fxz- + VyGyxa) 3x(x=a - + VyGyaa)

3.2

Translation of English to Quantified Wffs

translation scheme

Definition. A ttanslation schemefor the languageof predicatelogic consistsof a pairing of predicateletters with English predicate phrases and of names of predicatelogic with namesin English. We also include metavariables with the predicates and associated phrasesto indicatethe appropriateorder for namesand variables. Example. According to the ttanslationscheme a likes p LaP : a: Abigail , the sentence ' ' Abigail likes everything is translatedas "v'xLax.

Chapter3

Comment. It is possible to give several non-equivalent translation schemes for sentences of English , depending on how many places are assigned to the predicates.

Example. Using the translationscheme Fa.: a. is the fatherof Mary a: John, F is specified as a I -place predicate. Using this scheme,the sentence Johnis Mary' s father is translatedas Fa. Using the b"anslationscheme a is the father of p Fap : a: John b: Mary, F is specified as a 2-place predicate with the first position (occupiedby a ) correspondingto the subject of the phrase ' is the father of ' and the second(occupied by P) correspondingto its object. Using this scheme,the sentence Johnis Mary' s father is b"anslatedas Fab.

Chapter3

Comment. The choice of whether to represent English phrases with one-place or many-place predicates is dependent on the degree of structure that must be included in order for an argument to be analyzed adequately. In general, more detail is better than less detail , since arguments may be labeled invalid erroneously if insufficient detail is represented.

universals

Variantswhoselogical form is 'v'xFx include the following : Everythingis F. All things areF. Variantswhoselogical form is 'Q 'x(Fx - + Gx) include the following : Every F is a G. All Fs areGs. If it ' s an F, it ' saG . Everythingthat is F is G. Anything that is an F is a G. Any F is G. If somethingis an F, it is a G. Only Gs areFs.

Chapter3

There are several variants having the fonn

'v'x(Fx ~ - Gx) including these: No Fs are Gs. Not a singleF is G. Fs are neverGs. Every F is not G.

existentials

Variantswith the form 3xFx include the following : Somethingis F. Thereexistsan F. Thereis at leastone F. Variantshaving the form 3x( Fx & Ox) include the following : SomeFs areOs. At leastone F is O. Thereexistsan F that is G. Comment. Notice the difference between ttanslating ' ' ' ' Every F is G (equivalently All Fs are Gs ) and ' Some Fs are ' . In G the first case, an arrow is used in the scope of a universal quantifier . In the second, an ampersand is the appropriate connective in the scope of the existential quantifier .

3 Chapter Comment. When translating sentences of English without the use of the identity symbol, the distinction between ' Some F is G' ( ' At least one F is G' ) and ' Some Fs are G' ' At least two Fs are G' cannot be ( ) . We comment on the translation of ' at least represented n' below.

identity

Variantswith the form a=f3 include the following : a is p. a is (numerically) identical to p. a is the same(entity) as p. a and p areone and the same. a is the very sameindividual as p.

quantities

Numericalquantitiescanbe expressedusingthe quantifiers in conjunctionwith the identity symbol.

at least n

The existential quantifier expresses ' at least one' . Other numericalquantitiescan be expressedby asserting the existenceof non-identical objects. Thus, for example: At leasttwo 3xy x~y At leastthree 3xyz x~y & x~ ) & y~ ) The sentence' There are at least two dogs' may translated3x3y Dx & Dy ) & x~y) .

3 Chapter

exactly n

There are exactly n objects if there are at least n, and all objectsare identical to one or other of thosen. For example: 3x'v'y x=y Exactly one Exactly two 3xy (x~y & 'v'z(z=x v z=y 3xyz (x~y & x~z) & y~ ) & 'v'w w=x v w=y) v w=z Exactly three

at most II

Thereare at most n objectsif thereare exactly zero, or exactly one, etc., up to exactly n objects. For example, ' ' Thereareat most two dogs may be b' anslatedas: - 3xDx v (3x (Dx & Vy (Dy ~ y=x v 3xy (Dx & Dy) & x' *Y) & Vz(Dz ~ z=x v z=y This is equivalent to saying that there are not three distinct dogs, i.e.: - 3xyz Dx & (Dy & Dz & (x* y & (x* z & y* z ) Comment. There are many subtleties in the ttanslation of English quantifier phrases into the language of predicate logic . Such phrases often introduce ambiguity into the expressions of English . The exercises below illusttate some of the subtleties and ambiguities .

Chapter3

Exercise3.2

Give translation schemesand translatethe following sentencesof English into the language of predicate logic. If a sentenceis ambiguous, give all the reasonable translationsof it .

( 1- 22: Translate using one-place predicates only .) 1* All dogs are mammals. 2* Some sharks are ovoviviparous . 3* No fishes are endothermic. 4* Not all fishes are pelagic . * 5 Reptiles and amphibians are not endothermic. 6* Some primates and rodents are arboreal. 7* Only lagomorphs gnaw. * 8 Among spiders, only tarantulas and black widows are poisonous. 9* All and only marsupials have pouches. * 10 No fish have wings unless they belong to the family 11* 12* 13* 14* 15* 16* 17* 18* 19* 20 *

Exocoetidae. Some organisms are chordates and some organisms are molluscs , but nothing is both a chordate and a mollusc . None but phylogenists are intelligent . Animals behave normally if not watched. Animals behave normally only if not watched. Some sharks are pelagic fish , but not all pelagic fish are sharks. If Shamu is a whale and all whales are mammals, then Shamu is a mammal. No sparrow builds a nest unless it has a mate. No organism that is edentulous is a predator. All predators are not herbivorous. Not all predators are carnivorous.

Chapter3

21*

.

A mammalwith wmgs * 22 A mammalwith wIngs flying . (23 29: Try thesefirst with one-placepredicates,then with manyplacepredicates.) 23* Shamucan do everytrick. * 24 Shamucan do any trick. * 25 Shamucannotdo everytrick. 26* Shamucannotdo any trick. * If any whalecan do a trick , Shamucan. 27 * If everywhale can do a trick , Shamucan. 28 * If any whalecan do a trick , any whale can do a trick. 29 30Translations with many-placepredicates.) 57: ( * 30 Godzilla ate Bambi. * 31 SomethingateBambi. * 32 Godzilla ate something. 33* Bambi ateeverything. * 34 EverythingateBambi. * 35 Somethingate something. * 36 Somethingateeverything. 37* Everythingate something. * 38 Everythingateeverything. * 39 Everythingate itself. * 40 Somethingateitself. 41* Nothing ate itself. * 42 Somethingate nothing. * 43 Everyonesaidsomethingto everyone. * 44 Everyonesaid somethingto someone. 45* Everyonesaidnothing to someone. * No one saidanythingto anyone. 46 * 47 There is a reptile smaller than a cat but larger than a dog.

3 Chapter 48* 49* 50* 51* 52* 53* 54*

Somefishesswim slowerthan humans. Somefishesare smallerthan everymammal. Somewhaleseat only fast-moving fishes. Somewhalesdo not eat any fast-moving fishes. If anythingeatsfast-moving fishes, sharksdo. ' ' Jaguarstails are longer thanocelots tails. If an organismis symbiotic with a clown fish then it is a seaanemone. 55* The phalangesof birds are homologousto the phalanges of humanswhereasthe eyes of octopi are analogous but not homologousto the eyesof mammalsand birds. 56* Somewhaleseat more thanall fishes. * 57 There is a monkey who grooms all and only those . monkeyswho do not groom themselves Translations the ( involving identity symbol.) 58* one cheetahexists. Exactly * 59 Thereis only one Paris. 60* Bambi ateat leasttwo trees. 61* Bambi ateeverythingexcepthimself. * 62 Every dog hasexactly one tail. 63 G.odzilla ate Bambi, and somethingelseate Godzilla. 64 Bambi was not eaten by Godzilla but by something else. 65 Godzilla atenothing but Bambi. 66 Godzilla ateeverythingexceptBambi. 67 Only Bambi is afraid of Godzilla. 68 Nothing but Godzilla likes Bambi. 69 Thereis a fish that' s bigger than all the others. 70 Nobody likes somebodywho eats everything except Bambi.

Chapter3

3.3

Primitive Rule. . of Proof

. We introducesix newprimitiverulesof Comment , , universalintroduction proof: universalelimination identity elimination existential introduction existential , , . To allow introduction , andidentityelimination of the f1r8tfour of these succinctstatements , the notionsof universalization , and , existentialization . aredefined instance

universa lization

Definition. A UNIVERSALIZATION of a sentence with respectto a given nameoccurring in the sentence is obtainedby the following two steps: ( 1) Replace all occurrences of the name in the sentenceby a variable a , where a does not already occur in the sentence .

(2) Prefix'v'a to theopenfonnularesultingfrom step1. . Examples of Universalizations ~ (Fa Ga) include 'v'x(Fx ~ Gx) and'v'y(Fy ~ Gy).

3 Chapter Universalizationsof Faa include 'v'xFxx

andVyFyy.

lization

Definition. An EXISTENTIALIZATION of a sentence with respectto a given name occurring in the sentenceis obtainedby the following two steps: ( 1) Replace at least one occurrence of the name in the sentence by a variable a, where a. does not already occur in the sentence .

(2) Prefix 3a. to the openfonnula resulting; from step 1. Comment. Notice the difference between step 1 in the definition of universalization and step 1 in the definition of existentialization . Universalization requires replacement of all occurrences of the name by with a .

Examples. Existentializations ( Fa- + Oa) include 3x ( Fx - + Ox), 3x( Fa- + Ox), and3y ( Fy - + Oa) .

Chapter

Existentializationsof Faa include 3xFxx, 3xFax, and3yFya.

i DStance

of a universally or existentially Definition . An INSTANCE quantified sentence is the result of the following two steps :

( 1) Removethe initial quantifier. (2) In the open formula resulting from step 1, uniformly replace all occurrencesof the unbound variableby a name. Comment. This is called INSTANTIATING the sen-

teRce.Thenameis calledtheINSTANTIAL NAME. Examples. The sentence 'v'xFx hasinstances

Fa, Fb, Fc, etc. The sentence 3x(Fx & Gx) hasinstances ( Fa& Ga), ( Fb & Gb) , ( Fc & Gc), etc.

Chapter3

The sentence

3xVy(Fxy ~ Gy) hasinstances

Vy( Fay~ Gy), Vy(Fby~ Gy), etc. Exercise3.3.1* Pair wffs and below.

. I 11 .. . 111 . IV V VI VII ... VII1 . IX X

from the list of sentences lY appear in several pairs.

VxFax 3x(Fxa & VyGyxa)

3xFax Fab 3yVxFyx 3zx(Fxz& VyGyxa) VxyFxy VxFxa 3zx( Fxz& VyGyxz) Fha & VyGyba Comment . The primitive rules of proof for predicate logic include all the primitive rules from chapter 1. There are also introduction and elimination rules for the two quantifiersand for the identity symbol. Two of the new rules haveconditionsthat must be met for the applicationof the rules to be correct.

Chapter3

universal -e1im Given a universally quantified sentence (at line m), conclude any instance of it . Condition : Annotation : Assumption set: Also known as:

intro

None. mOVE sameas line m. Universal Instantiation .

. Examples (a) 1 ( 1) 'v'xFx 1 (2) Fa 1 (3) Ph

A I ' VE IVE

(b) 1 1

A 1 'v'E

(1) (2)

VyRyy Rbb

Given a sentence(at line m) containing at least one occurrenceof a name, conclude a universalizationof the sentencewith respectto that name. The name in question must not ' occur in any assumptions in m s assumptionset. m VI Annotation: sameas line m. set: Assumption Generalization. Universal as: known Also Condition:

Chapter3

VxFx Fb VxFx VyFy

A 1VE 2VI 2VI

(b) 1 1 3 3 1,3 1,3

'v'x(Fx~ Ox) Fa~ Oa 'v'xFx Fa Oa 'v'xOx

A I ' VE A 3 VE 2,4 - +E 5 VI

---

. Examples a () 1 (1) 1 (2) 1 (3) 1 (4)

Exampleof violation of the VI condition.

wrong !

(c) 1 2 1 1,2 1,2

(1) (2) (3) (4) (5)

'v'x(Fx- + Gx) Fa Fa- + Ga Ga 'v'xGx

A A

1 VE 2,3 - +E 4VI

Comment. Ordinarily we cannot conclude VxFx ' merely from Fa- the fact that one thing is F doesn t guarantee that everything is F ! The condition on VI ensures that we do not make this mistake. H the sentence Fa is b"ue, and furthennore would still be true no Inaner what the name denotes, then clearly everything is F , so we can conclude VxFx . When the condition on VI is met, then we are in such a situation : if we

Chapter3

prove Fa from assumptionsthat do not contain the name a and hence say nothing in particular about its referent, then we could just as well have used a different name, sayb, and provedPh. In fact, when the condition on VI is met, any proof of Fa can be turned into a proof of Ph just by replacing any involved occurrencesof the name a by the name b. This is sufficient to guaranteethat everything is F; hence, we can concludeVxFx. Given a sentence (at line m) containing at leastone occurrenceof a name, concludean existentializationof that sentencewith respectto that name. Condition: Annotation: .

None. m 31

Also known as :

ExistentialGeneralization.

-set m. asline : same Assumption Examples.

(b) 1 2 1 1,2 1,2 1,2

< m ~

(a) 1 1

---

existentialintro

Vx(Fx~ Gx) Fa Fa~ Ga Ga

Fa& Ga 3x(Fx& Gx)

A A

1VE 2,3 ~ E 2,4 &1 531

3 Chapter

(c) 1 1

existentialeUm

(1) (2)

VxFax : 3yVxFyx

A 131

Given a sentence(at line m) and an assumption(at line i ) that is an instanceof some existentially quantified sentencethat is present(at line k), concludethe given sentenceagain. Condition:

Annotation : Assumption set:

The instantial name at line i must not appearin the sentenceat line k or in the sentenceat line m. Also, it must not appear in any of the assumptions belonging to the assumption set at line m, other than the instancei itself. k , m3E (I) all assumptionsat line m other than i , and all assumptionsat line k.

Examples.

(a) 1 2 2 2 1

(1) (2) (3) (4) (5)

3xFx Fa Fav Oa 3x(Fxv Ox) 3x(Fxv Ox)

A A

2vl 331 1,4 3E(2)

3 Chapter

wrong !

(1) (2) (3) (4) (5) (6)

3x(Fxx- + P) Faa- + P 'v'xFxx Faa P P

A A A

3 VE 2,4 -+E 1,5 3E(2)

Examples violationof3Econdition. (a) 1 (1) 3xFx 2 (2) Fa 3 (3) Ga 2,3 (4) Fa& Ga 2,3 (5) 3x(Fx& Ox) 1,3 (6) 3x(Fx& Ox)

A A A

f(~~

r-

(b) 1 2 3 3 2,3 1,3

(b) 1 2 1

(1) (2) (3)

3xFx Fa Fa

(c) 1 2 2 1

(1) (2) (3) (4)

3xFax Faa 3xFxx 3xFxx

1,53E

1,2 3E

A A

231 1,3 3E(2)

Chapter3

. If all we know is that somethingis F, we are Comment not entitled to reasonas if we know what it is that is F. As in the case of VI , a use of 3E that meets the conditionsaboveand usesa certaininstantialnamecan be turned into a proof of the sameconclusionfrom the same assumptionsbut using any different instantial name. This showsthat the conclusiondoesnot rest on any assumptionsabout the actual identity of the thing that is said to exist. That is, if we apply 3E to 3xFx by discharging the assumedinstance Fa, the conditions ensurethat we do not mistakenly use any information about the referent of ' a' in particular . After all , 3xFx ' saysonly that somethingis Fit doesnt tell us which individual is F.

identity -intro

Concludeany sentenceof the form a=a.

Condition : Annotation: Assumptionset:

None. =1 Empty.

. Example c =c

=1

Comment. An identity statementof the form a=a, like a theorem, requiresno assumptionsto justify its assertion .

Chapter3

Given a sentence cI> ( at line m) containing a name a, and another sentence (at line n) that is an identity statement containing a and another name (3, conclude a sentence that is the result of replacing at least one occurrence of a in cI>with (3. Condition : Annotation : Assumption set : Also known as :

None. m, 1I =E The union of the assumption sets at lines m and n. Leibniz ' s Law , Substitutivity of Identity

Examples. - =-

Vx ( Fxa - + x = a) Fha Fha - + b=a b=a Vx ( Fxb - + x= b)

1,2 =E

A A

-.~II~

Fa& Ga b=:;a Fb & Ga Fb & Gb

2 1,2 1,2 (c) 1 2 1 1,2 1,2

Fa a=b Ph

- - -

( 1) (2) (3)

- - ~

identity -elim

A A

I ' VE 2,3- +E 1,4 =E

Chapter3

Comment. The role of identity elimination is not regardedas valid in all contexts. For instance, if Frank believesthat Mark Twain is a novelist then, even thoughTwain= Clemens, it doesnot follow that he believes Samuel Langhorne Clemens is a novelist (if , for " " " " example, he has heard the name Twain but never Clemens). For historical reasons , contextswhere the role fails, such as belief reports, are called intensional contexts in contrast to the extensionalcontextsprovidedby the ordinary predicateswhich the languagedevelopedin this chapteris intendedto represent.

Exercise3.3.2 Provethe following sequents , using the primitive rules of predicatelogic. You may also usederived sentential rules.

S87* S88* S89* S90 S91* S92* S93* S94 S95* S96* S97 S98 S99 SlOO SlOl * SlO2* SlO3

3x(Gx & - Fx), ~x(Gx ~ Hx) ~ 3x(Hx & - Fx) 3x(Gx & Fx), ~x(Fx ~ - Hx) ~ 3x- Hx ~x(Gx ~ - Fx), ~x(- Fx ~ - Hx) ~ ~x(Gx ~ - Hx) 3x(Fx & Ga), ~x(Fx ~ Hx) ~ Ga& 3x(Fx & Hx) ~x(Gx ~ 3y(Fy & Hy ~ ~x- Fx ~ - 3zGz ~x(Gx ~ Hx&Jx), ~x(Fx v - Jx~ Gx) ~ ~x(Fx~ Hx) ~x(Gx & Kx H Hx), - 3x(Fx & Gx) ~ ~x- (Fx & Hx) ~x(Gx~ Hx), 3x Fx&Gx) & Mx) ~ 3x(Fx& (Hx&Mx ~x(...(jxv- Hx), ~x Jx ~ Fx) ~ Hx) ~ - 3x(Fx & Gx) - 3x(- Gx & Hx), ~x(Fx ~ - Hx) ~ ~x(Fxv...(jx~ - Hx) ~x- (Gx & Hx), 3x(Fx & Gx) ~ 3x(Fx & - Hx) 3x(Fx & - Hx), - 3x(Fx & - Gx) ~ - ~x(Gx ~ Hx) ~x(Hx ~ Hx & Gx), 3x(- Gx & Fx) ~ 3x(Fx & - Hx) ~x(Hx ~ - Gx), - 3x(Fx & - Gx) ~ ~x- (Fx & Hx) ~x(Fx H GX) ~ ~xFx H ~xGx 3xFx ~ Vy(Oy~ Hy), 3x.Jx~ 3xGx~ 3x(Fx&Jx) ~ 3zHz 3xFxv 3xGx, ~x(Fx ~ Gx) ~ 3xGx

Chapter3

SI04 SIO5* SI06 SIO7* SIO7 SI09* SIlO SIII * S112* S113 S114 S115 S116 S117* S118 S119 S120 S121 S122 S123* S124 S125* S126 S127* S128 S129 S130*

'v'x(Fx ~ - Gx) ... - 3x(Fx & Gx) 'v'x(Fx v Hx ~ Gx & Kx), - 'v'x(Kx & Gx) ... 3x- Hx 'v'x(Fx & Gx ~ Hx), Ga& 'v'xFx ... Fa& Ha 'v'x(Fx H 'v'yGy) ... 'v'xFx v 'v'x- Fx 'v'y(Rl~ (3x<:Jx~ Oy ,'v'x(Gx ~ Hx), 'v'x(- Jx~ ....JIx ) ... 3x Jx ~ Fav 'v'x Gx 'v'x(Dx ~ Fx) ... 'v'z(Dz ~ ('v'y(Fy ~ Gy) ~ Gz 3xFxH 'v'y(FyVOy~ Hy), 3xHx, - 'v'z- Fz'" 3x(Fx&Hx) 'v'xFx'" - 3xGxH - (3x(Fx & Gx) & 'v'y(Gy ~ Fy "V'x(3yFyx ~ "V'zFxz) ~ "V'yx(Fyx ~ Fxy) 3x(Fx & "V'yOxy), "V'xy(Gxy~ Oyx) ~ 3x (Fx & "V'yOyx) 3x- "V'y(Oxy ~ Oyx) ~ 3x3y(Oxy & - Oyx) "V'x(Gx~ "V'y(Fy ~ Hxy , 3x(Fx & "V'z- Hxz) ~ - "V'xOx "V'xy(Fxy ~ Oxy) ~ "V'x( Fxx ~ 3y(Oxy & Fyx "V'xy(Fxy ~ - Fyx) ~ - 3xFxx "V'x3y(Fxy & - Fyx) ~ 3x- "V'yFxy "V'y(3x- Fxy ~ - Fyy) ~ "V'x(Fxx ~ "V'yFyx) 3xFxx ~ "V'xyFxy ~ "V'x(Fxx ~ "V'yFxy) a=b ~ b=a a=b & b=c ~ a=c a=b, ~ ~ a~ Fa & "V'x(Fx ~ x=a), 3x( Fx & Ox) ~ Oa "V'x x=x ~ 3xFx, "V'x(- Fx v Ox) ~ 3x(Fx & Ox) "V'x(Fx ~ Ox), "V'x(Ox ~ Hx), Fa & - Hb ~ a~b 3x Fx & "V'y( Fy ~ y=x & Ox), - Oa ~ - Fa 3x"V'y - Fxy ~ x=y) & Ox) ~ "V'x(- Ox- +3y(y~x & Fyx 3x(Px & ("V'y(Py ~ y=x) & Qx , 3x- (- Px v - Fx) ~ 3x( Fx & Qx) "V'x3yOyx, "V'xy(Oxy ~ - Oyx) ~ - 3y"V'x(x~y ~ Oyx)

3 Chapter

3.4

Sequents, Theorems, and Derived Rules of Proof

Exercise3.4.1 Provethe following sequents , using the primitive rules from chapter 3 and any of the primitive or derived rules from chapter1.

* S150 S151 S152 S153 S154 * S155 * S156 * S157 S158 S159 Sl60* S161

- ~xPx+ 3x- Px QuantifierExchange - 3xPx~~ ~x- Px QE - ~x- Px + 3xPx QE - 3x- Px ~~ ~XPX QE Confinement ~x(Px & Qx) ~~ ~XPX& ~ xQx Conf ~x(Px ~ Q) ~~ 3xPx~ Q Conf ~XPXv ~ xQx ~ ~x(Px v Qx) Conf 3xy(Px & Qy) ~~ 3xPx& 3xQx Conf 3x(Px v Qx) ~~ 3xPxv 3xQx Conf 3x(Px ~ Q) ~~ ~xPx~ Q Conf p ~ 3xQx ~~ 3x(P ~ Qx) Conf P ~ ~ xQx ~~ ~x(P ~ Qx)

Comment. The important quantifier -exchange rules establish that an initial tilde can always be moved to the (derived rule ) right of an adjacent quantifier , changing the quantifier from a universal to an existential (or vice versa) . Also ,

QE

a tilde that immediately follows an initial quantifier can be moved to the front of the sentence provided , again, that the quantifier is changed as just described. Although the above versions of the rules (S150- S153) involve quantifications of a simple formula , it is easily recognizable that the proofs of these sequents do not

Chapter3

dependon the simplicity of the quantifiedfonnula. ~ QUANrrD IER EXCHANGE(QE) maythusbe used asa derivedruleof proofasbelow.

---

. Example 1 2 1 1,2 1,2

3x- (Fx & Ox) 3xOx~ Vx(Fx & Ox) - Vx(Fx& Ox) - 3xOx Vx- Ox

Exercise3.4.2 Prove the following

A A

1QE 2,3 MTr 4QE

sequents , using any of the

sofar. primitiveor derivedrulesestablished T40 T41 T42 T43 T44 T45 T46 T47 T48 T49 T50 T51 T52 T53 T54 T55

I- ' Vx(Fx - + Ox) - + (' VxFx- + ' VxOx ) I- ' Vx(Fx - + Ox) - + (3xFx - + 3xOx) I- 3x(Fx v Ox) H 3xFxv 3xOx I- ' Vx(Fx & Ox) H ' VxFx& ' VxOx I- 3x(Fx & Ox) - + 3xFx& 3xOx I- ' VxFxv ' VxOx- + ' Vx(Fx v Ox) I- (3xFx- + 3xOx) - + 3x(Fx- + Ox) I- (' VxFx- + ' VxOx ) - + 3x(Fx - + Ox) ~ ' Vx(Fx H GX) v (' VxFxH ' VxGX ) ~ ' Vx(Fx H GX) v (3xFxH 3xGx) ~ - ' Vx(P & Fx) H (P - + - ' VxFx) ~ - 3x(P & Fx) H (P- + - 3xFx) ~ ' Vx(P v Fx) H (- P - + ' VxFx) ~ 3x(P v Fx) H (- P - + 3xFx) ~ ' Vx(Fx - + P) H (3xFx- + P) ~ - 3x(Fx - + P) H - (' VxFx- + P)

Chapter3

T56 T57 T58 T59* T60 T61 T62 T63

.- Vx ( Fx H P) - + (VxFx H P) .- Vx (Fx H P) - + (3xFx H P) .- (3xFx H P) - + 3x(Fx H P) .- (VxFx H P) - + 3x(Fx H P) .- Vx3y x=y .- Vx ( Fx H 3y(x=y & Fy .- Vx ( Fx H VY(X=y - + Fy .- Vxy ( Rxy H X=y) - + VxRxx

prenex fonD

Comment . The quantifier- exchangerules and the confinementrules (S154- S151) indicate that any sentencemay be convertedinto an equivalentsentencein which no connectiveis outsidethe scopeof , called a PRENEX any quantifier in the formula. Such a sentence sentence , has all its quantifiers in a row at the beginnin~ of the sentence .

Eyem. Q3.4.3

For each of the following , find a prenex equivalent and prove the

. equivalence 1* 2* 3* 4* 5*

v x(Px-+- VzRxz) 3y(Fy&Vz( Hyz& Jz 3xFxa-+- VyGyaa - VxFx-+- 3xHx - 3x( 3yFyx-+- - VzGzx) Find prenexequivalentsfor the other non-prenex sentences in this chapterand the next.

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Chapter 4 Models

4.1

Finite Interpretations and Expansions with OnePlace Predicates without Identity

finite

Definition. A FINITE INTERPRETATION for a set of symbolic sentences(containingone-placepredicates but no many-place predicates ) consists of three : components

interpretation

. UDiverse

.

A finite set of objects called the UNIVERSE or DOMAIN . The universemust contain at least one object.

predicate extensions

.

An EXTENSION for each of the predicatesin the sentences . Each extensionis a (possibly empty subset of the universecontainingthoseobjects ) to which the predicateapplies.

truth -value

.

TRUTH - VALUE SPECIFICADONS for the sentence letters in the sentences . Each of the sentenceletters is paired with the specification True or with the specificationFalse.

specifications

Comment . Such universe

an interpretation

is finite

is a finite set . In the rest of this ' ' will use as shorthand interpretation ' . interpretation

because

its

section , we for

' finite

Chapter4

evaluation

Comment. Given an interpretation for a set of sentences, it will be possible to detennine truth values for the sentencesin the set.

Example. Here is a conditional sentenceand an interoretationin

Vx(Fx v - Gx) ~ P v 3x(Gx & - Fx) U: { a,b,c} F: {a, b} G: {b} P is False In this interpretation the antecedent of the sentence is true since everything in the universe is either F or not G (a and b are both F, c is not G ) . The consequent of the conditional is false, since both disjuncts are false. (P is specified false. The existential wff is false because there is nothing in the extension of G that is not also in the extension of F.)

The conditionalis thereforefalse. Comment. The procedure for determining the buth values of sentences in an interpretation for them is given more precisely in section 4.2.

Chapter4

universal expausio D

Definition. The EXPANSION OF A UNIVERSAL WFF relative to a universeof n elementsconsistsof n conjuncts, where the nth conjunct is an instanceof the formula with the name of the nth elementin the universe as the instantial name. ( Werefer to this as a UNIVERSAL EXPANSION for short.) Comment . Sbictly speaking, all conjunctions have exactly two conjuncts. Expressionshaving the form (j) & 'I' & ... are unproblematic, however, becauseof the associativityof & (S40) . So, it is acceptableto use the notion of a conjunction with more than two conjuncts in the definition of a universal expansion. Likewise, becauseof the associativity of v (S41), we use the notion of a disjunct with more than two disjunctsin the definition of an existentialexpansionbelow. Example. The expansionof Vx (Fx ~ Gx) in the universe{ a} is ( Fa~ Ga) . In the universe{ a,b } its expansionis ( Fa~ Ga) & ( Fb ~ Gb) . In the universe{ a,b,c } its expansionis (Fa ~ Ga) & ( Fb~ Gb) & ( Fc ~ Gc), and so on.

Chapter4

existential . expansion

Definition. The EXPANSIONOF AN EXISTENTIAL WFF relative to a universeof n elements consistsof n disjuncts , wherethe nth disjunctis an instanceof the fonnulawith the nameof the nth element in the universeas the instantialname. (EXISTENTIAL EXPANSION for short.) Example. The existential 3x ( Fx & Ox) expandsto (Fa & Oa) v (Fb & Ob) v ... for the universe{ a,b, ... } .

overlapping quanti:fiers

. In caseswherequantifiersoverlap, expansion Comment may take severalsteps, startingwith the quantifier with the widest scope and then expanding those with narrower scope. Expansion is complete when no quantifiersremain. Example. In the universe{ a,b } 'Q 'x( Fx - + 3yGy) is first expandedto

(Fa- + 3yGy) & (Fb - + 3yGy), thento Fa- +(Gav Gb & (Fb - + (Gav Gb .

Chapter tnltb

values

of complex sentences

quantifters

Comment . The truth values of complex sentences in a given interpretation are determined as follows .

(i ) Construct the expansions of all universal and existential formulas , then assign truth values for the resulting quantifier -free sentences according to steps ii - iv below. The truth value of a quantified sentenceis the truth value of its expansion.

sentence

letters

(ii )

Sentence letters

have the b" uth values

directly

assigned to them in the interpretation .

predicates

(iii ) Formulas of the form Fa , where F is a predicate and a is a name, are b1Ie if the object a is in the extension of F and false otherwise.

connectives

(iv ) The b"uth values for the sentential connectives are determined according to the usual b"uth- functional rules for the connectives.

4 Chapter sentences

Exercise4.1.1

0 I* 00 * 11 000 III * iv * v* vi * vii * 000 VIII* IX* x* ExerCise 4.1.2

'v'xFx 3xFx & p

'v'xFx - + 3xGx 'v'x(Gx H P) v 'v'xHx Ha v 3xGx 3x(Fx v Hx) 'v'xFx H 3x(Fx& - Hx) - 'v'x(Fx & Gx) - 'v'x(Fx& - 'v'yGy) - ('v'xGx H 3x(Hx & - Fx Say whether the sentences in exercise 4 .1.1 are hUe in

the following interpretations:

U: {a}, F: {a}, G: { }, II: { }, P is False

U: {a,b}, F: {a}, G: {a,b}, H: { }, Pis True U: {a,b,c}, F: {a,b,c}, G: {a,b}, H: {b}, Pis False

Chapter4

4.2

Finite Countermodels for Arguments with OnePlace Predicates without Identity

model

Definition . A MODEL for a set of sentences is an interpretation in which all the sentences in the set are true.

countennodel

Definition. A COUNTERMODEL for a given argument is a model for the premises in which the conclusionis false. Comment. The idea behind a countermodel is the same as that behind using a truth table to demonstrate that an argument is invalid . The point is to demonstrate that it is possible for all the premises of an argument to be true and still have the conclusion turn out false. Thus , a countermodel is the predicate- logic analogue of an invalidating assignment (introduced in chapter 2) . Comment. Given an invalid sequent with one-place predicates and no many-place predicates, it is always possible to find a finite countermodel . Indeed, if the sequent contains n predicates, the universe of a countermodel need not have more than 2n elements, and will often have fewer . Comment. Expansions provide a convenient way of demonstrating that a given interpretation is acounter -

model for an argument.

100

Chapter4

Examples. (a) Give a countermodeland an expansion to show this sequentinvalid:

3xGxJ- P-+ "V'xGx Model:

U: {a,b } G: {a} P is True Expansions: The premise3xGx expandsto Ga v Gb with thesebUth assignments : TvF T The conclusionP - + 'v'xGx expandsto P - + (Ga & Gb) with thesebUth assignments : T - + (T & F) T -+ F F The premise is b1Ie and the conclusion is false in this interpretation , so the argument is invalid .

101

Chapter4

(b) Give a countennodeland an expansionto show this sequent invalid :

'o'xFx - + 'o'xGx .- Fm - + 3xGx

-

Model: U: m,a} F: m} G: } Expansion: The premise('V"xFx ~ 'V"xGx) expandsto Fm & Fa~ Gm & Ga T FFT The conclusion( Fm - + 3xGx) expandsto Fm - + GmvGa T FF FF The conclusion is false in this interpretationand the premiseis true; hence, this interpretationis acounterexamplefor the given sequent. Exercise4.2

Consbuct

countermodels

and expansions to show the

invalid. followingsequents i* ii * iii * iv * v*

VxFx ~ Vx Gx J- Vx(Fx ~ Gx) 3xFx ~ 3xGxJ- Vx(Fx ~ Gx) 3xFx & 3xGxJ- 3x(Fx& Gx) 3x(Fx v Gx) J- VxFx v VxGx 3x(Fx ~ Gx) J- 3xFx ~ 3xGx

102

Chapter4

vi * vii * ... Vlll * ix * x* xi * xii * ... Xlll * xiv * xv * xvi * xvii * xviii * xix * xx *

3x(Fx ~ 'v'xFx H 3xFx H 'v'xFx H 3xFx H

4.3

Finite Countermodels for Arguments with ManyPlace Predicate. without Identity

ordered pair

The notation (a, fi) denotes the ORDERED PAIR consistingof two objects namedby a and fi (a and fi may be the same). So long as the two objects are different objects, the orderedpair denotedby (a, fi) is different from the pair denotedby (fi,a ) .

Gx) ~ 'v'xFx ~ 'v'xGx 'v'xGx ~ 'v'x(Fx H GX) 3xGx ~ 'v'x(Fx H GX) P ~ 'v'x(Fx H P) P ~ 'v'x(Fx H P)

3x(Fx H P) ~ 3xFxH P 'xFx H P 3x(Fx H P) ~ "Q 'x(Hx - + Fx) 'x(Gx - + Hx) ~ "Q "Q 'x(Fx - + Gx), "Q 'x(Hx - + Gx) ~ 3x(Fx& Gx) "Q 'x(Fx - + Hx), "Q 'xGx, - "Q 'x(Fx - + Hx) ~ 3xHx - + 3x- Gx 3xFx H "Q "Q 'x(Gx v - Hx), 3x(Gx & Fx) ~ 3x- Hx "Q 'x(Fx & Gx - + Hx), 3x(Fx & Hx) ~ 3xGx 'x(Fx v Gx - + Hx) 3xFx, 3xGx, 3xHx ~ "Q - "Q 'xFx ~ "Q 'x- Fx 3x(Fx - + 3yGy) ~ 3xFx- + 3yGy

. The idea behind ordered pairs is easily Comment extendedto coverorderingsof more thantwo objects.

103

Cbapter4

ordered II - tuple

An ORDERED n - TUPLE , <~ a I' ..., an) ' consists of the n objectsnamedby ~ a I' ..., an. . As with orderedpairs, changingthe ordering Comment of a O ' a I' ..., an usually changesthe identity of the n tuple.

n - place

extensions

Definition. The EXTENSION OF AN n - PLACE PREDICATEis a set of orderedn-tupiesof objects . from theunverse Example. Given a universecontainingthe objectsa, b, andc, and a two-place predicate R, the set {(a,b) , (c,b), (a,a)} gives a possibleextensionfor R. In this example, the sentencesRab, Rcb, and Raa are true, while the sentencesRac, Rbc, Rba, Rca, Rbb, and Rcc are all false.

finite interpretation

Definition . A finite interpretation for a set of sentences containing one-place and many-place predicates consists of the following :

.

A finite universe, or domain. Extensionsfor all the predicatesappearingin the sentences .

104

Chapter4

.

Truth value specificationsfor the sentenceletters . appearingin the sentences

. Example Givena universe U: {a, b}

is constructed by first expanding the universal quantifier (sinceit haswider scope) to yield 3yFay & 3yFby. Eachexistentialis then expandedto yield (Faav Fab) & (Fba v Fbb). Comment . Thedefinition of aninterpretation for a setof sentences one , givenin section4.1, is just a containing placepredicates -placepredicates case of the definition for . special many countermodels: Comment . As before, a countennodel for a given is a model for the premises where the sequent conclusionis false.

. Example Thesequent 'v'x3yRxy ~ 3y'v'xRxy is invalid, asshownby thefollowinginterpretation : U: {a, b} R: {(a, b), (b, a)} Expansions: Premise

(Raav Rab) & (Rbav Rbb)

105

Chapter4

TvF

F v T & T Conclusion

( Raa& Rab) v (Rba& Rbb)

T & F F & T v F F

Exercise4. 3.1 Consb"Uct countennodels for the following invalid . sequents ' 1* " 11* " '* 111 iv * v* vi * vii * viii * ix *

3xFxx~ 'v'xyFyx 'v'y3xFxy ~ 3xFxx 'v'x3yFxy ~ 3x'v'yFxy Vx3y- Fxy, VxVy(Gxy ~ - Fxy) .- Vx3y- Gxy Vx(Fx ~ 3yGxy) .- VxVy(Fxv - Gxy) Vx3yVzVxyz .- 3y VxVzVxyz Vx- VyTxy '- Vx- 3yTxy 3xyz Fxy& Fyz) &~ FxzvFyx '- Vx3yFyx~ Vx- Fxx Vx3yFxy, 3x- VyGyx, 3xyFxy+-+3xy(Gyx & - Gxy) ~ 3y(Gxy v Gyx)

3x3yFxyH - 3xGxx,Vy3xGyx~ Vx- Fxx

106

Chapter4

Exercise4.3.2 Establish whether each of the following sequents is valid or invalid with either a proof or a countennodel .

. 1 .. 11 iii

. VI .. Vll ... Vlll IX X

.. Xll ... Xll1 . XlV XV . XVI XVll ... XVW XIX XX

'v'x(Fx ~ - Gx), 3x(Gx & - Fx) .- - 3xFxv 'v'x- Gx 'v'x(Fx ~ - Gx), 3x(Gx & Fx) .- - 3xFxv 'v'x- GX ; 3x(Fx ~ Gx) ~ 3x(Fx & - Hx) .- 3x(Fx& Gx) ~ - 'v'x(Gx ~ Hx) 3x(Fx ~ Gx) ~ 3x(Fx& - Hx) .- 3x(Fx& Gx) ~ 'v'x(Gx ~ Hx) 3x(Fx v F), P H - 3xGx, - 3x(Fx & Gx), 'v'x(Hx ~ - Fx & - Gx) .- P v 'v'x(Hx ~ P) 'v'x(Fx & Gx ~ Hx), 3xFx .- 3x(Gx ~ Hx) 'v'x(Fx & - Gx ~ Hx), 3xFx .- 3x(Hx ~ - Gx) .- 'v'xFx v 3x- Gx ~ - 3x(Fx v Gx) .- 'v'xFx v 3xGx~ 3x(Fx v Gx) 3x(Fx v Gx), 3xFx ~ 'v'xH:x, 3xGx~ - 3xHx .- - (3xHx & 3x- Hx) 3x(Fx v Gx), 3xFx ~ 'v'xHx, 3xGx~ - 3xHx .- - ('v'xHx v 'v'x- Hx) .- 3x(Bx & 'v'y(SxyH Syy .- 'v'xy(Fxy ~ Fyx) H 'v'xy(FxyH Fyx) .- 'v'xy(Fxy ~ Fxy) H 'v'xy( FxyH Fyx) 'v'xyz(Rxy&Rxz~ - Ryz) .- - 'v'xR.xx 'v'xyz(Rxy& Ryz~ - Rxz) .- 'v'x- Rxx 'v'xyz(Fxy& Fyz ~ Fxz), 'v'xy(Fxy ~ Fyx) ~ '-v'x3yFxy - + '-v'xFxx

Vxyz(Fxy& Fyz --+Fxz), Vx3y( Fxy--+Fyx) ~ Vx3yFxy - + "V'xFxx 3xVy Fxy ~ 3xVyz(Fxz--+Fzy) 3xVyFxy ~ 3x- Vyz(Fxz--+Fzy)

107

4 Chapter

4.4

Finite Countermodels with Identity

name

Definition. A NAME EXTENSION consists of a singleobject selectedfrom the universe.

extension

Comment. The inuoduction of identity statements requires greater care about the different roles played by namesin the languageof predicatelogic (which we refer to as the object language) and in the languagewe use to specify interpretations (referred to as the ) . Things may have more than one name metalanguage in the object languagebut each must have a unique metalinguistic name in the specification of an interpretation . To mark this distinction, in this section we shall use italicized letters as names in the metalanguage . To reducethe potential for confusion when specifying an interpretationfor a set of sentenceswe also recommendthe practice of not using italicized . versionsof lettersalreadyappearingin the sentences . Italicized (metalinguistic) names are not Comment of the languagewe are studying. Rather, they are part our names for the things denoted by names in the object language. It is important to bear in mind that although a thing may be namedby various namesin the object language, each thing has only one metalinguistic name. This will aid in the specification of interpretationsfor wffs containingthe identity symbol.

108 finite

Chapter4

Definition . A finite interpretation for a set of sentences containing one-place and many-place predicates as well as the identity symbol consists of the following : . A finite universe, or domain. . Extensions for all the predicates appearing in the sentences. . Extensions for the names appearing in the .

identity truth valuations

sentences. Truth value specifications for the sentence letters appearing in the sentences.

Identity statements of the form a. =J3are true if and only if the extension of a. is the extension of J3. Comment. Because an object never has more than one metalinguistic name, identity statements occurring in expansions are true if and only if they are of the form a. =a. . Example . The expansion of a sentence containing the identity symbol can only proceed given both the universeand the name extensions. Given the universe U : { c, d } and name extensions a: c b: c the expansion of the wff ' Vx(Fxa --+ x* b)

4 Chapter

109

requires first the replacementof the object language nameswith their metalanguage equivalents,yielding: 'Q 'x(Fxc - + x*,"). Then expansionproceedsby the normal method, to yield: (Fcc - + c:#c) & (Fdc - + d:#:c) . Finally, given the predicateextension F: {(c, d) , (d, c) } the conjunctioncan be seento be hUegiven the falsity of the antecedentof the left conditional and the buth of both antecedentand consequenton the right.

Exercise4.4 " I* "" 11 """ 111 iv v* " VI "" Vll """ V111 ix * x

. Constructcountermodelsfor the following sequents = a b, c=d J- a=c Fa, a~b J- - Fb ' Vx3yx=y J- 3y' Vxy=x 3x(x~ - + Fx), a=b J- Fb 3xy Fx & Fy) & x ~ y ) J- ' VxFx ' Vxy( Fx & Gy - + x=y) J- - 3x(Fx & Gx) 3x (x~ - + Fx v Gx) J- 3x( Fx v Gx - + x~ ) ' Vxy(Fxy - + y=x) J- 3xFxx 3x' Vy(Fxy H x:#y) J- ' Vxyz Fxy & Fxz) - + y=z) ' Vxy Fx & Fy) & x~y) J- ' Vxy (Fx & Fy) & Fz) & X:l:y & y~ ) & x~

110

Chapter

4.5

InfiniteCountermodels

infinite countennodel

. Sometimesan invalid argumentcannot be Comment showninvalid by meansof afinite countermodel. Such casesrequire an INFINITE OOIJNTERM<XEL , i.e., one with an infinite numberof objectsin its domain.

Comment. A fonnal b"eabnentof infinite setsrequires an advancedcourse in set theory. Nonethelessit is possibleto exploit knowledgeaboutsetsof numbersto construct counterexamplesfor invalid argumentsthat requireinfinite models. The wffs of predicatelogic can be given interpretations in tenDS of arithmetical relationships in infinite domains such as the natural numbersor the setof positive and negativeintegers.

For ease of exposition we will take the natural numbers (0 , 1,2,3, etc.) as the infinite domain to be used. (This set is denoted N .) Note also that we cannot use expansions to construct countermodels, since an expansion for an infinite number of objects would involve infinitely long sentences.

111

Chapter4

numerical countennodel

to COUNTERMODEL Definition . A NUMERICAL an argument is a countermodel whose universe is N .

Example. 'v'xyz(Rxy & Ryz ~ Rxz), 'v'xy(Rxy ~ - Ryx), 'v'x- Rxx, 'v'x3yRxy ~ 3y'v'xRxy

Model: U: N R: {(m,n) suchthatm
112

Chapter4

'

premise and it doesn t bear R to the second object (secondpremise) , and since the first object bears R to this third object as well as the second (first premise) , the third doesn' t bear R to the first (second premise) . But this third object bears R to something (fourth premise) , hence a fourth object must be present, and so forth . Thus, the four premises require an infinite universe.

Exercise4.5.1 Return to any invalid sequentof this chapter, and give a numericalcountermodelfor it. (Of course, up to this point infinite models were not necessaryfor demonsttating invalidity, but they arepossible.) Exercise4.5.2 Give numerical countermodels to the following sequents .

" 1* ""* 11 """* 111 iv * v* vi *

Vxyz(Fxy & Fyz- + Fxz), Vx3yFxy ~ 3xFxx Vx3yVz(Fxy & (Fyz - + Fxz ~ 3xFxx Vx3yFxy, Vxyz(Fxy & Fyz- + Fxz), Vx- Fxx ~ Vxy(Gx & - Gy - + Fxy v Fyx) Vx3yz(Fxy & Fzx), Vxyz(Fxy & Fyz- + Fxz) ~ 3xy(Fxy & Fyx) Vx- Fxx, Vx3yVz(Fxy & (Fyz- + Fxz ~ Vxyz(Fxy & Fyz - + Fxz) Vxyz(Gxy & Gyz- + Gxz), Vxy(Gxy - + - Gyx), Vx3yGyx, Vx(x~ - + Gxa) ~ 3yVx(x~y - + Gyx)

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Answers to Selected Exercises Note : In almost all cases the answers given are not the only con-ect answers possible .

Chapter 1 Exerdse1.1 i ii iii iv v vi vii viii ix x

False False False True False True False True True False

Exerdse 1. 2.1 i ii iii iv v vi vii viii ix x xi xii xiii xiv xv

Atomic Sentence Not wff Not wff Conditional: AntecedentA ; ConsequentB Not wff Conditional: AntecedentA ; Consequent( 8 - + C) Conditional: Antecedent( P& Q) ; ConsequentR Disjunction: Left disjunct (A & B); Right disjunct (C - + ( 0 H 0 Negation to be a disjunction Not wff ; requiresouter parentheses Not wff Not wff Conjunction: Left conjunct - ( P & F); Right conjunct( P H (Q V - Q Biconditional: Left side - 8 v P) & C); Right side 0 v - 0 ) - + H) Not wff

114

Answersto Chapter 1 Exercises

Exerdse1.2.2 iv vi vii viii xiii xiv

A - +B A - + (8 - + C) P& Q -+R (A & B) v (C -+ ( D+-+0 - (P& P) v (p +-+Q v - Q) - 8 v P) & C) +-+D v - 0 -+ H

Exerdse 1.2. 3 i ii ill iv v vi vii viii ix x

: (PH (- Q v R Unambiguous : P v Q) -+ ( R& S Unambiguous : (Pv Q) -+ R) H S) Unambiguous Ambiguous Ambiguous Ambiguous : P & Q) H (- R v S Unambiguous Ambiguous : P - + (Q & - R H - S v T) -+ U Unambiguous Ambiguous

Exercise 1.3

1 2 3 4 5 6 7 8 9 10 11 12 13 14

P& - Q - P-+ - T P-+T T - +P - P v T(or - T - + - F) ( T -+ P) -+ - U (Q & - g) -+ R - (Pv R) -+ - T - T v (Pv R) (or thesameas8) (P& R) -+T T & - (Pv R) R - + (Q -+ P) T -+ U - T -+ - (Pv R)

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Answers to Chapter 1 Exercises

Answersto Chapter. I Exercises

116

3 1 1,2 1,2,3 1,3

- P H Q. - Q v R .- - P -+ R - PHQ -P - QvR - P -+ Q Q R - P -+ R

---

iv 1 2

A A

A tHE 2,4 -+E 3,5 v E 6 -+1(2)

Exerdse 1.4.2

1,3 1,2,3 53 1

2 3 3 1,3 1,3 1,2,3 S4 I 2 3 1 1 1 1 1,2 1,2,3

P v - R, - R -+ 5, - P ..- 5 Pv - R - R -+ 5 -P -R 5

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P -+ - Q, - Q v R -+ - S, P & T ~ - 5 P -+ - Q (I) - Q v R -+ - 5 (2) P& T (3) P (4)

(5) (6) (7)

-Q - QvR -5

P& (Q & R), P& R -+ - 5, 5 v T J- T P& (Q & R) (1) 2 P& R -+ - 5 () SvT (3) P (4) (5) Q& R R (6) P& R (7) -S (8) T (9)

A A A

1,3 vE 2,4 -+E

A A A 3 &E 1,3 - +E 5 vI 2,6- +E

A A A l &E l &E 5 &E 4,6 &1 2,7-+E 3,8 v E

Answersto Chapter Exercises

117

- P, R v - PH Pv Q ~ Q -P (I ) v - PH Pv Q 2 R () Rv - P (3) Rv - PHPV Q (4) Pv Q (5) (6) Q

A A 1vI 2HE 3,4 -+E 1,5 vE

59 1 2 3 3 1,3 1,3 3 1,3 3 2,3 1,2,3

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A A A 3 &E 1,4 -+E 5 &E 3 &E 6,7 &I 4vl 2,9 - +E 8, 10&1

Exerdse1.5.1

IS-

1I Q. Q

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P -i ~ - - p

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SII

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87 1 2 I 2 1,2 1,2

118

Answersto Chapter I Exercises

- P-+Q, - Q .- P - P-+Q (1) -Q (2) -P 3 () (4) Q P (5)

1 2 3

1,3 1,2 Sl6 I

2 3

1,3 1,2,3 1,2 Sl7 I 2 I 518 1 2 3 1,2 1

1,3 - +E 2,4 RAA (3)

A A A

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A A A

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A A

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514

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4,7 &I

(b) 1 1 1 4 1,4 1

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A l &E l &E A 2,4 vE 3,5 RAA (4)

8 1 1

(b) 1 2 1

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P& - Qr - (p-+Q) P& - Q (1) P-+Q (2) P (3)

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122

I HE 2,3 HI

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(3) (4)

P& (Q & R) -i~ (P& Q) & R P& (Q& R) ~ (P& Q) & R P& (Q& R) ( 1) P (2) (3) Q& R (4) Q R (5) P& Q (6) (P& Q) & R (7)

A I &E I &E 3 &E 3 &E 2.4 &1 5.6 &I

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Answersto Chapter 1 Exercises

1,3,5 1,3,5 1,2,5 1,2,5 1,2,5 1,2 1,2 1,2 1

Pv (Q v R) -il - (Pv Q) v R Pv (Q v R) I- (Pv Q) v R P v (Q v R) (1) - P v Q) v R) 2 () -P 3 () 4 () Qv R -Q 5 ) ( R 6 () (Pv Q) v R (7) P 8 () Pv Q (9) (Pv Q) v R (10) (11) Q Pv Q ( 12) (Pv Q) v R ( 13) (Pv Q) v R (14)

A A A

1,3 vE A 4,5 vE 6vI 2,7 RAA(3) 8 vI 9vI 2,10RAA(5) 11vI 12vI 2,13RAA(2)

123

Answersto ChapterI Exercises

(b) I 2 3 3 2 6 6 6 2 1,2 1,2 1,2 1,2 1

(Pv Q) (I ) (2) (3) (4) (5) (6) (7) (8) (9) (10) (II ) (12) (13) (14)

542

P& (Q v R) -il - (P& Q) v (P& R) P& (Q v R) I- (P& Q) v (P& R) P& (Q v R) (I ) - P & Q) v (p& R 2 () P (3) (4) Qv R (5) Q P& Q (6) (7) (p& Q) v (p& R) -Q (8) R (9) P& R (10) (II ) (P& Q) v (P& R) 12 ( ) (p& Q) v (p& R)

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(a) 1 2 1 1 5 1.5 1.5 1.2 1.2 1.2 1.2 1

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A A A 3 vI 2,4 RAA (3) A 6vI 7vI 2,8 RAA (6) 1,9 v E 5, 10v E 11v I 12v I 2,13RAA (2)

A A [forRAA] l &E l &E A [forRAA] 3,5 &I 6vI 2,7 RAA(5) 4,8 v B 3,9 &1 10v I

2,11RAA(2) A A [for RAA] A [for RAA] 3 &E

, I Exercises Answersto Chapter

124 - (P& Q) P& R P P - (Q v R) -Q Q - (P& Q) P& R R QvR Q QvR QvR P& (Q v R)

2,4 RAA(3) 1,5 vE 6&E 2,7 RAA(2) A [forRAA] A 3&E 10,11RAA(3) 1,12vE 13&E 14vi 9,15RAA(10) 16vi 9,17RAA(9) 8,18&1

(a) I I I 4 5 6 1,6 1,5 1,5 1,5 1,4 1,4 1,4 1,4 I

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A 1 HE 1 HE A A A 3,6 -+E 5,7 RAA (6) 5,8 &1 9 vi 4,10RAA (5) 2,11-+E 11,12&1 13vi 4, 14RAA (4)

(b) 1 2

(P& Q) v (- P& - Q) .- PH Q ( 1) (P& Q) v (- P& - Q) P (2)

A A

2 1,2 1,2 1 9 10 3 10 1,10 1,10 1,10 1,9 1,9 1 1

(5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18) (19)

Exerdse1.5.2 553

~~~.~ 1Q

125

Answers to Chapter I

(3) (4) (5) (6) (7) (8) (9) ( 10) (II ) (12) (13) ( 14) ( IS) ( 16) (17) ( 18)

sss

(- Pv Q) & R, Q-+SIP -+(R-+S) (I ) (- PvQ) &R (2) Q -+ S p (3) R (4) - PvQ (5) (6) Q s (7) R-+S (8) P-+ (R-+ S) (9)

-

3 4 4 3 1,3 1,3 1,2 1 11 12 12 11 I ,ll I ,ll 1 1

I 1,3 1,2,3 1,2,3 1,2 556 1 2 3 1 1

1,2 7 1,2,7 1,2,7 1,2,3

Q& R, Q -+P v S, - (S& R) ~ P (I ) Q& R 2 () Q -+ Pv S

(3) (4) (5) (6) (7) (8) (9) (10)

- (5 & R) Q R PvS -P S S& R P

A A 4&E 3,5 RAA(4) 1,6 vE 7&E 2,8 RAA(3) 9 -+1(2) A [for-+1] A [forRAA] 12&E 11,13RAA(12) 1,14vE 15&E 16-+1(II ) 10,17HI

A A A A

l &E 3,5 vE 2,6 -+E 7 -+1(4) 8 -+1(3)

A A A l &E l &E 2,4 - +E A [for RAA] 6,7 vE 5,8 &1 3,9 RAA (7)

126

Answersto Chapter 1 Exercises

Exerdse 1.5.3

i ii ii iv v vi vii viii ix x

Form Trans HS v Comm Dist&Iv DN Neg~ v~ TC DM SimOil

Substitution (P/ R; Q/S) (p/- P; Q/Q v R; RlS) (pIP& Q; Q/ R) (pIPv Q; Q/- R; RI- S) (P/ Rv S) (pIPv R; Q/S) (pIP;Q/Q v R) (p/- (P& Q) ; Q/ R) (pIP& Q; Q/ R& S) (pIP;Q/ Rv S; RlQ& R)

Exerdse l .5A

_ M

S66 1 2 3 1,3 1,3 1 7 1,7 1,7

1 1 4 4 1.4

PHQr

( 1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

PvQ -+P& Q PHQ PVQ P Q P& Q P-+P& Q Q P P& Q Q -+P& Q P& Q PVQ -+P& Q

P+-+Q -i.- - (Pv Q) v - (- Pv - Q) P+-+Q '- - (Pv Q) v - (- Pv - Q) P+-+Q (1) P-+Q (2) (3) Q -+ P (4) PvQ - P -+Q 5 () (6) Q

A A [for -+1) A 1,3 BP 3,4 &I 5 - +1(3) A 1,7 BP 7,8 &I 9 -+1(7) 2,6,10SiI DOil 11-+1(2)

A tHE tHE A 4v - +

2,5Spec Dll

. I Exercises Answersto Chapter

(7) (8) (9) (10) (11) (12)

(b) I

- (pv Q) v - (- p v - Q) J- PH Q - (Pv Q) v - (- Pv - Q) (1) P (2) (3) PvQ - (- Pv - Q) (4) P& Q (5) (6) Q P-+Q (7) (8) Q (9) PvQ - (- Pv - Q) (10) 11 P& ( ) Q P ( 12) ( 13) Q -+ P 14 P+-+Q ( )

2 2

1,2 1,2 1,2 1 8 8 1,8 1,8 1,8 1 1

-

S73

6 7 7

6 I

(P& Q) (1) (2) (3) (4) (5) (6) (7) (8) (9) ( 10) ( 11) (12) (13)

v (Rv S) I- p & Q) v R) v S (p& Q) v (Rv S) P& Q (p& Q) v R p & Q) v R) v S P& Q-+ p & Q) v R) v S Rv S R (p& Q) v R R -+ (p& Q) v R S-+ S P & Q) v R) v S R v S-+ P & Q) v R) v S P & Q) v R) v S

~>tf ~ ~

- Q -+P P P& Q - (- P v - Q) P v Q -+ - (- P v - Q) - (Pv Q) v - (- P v - Q)

4 1,4 1,4 1,4 1 1

127

V

+

[

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H J

d

H

U

t

J ~ '

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If, ~t p

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~ ~

IH

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t

d b

b

b

HLS I

l

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9 9

H H

S

9

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9

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.

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(

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31 VI 31

S

9 9

+

l

v v v

d S

S

S

' ' 9LS I

8

Answers to Chapter 1 Exercises

Answersto Chapter1 Exercises

879 I 2 3 4 5 5 4 4 ,5 1 ,4 , 5 1, 3 ,4 , 5 1 , 3 ,4 , 5 1 , 3 ,4 , 5 1 , 3 ,4 , 5 1 , 3 ,4 , 5 15 15 15 15 15

1 , 3 ,4 , 5 1 , 2 , 3 ,4 , 5 1 , 2 , 3 ,4 , 5 1 , 3 ,4 , 5 1 , 2 , 3 ,4 1, 2 ,3

880 I 2 3 3 6 6 6

129

(PH - Q) - + - R, (- P & S) v (Q & T) , S v T -+ R ~ Q -+ P A ( 1) (PH - Q) -+ - R 2 v P & S & A ( ) ( ) (Q T) S v T -+ R A (3) A [for -+1] (4 ) Q -p A [for RAA] (5 ) P- + - Q 5FA (6 ) - Q - +P 4FA (7) PH - Q 6,7 HI (8) -R 1,8 -+E (9 ) v 10 5 3,9 MTr ( ) ( T) -5 & -T 10DM ( 11 ) -5 II &E ( 12 ) Pv - 5 12vl ( 13 ) - (- P & - - 5) ( 14 ) 13DM - P& 5 A ( 15 ) -P 15&E ( 16 ) 5 15&E ( 17 ) --5 17DN ( 18 ) - P& - - 5 16,18&1 ( 19 ) - P& 5 - + - P& - - 5 19-+1( IS) ( 20 ) - (- P & 5) 14,20MTr ( 21 ) 2,21v E ( 22 ) Q& T T 22&E ( 23 ) -T II &E ( 24 ) P 23,24RAA (5) ( 25 ) 25- +1(4) ( 26 ) Q -+ P - s v (S & R), (S -+ R) - + P .- P - S v (S & R) ( 1) (2) (S -+ R) -+ P -S (3) S -+ R (4) - S -+ (S -+ R) (5) S& R (6) R (7) S- + R (8) S & R -+ (S -+ R) (9)

A A

A [for-+1] 3FA 4 -+1(3) A 6&E 7TC 8 -+1(6)

(

'

Vd

(

V

(

tl tl tl

'

NQtl ~

.

tl tl

H

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1

~

(

L

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,

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L '

V

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dw UW WO6 v v

b

L L II II

'

t

v v

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s SA

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3 3 3

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Answersto Chapter. 1 Exercises

131

Answersto Chapter1 Exercises

1,2,3 1,2

(22) (23)

8 -+ R R v - T -+ (8 -+ R)

S83 1 2 2 4 2 2,4 1,2,4 2 1,2,4 1,2 1 1

P& Q -+ R v SJ- (P-+R) v (Q -+ S) P& Q -+RvS (1) - (P-+ R) 2 () P& - R (3) (4) Q P (5) 6 P& Q () 7 Rv S () -R 8 () 9 S () (10) Q -+ S - (P-+ R) -+ (Q -+ S) (11) (12) (P-+ R) v (Q -+ S)

S84 I 2 I I 2 2 2 2 I 1,2 1,2 I I 2 1,2 1,2 1,2 1,2

(p-+Q) & (R -+P), (pv R) & - (Q& R) .- (p& Q) & - R A (I ) (p-+ Q) & (R-+ P) v R & & R A t2) (p ) (Q ) -+ P Q I &E (3) R-+ P I &E (4) PvR 2&E (5) - (Q & R) 2 &E (6) - Qv - R 6DM (7) - P-+ R 5 v-+ (8) - P-+ - R 4 Trans (9) --P 8.9 IA (10) P 10DN (II ) R -+Q 3,4 HS ( 12) - Q -+ - R 12Trans ( 13) -+ v-+ 14 R 7 ( ) Q -R 13,14SpecDll (IS) 3,11-+E (16) Q P& Q 11,16&I ( 17) 15,17&I ( 18) (p& Q) & - R

3,6,21SimOil 22-+1(3)

A A [for - +1] 2 Neg-+ A [for - +1] 3 &E 4,5 &I 1,6 - +E 3 &E 7,8 vE 9 -+1(4) 10- +1(2) 11v-+

132

S85

Answersto Chapter 1 Exercises

P& Q -+ (RvS) & - (R& S), R & Q -+ S, (I) (2 )

3

(3)

4

(4 )

5

(5)

4 ,5

(6 )

1,4 , 5

(7 )

1,4 , 5

(8)

1 ,4 , 5

(9 )

1 ,4 , 5

( 10 )

II

( II )

5 , 11

( 12 )

2 , 5 , 11

( 13 )

2 ,5

( 14 )

1 , 2 ,4 , 5

( 15 )

1 , 2 ,4 , 5

( 16 )

1 , 2 , 3 ,4 , 5 ( 17 ) 1 , 2 , 3 , 4 , 5 ( 18 ) 5

( 19 )

5

( 20 )

1 , 2 , 3 , 4 , 5 ( 21 ) 1 , 2 , 3 ,4 , 5 ( 22 ) 1 , 2 , 3 ,4

( 23 )

1 ,2 , 3

( 24 )

s -. R & Q) v (- R & - Q v - Pr p -. - Q P& Q - . ( RvS) & - ( R& S) A R & Q -. S A S -. R & Q) v (- R & - Q v - P A p A [for -+1] A -[for RAA] Q

P& Q (Rv S) & - (R& S) RvS - (R& S) R -+ - S R R& Q S R-+S -R S

R & Q) v (- R & - Q v - P ( R& Q) v (- R& - Q) - R - +Q - (- R & - Q) R& Q R -Q P -+ - Q

4.5 &I 1.6 -+E 7&E 7&E

9 Neg-+ A 5, 11&1 2, 12-+E 13- +1( II ) 10,14IA 8,15vE 3, 16-+E 4, 17v E 5TC 19Neg- + 18,20v E 21&E 15,22RAA (5) 22 -+1(4)

Exerdse 1.6.1

Tl 1

I- P-+P (I ) p -+p (2)

T2 I- P v - P (i) primitiverolesonly - (Pv - f ) 1 (I) 2 P (2) 2 Pv - P (3)

A 1 - +1 ( I )

A [forRAA] A [forRAA] 2vI

133

Answersto ChapterI Exercises

I I

(4) (5) (6)

-p Pv - P Pv - P

IQ t IQ

---

( ii ) derived rules allowed I

1,3 RAA (2) 4vI 1,5 RAA ( 1) A 1 - +1 ( 1) 2v - +

A [for - +1] A [for -+1] 1 -+1(2) 3 -+1( 1)

(ii ) derivedmlesallowed 1 P ( 1) 1 2 ( ) Q -+ P P -+ (Q - + P) (3)

A ITC 2 -+1( I )

I- (P-+Q) v (Q -+ P) T5 i ( ) primitiverolesonly - P -+Q) v (Q -+ P I (I ) 2 P-+Q (2) 2 (P-+Q) v (Q -+ P) (3) - (P-+Q) I (4) -P 5 (5) 6 P (6) -Q 7 (7) 5,6 8 () Q P-+Q 5 (9) P I ( 10) II (II ) Q I ( 12) Q -+P I (13) (P-+Q) v (Q -+ P) (14) (P-+Q) v (Q -+P)

A [forRAA] A [forRAA] 2vl 1,3 RAA(2) A [forRAA] A [for-+1] A [forRAA] 5,6 RAA(7) 8 -+1(6) 4,9 RAA(5) A [for-+1] 10-+1( II ) 12vl 1,13RAA(I )

---

t- P-+ (Q -+P) T4 i( ) primitiverulesonly I P 2 Q I Q -+P P-+ (Q -+ P)

r - (P+-+Q) +-+(- P+-+Q) T8 i( ) primitiverulesonly - (p+-+Q) I (I ) -P 2 (2) 3 3 () Q - Q -+ - P 2 (4) 5 P (5) -P 2,3 (6) 2,5 (7) Q 2 P-+Q (8) - P-+ - Q 3 (9) 10 10 ( ) Q -P II (II ) -Q 3,11 (12) 3,10 (13) P 3 (14) Q -+P 2,3 15 P+-+Q ( ) 1,2 ( 16) Q - P-+Q I (17) 10 18 P -+Q ( ) 5 (19) Q -+P 5,10 (20) P+-+Q -P 1,10 (21) I (22) Q -+ - P - P+-+Q I (23) - (p+-+Q) -+ (- P+-+Q) (24) - P+-+Q 25 25 ( )

2 -+1(3) A [for-+1] 3.4 -+E 5.6 RAA(3) 7 -+1(5) 3 -+1(2) A [for-+1] A [forRAA] 9.11-+E 10.12RAA(11) 13-+1(10) 8.14HI 1.15RAA(3) 16-+1(2) 10-+1(5) 5 -+1(10) 18.19HI 1.20RAA(5) 21-+1( 10) 17.22HI 23-+1( 1) A [for-+1]

--

11

rulesallowed (ii) derived - P -+ Q) v (Q -+ P 1 (I ) - (p-+Q) & - (Q -+P) I (2) - (p-+ Q) 1 3 () - (Q -+ P) 1 4 () 1 P& - Q (5) 1 (6) Q& - P 1 P (7) -P 1 (8) 9 () (P-+ Q) v (Q -+P)

Answersto Chapter Exerci~

-~

134

-~ NNN r

135

Answers to Chapter 1 Exercises

2S 25 28 28 28 10,28 10,25 25,28 5,28 25,28 25,28 25

- P- +Q Q -+ - P PHQ P-+ Q Q -+P P -P -Q Q -p Q - (P+-+Q) (- P +-+Q) - + - (P+-+Q) - (P+-+Q) +-+(- P +-+Q)

25HE 25HE

A [forRAA] 28HE 28HE 10,30-+E 10,27-+E 31,32RAA(10) 5,29-+E 33,34RAA(5) 26,35-+E 33,36RAA(28) 37-+1(25) 24,38HI

J- P ~ Q) ~ P) ~ P 1' 9 rulesonly (i) primitive 1 (P~ Q) ~ P (1) -P 2 (2) P~ Q 3 (3) P 1.3 (4) - (P~ Q) 1.2 (5) P 6 6 () -Q 7 (7) 2.6 (8) Q P~ Q 2 (9) P 1 (10) P ~ Q) ~ P) ~ P (11)

A [for-+1] A [forRAA] A 1,3 -+E 2,4 RAA(3) A [for-+1] A [forRAA] 2,6 RAA(7) 8 -+1(6) 5,9 RAA(2) 10-+1(1)

rulesallowed (ii) derived - (p-+Q) -+P) -+ P) 1 (1) 1 P -+ Q) -+P) & - P (2) 1 (3) (p-+ Q) -+P -P 1 (4) - (P-+ Q) 1 (5) P& - Q 1 (6) P 1 (7) (8) P -+ Q) -+ P) -+ P

1Neg-+ 2& 2& 3,4 M1T 5 Neg-+ 6& 4,7 RAA ( 1)

Answersto Chapter 1 Exercises

I- (p - + Q) v (Q -+ R) TI0 (i) primitiverulesonly - P -+ Q) v (Q -+ R 1 ( 1) 2 (2) (p -+ Q) 2 (3) (p -+ Q) v (0 -+ R) - (p -+ Q) 1 (4) 5 5 ( ) 'Q -R 6 (6) 7 P (7) 5 P -+ Q (8) 1,5 R (9) 1 ( 10) Q -+ R 1 ( 11) (P-+Q) v (Q -+R) 12 ( ) (P-+Q) v (Q -+R)

---

-

(ii ) derivedrulesallowed - (P-+Q) 1

P& - Q -Q Q -+R - (P-+ Q) -+(Q -+R) (P-+Q) v (Q -+R)

roles (ii) alternative proofusingderived 1 (I ) Q 1 P-+ Q (2) 1 (3) (P-+Q) v (Q -+R) 4 () Q -+ (P-+Q) v (Q-+R) -Q s (5) s 6 () Q -+R 5 7 () (P-+Q) v (Q -+R) - Q -+ (p-+Q) v (Q -+R) (8) (9) (P-+Q) v (Q -+R)

A [for RAA] A 2vl 1,3 RAA (2) A [for - +1] A [for RAA] A 5 -+1(7) 4,8 RAA (6) 9 - +1(5) 10vl I , ll RAA ( 1) A 1Neg- + 2 &E 3FA 4 - +1( I ) 5v -+ A ITC 2vl 3 -+1( I ) A 5FA 6vl 7 - +1(5) 4,8 SpecDll

137

Answers to Chapter 1 Exercises

.- (PHQ) H (- PH - Q) TII rulesonly i ( ) primitive I I PHQ () P-+Q I (2) I (3) Q -+P -P 4 (4) 5 (5) Q P 1,5 (6) -Q 1,4 (7) - P-+ - Q I (8) -Q 9 (9) P 10 ( 10) 1,10 (II ) Q -P 1,9 ( 12) - Q -+ - P I (13) - PH - Q I (14) (PH Q) -+ (- PH - Q) (IS) - PH - Q 16 16 ( ) - P-+ - Q 16 (17) - Q -+ - P 16 (18) P 19 ( 19) -Q 20 (20) -P 16,20 (21) 16,19 (22) Q P-+Q 16 (23) 24 Q (24) -P 25 (25) -Q 16,25 (26) P 16,24 (27) 16 Q -+P (28) 16 PHQ (29) 30 (- PH - Q) -+ (PH Q) ( ) 31 (PH Q) H (- PH - Q) ( ) rulesallowed (ii) derived I I PHQ () 2 I () QHP - PH - Q I (3)

A [for - +1] IHE IHE A [for -+1] A [for RAA] 3,5 - +E 4,6 RAA (5) 7 -+1(4) A [for -+1] A [for RAA] 2,10-+E 9,11RAA ( 10) 12- +1(9) 8,13HI 14-+1( I ) A [for - +1] 16HE

16HE A [for-+1] A [forRAA] 18,20-+E 19,21RAA(20) 22-+1( 19) A [for-+1] A [forRAA] 17,25-+E 24,26RAA(2S) 27-+1(24) 23,28HI 29-+1(16) IS,30HI A 1 Comm 2 Bitrans

138

-"-

Answers to Chapter 1 Exercises

5 5 5

(PH Q) - + (- P H - Q) - PH - Q QHP PHQ (- P H - Q) -+ (PH Q) (PH Q) H (- PH - Q)

T12 ... (- P-+ Q) & (R-+ Q) +-+(p-+ R) -+ Q i ( ) primitiverulesonly 1 ( 1) (- P-+Q) & (R-+Q) - P-+Q 1 (2) 1 R -+Q (3) 4 P-+R (4) -Q 5 (5) 6 P (6) 1,6 (7) Q 1,5 P (8) 1,4,5 R (9) 1,4,5 (10) Q 1,4 (11) Q 1 (12) (p-+ R) -+ Q (13) (- P-+ Q) & (R-+ Q) -+ P -+R) -+ Q) 14 (14) (p-+ R) -+ Q -P 15 (15) 16 P (16) -R 17 (17) 15,16 (18) R 15 P-+R (19) 14,15 (20) Q - P-+Q 14 (21) 22 R (22) 22 P-+R (23) 14,22 (24) Q 14 R -+Q (25) 14 (26) (- P-+Q) & (R-+Q) 27 ( ) P -+ R) -+Q) -+ (- P-+Q) & (R-+Q) (28) (- P-+Q) & (R-+Q) +-+(p-+ R) -+Q

3 -+1(2) A 5 Bitrans 6 Comm 7 -+1(5) 4,8 HI

A l &E l &E A [for -+1] A [for RAA]

A 2.6 -+E

5,7 RAA (6) 4,8 -+E 3,9 -+E 5,10RAA (5) 11-+1(4) 12- +1( 1) A A A A 15,16RAA ( 17) 18-+1( 16) 14,19- +E 20 -+1( 15) A 22- +1( 16) 14,23 -+E 24- +1(22) 21,25&J 26-+1( 14) 13,27HI

139

Answersto Chapter1 Exercises

rulesallowed (ii) derived I (I ) (- P-+Q) & (R-+Q) - P-+ Q I (2) R -+Q I (3) P-+ R 4 (4) 1,4 P-+ Q 5 () 1,4 (6) Q I (7) (P-+ R) -+Q 8 (- P-+R) & (R-+Q) -+ P -+ R) -+Q) () 9 (9) (P-+ R) -+Q -P 10 10 ( ) II P -+ R 10 ( ) 9,10 (12) Q - P-+ Q 9 (13) 14 R (14) P-+ R 14 (15) 9,14 (16) Q R-+ Q 9 (17) 18 9 ( ) (- P-+Q) & (R-+ Q) P -+R) -+Q) -+ (- P-+Q) & (R-+Q) (19) (20) (- P-+Q) & (R-+Q) H (P-+R) -+Q TI3 1 1 4 4

T14 1 1 4 5 4,5

J- PHP (I ) (2) (3) (4) (5) (6) (7)

&P P P& P P-+P& P P& P P P& P-+ P PHP & P

t- PH PV P P ( 1) 2 PV P () P-+PV P 3 () P VP (4) -P 5 () P 6 ()

A l &E l &E A 3,4 US 2,5 SpecOil 6 -+1(4) 7 -+1(1) A A 10FA 9,11-+E 12-+1(10) A 14TC 9,15-+E 16-+E (14) 13,17&1 18-+1(9) 8,19HI

A 1,1 &I 2 - +1( 1) A 4& 5 -+1(4) 3,6 HI

A 1 vI

2 -+1( I ) A A [forRAA] 4,5 vE

140

Answersto Chapter 1 Exercises

(7) (8) (9)

p PvP -+P PHPVP

-NNN r-

TI7 t- (p+-+Q) & (R+-+S) -+ (p v R +-+Q v S) (i) primitive .rulesonly 1 (I ) (p+-+Q) & (R+-+S) I (2) PHQ I P-+ Q (3) I 4 () Q -+P I RHS (5) I R-+S (6) I S-+R (7) S PvR 8 () - (Q v S) 9 (9) 10 (10) Q 10 (11) QvS -Q 9 (12) 13 P (13) 1,13 (14) Q -P 1,9 (15) I ,S,9 16 R ( ) I ,S,9 S (17) I ,S,9 (18) QvS I ,S (19) QvS I PvR -+QvS (20) 21 21 ( ) QvS - (Pv R) 22 (22) P 23 (23) 23 Pv R (24) -P 22 (25) 26 (26) Q 1.26 P 1.22 -Q 1,21,22 S 1,21,22 R 1,21,22 PvR 1,21 PvR

5,6 RAA (5) 7 -+1(4) 3,8 HI

A I &E 2 +-+E 2 +-+E I &E SHE SHE A [for -+1] A [for RAA] A 10vi 9,11RAA ( 10) A 3,13- +E 12,14RAA ( 13) 8,IS vE 6,16- +E 17vi 9,18RAA (9) 19-+1(8) A A A

23vI 22,24RAA(23) A 4,26-+E 25,27RAA(26) 21,28vE 7,29-+E 30vI 22,31RAA(22)

141

Answers to Chapter I Exercises

QvS -+ PvR PVRHQVS

(PH Q) & (RH S) -+ (Pv R H Q v S)

-

1

1,4 1 9 1 1 1,9 1 1

---

(ii ) derived rules allowed 1 (p H

Q) & ( RH S)

-

(33) (34) (35)

1 1

PHQ RHS PvR P-+ Q R -+ S QvS PvR - + QvS

QvS Q -+P S-+R PvR QvS - +PvR PVRHQVS (PH Q) & ( RH S) -+ (Pv R H Q v S)

32-. 1(21) 20,33HI 34-. 1(1) A I &E l &E A 2HE 3HE 4,5,6 CornDll 7 - +1(4) A 2 +-+E 3 +-+E 9,10,11CornDll 12-+1(9) 8,13HI 14-+1( 1)

-

~ (PH Q) -+ R -+ P) H ( R- + Q & P -+ R) H (Q -+ R T19 (i ) primitiverulesonly PH Q A [for - +1] ( 1) P- + Q IHE (2) 3 -+ P IHE ( ) Q 4 R + P A ( ) [for - +1] R A [for - +1] (5) 4,5 P 4,5 -+E (6) 1,4,5 2,6 - +E (7) Q 1,4 R -+ Q 7 - +1(5) (8) 1 9 + P + -+ 8 -+1(4) ( ) (R ) ( R Q) 10 R - +Q A ( 10) 11 R A ( 11) 10,11 ( 12) 10,11-+E Q 1,10,11 ( 13) P 3,12-+E 1,10 R -+ P 13-+1( 11) ( 14)

1 Exercises to Chapter Answers

142 1 1 17 18 1,18 1,17,18 1,17 1 23 24 1,24 1,23,24 1,23 1 1 1

(15) (16) (17) (18) (19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) (30) (31)

(R-+Q) -+ (R-+P) (R-+P) H (R-+Q) P-+R Q P R Q -+R (P-+R) -+ (Q -+ R) Q -+R P Q R P-+R (Q -+R) -+ (P-+R) (P-+R) H (Q -+R) R -+ P) H (R-+Q & P -+R) H (Q -+R (PH Q) -+ R -+ P) H (R-+Q & P -+R) H (Q-+ R

14-+1(10) 9,15HI A A 3,18-+E 17,19-+E 20-+1(18) 21-+1( 17) A A 2,24-+E 23,25-+E 26-+1(24) 27-+1(23) 22,28HI 16,29&I 30-+1(1)

A tHE tHE A 2,4 US 5 -+1(4) A

3,7 US 8 -+1(7) 6,9 HI A 3,11US 12-+1(11) A 2,14HS IS-+1(14) 13,16HI

V

1

3 3

V

b b d d

IA91

( gA g I

( ( IH (

V V

v

1

( 3 3 1

( (

V

V

HHIHHI

~ >51

~. c~: 5I1Q j ~ >Q: ~ >51 -. ~ . : > I (

Answersto Chapter1 Exercises

, 1 Exercises Answersto Chapter

144 -R -+P -R -+Q

~~~ ~IU 0~tu5It 0: g t c t ~ .:

4v ~ 2,5 as 6v ~ 7 ~ I (4) A 9v ~ 3, 10 as 11 v ~ 12 ~ I (9) 8, 13 HI 14 ~ I ( 1)

RvQ RvP - + RvQ RvQ - R -+ Q - R -+ P RvP RvQ -+RvP R V P H Rv Q (PH Q) -+ ( RV PH R V Q)

c c

(5) (6) (7) (8) (9) (10) (11) (12) (13) ( 14) ( 15)

.

4 1,4 1,4 1 9 9 1,9 1,9 1 1

A [for -+1] A [for - +1] A [for RAA] A 2,4 - +E 3,5 RAA (4) A 1,7 - +E 6,8 RAA (7)

A tOyI 3.11vE 12-+1( 10) 9.13RAA(3) 14-+1(2) IS-+1( 1) A [for-+1] A [for-+1] A [forRAA] A 18.20-+E 19.21RAA(20) A 17.23-+E

Answersto ChapterI

Exercises

17,18,19(25) - (Q ~ P) 26 (26) Q 26 (27) Qv P 19,26 (28) P 19 29 ( ) Q~ P 17,18 (30) Q 17 31 ( ) (P~ Q) ~ Q 32 ( ) Q ~ P) ~ P) ~ P ~ Q) ~ Q) (33) (P~ Q) ~ QH (Q ~ P) ~ P

145

22.24RAA(23)

A 26vI 19,27vE 28-+1(26) 25,29RAA( 19) 30-+1(18) 31-+1(17) 16,32HI A 1Neg- + 2 &E 2 &E 4 Neg-+ 5 &E 5 &E 7FA

3,8 -+E 6,9 -+E 7,10RAA( 1) A 12Neg-+ 13&E 13&E 15Neg-+ 16&E 16&E 18FA 14,19-+E 17,20-+E 18,21RAA(12) 11,22HI

-~

Answersto Chapter 1 Exercises

J- (P& 0) v (R& S) H (Pv R) & (Pv S & Q v R) & (Q v S (i) primitiverulesonly A [for -+1] I (p& Q) v (R& S) (I ) A [for RAA] v R 2 2 (P ) () A [for RAA] v 3 (P S) (3) A 4 P 4 () A P& Q 5 (5) 5 &E P 5 (6) - (P& Q) 4,6 RAA (5) 4 (7) 1,7 vE R& S 1,4 (8) 8&E R 1,4 (9) 8&E S 1,4 (10) vI 9 Pv R 1,4 (11) 2 P 1,2 , 11RAA (4) ( 12) vI 12 Pv R 1,2 ( 13) 2, 13RAA (2) Pv R I (14) 10vI Pv S 1,4 ( 15) 3,15RAA (4) P 1,3 ( 16) 16v I Pv S 1,3 (17) 3 PvS I 18 ,17RAA (3) ( ) 14 v v R & S I 19 ,18&1 (P ) (P ) ( ) A 20 [for RAA] (Q v R) (20) A [for RAA] 21 (Q v S) (21) -Q A 22 (22) 5 &E 23 5 ( ) Q - (P& Q) 22,23RAA (5) 22 (24) 1,24 v E R & S 25 1,22 ( ) 25&E R 1,22 (26) 25&E S 1,22 (27) 26vI v R 1,22 28 ( ) Q 20,28RAA(22) 1,20 (29) Q 28vI v R 30 1,20 ( ) Q 20,30RAA(20) v R I 31 ( ) Q 27vI v 1,22 (32) Q S 21,32RAA(22) 1,21 33 ( ) Q 33vI 1,21 (34) Qv S 21,34RAA(21) I (35) Qv S 30,35&I I (Q v R) & (Q v S) (36)

V

(

(

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mleg

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---

I I I p I ~

If

H +

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S (

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( (

[

d

V V

' S

I

S

I

( 3A 3AI

d l

l

(

l

I

( 3A 3AI

+

IA

(

( (

( (

(

I (

' s ' SAd

( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (

Ot

Ltl

ChapterI Exercises

-

148

15 15 15 15 15 15 15 15 15 15 15 15 15

Answersto Chapter I Exercises

(7) (8) (9) (10) ( 11) ( 12) ( 13) ( 14)

(IS) (16) (17) (18) (19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29)

PvR Pv S (Pv R) & (Pv S) Qv R Qv S (Q v R) & (Q v S) P v R) & (Pv S & Q v R) & (Q v S (P& Q) v (R& S) -+ P v R) & (Pv S & Q v R) & (Q v S P v R) & (Pv S & Q v R) & (Q v S P ( v R) & (Pv S) (Q v R) & (Q v S) PvR PvS QvR QvS (Pv R) & (Q v R) (Pv S) & (Q v S) (P& Q) v R (P& Q) v S P & Q) v R) & P & Q) v S) (P& Q) v (R& S) P v R) & (Pv S & Q v R) & (Q v S -+

(P& Q) v (R& S) (P& Q) v (R& S) H

5&E 6&E 7,8 &1 5&E 6&E 10,11&1 9,12&1 13-+1(1) A 15&E 15&E 16&E 16&E 17&E 17&E 18,20&1 19,21&1 22Dist 23Dist 24,25&1 26Dist 27-+1( 15) 14,28HI

P v R) & (Pv S & Q v R) & (Q v S T31

J- (Pv Q) & (Rv S) H P & R) v (P& S v Q & R) v (Q & S

(i) primitiverulesonly 1 A [for -+1] (I ) (Pv Q) & (R v S) - (P& R) v (P& S v Q& R) v (Q& S ) A [for RAA] 2 (2) 3 P& R A (3) 3 3 vI (4) (P& R) v (P& S) 3 4vI (5) P & R) v (P& S v Q& R) v (Q & S - (P& R) 2 2,5 RAA (3) (6) 7 P& S A (7)

149

Answersto Chapter1 Exercises

tf' tfr

M M

(35) (36) (37) (38) (39) (40) (41) (42) (43) (44) (45)

Q& R) v (Q& S

Q & R) v (Q & S

7 vI 8 vI 2,9 RAA (7) A

'--

(P& R) v (P& S) P & R) v (P& S v - (P& S) Q& R (Q& R) v (Q & S) P & R) v (P& S v - (Q& R) Q& S (Q & R) v (Q & S) P & R) v (P& S v - (Q & S) PvQ RvS P R P& R -R S P& S -P Q Q& R -R S Q& S P & R) v (P& S v (Pv Q) & (R v S) -+ P & R) v (P& S v P & R) v (P& S v - (pvQ) (P& R) v (P& S) P& R P PvQ - (P& R) P& S P PvQ - P& R) v (P& S

-->-

(8) (9) (10) (II ) (12) (13) (14) (15) (16) (17) (18) (19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) (34)

Q & R) v (Q & S I &E l &E A

Q& R) v (Q & S Q& R) v (Q& S Q & R) v (Q& S

A 21,22&1 6,23RAA(22) 20,24vE 21,25&1 10,26RAA(21) 19,27vE 22,28&1 14,29RAA(22) 20,30vE 28,31&1 18,32RAA(2) 33-+1(1)

cS ~~ .~ ~~-

35 36 37 38 38 38 36

-

-

7 7 2 II II II 2 15 15 15 2

A for - +1

150

f. ~ ~

Q & R) v (Q & S Q& R Q PvQ - (Q & R) Q& S Q PvQ PvQ - (Rv S) (Q & R) v (Q & S) Q& R R

~-

>~ ~r--

36,53RAA (36) A [for RAA] A A 57&E 58 vI 55,58RAA (57) 56,00 v E 61&E 62 v I 55,63RAA (56) 35,64 v E A 66&E 67 vI 55,68 RAA (66) 65,69 v E 70&E RvS 71 vI RvS 55,72 RAA (55) v v & S 54,73&1 (P Q) (R ) P & R) v (P& S v Q & R) v (Q & S -+ 74--+1(35)

~>r

(46) (47) (48) (49) (50)

-~--

35,36 47 47 47 36 35,36 35,36 35,36 35 55 56 57 57 57 55 55,56 55,56 55,56 55 35,55 66 66 66 55 35,55 35,55 35,55 35 35

Exercises

~ -

Answers to Chapter

(Pv Q) & (Rv S) (Pv Q) & (Rv S) H P & R) v (P& S v Q& R) v (Q & S (ii ) derivedrulesallowed 1 ( 1) (Pv Q) & (R v S) 1 (2) (P& ( Rv S v (Q & ( Rv S 3 P & ( Rv S) (3)

34,7SHI

A 1 Dist A

151

19 19

t~>0-~~>(-

PvQ (Pv Q) & (Rv S) (Q& R) v (Q & S) -+ (Pv Q) & (Rv S) (Pv Q) & (Rv S) P & R) v (P& S v Q & R) v (Q& S -+

11

(28) T32

(P& R) v (P& S) P& (Rv S) -+ (P& R) v (P& S) Q& (Rv S) (Q& R) v (Q & S) Q & (Rv S) -+ (Q& R) v (Q & S) P & R) v (P& S v Q & R) v (Q& S (Pv Q) & (Rv S) -+ P & R) v (P& S v Q & R) v (Q & S P & R) v (P& S v Q & R) v (Q & S

~~~-~~(-

-0

11 12 12 12 12 12 12

(4) (5) (6) (7) (8) (9) ( 10)

----

\

Answersto Chapter1 Exercises

(Pv Q) & (Rv S) (Pv Q) & (Rv S) H P & R) v (P& S v Q& R) v (Q& S

3 Dist 4 -+1(3) A 6 Dist 7 -+1(6)

A A 12Dist 13&E 13&E 14vl 15.16&I 17-+1( 12) A 19Dist 20&E 20&E

21vI 22.23&I 24-+1(19) 11.18.25SimDil 26-+1(11) 10.27HI

.- (P-+Q) & (R-+ S) H - P& - R) v (- P& S v Q & - R) v (Q & S

(i) primitiverulesonly 1 (1) (p-+Q) & (R-+ S) - (- P& - R) v(- P& S v Q& - R) v(Q & S 2 (2) 1 P-+Q (3) 1 4 R -+ S () 5 5 () (- P& - R) v (- P& S)

A [for-+1] ) A [forRAA] 1&E 1&E A

-

Answersto Chapter I Exercises

- P& - R) v (- P& S v Q & - R) v (Q& S - - P& - R) v (- P& S (Q& - R) v (Q& S) - P& - R) v (- P& S v Q & - R) v (Q & S - Q& - R) v (Q & S (- P& - R) (- P& - R) v (- P& S) - (- P& - R) - P& S (- P& - R) v (- P& S) - (- P& S) Q& - R (Q& - R) v (Q& S) - (Q & - R) Q& S (Q & - R) v (Q & S) - (Q& S) -P

-r

~

5 2 8 8 2 11 11 2 14 14 2 17

-~-

152

& -R

37 38 39 40 40 39 43

R S - P& S P Q Q& - R R S Q& S - P& - R) v (- P& S v Q& - R) v (Q & S (P-+Q) & (R-+ S) -+ - P& - R) v (- P& S v Q & - R) v (Q& S - P& - R) v (- P& S v Q & - R) v (Q & S P -Q Q& - R Q - (Q & - R) Q& S

5 vI 2,6 RAA(5) A 8 vI 2,9 RAA(8) A 11vI 7,12RAA(11) A 14vI 7,15RAA(14) A 17vI 10,18RAA(17) A 20vI 10,21RAA(20) ARAP 23,24&1 13,25RAA(24) 4,26-+E 23,27&1 16,28RAA(23) 3,29-+E 24,30&1 19,31RAA(24) 4,32-+E 30,33&1 22,34RAA(2) 35-+1( 1) A [for-+1] A [for-+1]

A [for RAA] A 4O&E 39,41 RAA (40) A

153

Answersto Chapter1 Exercises

43 (44) 39 (45) 46 (46) 39,46 (47) 39 (48) 37,39 (49) 50 (50) 50 (51) 38 (52) 37,38,39(53) 37,38,39(54) 37,38 (55) 37 (56) 57 (57) 58 (58) 59 (59) 59 (60) 58 (61) 62 (62) 58,62 (63) 58,62 (64) 57,58 (65) 37,57,58(66) 67 (67) 67 (68) 57 (69) 37,57,58(70) 37,57,58(71) 37,57 (72) 37 (73) 37 (74) (75) (76)

43 &E

39,44RAA(43) A 42,46 v B

45,47 RAA (46) 37,48 v E A 50&E 38,51 RAA (50) 49,52 v E 53&E 38,54RAA (39) 55 ~ I (38) A [for ~ I] A [for RAA] A 59&E 58,60RAA (59) A 61,62 v E 63&E 57,64RAA (62) (- P & - R) v (- P & S) - P& - R -R - (- P & - R) - P& S S S R -+ S (P-+ Q) & ( R-+ S)

37,65vE A 67&E

57,68RAA(67) 66,69vE 70&E 58,71RAA(58) 72-+1(57) 56,73&1 - P& - R) v (- P& S v Q& - R) v (Q & S -+ 74-+1(37) (P-+ Q) & (R-+ S) 36,7SHI (P-+Q) & (R-+ S) H - P& - R) V (- P& S V Q& - R) V (Q& S

(ii ) derivedrulesallowed 1 (p - + Q) & ( R-+ S) ( 1) P -+ Q 1 (2)

A l &E

154

~ 16 16 16 16 16 16 16

(31)

0 -_ ~-

~

-.~Ai> f -

~- - 1~ > f ~

~ f-

~ f --

i

- ~~

0~

( 16) ( 17) (18) (19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) (30)

f

16 17 17

(3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)

~i

- -

1 1 1 1 1 8 8

Answersto Chapter Exercises

15,30HI (P-+ Q) & (R-+ S) +-+ - P& - R) v (- P& S v Q & - R) v (Q & S

155

Answers to Chapter 2 Exercises

Chapter 2

~O

Exerdse2.1

PQ

PQ TT TF FT FF

""

QR TT TF FT FF TT TF FT FF

AO ~

>

iii PQ TT TF FT FF

Iu! t Q-t -A

Pv (- PvQ ) TF T TF F TT T TT T

(P& Q H Q) -+ (Q -+P) T T T T F T T T F F T F F T T T

ii - {P& Q) v P TTFT T TFTFT FTTFT FFTFT

iv (pv Q) v (- P& Q) T TF F T TF F T TT T F FT F vi RH - Pv (R& Q) TFT T TF F F FFF F TF F F TTT T FTT F TTT F FTT F viii PQ TT TF FT FF

(PH - Q) H (- P H - Q) FF F F TF TT FF FT TF FT FF FT FT TT

Answersto Chapter2 Exercises

156

ix PQR TTT TTF TFT TFF FTT FTF FFT FFF

(p +-+Q) +-+(pv R-+ (- Q-+ R T T T T F T T T T F T T F T T T T F T T FTF F F F T TFT F F F T F T T T T T T T T T F T T F

x P Q R S (P& Q) v ( R& S) -+ (P& R) v (Q & S) TT T T T T T T T T T TT T F T T F T T T F TTFT T T F T F T T TTFF T T F F F F F TF T T F T T T T T F TF T F F F F T T T F TF F T F F F T F F F TF F F F F F T F F F FT T T F T T T F T T FT T F F F F T F F F FTFT F F F T F T T FTFF F F F T F F F FF T T F T T F F F F FF T F F F F T F F F FF F T F F F T F F F FF F F F F F T F F F Exerdse2.2

A [forRAA] l &E 2,3 BP l &E 4,5 RAA(2)

-

' c>0

<~c2 .i Q

N~

.

~

>..

(~

~ tI

0~EtI a01 {~ 0I~~an. ~>E~ ac~ ~~

NN

~-

d d (

d

~ ~~tI ----fMMOI -01

.>

-

.cS. t ~

~EQ>0~0 tI~ OIWA 6~~tI 0I~E0I iE

M ~~ .(la

Q

~f

-

-Ir

S

01

Chapter

2 Exercises Answersto Chapter

o0

1

~:~~ t~>Ia -

5 5

P-+ (- P-+ R) Q - Q -+R Q -+ (- Q -+R) (- P-+ R) v (- Q -+R)

3 -+1(2) A [for-+1] SPA 6 -+1(5) 1,4,7 CornOil

~cS .~ -

-~-

158

VALID (1) (2) (3) R& - P (4) 5 R () 6 () Q (7) QvR - (Q v R) (8) (9) (QH R) xiv 1 2 3 4 1,2 1,2,4 1,2,3

VALID (1) (2) (3) (4) (5) (6) (7)

xv

INVALID P:F Q:F

P-+ (Q & R -+ S) P -S Q& R Q & R -+ S S - (Q& R)

R:T

A A A A [for RAA] 1,2 -+E 4,5 - +E 2,6 RAA (4)

159

,

~'

Answersto Chapter2 Exercises

INVALffi

P:T

Q:F

R:F

S:F

T:F

VALID

(1) (2) (3) (4) (5) (6) (7) (8)

4 4 3,4 8 8 (9) 2,8 (10) 1,2,8 (II ) 1,2,3,4,8 (12) 1,2,3,4,8 (13) 1,2,3,4 (14) 1,2,3 ( 15) 1,2,3 (16) xviii

Q -+ (P- +R & - Q) - Q -+ - (TvV ) U & SHP - (S -+ - V) S& U U& S P T TvV Q P-+R & - Q

A A A

A [for -+1] 4 Neg- + 5 Comm 3.6 BP A [for RAA] 8 vI 2.9 M1T 1.10-+E 7.11-+E 12&E 10.13RAA (8) 14-+1(4) 15v- +

~~I0

1 2 3 4

- (5 -+ - U) -+ - T (5 -+ - U) v - T

INVALID Q:F

R:F

S:F

T:F

i ii iii iv v vi vii viii ix x xi xii xiii 1

P:F PiT P:T P:F PiT P:F PiT P:F PiT P:T PiT P:F VALID ( 1)

Q:T Q:F Q:T Q:F Q:F Q:F Q:F Q:F Q:T Q:F Q:F Q:T

R:F RiT R:T R:F RiT R:T R:F RiT R:F R:F

ri5ri5 f

Exerdse 2.4.2

P - + (- Q - + - R & - 8 )

T:T T:F T:T TiT

U:F

V:T

W:F

Answers to Chapter2 Exercises

xiv xv xvi xvii I

2 1 1 1 1 1,2 1,2 1,2 1,2 -

xviii

1,2 1,2 1,2 1,2 1

-

I~ : CI

2 3 4 1,4 1,3,4 1,3,4 1,3,4 1,3,4 1,3,4 1,3,4 1,2,3

-Ntf -

160

A A A 1,4 -+E 3,5 -+E 6 &E 6 &E 7FA 8FA 9, 10HI

- Q -+ - R & - S -R& -S -R -S R -+S S-+R RHS -p

P:F PiT PiT VALID (1) (2) (3) (4) (5) (6) (7) (8) (9) ( 10) VALW (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

Q:T Q:F Q:F

RiT RiT RiT

2,11RAA(4)

S:F S:F S:F

T:F T:F

U:F

- (p-+ - Q & R) - RH - P P& - (- Q& R) P - (- Q & R) - - Qv - R R --Q Q P& Q

1Neg-+ 3&E 3&E 50M 2,4 BT 6,7 vE SON 4,9 &1

(p -+ Q) & (- Q - + P & R) -+ (5 v T - + - Q) Q P- + Q - Q -+ P& R (p -+ Q) & (- Q - + P & R) 5vT -+ - Q - (S v T) - S& - T - (- S -+ T) Q -+ - (- S -+ T)

A A 2TC 2FA 3,4 &1 1,5 -+E 2,6 MTr 7DM 8Neg-+ 9 -+1(2)

A A

Answersto Chapter3 Exercises

8 9 8,9 8,9 8

xx 1 2 1,2 1 1 1 1

P v - Q) -+ (- P & - Q -+ - P -P Pv - Q -Q - P& - Q (Pv - Q) -+ (- P & - Q) - P -+ P v - Q) -+ (- P & - Q P v - Q) -+ (- P & - Q +-+- P

VALill (I ) (2) (3) (4) (5) (6) (7)

QH - Q Q -Q -Q Q QV (PH - P) PH - P

Pv - Q -+-P& -Q P Pv - Q -P& -Q -P -P

A [for-+1] A [forRAA] 2vl 2,3 -+E 4&E 2,5 RAA(2) 6 -+1(1)

4,6 v B

Chapter 3

.

. .

.

Exerdse 3.1.1 Not a wff Not a wff Universal Existential Not a wff Not a wff Universal Not a wff ( but acceptable biconditional abbreviation given the dropping convention )

-

I 2 2 1,2 1,2 I

VALill (I ) (2) (3) (4) (5) (6) (7) (8) (9) (10) (II ) (12) ( 13) ( 14)

-c2c2 .tt~ ~

xix

161

162 ix x Xl xii xiii xiv xv xvi xvii xviii xix xx xxi xxii xxiii xxiv xxx

Answersto Chapter3 Exercises Negation Not a wff (but acceptableconditional abbreviationgiven the parenthesis droppingconvention) Not a wff (but acceptableconditional abbreviation given the parenthesis convention ) dropping Negation Not a wff Existential Not a wff Atomic sentence Not a wff Not a wff Not a wff Universal Not a wff Not a wff Negation Biconditional Not a wff

Exercise 3.1.2 Fonnula Open i Fz ii None Gcax ill iv

v vi vii

Gxy Gyx (Gxy& Gyx) 'v'y(Gxy& Gyx) Gxy Hy Ax Fxx Fxy Hxyz Jz (Hxyz& Jz) 'v'z(Hxyz& Jz) ( Fxy- + 'v'z( Hxyz& Jz 'v'y( Fxy-+ 'v'z( Hxyz& Jz

ExampleWFF 3zFz

3y- VxGxy - Vy- Hy VxAx 3xFxx

Answers to Chapter,3 Exercises

viii

xi xii xiii xiv xv

163

Fxx Fxy

'v'yFxy Hz Ix ( Hzv Ix ) 3z( Hzv Ix ) - 3z( Hz v Ix ) Fxx (Hav Fxx) - (Hav Fxx) None Fx Fx Fyyy (Fyyy& P) Fzx Hxyz (FzyH Hxyz)

'v'xFx

Exercise 3.1. Translationschemeis providedonly whereit is not obvious. alt: indicatesan alternative, logically equivalentttanslation. . amb: indicatesnon- equivalentrenderingof an ambiguoussentence inc ! indicatesa common, but incorrectanswer.

I inc! 2 inc! 3 all: inc!

alt: inc!

Vx( Dx-+ Mx) Vx( Dx& Mx) 3x(Sx& Ox) 3x(Sx- + Ox) Vx(Fx-+ - Ex) - 3x(Fx& Ex) - Vx(Fx-+ Ex) - Vx(Fx-+ Px) 3x(Fx& - Px) 'v'x( Rx - + - Ex) & 'v'x(Ax - + - Ex) 'v'x( Rx v Ax - + - Ex) 'v'x( Rx & Ax - + - Ex)

164

inc! inc!

7 inc! 8 alt: 9 alt: 10 alt: 11 12 alt: inc! 13 14 15 alt: 16 17 18 alt: 19 arnb: 20 21 22 23-29

Answersto Chapter3 Exercises

3x(Px& Ax) & 3x(Rx & Ax) 3x Px& Rx) & Ax) 3x Pxv Rx) & Ax) 'v'x(Ox -+ Lx) 'v'x(Lx -+ Ox) 'v'x(Sx- + (Px -+ Tx v Bx 'v'x(Sx& Px -+ Tx v Bx) 'v'x(Mx H Px) 'v'x Mx -+ Px) & (Px -+ Mx 'v'x(Fx- + - Wx vEx ) 'v'x(Fx& Wx -+ Ex) (3x(Ox & Cx) & 3x(Ox & Mx & - 3x(Cx & Mx ) 'V'x(lx - + Px) 'V'x(- Px - + - Ix ) 'v'x(Px - + Ix ) 'v'x(Ax - + (- Wx - + Nx 'v'x(Ax - + (Nx - + - Wx

3x(Sx& (Px& Fx & - Vx(Px& Fx-+ Sx) 3x(Sx& (Px& Fx & 3x Px& Fx) & - Sx)

Wa& 'v'x( Wx-+ Mx) -+ Ma 'v'x(Sx-+ (- Nx v Mx 'v'x(Ox & Ex -+ - Px) - 3x Ox & Ex) & Px) 'v'x(Px -+ - Hx) - 'v'x(Px - + Hx) [possiblereadingin someregional dialects ofEnglish ] - 'v'x(Px - + Cx) 'v'x(Mx & Wx -+ Bx) 3x Mx & Wx) & Fx) TranslationschemeI using single-placepredicatesonly : Ta : a is a trick Wa : a is a whale Sa: Shamucan do a Ca: a can do a trick s: Shamu Note: Strictly we shouldusea letter from a-d for a name, but the useof s for . Shamuis more perspicuous

165

Answersto Chapter3 Exercises

'v'x(Tx-+ Sx) 23 'v'x(Tx-+ Sx) 24 - 'v'x(Tx -+ Sx) 25 'v'x(Tx-+ - Sx) 26 : - Cs thatis notlogicallyequivalent alternative translation 3x(Wx & Cx) -+ Cs 27 . Notedifferencein scope 'v'x( Wx& Cx -+ Cs) . Lessnaturalreading ambi: 'v'x( Wx-+ Cx) -+ Cs ambii: If anywhalecandoa trick, Shamucandothatsametrick. Thisreadingis notexpressible only. usingsingle-placepredicates Notescopeagain. 'v'x( Wx-+ Cx) -+ Cs 28 3x( Wx& Cx) -+ 'v'x( Wx-+ Cx) 29 . Thisreadingis lessnatural amb 'v'x( Wx-+ Cx) -+ 'v'x( Wx-+ Cx) 23-29

TranslationSchemen lI~ing many a is a hick Ta : : a can do p Cap a is a whale Wa : s: Shamu

23 24 25

-

place predicates

'v'x( Tx-+ Csx) 'v'x( Tx-+ Csx) - 'v'x(Tx-+ Csx) 3x(Tx & - Csx) alt: 'v'x( Tx-+ - Csx) 26 27 3xy Wx & Ty) & Cxy) - + 3z(Tz& Csz) ! 'v'xy ( Wx& Ty) & Cxy) - + 3z(Tz& Csz alt: Scope amb-i: 'v'x3y(Wx -+ Ty & Cxy) -+ 3z(Tz & Csz) amb-ii: 'v'xy ( Wx&Ty) & Cxy) -+ Csy) . scheme with thepreviousb' anslation Thisis theambiguous readingnotexpressible 'v'x3y( Wx-+ (Ty & Cxy -+ 3z(Tz& Csz) 28 'v'x( Wx-+ 3y(Ty & Cxy -+ 3z(Tz & Csz) alt: 29 3xy Wx & Ty) & Cxy) - + 'v'x( Wx-+ 3y(Ty & Cxy amb: 'v'x3y( Wx-+ ( Ty& Cxy -+ 'v'x( Wx-+ 3y(Ty& Cxy

30 31

Agb 3xAxb

166 32 33 34 35 36 amb: 37 amb: 38 39 40 41 alt: 42 alt:

Answersto Chapter3 Exercises

3xAgx VxAbx VxAxb 3xyAxy 3xVyAxy Vy3xAxy Vx3yAxy 3yVxAxy VxyAxy VxAxx 3xAxx Vx- Axx - 3xAxx 3xVy- Axy 3x- 3yAxy

Translation Scheme for 43-46. SaJiy; a saidIi to 'Y fa : a is a person 43 amb: 44 45 46

Vx(Px-+ 3yVz(Pz-+ Sxyz Vxy(Px& Py -+ 3zSxzy Vx(Px-+ 3yz(Pz& Sxyz Vx(Px-+ 3y(Py& - 3zSxzy Vxy(Px& Py-+ - 3zSxzy)

47 48 alt: amb: 49 50 amb: 51 52 amb: 53

3xyz Rx & (Cy & Sxy & ( Dz& Lxz 3x(Fx& Vy(Hy -+ Sxy 3xVy(Fx& (Hy -+ Sxy 3x(Fx& 3y( Hy& Sxy 3x(Fx& Vy( My-+ Sxy 3x( Wx& Vy(Fy& Exy -+ My 3x( Wx& Vy(Exy-+ Fy & My 3x( Wx& Vy(Fy& My -+ - Exy 3xy My & Fy) & Exy) -+ Vx(Sx- + 3y My & Fy) & Exy 3xy My & Fy) & Exy) -+ 3x(Sx& 3y My & Fy) & Exy ' Vwxyz Jw & Txw) & (Oy & Tzy) - + Lxz) [ Tali: a is Ii s tail]

Answers to Chapter 3 Exercises -- alt :

54 alt: 55 56

167

'v'wxyz (Jw & Tx) & Bxw) & Oy & Tz) & Bzy) -+ Lxz) (Bali: a belongs to Ii] 'v'x(3y(Cy & Sxy) -+ Ax) 'v'xy(Cy & Sxy-+ Ax) 'v'uvxy( Bu & Pvu - + (Hx & Pyx - + Mvy & 'v'uvwx(Ou & Evu - + Mw v Bw) & Exw - + Avx & - Mvx Ambiguous. i. The amounteatenby somewhalesis more than the amounteatenby any fish. Translationscheme: a is an amount(of food) Au : a is a fish Fa: : a eatsfi amount(of food) Eafi : a is greaterthan fi Gafi 3x( Wx& 3y Ay & Exy) & 'v'zw( Fz& Aw & Ezw - + Gyw ) ii . The amounteatenby somewhales is more than the amounteatenby all the fishescombined. Addition to translationscheme: the amounteatenby all the fishescombined a:

3xy Wx & Ay) & ( Exy& Gya 3x( Mx& 'v'y( My-+ (GxyH - Gyy ) ( Usingidentity) 3x( Cx & 'v'y(Cy -+ y =x 58 3x x=p 59 alt : 3x( x=p & 'v'y ( y=p - + y=x 60 3xy (Tx& Ty) & x * y) & (Ebx& Eby 'v'x( x~ -+ Exb) & - Ebb 61 'v'x( Dx -+ 3y Ty & Byx) & 'v'z(Tz& Bzx-+ y=z 62 Exerdse 3.3.1 i is an instanceof v ii is an instanceof vi ii is an instanceix

Answers to Chapter3 Exercises

168 iv is an instanceof i iv is an instanceof ill x is an instanceof ii Exerdse 3.3. 2

3x(Gx & - Fx), 'v'x(Gx -+ Hx) ~ 3x(Hx & - Fx) 3x(Gx & - Fx) ( 1) 'v'x(Gx -+ Hx) (2) 3 - + Ha Ga ( ) 4 Ga & - Fa ( ) Ga (5) Ha (6) - Fa (7) Ha& - Fa (8) 3x(Hx & - Fx) (9) 3x(Hx & - Fx) ( 10)

S88 1 2 2

3x(Gx& Fx). '9'x(Fx-+ - Hx) .- 3x- Hx 3x(Gx & Fx) (I ) 'v'x(Fx-+ - Hx) (2) Fa-+ - Ha (3) Ga& Fa (4) Ga (5) Fa (6) - Ha (7) 3x- Hx (8) 3x- Hx (9)

A A 2VE A [for 3E on I ] 4 &E 4 &E 3.6 ~ E 731 1.8 3E (4)

'v'x(Gx-+ - Fx), 'v'x(- Fx-+ - Hx) J- 'v'x(Gx-+ - Hx) 'v'x(Gx-+ - Fx) (1) 2 'v'x(- Fx-+ - Hx) () Ga-+ - Fa (3) - Fa-+ - Ha (4) Ga-+ - Ha (5) 'v'x(Gx-+ - Hx) (6)

I ' VE 2 ' VE 3,4 US S' v1

NN ~~

4 4 4

1 2 1 2 1,2 1,2

<~~ m ~0=-~

S87 1 2 2 4 4 2,4 4 2,4 2,4 1,2

169

Answersto Chapter3 Exercises

S91 1 1 3 3 5 6 1,6 8 8 3,8 1,3,6 1,3,5 1,3 1

'Vx(Gx-+3y( Fy& Hy ~ 'Vx-Fx-+ - 3zGz 'Vx(Gx-+3y(Fy& Hy (I ) Ga-+3y( Fy& Hy) (2) 'Vx-Fx (3) - Fa (4) 3zGz 5 () Ga (6) (7) 3y(Fy& Hy) Fa& Ha (8) Fa (9) - 3zGz (10) - 3zGz (II ) - 3zGz 12 ( ) - 3zGz 13 ( ) 'Vx-Fx-+ - 3zGz ( 14)

S92 1 2 1 2 5 5 2,5 1,2,5 1,2,5 1,2 1,2

'v'x(Gx -+ Hx & Jx), 'v'x(Fxv - Jx - + Gx) I- 'v'x(Fx- + Hx) 'v'x(Gx - + Hx & Jx) (I) 'v'x(Fxv - Jx - + Gx) (2) Ga- + Ha& Ja (3) 4 Fav - Ja-+ Ga ( ) Fa 5 ( ) Fav - Ja (6) Ga (7) Ha& Ja (8) Ha (9) Fa-+ Ha ( 10) 'v'x(Fx-+ Hx) ( II )

593 1 2 3 1 3 1,3 1,3

'v'x(Gx & Kx H Hx), - 3x(Fx& Gx) .- 'v'x- (Fx& Hx) 'v'x(Gx & Kx H Hx) (I ) - 3x(Fx& Gx) (2) Fa & Ha 3 ( ) 4 KaHHa Ga& ( ) Ha 5 ( ) Ga& Ka (6) Ga (7)

"\ t ~

A A 1'v'E 2'v'E A [for-+1] 5 vI

I ' VE 3&E 4,5 BP 6&E

~

A IVE A [for-+1] 3VE A [forRAA] A [for3Eon5] 2,6 -+E A [for3Eon7] 8&E 4,9 RAA(5) 7,103E(8) 5,113E(6) 5,12RAA(5) 13-+1(3)

170

895 1 2

3 4 4 4 1

3&E 7,8 &1 931 2,10RAA(3) 11'v1

A A

A [forRAA] A [for3Eon3] 4&E 4&E I ' VE 6,7 vE 5TC 2'v'E 9,10~ E 8,11RAA(3) 3,123E(4) 3,13RAA(3)

-

NNN ~ .

-

1,4 4

Fa Fa& Oa 3x(Fx& Ox) - (Fa& Ha) 'v'x- (Fx& Hx)

-

(8) (9) ( 10) (II ) (12)

MM ~ -I~~~~~gl~ ~ ~i -. i ~-. Im ~g ~0~ -

3 1,3 1,3 1,2 1,2

Answersto Chapter Exercises

A 2'v'E A [for-+1] A [forRAA] 5 Neg-+ 631 1,7 RAA(5) 3,4,8 SiIDDll 9 -+1(4) 10' v1

--

~

N

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m T ~~

f-s

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o~

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~

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MM

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-

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Fa H Ga Ga 'v'xGx 'v'xFx - + 'v'xGx 'v'xGx

N ~ ~.'..1-_(- m c

-~ ~0~-. m ~I~~

~>i :' ~~~i m ca ~.m ~g~.~-i

-N- mm ~~t ~~ ~

Answers to Chapter

'v'xGx - + 'v'xFx

'v'xFXH

Answersto Chapter. 3 Exercises

172 5

(

5

(

4

(

4

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1

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1

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4,6 RAA (5) 7 vI 3,8 ~ E 9 Comm 10VI 2,11RAA (4)

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'v'x(Dx-+Fx) r 'v'z(Dz-+ ('v'y(Fy-+ Gy) -+Gz 'v'x(Dx-+Fx) (I ) Da (2)

A A [for-+1] A [forRAA] A 431 3,5 RAA(4) 6VI 2,7 RAA(3) A [for3Eon8] I ' VE 9,10BT 8,113E(9) A [forRAA] A [for3Eon13] 10,14BP 13,153E( 14) 12,16RAA(13) A [forRAA] 1831 17,19RAA(18) 20VI 21-+1(2) 22v-+

A A

173

Answersto Chapter3 Exercises

(3) (4) (5) (6) (7) (8) (9) ( 10)

SIll 1 2 3 3 5 5 5 3 2

11 12 1 1 1 1,12 1,12 1,12 1,11 1 1 1

'v'xFx I- - 3xOx+-+- (3x(Fx& Ox) & 'v'y(Oy - + Fy 'v'xFx - 3xOx 3x(Fx& Ox) & 'v'y(Oy -+ Fy) 3x(Fx& Ox) Fa& Oa Oa 3xOx 3xOx - (3x(Fx& Ox) & 'v'y(Oy -+ Fy - 3xOx-+ - (3x(Fx& Ox) & 'v'y(Oy -+ Fy 3xOx Oa Fa Oa- + Fa 'v'x(Ox -+ Fx) Fa& Oa 3x(Fx& Ox) 3x(Fx& Ox) & 'v'x(Ox -+ Fx) 3x(Fx & Ox) & 'v'x(Ox -+ Fx) 3xOx-+ 3x(Fx& Ox) & 'v'x(Ox -+ Fx) - (3x(Fx& Ox) & 'v'x(Ox - + Fx -+ - 3xOx - 3xOx+-+- (3x(Fx& Ox) & 'v'y(Oy - + Fy

A A [for-+1] A [forRAA] 3&E A [for3Eon4] 5&E 631 4,7 3E(5) 2,8 RAA(3) 9 -+1(2) A [for-+1] A [for3Eon11] 1'v'E 13TC 14VI 12,13&1 1631 15,17&1 11,183E(12) 19-+1(11) 20Trans 10,21HI

Sl 12 1 2 2

'v'x(3yFyx-+ 'v'zFxz) ~ 'v'yx( Fyx- + Fxy ) 'v'x( 3yFyx-+ 'v'zFxz) ( 1) Fab (2) (3) 3yFyb

A A [for-+1] 231

'v'y( Fy - + Gy) Da - + Fa Fa - + Ga Da - + Ga Ga 'v'y( Fy - + Gy) - + Ga Da - + ('v'y( Fy - + Gy) - + Ga) 'v'z(Dz - + ('v'y( Fy - + Gy) - + Gz

-

--

3 1 3 1,3 1,2,3 1,2 1 1

A IVE 3 ' v' E 4,5 HS 2,6 -+E 7 -+1(3) 8 -+1(2) 9 VI

(

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Chapter

175

Answersto Chapter3 Exercises

S127*

I 2 3 3 3 6 3 3,6 3 3,6 2,3 1,2 * S130 1 2 3 4 1 6 2 2 2,6 4 2,4,6 2,4,6 2,4,6 2,4,6 1,2,4 1,2,3 1,2

3x Fx & 'v'y( Fy-+ y=x & Ox), - Oa'- - Fa 3x Fx & 'v'y(Fy-+ y=x & Ox) ( 1) - Oa 2 ( ) 3 ( ) ( fb & 'v'y(Fy-+ y=b & Ob Fb& 'v'y(Fy-+ y=b) (4) 'v'y(Fy-+ y=b) (5) Fa (6) Fa-+ a=b (7) a=b (8) Ob ,(9) Oa ( 10) - Fa ( 11) - Fa 12 ( )

(I)

' v' x3y Gyx

(2 )

' v' xy ( Gxy

(3 )

3y ' v' x ( x : #y - + Gyx )

(4 )

' v' x ( x ~

(5 )

3yGya

(6 )

Gba

(7 )

' v' y ( Gay

- + - Gyx )

- + Gax )

- + - Gya )

- + - Gba

Gab

SVE 6,7 -+E 3&E 8,9 =E 2,10RAA(6) I ,ll 3E( 3)

- + - Gyx ) r - 3y ' v' x ( x : #y - + Gyx )

' v' x3y Gyx , ' v' xy ( Gxy

(8)

A A A 3 &E 4 &E A

(9 )

- Gab

( 10 )

~

( II )

b=a

( 12 )

Gaa

( 13 )

- Gaa

( 14 )

-

3y ' v' x ( x : #y - + Gyx )

( IS )

-

3y ' v' x ( x : #y - + Gyx )

( 16 )

-

3y ' v' x ( x : #y - + Gyx )

( 17 )

-

3y ' v' x ( x : #y - + Gyx )

- + Gab

A A

A [for RAA] A [for 3E] I' VE A [for 3E] 2 ' VE 7 ' VE 6,8 ~ E 4 ' VE 9,10MTr 6,11=E 9,11=E 12,13RAA (3) 5,143E (6) 3,153E (4) 3,16RAA (3)

176

Answers to Chapter 3 Exercises

Exercise 3.4.1 .VxPx -1.- 3x - Px

~~ -l ~.m

5150 (a) 1 2 3 3 2 2

(b) 1 2 3 2 3 1

A A A 331 2,4 RAA (3) SVI 1,6 RAA (2)

A A A

2' VE 3,4 RAA(2) 1,5 3E(3)

'v'x(Px-+Q) -it- 3xPx-+ Q

(a) I 2 I 4 1,4 1,2 1

'v'x(Px-+Q) t- 3xPx-+ Q 'v'x(Px-+Q) 3xPx Pa-+ Q Pa Q Q 3xPx-+Q

3,4 -+E 2,5 3E(4) 6 -+1(2)

3xPx-+Q ~ 'v'x(Px-+Q) 3xPx-+Q (I ) Pa (2) 3xPx (3)

A A 231

(b) 1 2 2

---

SISS

A A I ' VE A

177

Answers to Chapter 3 Exercises

7 7 7

~ >; 0~. ~-~ 0~

5156 1 2 2 2 2

( 1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

Q Pa- + Q 'v'x(Px-+ Q)

1.3 -+E 4 -+1(2) 5 VI

-

(4) (5) (6)

1,2 1 1

'v'xPx

A A

2VE 3 vI 4 VI 5 -+1(2) A 7 ' v' E 8 vi 9 ' v1

10-+1(7) 1,6,11SimDll

A A A 3 &E 3 &E 431 531 6.7 &1 2.8 3E (3) 1.9 3E (2)

(b) 1 1 1 4

3xPx& 3xQxI- 3xy(Px& Qy) 3xPx& 3xQx 3xPx 3xQx Pa

A l &E l &E A

---

(a) 1 2 3 3 3 3 3 3 2 1

3xy(Px& Qy) -i~ 3xPx& 3xQx 3xy(Px& Qy) ~ 3xPx& 3xQx (I ) 3xy(Px& Qy) 2 () 3y(pa& Qy) Pa& Qb 3 () Pa (4) (5) Qb 3xPx (6) 3xQx (7) 3xPx& 3xQx 8 () 3xPx& 3xQx (9) 3xPx& 3xQx ( 10)

5157

178

Answers to Chapter 3 Exercises

5 4,5 4,5 4,5 1,4 1

(5) (6) (7) (8) (9) ( 10)

SI60

P-+3xQx~J- 3x(P-+ Qx)

(a) 1 2 3 3 2 2 2 1,2 2 10 10 1,2 1

P-+3xQxJ- 3x(P-+Qx) P-+3xQx ( 1) - 3x(P-+ Qx) 2 () P -+ Qa (3) 3x(P-+ Qx) (4) - (p -+ Qa) (5) - Qa 6 P& () P (7) (8) 3xQx - Qa (9) ( 10) Qa ( 11) ( 12) ( 13)

(b) 1 2 3 2,3 2,3 3 1

3x(P-+ Qx) ~ P -+ 3xQx 3x(P-+ Qx) P P -+ Qa

---

mmm ~~t 666

Qb Pa& Qb 3y(pa& Qy) 3xy(Px& Qy) 3xy(Px& Qy) 3xy(Px& Qy)

Qa 3xQx p ~ 3xQx p ~ 3xQx

A 4,5 &1 631 731 3,8 3E(5) 2,9 3E(4)

A A

A 331 2,4 RAA (3) 5 Neg-+ 6 &E

1,7 -+E 6&E A 9,10RAA(2) 8,113E( 10) 2,12RAA(2)

A A A 2,3 - +E 431 S-+1(2) 1,6 3E (3)

Exercise 3.4. 2

T40 1 2

.- 'v'x( Fx - + Gx) - + ('v'xFx - + 'v'xGx)

(1) (2)

'v'x(Fx-+Ox) VxFx

A [for - +1] A [for - +1]

179

Answers to Chapter 3 Exercises

I 2 1,2 1,2 1

(3) (4) (5) (6) (7) (8)

T42 I 2 2 2 2 2 2 2 2 10 2,10 10 1

r 3x(Fxv Ox) H 3xFxv 3xOx 3x(Fxv Ox) - (3xFxv 3xOx) - 3xFx& - 3xOx - 3xFx - 3xOx 'v'x- Fx 'v'x- Ox - Fa - Oa Fav Oa Oa 3xFxv 3xOx 3xFxv 3xOx 3x(Fxv Ox) -+ (3xFxv 3xOx) 3xFxv 3xGx 3xFx Fa Fav Ga 3x(Fxv Gx) 3x(Fxv Gx) 3xFx-+ 3x(Fxv Gx)

---

-

Fa - + Ga Fa Ga VxGx 'V'xFx - + 'V'xGx 'v'x( Fx - + Gx) - + ('v'xFx - + 'v'xGx)

15 16 17 17 17 16 22 23 23 23 22 15

(22) (23) (24) (25) (26) (27) (28)

3xGx Ga Fav Ga 3x(Fxv Gx) 3x(Fxv Gx) 3xGx-+ 3x(Fxv Gx) 3x(Fxv Gx)

i 've 2VE 3,4 -+E S' v1 6 -+1(2) 7 -+1( I )

A [for-+1] A [forRAA] 20M 3 &E 3 &E 4QE 5QE 6VE 7VE A [for 3E on I ] 8,10vE 9,11RAA (2) 1,123E( 10) 13-+1( I ) A [for - +1] A A [for 3E on 16] 17v I 1831 16,193E ( 17) 20-+1( 16) A A [for 3E on 22]

23vI 2431

22,253E(23) 26-+1(22) Dll 15,21,27Sirn

180

Answers to Chapter 3 Exercises

(29) (30) T46 1 2 2 2 2 2 2 1,2 9 2 9 1,2 1

T59 1 2 2 2 5 1,5 1,5 1 9 1,9 1,2 1,2 1,2 1,2 1,2 1,2 1

3xFxv 3xOx-+ 3x(Fxv Ox) 3x(Fxv Ox) H 3xFxv 3xOx

J- (3xFx-+ 3xGx) -+ 3x(Fx-+ Gx) 3xFx-+ 3xGx ( 1) - 3x(Fx-+ Gx) (2) 'v'x- (Fx- + Gx) (3) (4) (5) (6) (7) (8) (9) ( 10) ( 11) ( 12) ( 13) ( 14)

( 1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) ( 13) ( 14) ( 15) (16) (17)

- 3x(FxH P) 'v'X- (FxH P) - (FaH P) P 'v'xFx Fa P -+ Fa Fa-+ P FaHP - (Fa-+ P) Fa& - P FaP 'v'xFx P 3x(FxH P)

28 -+1( 15) 14,29HI

A [for - +1] A [for RAA]

2QE 3'v'E 4 Neg~ 5&E 631 1,7 -+E A [for3Eon8] 5&E 9,10RAA(2) 8,113E(9) 2,12RAA(2) 13-+1( I )

A A

2QE 3 ' v' E A 1,5 SP 6 ' v' E 7 -+1(5) A 8,9 HI 4,10RAA (9) II Neg-+ 12&E 12&E 13VI 1, 15BP 14,16RAA (2)

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181

Answers to Chapter 4 Rxerci ~ ~

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Answers to Chapter 4 Exercises

- (Fa& Ga& fb & Gb& Fc& Gc) - (Fa& - Ga) - Fa & - (Ga& Gb & (Fb& - (Ga& Gb )

- Fa & - (Oa&Ob&Gc & (Fb&--(Oa& Ob&Gc & (Fc& - (Oa&Gb&Gc ) - (OaH .Ha& - Fa) - (Oa& ObH (Ha& - Fa) v ( Hb& - fOb - (Oa& Ob& GcH (Ha& - Fa) v ( Hb& - fOb ) v (Hc & - Fc

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Exerdse 4.1.2

Exerdse 4.2

U:{a,b} F:{a}

G:{ }

Fa& fb -+ Ga& Gb.- (Fa-+ Ga) & (fb -+ Gb) U: {a, b} F:{a}

G:{b}

Fav fb -+ Gav Gb.- (Fa-+ Ga) & (fb -+ Gb) Samemodelasii (Fav fb ) & (Gav Gb) .- (Fa& Ga) v (fb & Gb) Samemodelasi (Fav Ga) v (fb v Gb) .- (Fa& fb ) v (Ga& Gb)

183

Answers to Chapter 4 Exercises v

Same model as i

(Fa-+Ga) v (Fb-+Gb) r (Fav fb ) -+ (Gav Gb) U:{a,b} F:{a,b} G:{a} (Fa-+ Ga) v (fb - + Gb) I- (Fa& fb ) -+ (Ga& Gb) Samemodelasi Fa& FbH Ga& Gbl- (FaH Ga) & (fbH Gb) Samemodelasii Fav fb H Gav GbI- (FaH Ga) & (fb H Gb) U:{a,b} F:{a}

P is FALSE

(Fa& fb ) H PI - (FaH P) & ( fbH P) U:{a,b} F:{a}

Pis TRUE

(Fav fb ) H PI- (FaH P) & (fbH P) Samemodelasix (FaH P) v (fbH P) I- (Fav fb ) H P Samemodelasx (FaH P) v (fbH P) I- (Fa& fb ) H P xiii

U: {a}

F:{ }

G:{ }

H: {a}

Fa-+ Ga, Ga-+ HaI- Ha-+ Fa F:{ } H: { } G:{ } Fa-+ Ha, Ha-+ GaI- Fa& Ga U: {a}

U: {a, b} F:{a}

G:{a,b} H: {b}

Fa v fObH Oa & Ob, - Fa - + Ha) & ( Fb- + Hb ~ Ha v Hb - +- Oa v - Gb

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185

Answersto Chapter4 Exercises vii

U:{a, b} T:{(a, b)}

- (Taa& Tab) & - (Tba& Thb) ..- - (Taav Tab) & - (Thav Thb) viii

U:{a.b,c}

F:{(a.b), (h,c),(a.a),(c,a)}

ix

U:{a.b,c}

F:{(a.b), (h,c), (c,a)}

G:{(a,b)}

x

U:{a.b}

F:{(a,a)}

G:{(a,b), (b,a) }

Exerdse4.4 i

U: {m, n}

a: m

b: m

c: n

d: n

m= m, n= n ~ m= n

F: {a, b} U: {a, b, c} [ Fa & Fa) & ~ ) v Fa & Fb) & a$b) v Fa & Fc) & a#C)] v [ Fb & Fa) & ~ ) v Fb & Fb) & ~ ) v Fb & Fc) & ~ )] v [ Fc & Fa) & ~ ) v Fc & Fb) & ~ ) v Fc & Fc) & t * )] I- Fa& Fb& Fc

F: {(a, b), (a. c) } U: {a, b,c } Faa H ~ ) & ( FabH ~ ) & ( FacH ~ v Fha H b:I:a) & ( FbbH b#b) ( FbcH b#c)) v ( FcaH c:#:a) &: ( FcbH ~ ) ( FccH t * )) t& & & & & & & &

( Faa& Faa - + a=a) & ( Faa& Fab - + a=b) & ( Faa& Fac - + a=c) ( Fab& Faa - + b=a) & ( Fab& Fab - + b=b) & ( Fab& Fac - + b=c) ( Fac& Faa - + c=a) & ( Fac& Fab - + c=b) & ( Fac& Fac - + c=c) ( Fha& Fha - + a=a) & ( Fha& Fbb - + a=b) & ( Fha& Fbc - + a=c) ( Fbb& Fha - + b=a) & ( Fbb& Fbb - + b=b) & ( Fbb& Fbc - + b=c) ( Fbc& Fha - + c=a) & ( Fbc& Fbb - + c=b) & ( Fbc& Fbc - + c=c) ( Fca& Fca - + a=a) & ( Fca& Fcb - + a=b) & ( Fca&Fcc - + a=c) ( Fcb& Fca - + b=a) & ( Fcb& Fcb - + b=b) & ( Fbb& Fbc - + b=c) ( Fcc& Fca - + c=a) & ( Fcc& Fcb - + c=b) & ( Fbc& Fbc - + c=c)

Answers to Chapter 4 Exercises

186

Exerdse 4.5.2

'v'xyz(Fxy& Fyz-+ Fxz), 'v'x3yFxy r 3xFxx U: N. F: {(m,o) : m
'v'x3y'v'z(Fxy& (Fyz- + Fxz ~ 3xFxx U: N. F: {(m,o) : m
Vx3yFxy, Vxyz(Fxy& Fyz- + Fxz), Vx- Fu r Vxy(Gx & - Gy -+ Fxyv Fyx) U: N. F: ({m,n) : n is anevennumbergreaterthanm} G: (m : m is even} I st premise :T 'Foreachnumberthereis anevennumberthatis .' ) greater ( :T 2ndpremise ' z is anevennumbergreaterthan ( If Y is anevennumbergreaterthanx, and ' x. than number z is an even ) greater y, then :T 3rdpremise ' ' ( No numberis anevennumbergreaterthanitself. ) : F. Conclusion ' ( If x is evenandy is odd, theneitherx is anevennumbergreaterthan y or y is anevennumbergreaterthanx.' )

Answersto Chapter4 Exercises Vx3yz( Fxy & Fzx), Vxyz( Fxy & Fyz - + Fxz) ~ 3xy( Fxy & Pyx) IJ~NF: {{m, n) : either m and n are even and m n , or m is odd and n is even. }

'v'x- Fxx, 'v'x3y'v'z( Fxy &; ( Fyz - + Fxz J- 'v'xyz( Fxy &; Fyz - + Fxz) U: N F: {(m, n) : 3k( k>O&; (n=2k(m+ l ) - 1 or n=2k(m+ l ) } 'v'xyz(Oxy &; Oyz - + Oxz), 'v'xy(Oxy - + - Gyx), 'v'x3y Gyx, 'v'x(x~ - + Ou ) J- 3y'v'x(x* y - + Oyx) U: N G: {{m, n) : m > n } a: zero

187

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