Metatheoryfor truth-functional logic

Tim ButtonUniversity of Cambridge

In 2005, P.D. Magnus released the rst version of his open-source logic text-book, forallx. In 2012, Tim Button adapted this for a rst-year course at theUniversity of Cambridge. Metatheory takes up exactly where the Cambridgeimplementation of forallx leaves o.

Tim would, once again, like to thank Magnus for making forallx freely avail-able. With this in mind, Metatheory is released under the same open-sourcelicense as forallx.

Tim would also like to thank Brian King and Mat Simpson for comments andcorrections. This version (released July 2014) corrects minor typos from theprevious version.

20132014 by Tim Button. Some rights reserved. This book is oered undera Creative Commons license (Attribution-ShareAlike 3.0).

You are free to copy this book, to distribute it, to display it, and to make derivative works,under the following conditions: (a) Attribution. You must give the original author credit.(b) Share Alike. If you alter, transform, or build upon this work, you may distribute theresulting work only under a license identical to this one. For any reuse or distribution,you must make clear to others the license terms of this work. Any of these conditions can bewaived if you get permission from the copyright holder. Your fair use and other rights arein no way aected by the above. This is a human-readable summary of the full license,which is available on-line at http://creativecommons.org/licenses/by-sa/3.0/

Typesetting was carried out entirely in LATEX2". The style for typesettingproofs employs forallx.sty by P.D. Magnus, University at Albany, State Uni-versity of New York; Magnuss le is based upon tch.sty (v0.4) by PeterSelinger, University of Ottawa.

Contents

1 Induction 31.1 Stating and justifying the Principle of Induction . . . . . . . . 31.2 Arithmetical proofs involving induction . . . . . . . . . . . . . 41.3 Strong Induction . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Induction on length . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Unique readability . . . . . . . . . . . . . . . . . . . . . . . . . 9Practice exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Substitution 122.1 Two crucial lemmas about substitution . . . . . . . . . . . . . . 122.2 Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Practice exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Normal forms 203.1 Disjunctive Normal Form . . . . . . . . . . . . . . . . . . . . . 203.2 Proof of DNF Theorem via truth tables . . . . . . . . . . . . . 213.3 Proof of DNF Theorem via substitution . . . . . . . . . . . . . 223.4 Cleaning up DNF sentences . . . . . . . . . . . . . . . . . . . . 273.5 Conjunctive Normal Form . . . . . . . . . . . . . . . . . . . . . 27Practice exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Expressive adequacy 304.1 The expressive adequacy of TFL . . . . . . . . . . . . . . . . . 304.2 Individually expressively adequate connectives . . . . . . . . . . 324.3 Failures of expressive adequacy . . . . . . . . . . . . . . . . . . 33Practice exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5 Soundness 375.1 Soundness dened, and setting up the proof . . . . . . . . . . . 375.2 Checking each rule . . . . . . . . . . . . . . . . . . . . . . . . . 38Practice exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 Completeness 436.1 Completeness dened, and setting up the proof . . . . . . . . . 436.2 An algorithmic approach . . . . . . . . . . . . . . . . . . . . . . 446.3 A more abstract approach: polarising sentences . . . . . . . . . 546.4 The superior generality of the abstract proof . . . . . . . . . . 58Practice exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

iii

Introduction

This book is an introduction to the metatheory of truth-functional logic, alsoknown as the propositional calculus.

This book does not itself contain an account of truth-functional logic, butinstead assumes a prior understanding. More specically, this book assumesa thorough understanding of the syntax, semantics and proof-system for TFL(truth-functional logic) that is presented in forallx :Cambridge 2013-4 (forbrevity, I henceforth refer to that book as fx C). fx C can be obtained, freeof charge, at www.nottub.com/forallx.shtml. There is nothing very surprisingabout the syntax or semantics of TFL, and the proof-system is a fairly standardFitch-style system.

This book does not, though, presuppose any prior understanding of(meta)mathematics. Chapter 1 therefore begins by explaining induction fromscratch, rst with mathematical examples, and then with metatheoretical ex-amples. Each of the remaining chapters presupposes an understanding of thematerial in Chapter 1.

One idiosyncrasy of this book is worth mentioning: it uses no set-theoreticnotation. Where I need to talk collectively, I do just that, using plural-locutions.Although this is unusual, I think Metatheory benets from being free of un-necessary epsilons and curly brackets. So, where some books say:

fA1;A2 ; : : : ;Ang is inconsistentto indicate that no valuation makes all of A1;A2; : : : ;An true, I say:

A1;A2; : : : ;An are jointly inconsistentI use upper-case Greek letters for plural terms, so I might equally say:

are jointly inconsistentInconsistency is therefore a (collective) property of sentences, rather than aproperty of sets (of sentences). Semantic entailment also comes out as a (col-lective) property of sentences. To indicate that every valuation that makes allof true also makes C true, we write:

C

where, of course, are some sentences. If there is a valuation that makes allof true but C false, we write:

2 C

1

Contents 2

The corresponding notions from proof-theory are also properties of sentences.Thus we write:

` Cto indicate that there is a TFL-proof which starts with assumptions among and ends with C on no further assumptions. If there is no such proof, we write:

0 C

A special case of this notation is used when there is a TFL-proof which startswith assumptions among and ends with ? (i.e. with an absurdity). In thissituation, we say that ` ?, or that are jointly contrary. (I should note that? is not a primitive 0-place connective of TFL. Rather, ? shows up only inTFLs proof-theory. For more details, see fx C26.7)

Induction 1

This book proves a bunch of results about formal logical systems. Specically,it proves results about truth-functional logic(s): that is, logics where all ofthe connectives are truth functional. More specically still, it proves resultsabout TFL, a particular truth-functional logic whose syntax, semantics andproof-system are detailed in fx C.

When we prove results about formal logical systems, we inevitably have torely upon some background logical reasoning. This reasoning need not, itself,be completely formalised. Nonetheless, the logical principles that we shallinvoke informally like modus ponens, or universal generalisation should allfeel familiar from their formalised counterparts.

However, when proving results about formal logical systems, we also relyupon mathematical reasoning. And there is an important mathematical prin-ciple which we shall use frequently: induction.1 Indeed, almost every result inthis book will invoke induction, in some form or other. So, the task of this rstchapter is to familiarise ourselves with induction.

1.1 Stating and justifying the Principle of InductionI mentioned that induction is a mathematical principle. In its plain vanillaform, it governs the behaviour of the natural numbers. These are all the nite,whole, non-negative numbers: 0, 1, 2, 3, and so on. (In what follows, wheneverI say number, I always mean natural number.) Here is the principle:

Principle of Induction. Let be any property such that both: 0 has ; and for every natural number n: if n has , then n+ 1 has .

Then every natural number has .

Here is an intuitive way to think about the Principle of Induction. Imagine thatwe had innitely many dominoes one for every number balanced on theirends. Whenever we prove n has , we get to knock down the correspondingdomino. Our aim is to knock down all of them. But there are too many of themfor us to hope to knock them down one at a time. Fortunately, dominoes have anice property: if we can arrange them in a line , so that when a domino falls itknocks over the next in line, then we can knock all the dominoes down just by

1NB: mathematical induction must be sharply distinguished from the kind of empiricalinduction that David Hume discussed.

3

1. Induction 4

knocking the rst one down. To knock down the rst domino corresponds withproving that 0 has . And to arrange the dominoes in this way correspondswith proving the conditional: if n has then n+ 1 has .

Of course, this is just an analogy, and a spatial one at that. Why, then,should we believe the Principle of Induction? Here is an argument in its favour.Suppose you have proved both that 0 has and also that, for every n, if n has then n+ 1 has . Now, pick any number, m. Given what you have proved,you would be justied in writing down all of the following:

0 has if 0 has , then 1 has if 1 has , then 2 has ...if m 1 has , then m has

It would now follow, by m-many applications of modus ponens, that m has .And since m was arbitrary, this point generalises: every number has .

I doubt this line of argument will convince the determined sceptic.2 And,in the end, the question Why believe in the Principle of Induction? is a deepquestion in the philosophy of mathematics. That question is interesting, but itis a question for another day. For the purposes of this book, we shall take thePrinciple of Induction for granted.

1.2 Arithmetical proofs involving inductionTo invoke induction in a proof, we must rst identify some property, , thatconcerns us. We call this the induction property. We then do two things:

Prove the base case: we need to show that 0 has . Often, this is a fairlystraightforward task.

Prove the induction case: we need to show that, for any n, if n has , sodoes n + 1. This is normally the harder step. We typically proceed bychoosing some arbitrary n, and assuming for induction that n has .If, on just this basis, we can show that n+1 has , then we have provedthe conditional: if n has , then n+ 1 has . And since n was arbitrary,this holds for all n.

To illustrate the strategy, I shall oer a few simple examples of the use ofinduction, in establishing some arithmetical results. (Just to be clear: I amnot going to make any use of the arithmetical results themselves; I just wantto show how induction is used in proofs.) So, for example, suppose we want toprove a claim that should be very obvious:

Proposition 1.1. Every number is either even or odd. That is: for everynumber n, either there is a number k such that n = 2k (i.e. n is even), or thereis a number k such that n = 2k + 1 (i.e. n is odd).

I shall prove this Proposition by induction. The induction property will be xis either odd or even. It suces to prove the base case and the induction case;then all we need to complete the proof is to appeal to induction. Here goes:

2It is worth spending some time asking yourself: Why wont this convince the sceptic?

1. Induction 5

Proof. Evidently, 0 = 2 0. So 0 has the induction property.Now pick an arbitrary number, n, and suppose (for induction) that n has

the induction property. There are two cases to consider Suppose n is even, i.e. n = 2k, for some number k. Then n+1 = 2k+1,i.e. n+ 1 is odd. Suppose n is odd, i.e. n = 2k + 1, for some number k. Then n + 1 =2k + 2 = 2(k + 1), i.e. n+ 1 is even.

Either way, n+1 has the induction property. So if n has the induction property,then so does n+ 1.

This completes the proof of Proposition 1.1 by induction. In addition to the use of induction, a couple of things should strike you aboutthis proof. In particular, you should note that it is not a formal proof in anyparticular deductive system. It is a bit of informal, mathematical, reasoning,written in mathematised English. That said, it is perfectly rigorous. (Thesymbol at the end of the proof just indicates that the proof is complete;other texts use other symbols, e.g. QED or .)

To cement ideas, let me oer two more examples of proofs by induction:

Proposition 1.2. 20 + 21 + 22 + : : :+ 2n = 2n+1 1, for any number n.Proof. Evidently 20 = 1 = 20+1 1. So 0 has the induction property.

Now, pick any arbitrary number, k, and suppose (for induction) that20 + 21 + 22 + : : :+ 2k = 2k+1 1

Then we have:20 + 21 + 22 + : : :+ 2k + 2k+1 = 2k+1 1 + 2k+1

= (2k+1 2) 1= 2k+2 1

So the induction property holds for k + 1 as well.

Proposition 1.3. 0 + 1 + 2 + : : :+ n = n2+n2 , for any number n.

Proof. Evidently 0 = 02+02 . So 0 has the induction property.Now, pick any arbitrary number, k, and suppose (for induction) that

0 + 1 + 2 + : : :+ k =k2 + k

2

Then we have:

0 + 1 + 2 + : : :+ k + (k + 1) =k2 + k

2+ (k + 1)

=k2 + k + 2(k + 1)

2

=(k2 + 2k + 1) + (k + 1)

2

=(k + 1)2 + (k + 1)

2

1. Induction 6

So the induction property holds for k + 1 as well.

We could keep going with examples from arithmetic, but these illustrate theuse of induction, and give some sense of how powerful it is.

1.3 Strong InductionNow let me introduce a second inductive principle:

Principle of Strong Induction. Let be any property. Suppose that, forevery number n, if all the numbers less than n have , then n itself has .Then every number has .

Why should we believe this? In fact, it follows directly from the original Prin-ciple of Induction, as I shall now show:3

Proof. Suppose that, for every number n, if all the numbers less than n have, then n itself has . We aim to show that this entails that every number has (thereby vindicating the Principle of Strong Induction).

To do this, we shall dene a new property, , as follows:every number less than x has

We shall now prove by (ordinary) induction that every number has .Base case: 0 has , vacuously, since 0 is the smallest number.Induction case: Suppose, for (ordinary) induction, that n has . That isto say that every number less than n has . So, by the supposition atthe start of this proof, n has . It follows that every number less thann+ 1 has , i.e. that n+ 1 has .

It follows, by (ordinary) induction, that every number has . And since everynumber has , and for every number there is a larger number, every numberhas . Hence the use of Strong Induction is vindicated.

It is worth illustrating the power of the Principle of Strong Induction, with anarithmetical example:4

Proposition 1.4. Every number greater than 1 has a prime factorisation (i.e.it can be expressed as the product of one or more prime numbers).

Proof. Fix n, and suppose for (strong) induction that every number less thann has the induction property (i.e. the property that if x > 1 then x has a primefactorisation). There are now three (exhaustive) cases to consider:

Case 1: n 1. Then n obviously has the induction property.Case 2: n is prime. Then n obviously has the induction property.

3And it entails the original Principle too, though I shall not show this.4This is one half of the Fundamental Theorem of Arithmetic; the other half is that prime

factorisations are unique.

1. Induction 7

Case 3: n > 1 and n is not prime. Then n = ab with both n > a > 1and n > b > 1. By assumption, a and b have prime factorisations,i.e. a = p1 : : : pj and b = q1 : : : qk, for some prime numbersp1; : : : ; pj ; q1; : : : ; qk. So now n = p1 : : : pj q1 : : : qk, i.e. n hasa prime factorisation.

So, in every case, n has the induction property on the assumption that everysmaller number has the induction property. Hence, by strong induction, everynumber has the induction property.

Note that we needed to use strong induction here, because in the case wheren = ab, we know nothing about a and b except that n > a > 1 and n > b > 1.

1.4 Induction on lengthI now want to move on from arithmetical propositions to claims about TFL.By the end of this book, we will have proved some rather sophisticated results.But we need to start at the shallow end. In particular, we shall start by provinga syntactic result, which should already be familiar:

Proposition 1.5. In any TFL sentence, there are exactly as many instancesof ( as instances of ).

When I mention TFL sentences here I have in mind the the objects thatwere dened recursively in fx C6. In particular, then, I shall not allow us torely upon the convenient bracketing conventions that we introduced later infx C; so, we cannot, for example, allow for arbitrarily long conjunctions, andwe cannot drop the outermost brackets of a sentence. To save space, in whatfollows I shall speak of sentences rather than TFL sentences as given by therecursive denition of fx C6.

Now, we want to prove Proposition 1.5 by induction. But there is an im-mediate obstacle in our path. The property mentioned in Proposition 1.5 isx has as many instances of ( as of ), and that is a property of sentences,rather than a property of numbers. So, if we are to prove this Proposition usinginduction, we need a way to associate sentences with numbers.

The easiest way to associate sentences with numbers is just by consideringtheir length. More precisely, by the length of a sentence, we shall mean thenumber of instances truth-functional connectives (i.e. :, ^, _, ! and $)that the sentence contains.5 Thus, for example:

B has length 0 A234 has length 0 :(B _ C4) has length 2 ::B has length 2 :::B has length 3

Having dened the notion of the length of a sentence, we could now proveProposition 1.5 by (strong) induction on numbers, considered as lengths of

5Many books refer to this as the complexity of a sentence.

1. Induction 8

sentences. But things will go easier for us if we introduce yet another inductiveprinciple that relates more directly to TFL:

Principle of Strong Induction on Length. Let be any property. Supposethat, for any sentence, A , if every shorter sentence has , then A has . Thenevery sentence has .

This principle follows directly from (ordinary) strong induction:Proof. Suppose that, for any sentence, A , if every shorter sentence has , thenA has . We aim to show that this entails that every sentence has (therebyvindicating the Principle of Strong Induction on Length).

To do this, we shall dene a new property of the numbers, , thus:every sentence of length x has

Pick any arbitrary number, n, and suppose for (strong) induction that everynumber less than n has . That is, every sentence of length < n has . So,by supposition, every sentence of length n has . So n has . It follows, bystrong induction, that every number has .

Since every sentence has some number as its length, it now follows thatevery sentence has , as required. So, armed with the notion of the length of a sentence, and the Principle ofStrong Induction on Length, we can now prove Proposition 1.5:Proof of Proposition 1.5. Let A be any sentence, and suppose for (strong) in-duction (on length) that every shorter sentence has exactly as many instancesof ( as of ). We shall show that A has exactly as many instances of ( as of), by considering all the dierent forms that A might have. (These six casesare given to us by the recursive denition of the notion of a sentence; for areminder of the denition, see fx C6.)

Case 1: A is atomic. Then A contains no brackets, so it has exactly asmany instances of ( as of ).Case 2: A is :B, for some sentence B. Clearly, B is shorter than A . Bysupposition, then, B contains exactly as many instances of ( as of ).Hence A does too.Case 3: A is (B ^ C ), for some sentences B and C . Clearly, both Band C are shorter than A . By supposition, then, B contains exactly asmany instances of ( as of ), say i; and C contains exactly as manyinstances of ( as of ), say j. Hence A contains i + j + 1 instances of(, and i+ j + 1 instances of ).Case 4: A is (B _ C ), for some sentences B and C . Exactly similar tocase 3.Case 5: A is (B ! C ), for some sentences B and C . Exactly similar tocase 3.Case 6: A is (B $ C ), for some sentences B and C . Exactly similar tocase 3.

There are no other forms that A could take. Hence, for any sentence, A , ifevery shorter sentence has exactly as many instances of ( as of ), then sodoes A . The result now follows by strong induction on length.

1. Induction 9

Now, I hope that it was obvious that the Proposition was true, before we setabout proving it. The purpose of this proof was not, then, to convince youof its truth. Rather, I gave you this proof primarily in order to illustrate thePrinciple of Strong Induction on Length. Moreover, this proof exhibits threefeatures that are worth mentioning explicitly.

First, even though the result was obvious, it took quite a lot of eort to proveit. This is fairly typical of metalogical results: showing them with sucientrigour can be hard!

Second, we teamed an inductive principle with a recursive denition. Thisis fairly typical of metalogical results. Indeed, it is hardly surprising thatinduction and recursion should be best of friends. Recursive denitions tell youhow to build objects up, stage-by-stage; proofs by (strong) induction involveshowing that something holds at later stages if it has held at every stage so far.

Third, when we broke the proof down into cases, cases 36 were a bit rep-etitious. Indeed, all four of our two-place connectives have the same syntacticbehaviour, and that is all we cared about during the course of the proof. Sowe could simply have rolled cases 36 together in something like the following:Case 3: A is (B C C ), for some two-place connective C and sentences B and

C . Clearly, both B and C are shorter than A . By supposition, then, Bcontains exactly as many instances of ( as of ), say i; and C containsexactly as many instances of ( as of ), say j. Hence A contains i+j+1instances of (, and i+ j + 1 instances of ).

saving ourselves some time and eort.6 Generally, if we need to do a proofby cases and we often shall we should pause before ploughing in, to thinkabout exactly what the relevant cases are.

1.5 Unique readabilityI shall close this chapter by proving two more results about the syntax of TFL.

When A and B are expressions, say that B is an initial segment of Ai A can be obtained by concatenating some (possibly no) symbols to the(right-hand) end of B . For example:

: is an initial segment of ::A and an initial segment of :(P ^ Q),but not an initial segment of (:P ! Q).

(P ^:Q) is an initial segment of (P ^:Q), but not an initial segmentof ((P ^ :Q) _R)

Here is a cute result about initial segments.

Lemma 1.6. For any sentence A , the only sentence which is an initial segmentof A is A itself.

Proof. Let A be any sentence, and suppose (for induction) that every shortersentence, B , has the property that the only sentence which is an initial segmentof B is B itself (this is the induction property). There are three cases:

6Note that C, like A , is a symbol of the metalanguage, not the object language. For areminder of the distinction, fx C7.

1. Induction 10

Case 1: A is atomic.. Then the only initial segment of A is A itself.Case 2: A is :B, for some sentence B. Let A be any sentence that isan initial segment of A . So A is :B, for some sentence B. Evidently,B is an initial segment of B . But B and B are both sentences that areshorter than A ; so by the induction hypothesis, B must just be B itself.Hence A is A itself, and so A has the induction property.Case 3: A is (B C C ), for some sentences B and C and some two-placeconnective C. Let A be any sentence that is an initial segment of A .Evidently, A starts with a (; so, by the recursion rules, A must be(B J C ), for some sentences B and C and some two-place connectiveJ. Now, exactly as in Case 2, we can see that B must just be B itself(since B is an initial segment of B). Consequently, C and J must bethe same connective. And so, by a similar argument, C must just be Citself (since C is an initial segment of C ). Hence A is A itself, and soA has the induction property.

This completes the strong induction on length.

Considered on its own, this result is not very thrilling. But I called it a lemmabecause it will be used to prove a result that is suciently important to becalled a theorem, and to have its own name.

Before I state and prove the Theorem, I shall try to motivate it. When Iintroduced the idea of the main logical operator of a sentence (in fx C6),I described it as the symbol that was introduced last, when constructing thesentence by the recursion rules. However, I gave you no guarantee that therewas only one way to construct a sentence using those rules. Consequently, Igave you no guarantee that the main logical operator is unique. And, at thatpoint in your lives, we were not in a position to guarantee this result. Now,though, we are; and here is that overdue guarantee.

Theorem 1.7. Unique Readability Theorem. For every sentence A ,exactly one of the following obtains:

1. A is an atomic sentence; or2. there is a unique sentence, B , such that A is :B ; or3. there are unique sentences, B and C , and a unique two-place connective,C, such that A is (B C C ).

Proof of Theorem 1.7. Let A be any sentence. It is clear that at most one of(1), (2) or (3) obtains, since each requires that A start with a dierent symbol.It remains to show that at least one of them obtains. There are three cases toconsider:

Case 1: A is atomic. i.e. case (1) obtains.Case 2: A is :B, for some sentence B. Then it is clear that there is ex-actly one sentence that is formed by removing the negation, i.e. B isunique. So case (2) obtains.Case 3: A is (B C C ), for some sentences B and C and some two-placeconnective C. To establish that case (3) obtains, we need to show thatthis decomposition is unique. So suppose that there are also sentences,B and C , and a two-place connective, J, such that A is (B J C ).

1. Induction 11

We can use exactly the same technique as in the proof of Case 3 ofLemma 1.6 to show that B is B, that C is J and that C is C ; theonly dierence is that we now invoke Lemma 1.6 itself (rather than aninduction hypothesis). Hence, case (3) obtains.

This uniqueness really matters. It shows that there can be no syntactic am-biguity in TFL, so that each sentence admits of one (and only one) reading.Contrast one of the most famous newspaper headlines of all time: 8th Armypush bottles up German rear.

Practice exercisesA. Fill in the details in Case 3 of the proof of Lemma 1.6, and in Case 3 ofthe proof of Theorem 1.7.

B. Prove the following, by induction on length: for any sentence A , everyvaluation which assigns truth values to all of the atomic sentences in A assignsa unique truth value to A .

C. For any sentence A , let: aA be the number of instances of atomic sentences in A ; bA be the number of instances of ( in A ; cA be the number of instances of two-place connectives in A .

As an example: if A is (:B ! (B ^ C)), then aA = 3, bA = 2 andcA = 2. Prove the following, by induction on length: for any sentence A :aA 1 = bA = cA .

D. Suppose a sentence is built up by conjoining together all of the sentencesA1; : : : ;An, in any order. So the sentence might be, e.g.:

(A5 ^ ((A3 ^A2) ^ (A4 ^A1)))

Prove, by induction on the number of conjuncts, that the truth value of theresulting sentence (on any valuation which assigns a truth value to all ofA1; : : : ;An) will be the same, no matter how, exactly, they are conjoined. Thenprove that the same is true if we consider disjunction rather than conjunction.Which notational conventions that were introduced in fx C are vindicated bythese proofs?

Substitution 2

In this chapter, we shall put our new familiarity with induction to good work,and prove a cluster of results that connect with the idea of substituting onesentence in for another.

Before going further, we should state the idea of substitution a bit moreprecisely. Where A , C and S are sentences, A [S/C ] is the result of searchingthrough A and nding and replacing every occurrence of C with S . As anexample, when A is (Q ^ (:P _ (Q $ R))), when C is Q and when S is:(D ! B), then A [S/C ] is (:(D ! B) ^ (:P _ (:(D ! B)$ R))).

All of the results in this chapter employ this surprisingly powerful idea.

2.1 Two crucial lemmas about substitutionGiven your familiarity with TFL, here is a line of thought which should strikeyou as plausible. Suppose you have a valid argument in TFL. Now, run throughthat argument, uniformly substituting every instance of some atomic sentencethat occurs within that argument, with some other sentence. The resultingargument is still valid.

It is worth realising that this line of thought is not generally plausible innatural languages. For example, here is a valid argument in English:

Isaac is an eft; so Isaac is a newt.Its validity, plausibly, connects with the meanings of its component words.But recall that the semantics of TFL is purely extensional, and knows nothingabout meanings. Its only dimension of freedom is over how we assign truthand falsity to components of sentences. And uniform substitution of the sortmentioned in the previous paragraph cannot, in a relevant sense, increase ouroptions for assigning truth values to such components.

But to convince ourselves that this line of thought is correct, we of courserequire a rigorous proof. Here it is.

Lemma 2.1. For any atomic sentence C and any sentences A1; : : : ;An+1; S :if A1; : : : ;An An+1, then A1[S/C ]; : : : ;An[S/C ] An+1[S/C ].

Proof. Suppose A1; : : : ;An An+1. Suppose also that v is a valuation whichmakes all of A1[S/C ]; : : : ;An[S/C ] true. Let w be a valuation which diers fromv (if at all) only by assigning a value to C as follows: w makes C true i vmakes S true. I now make a claim:

Claim: for any sentence A : w makes A true i v makes A [S/C ] true.

12

2. Substitution 13

I shall prove the Claim by induction on the length of A . Suppose, for induction,that the Claim is true of every shorter sentence. Now there are three cases:

Case 1: A is atomic. Then either A is C , in which case A [S/C ] is just S ;or A [S/C ] is just A . Either way, the Claim holds of A .Case 2: A is :B, for some sentence B. Then, by assumption, w makesB true i v makes B [S/C ] true. By considering the truth-table for :, itfollows that w makes :B true i v makes :B [S/C ] true.Case 3: A is (B C C ), for some sentences B and C and some two-placeconnective C. This is much like Case 3; I leave the details as an exercise.

This proves the Claim, by induction. It follows from the claim that w makesall of A1; : : : ;An true. And, since A1; : : : ;An An+1, we see that w makesAn+1 true. So, appealing to the Claim again, v makes An+1[S/C ] true.

Lets see this Lemma in action. Consider these three claims:(:A! (A _B)) (A _B)

(::A! (:A _ :B)) (:A _ :B)(:(P $ :Q)! ((P $ :Q) _ :(B ^ (C $ D)))) ((P $ :Q) _ :(B ^ (C $ D)))

The rst of these is easy to check using a truth table; the second is marginallyharder; and the third is a pain! But if we recognise that the second and thirdare the result of uniform substitution into the rst, then we can use Lemma2.1 to save ourselves a lot of work.

There is a deeper point here. Lemma 2.1 vindicates the practice of describ-ing an entire argument structure as valid. For example, we can say that thefollowing is a valid pattern of entailment:

(:A ! (A _ B)) (A _ B)

We know already that at least one instance of this pattern of entailment isvalid. Lemma 2.1 then tells us that the entailment will hold, regardless of whatsentences we use for A and B . Lemma 2.1 therefore underscores the purelystructural nature of entailments, in our semantics for TFL.

Lemma 2.1 concerns tautological entailment. We shall also want to havea result to hand which concerns tautological equivalence. It will help us if weintroduce the symbol to symbolise tautological equivalence. So we write:

A B

to symbolise that A and B are tautologically equivalent; or, in other words,that A B and B A . Using this notation, we can prove a nifty little result:

Lemma 2.2. For any sentences A ;C and S : if C S , then A A [S/C ].

Proof. By an induction on the length of A ; I leave this as an exercise.

In a slogan: substitution of tautological equivalents preserves tautologicalequivalence. We shall rely upon this frequently, in what follows.

2. Substitution 14

2.2 InterpolationThe next result uses the idea of substitution to prove something rather deep.When working with an argument, one can get the idea that there is an activecore to the argument, which should be distinguished from some of the irrelevantsurrounding matter. The next result gives some formal content to this claim

I shall start with a little result which might be put as telling us the following:if I am to draw a conclusion from some claim, there had better be somethingin the original claim that my conclusion builds upon. More precisely:

Lemma 2.3. Suppose A B , that A is not a contradiction, and that B isnot a tautology. Then at least one atomic sentence occurs in both A and B .

Proof. Supposing A is not a contradiction, there is a valuation, v, that makesA true. Supposing B is not a tautology, there is a valuation, w, that makesB false. Supposing also A and B have no atomic sentences in common, thenthere is a valuation, x, which assigns the same values as v to all atomic sentenceletters in A , and the same values as w to all atomic sentence letters in B . Byhypothesis, x makes A true and B false. So A 2 B .

Our main result is going to employ a (fairly lengthy) induction to extendLemma 2.3 rather radically. First, we need a denition. Suppose that A B .Then an interpolant linking A to B is any sentence I meeting all three ofthese conditions:(int1) A I(int2) I B(int3) every atomic sentence in I occurs both in A and in BOur result guarantees the existence of interpolants.

Theorem 2.4. Interpolation Theorem. Suppose A B , that A is nota contradiction, and that B is not a tautology. Then there is an interpolantlinking A to B .

Proof. Our proof consists in an (ordinary) induction on the number of atomicsentences that occur in A but not B .

In the base case, there are no atomic sentences in A that are not also inB . Hence we can let A itself be our interpolant, since A A vacuously, andA B by hypothesis.

Now x n, and suppose (for induction) that we can nd an interpolantwhenever there are n atomic sentences that occur in the premise but not theconclusion of the entailment. Additionally, suppose that A B , that A isnot a contradiction, that B is not a tautology, and that there are n+1 atomicsentences that occur in A but not in B . Lemma 2.3 tells us that there is atleast one atomic sentence, C , common to both A and B . Now, where:

D is any sentence that occurs in A but not in B > is a tautology which contains no atomic sentence apart from C , e.g.(C ! C ) ? is a contradiction which contains no atomic sentence apart from C ,e.g. (C ^ :C )

2. Substitution 15

I make the following claim:Claim 1: (A [>/D] _A [?/D]) B

To prove Claim 1, let v be any valuation that makes (A [>/D] _ A [?/D]) true.If v makes A [>/D] true, let w dier from v just by making D true; otherwise,let w dier from v just by making D false. Either way, w makes A true, sow makes B true too (since A B). Finally, since D does not occur in B , vmakes B true too. This establishes the Claim.

Next, because there were n + 1 atomic sentences in A but not B , and Dwas one of these atomic sentences, there must be exactly n atomic sentences inA [>/D] _ A [?/D] but not B . Hence, by Claim 1 and the induction hypothesis,we can nd an interpolant, I , linking (A [>/D] _A [?/D]) to B , i.e.:

1. (A [>/D] _A [?/D]) I2. I B3. every atomic sentence in I occurs in both (A [>/D] _A [?/D]) and in B

I shall now show that I is an interpolant linking A itself to B . Point (2), above,tells us that I B . Moreover, given point (3), it is obvious that every atomicsentence in I occurs in both A and in B . So it remains to show that A I .Given that (A [>/D]_A [?/D]) I , it suces, by the transitivity of entailment,to prove the following claim:

Claim 2. A (A [>/D] _A [?/D])To prove Claim 2, let v be any valuation that makes A true. If v makes Dtrue, then since > is a tautology, v makes A [>/D] true. If v makes D false,then v makes A [?/D] true. Either way, any valuation that makes A true makes(A [>/D] _A [?/D]) true. This establishes Claim 2.

Hence I is an interpolant linking A to B . This completes the proof of theinduction case. The entire Theorem now follows by induction.

This is easily the longest proof that we have seen so far. So it is worth pausing,to make sure both that you understand every step in the proof, and also howthe proof is put together.

In fact, this proof merits particular scrutiny since, implicitly, the proofdoes more than just prove the Theorem. The Interpolation Theorem statesthat, when A B , an interpolant exists. If this were all we knew, then wemight have no idea about how to nd an interpolant. But, viewed correctly,the proof yields an algorithm for nding an interpolant. An algorithm is astep-by-step process, followed thoughtlessly (i.e. without any need for insight,just by implementing very simple little rules), that is guaranteed to terminatein nitely many steps. I shall describe the algorithm for nding an interpolant;and having done so, it should be clear how it relates to the proof.

Given that A B , linking A to B as follows:Step 1: Designate A as the input sentenceStep 2: Let D be the rst atomic sentence (alphabetically) that appears in

the input sentence, but that does not appear in B . Rewrite the inputsentence twice, once uniformly substituting a tautology in place of Dand once uniformly substituting a contradiction in place of D. Let thedisjunction of these two rewritten sentences be your new input sentence.

2. Substitution 16

Step 3: Repeat step 2, until there are no atomic sentences that appear in theinput sentence but not in B . At that point, stop. The last sentence youwrote down is an interpolant linking A to B .

It is worth seeing this in action. Suppose we start with the entailment:

((A ^ (:B _ C)) ^ (:C ^D)) ((B _D) ^ (:B _ E))

Running the algorithm, our initial input sentence is:

((A ^ (:B _ C)) ^ (:C ^D))

We would rst eliminate A, obtaining (using > to abbreviate some tautologycontaining B, and ? to abbreviate some contradiction containing B):

(((> ^ (:B _ C)) ^ (:C ^D)) _ ((? ^ (:B _ C)) ^ (:C ^D)))

This would become our new input sentence; from which we would eliminate C:

((((> ^ (:B _ >)) ^ (:> ^D)) _ ((? ^ (:B _ >)) ^ (:> ^D))) _(((> ^ (:B _ ?)) ^ (:? ^D)) _ ((? ^ (:B _ ?)) ^ (:? ^D))))

And this is our interpolant! But, as the example illustrates, the output ofthis algorithm can be quite a mouthful. To simplify it, we can apply someclean-up rules. Where > abbreviates any tautology, and ? abbreviates anycontradiction, note, for example, that:

> _A >? ^A ?:> ?:? >

Using these equivalences and Lemma 2.2, we can simplify the subsentences ofour interpolant.1 First, applying the rules for negation:

((((> ^ (:B _ >)) ^ (? ^D)) _ ((? ^ (:B _ >)) ^ (? ^D))) _(((> ^ (:B _ ?)) ^ (> ^D)) _ ((? ^ (:B _ ?)) ^ (> ^D))))

Next, applying the rules for conjunction and disjunction:

((((> ^>) ^ ?) _ (? ^?)) _ (((> ^ :B) ^D) _ (? ^D)))

And, then, applying them again and again, we get, successively:

((? _?) _ ((:B ^D) _ ?))(? _ (:B ^D))

(:B ^D)

Which is much simpler!1Recall that a sentence A is a subsentence of B i A is a part of B. We here read part

in a generous sense, so that every sentence is a part of itself.

2. Substitution 17

2.3 DualityThe nal result of this chapter will use substitution to explore a very powerfulidea. To explain the idea, recall the following equivalences, known as the DeMorgan Laws:

(A ^ B) :(:A _ :B)(A _ B) :(:A ^ :B)

These indicate that there is a tight symmetry between conjunction and dis-junction. The notion of duality is a generalisation of this tight symmetry.And in fact, it is worth introducing it with complete generality at the outset.

In TFL, our only primitive connectives are one-place (i.e. :) and two-place (i.e. ^, _, ! and $). But nothing stops us from introducing three-,four-, or ve-place connectives; or, more generally, n-place connectives, for anynumber n we like. (We shall, however, always assume that our connectives aretruth-functional.)

So, let C be an n-place connective. The dual of C, which I shall genericallydenote by underlining it, i.e. C, is dened to have a characteristic truth tableas follows:

C(A1; : : : ;An)

:C(:A1; : : : ;:An)

This is the obvious generalisation of the De Morgan Laws to many-place con-nectives.2 And it is immediate, on this denition, that ^ is the dual of _,and that _ is the dual of ^. I encourage you also to check that " is the dualof #, and vice versa; and that : is its own dual.

Inspired by these examples, we might correctly conjecture the following:

Lemma 2.5. The dual of a dual is always the original connective; that is, Cis the same connective as C.

Proof. I start by noting:

C(A1; : : : ;An)

:C(:A1; : : : ;:An) by denitionC(:A1; : : : ;:An) :C(::A1; : : : ;::An) by Lemma 2.1:C(:A1; : : : ;:An) ::C(::A1; : : : ;::An) by the denition of ::C(:A1; : : : ;:An) C(A1; : : : ;An) by Lemma 2.2

Next, observe that, by denition:

C(A1; : : : ;An)

:C(:A1; : : : ;:An)

Hence, by the transitivity of entailment:

C(A1; : : : ;An)

C(A1; : : : ;An)

as required. 2Hence some books talk of the De Morgan dual, rather than simply of the dual.

2. Substitution 18

We shall use the same kind of proof-technique in what follows, to great eect.First, we need a little more notation, connecting duality with negation.

Given any sentence, A , we say A is the result of replacing every connectivein A with its dual, and A is the result of replacing every atomic sentence inA with the negation of that atomic sentence. So, when A is the sentence(A _ (:B ^ C)), A is (A ^ (:B _ C)) and A is (:A _ (::B ^ :C)).

Theorem 2.6. A :A , for any sentence A .Proof. Let A be any sentence; we suppose, for induction on length, that everyshorter sentence B is such that B :B . There are two cases to consider:

Case 1: A is atomic. Then A is just A itself; and :A is just ::A ; andwe know that A ::A .Case 2: A is C(B1; : : : ;Bn), for some n-place connective C. So by hy-pothesis, Bi

:Bi for each Bi. Now observe that:

C(B1; : : : ;Bn)

C(B1; : : : ;Bn) by denition

C(:B1; : : : ;:Bn) by Lemma 2.2

:C(::B1; : : : ;::Bn) by denition of C

:C(B1; : : : ;Bn) by Lemma 2.2

:C(B1; : : : ;Bn) by denition

and hence, given the transitivity of entailment, A :A .This completes the induction on length.

And we can now obtain a very elegant result:

Theorem 2.7. A B i B A , for any sentences A and B . Hence, inparticular, A B i A B .

Proof. Left to right. Suppose A B . Then A B , by Lemma 2.1. So:B :A . So B A , by Theorem 2.6 and Lemma 2.2.

Right to left. Suppose B A . Then, by the preceding, A B . HenceA B , by Lemmas 2.5 and 2.2.

This ends my theorem-proving. But to close the chapter, I want to oer somesketch some reasons for thinking that duality is pretty great.

Illustration 1. You may have already noticed the following equivalences:

(A ^ (B _ C )) ((A ^ B) _ (A ^ C ))((A _ B) ^ C ) ((A ^ C ) _ (B ^ C ))

If not, then you can test these using a truth table. But, having done so,Theorem 2.7 immediately yields two further equivalences for free:

(A _ (B ^ C )) ((A _ B) ^ (A _ C ))((A ^ B) _ C ) ((A _ C ) ^ (B _ C ))

2. Substitution 19

These four equivalences are the Distribution Laws. We shall use them repeat-edly in the next chapter.

Illustration 2. Here is a playful way to think about negation, due to FrankPlumpton Ramsey.3 Suppose that we allow ourselves to represent a sentencesnegation by ipping the entire sentence upside-down, rather than by prexingthe sentence with :. With this notation, and supposing ~ to be an n-placeconnective, we would have:

~(A1; : : : ;An) :~(:A1; : : : ;:An) the original denition

~(A1; : : : ;An) replacing the interior negations

~(A1;

: : :;An) replacing the outer negation

and with a little imagination concerning punctuation (commas and brackets),we see that the same eect could be achieved more simply by merely invertingthe connective itself, i.e. we would represent the dual of ~ by ~. A fullyperspicuous notation would, then, represent connectives and their duals bysymbols which are mirror-images of each other. But that is exactly how wedo represent conjunction and disjunction!4 Note, too, that if we adopted thisnotation, then Lemma 2.5 would be visually immediate. And this suggests(correctly) that the success of our Ramsey-inspired notation system dependsonly upon Lemma 2.2 and the equivalence: ::A A . Once we realise this,we can see that all of our results about duality are extremely general: theyhold for any extensional language which respects the equivalence ::A Aand in which Lemmas 2.1 and 2.2 hold.

Illustration 3. An immediate consequence of our results so far is

A :A

I leave the proof of this consequence as an exercise. But the consequenceprompts a subtle thought. Typically, when I hear you say London is thecapital of the UK, or whatever, I take you to mean that London is the capitalof the UK, and to be asserting this. What if, instead, I took you to mean thatLondon is not the capital of the UK, and to be rejecting this? More generally,where before I took you to be asserting that A , I now take you to be rejectingthat A . Would anything be wrong with my reinterpretation? If not: whyshould I assume that you mean and, rather than or, by the word and?5

Practice exercisesA. Complete the inductions on length mentioned in the proofs of Lemmas 2.1and 2.2.

B. Prove that A :A , for any sentence A .

3Facts and Propositions (1927, Proceedings of the Aristotelian Society SupplementaryVolume 7.1).

4There is something, then, to be said for representing negation with a horizontally sym-metric . But I have no great ideas for how to represent (bi)conditionals.

5For discussion of all this, see Gerald Massey, The Indeterminacy of Translation,pp.329. (1992, Philosophical Topics 20.1) and Hilary Putnam, Replies, pp.399402 (1992,Philosophical Topics 20.1).

Normal forms 3

In this chapter, I prove two normal form theorems for TFL. These guaranteethat, for any sentence, there is a tautologically equivalent sentence in somecanonical normal form. Moreover, I shall give methods for nding these normal-form equivalents.

3.1 Disjunctive Normal FormSay that a sentence is in disjunctive normal form i it meets all of thefollowing conditions:(dnf1) No connectives occur in the sentence other than negations, conjunc-

tions and disjunctions;(dnf2) Every occurrence of negation has minimal scope (i.e. any : is imme-

diately followed by an atomic sentence);(dnf3) No disjunction occurs within the scope of any conjunction.(For a reminder of the denition of the scope of a connective, see fx C6.) So,here are are some sentences in disjunctive normal form:

A(A ^B) _ (A ^ :B)

(A ^B) _ (A ^B ^ C ^ :D ^ :E)A _ (C ^ :P234 ^ P233 ^Q) _ :B

Note that I have here broken one of the maxims of this book (see 1.4) andtemporarily allowed myself to employ the relaxed bracketing-conventions thatallow conjunctions and disjunctions to be of arbitrary length (see fx C10.3).These conventions make it easier to see when a sentence is in disjunctive normalform. I shall continue to help myself to these relaxed conventions, withoutfurther comment, until 3.3.

To further illustrate the idea of disjunctive normal form, I shall introducesome more notation. I write A to indicate that A is an atomic sentencewhich may or may not be prefaced with an occurrence of negation. Then asentence in disjunctive normal form has the following shape:

(A1 ^ : : : ^ Ai) _ (Ai+1 ^ : : : ^ Aj) _ : : : _ (Am+1 ^ : : : ^ An)We now know what it is for a sentence to be in disjunctive normal form. Theresult that we are aiming at is:

Theorem 3.1. Disjunctive Normal Form Theorem. For any sentence,there is a tautologically equivalent sentence in disjunctive normal form.

20

3. Normal forms 21

Henceforth, I shall abbreviate Disjunctive Normal Form by DNF.

3.2 Proof of DNF Theorem via truth tablesMy rst proof of the DNF Theorem employs truth tables. I shall rst illustratethe technique for nding an equivalent sentence in DNF, and then turn thisillustration into a rigorous proof.

Lets suppose we have some sentence, S , which contains three atomic sen-tences, A, B and C. The very rst thing to do is ll out a complete truthtable for S . Maybe we end up with this:

A B C ST T T TT T F FT F T TT F F FF T T FF T F FF F T TF F F T

As it happens, S is true on four lines of its truth table, namely lines 1, 3, 7and 8. Corresponding to each of those lines, I shall write down four sentences,whose only connectives are negations and conjunctions, where every negationhas minimal scope:

A ^B ^ C which is true on line 1 (and only then) A ^ :B ^ C which is true on line 3 (and only then) :A ^ :B ^ C which is true on line 7 (and only then) :A ^ :B ^ :C which is true on line 8 (and only then)

But if I now disjoin all of these conjunctions, like so:

(A ^B ^ C) _ (A ^ :B ^ C) _ (:A ^ :B ^ C) _ (:A ^ :B ^ :C)

I have a sentence in DNF which is true on exactly those lines where one of thedisjuncts is true, i.e. it is true on (and only on) lines 1, 3, 7, and 8. So thissentence has exactly the same truth table as S . So I have a sentence in DNFthat is tautologically equivalent to S . Which is exactly what I required!

Now, the strategy that I just adopted did not depend on the specics of S ;it is perfectly general. Consequently, we can use it to obtain a simple proof ofthe DNF Theorem.

Proof of DNF Theorem via truth tables. Pick any arbitrary sentence, S , andlet A1; : : : ;An be the atomic sentences that occur in S . To obtain a sentencein DNF that is tautologically equivalent S , we consider S s truth table. Thereare two cases to consider:

Case 1: S is false on every line of its truth table. Then, S is a contradic-tion. In that case, the contradiction (A1 ^ :A1) S , and (A1 ^ :A1)is in DNF.

3. Normal forms 22

Case 2: S is true on at least one line of its truth table. For each line i ofthe truth table, let Bi be a conjunction of the form

(A1 ^ : : : ^ An)

where the following rules determine whether or not to include a negationin front of each atomic sentence:

Am is a conjunct of Bi i Am is true on line i:Am is a conjunct of Bi i Am is false on line i

Given these rules, a trivial proof by induction shows that Bi is trueon (and only on) line i of the truth table which considers all possiblevaluations of A1; : : : ;An (i.e. S s truth table).Next, let i1; i2; : : : ; im be the numbers of the lines of the truth table whereS is true. Now let D be the sentence:

Bi1 _ Bi2 _ : : : _ BimSince S is true on at least one line of its truth table, D is indeed well-dened; and in the limiting case where S is true on exactly one line of itstruth table, D is just Bi1 , for some i1.By construction, D is in DNF. Moreover, by construction, for each linei of the truth table: S is true on line i of the truth table i one of Dsdisjuncts (namely, Bi ) is true on, and only on, line i. (Again, this isshown by a trivial proof by induction.) Hence S and D have the sametruth table, and so are tautologically equivalent.

These two cases are exhaustive and, either way, we have a sentence in DNFthat is tautologically equivalent to S .

So we have proved the DNF Theorem. Before I say any more, though, I shouldimmediately ag that I am hereby returning to the austere denition of a (TFL)sentence, according to which we can assume that any conjunction has exactlytwo conjuncts, and any disjunction has exactly two disjuncts.1

3.3 Proof of DNF Theorem via substitutionI now want to oer a second proof of the DNF Theorem.

At this point, you might reasonably ask: Why bother with two proofs?Certainly one proof is sucient to establish a result. But there are severalpossible reasons for oering new proofs of known results. One is that a secondproof might oer more insight (or a dierent insight) than the original proof.Another is that the proof-technique might be useful in other ways: it mighthelp you to prove other results (that you have not yet proved). Another issheer curiosity. All three reasons apply here.

The second proof of the DNF Theorem is very dierent in style from therst. Rather than going via truth tables, we shall employ Lemma 2.2. This

1NB: if you did the practice exercises for chapter 1, you will know how to justify theseconventions rigorously; and you will also, essentially, have performed the two trivial proofsby induction mentioned in the preceding proof.

3. Normal forms 23

allows us to take a sentence, and replace its subsentences with tautologicallyequivalent subsentences, thereby obtaining a sentence that is tautologicallyequivalent to what we started with. By choosing cunning substitutions, re-peated applications of Lemma 2.2 will allow us to reach a DNF equivalent.

I say that we shall choose the substitutions cunningly. In fact, we shallproceed by three lemmas. Each lemma is proved by induction. Each lemmacorresponds to one of the three clauses (dnf1)(dnf3) of the formal denitionof a DNF sentence (see 3.1). Each lemma brings us closer to DNF. Indeed most excitingly of all each lemma yields an algorithm for pushing a sentencetowards DNF (for a reminder about what an algorithm is, see 2.2).

I shall start with condition (dnf1). This tells us that any DNF sentencemust contain only negations, conjunctions and disjunctions. So our very rsttask is to show how to generate an equivalent sentence of this form.

Lemma 3.2. For any sentence, there is a tautologically equivalent sentencesatisfying (dnf1).

Proof. Let A be any sentence, and suppose, for induction on length, that forevery shorter sentence, B , there is a sentence B which satises (dnf1), andsuch that B B. There are now six cases to consider:

Case 1: A is atomic. Then A already satises (dnf1), and A A , sothere is nothing to prove.Case 2: A is :B, for some sentence B . Then A :B, by Lemma 2.2,and :B satises (dnf1).Case 3: A is (B _ C ), for some sentences B and C . Then A (B _C ), by Lemma 2.2, and this satises (dnf1).Case 4: A is (B ^ C ), for some sentences B and C . Then A (B ^C ), by Lemma 2.2, and this satises (dnf1).Case 5: A is (B ! C ), for some sentence B and C . In this case:

A (B ! C ) given

(B ! C) by Lemma 2.2

(:B _ C ) by considering truth tablesAnd this last sentence satises (dnf1).Case 6: A is (B $ C ), for some sentence B and C . Then, as in Case 5,A ((B ^ C ) _ (:B ^ :C )), which satises (dnf1).

This completes the induction.

It is worth noting that this proof doesnt simply tell us that an equivalentsentence exists; it implicitly provides us with an algorithm for eliminatingoccurrences of conditionals and biconditionals from a sentence, one at a time,whilst preserving tautological equivalence. (Compare the discussion of theproof of the Interpolation Theorem; see 2.2.) In particular, the algorithmjust tells us to rewrite our sentence repeatedly, relying on Lemma 2.2 and thefollowing general equivalences:

(A ! B) (:A _ B)(A $ B) ((A ^ B) _ (:A ^ :B))

3. Normal forms 24

until no (bi)conditionals remain. For example, if we start with the sentence:((:A _ (:B ^ C))! :(D $ E))

then we rst get:(:(:A _ (:B ^ C)) _ :(D $ E))

and then:(:(:A _ (:B ^ C)) _ :((D ^ E) _ (:D ^ :E)))

which is a bit closer to DNF.But there is still some way to go. In particular, condition (dnf2) tells us

that in any DNF sentence, every occurrence of negation has minimal scope. Soour next task is to prove:

Lemma 3.3. For any sentence satisfying (dnf1), there is a tautologicallyequivalent sentence satisfying both (dnf1) and (dnf2)Proof. Let A be any sentence satisfying (dnf1) and suppose, for induction onlength, that for every shorter sentence B , if B satises (dnf1) then there is asentence B satisfying (dnf1) and (dnf2) such that B B. There are nowthree cases to consider:

Case 1: A is atomic. Then there is nothing to prove, since A contains nonegations.Case 2: A is either (B ~ C ), for some two-place connective ~. In thiscase, A (B ~ C ), which satises (dnf1) and (dnf2).Case 3: A is :B, for some sentence B. Then there are four subcases:3a: A is :B, for some atomic sentence B. Then A itself satises (dnf1)

and (dnf2).3b: A is ::C , for some sentence C . Then A C , which satises

both (dnf1) and (dnf2).3c: A is :(C _D), for some sentences C and D. Then clearly A

(:C ^:D). Moreover, since :C is shorter than A , by the inductionhypothesis there is some E satisfying (dnf1) and (dnf2) such thatE :C ; similarly, there is some F satisfying (dnf1) and (dnf2)such that F :D. Hence A (E ^ F ), which satises (dnf1)and (dnf2).

3d: A is :(C ^D), for some sentences C and D. Exactly like Subcase3c, except that we start by noting that A (:C _ :D).

This completes the induction on length.

As with the previous result, this Lemma implicitly provides us with an algo-rithm. This time, it allows us to drive negations deep into a sentence, so thatthey have minimal scope. In particular, we repeatedly apply these equivalences:

::A A:(A _ B) (:A ^ :B):(A ^ B) (:A _ :B)

In more detail, heres an algorithm based on these rules:

3. Normal forms 25

Step 1: Remove all the double-negations in the sentence (working from left toright in the sentence).

Step 2: Starting from the left, look for the rst instance where we can apply oneof the De Morgan laws, and apply it, pushing negations inside brackets.

Step 3: Repeat steps 12, until no further change can occur.It might help to see this algorithm in action, on the example from before:

(:(:A _ (:B ^ C)) _ :((D ^ E) _ (:D ^ :E)))((::A ^ :(:B ^ C)) _ :((D ^ E) _ (:D ^ :E)))

((A ^ :(:B ^ C)) _ :((D ^ E) _ (:D ^ :E)))((A ^ (::B _ :C)) _ :((D ^ E) _ (:D ^ :E)))

((A ^ (B _ :C)) _ :((D ^ E) _ (:D ^ :E)))((A ^ (B _ :C)) _ (:(D ^ E) ^ :(:D ^ :E)))((A ^ (B _ :C)) _ ((:D _ :E) ^ :(:D ^ :E)))((A ^ (B _ :C)) _ ((:D _ :E) ^ (::D _ ::E)))((A ^ (B _ :C)) _ ((:D _ :E) ^ (D _ ::E)))((A ^ (B _ :C)) _ ((:D _ :E) ^ (D _ E)))

We are really very close to DNF! All that remains is to meet condition (dnf3),i.e. to ensure that no disjunction occurs within the scope of any conjunction.So our nal task is to prove:

Lemma 3.4. For any sentence satisfying (dnf1) and (dnf2), there is a tau-tologically equivalent sentence in DNF.

Proof. We shall almost do a (strong) induction on length. However, the onlymeasure of length that we care about is the number of occurrences of dis-junctions in a sentence. More precisely, then, the disjunctive-length ofa sentence is the number of occurrences of _ in the sentence, and we shallproceed by (strong) induction on disjunctive-length.2

Let A be any sentence satisfying (dnf1) and (dnf2). For induction, sup-pose that, for every sentence B which satises (dnf1) and (dnf2) and isdisjunctively-shorter than A (i.e. that contains fewer disjunctions than A),there is a DNF sentence B which is exactly as disjunctively-long as B andsuch that B and B. There are now four cases to consider:

Case 1: A is atomic. In this case, A itself is in DNF and is (trivially)exactly as disjunctively-long as A .Case 2: A is :B. In this case, B must be atomic, hence :B is in DNFand is (trivially) exactly as disjunctively-long as A .Case 3: A is (B _ C ). Then A (B _ C ), which is in DNF and, byassumption, is exactly as disjunctively-long as A .Case 4: A is (B ^ C ). There are three subcases to consider, dependingupon the form of B and C :

2I leave it as an exercise, to check that the required Principle of Strong Induction onDisjunctive-Length is legitimate; compare the justication of the Principle of Strong Induc-tion on Length in 1.4.

3. Normal forms 26

4a: B contains a disjunction. In that case, since B is in DNF, B is(B1 _B2), for some B1 and B2 in DNF that are both disjunctively-shorter than B. Next, observe:

(B ^ C ) ((B1 _ B2) ^ C )

((B1 ^ C ) _ (B2 ^ C ))Now, (B1 ^ C ) is disjunctively-shorter than A ; so by hypothesis,there is some DNF sentence D which is exactly as disjunctively-long as (B1 ^ C ) and such that D (B1 ^ C ). Similarly, thereis some DNF sentence E which is exactly as disjunctively-long as(B2 ^ C ) and such that E (B2 ^ C ). So A (D _E), which isin DNF and exactly as disjunctively-long as A .

4b: C contains a disjunction. Exactly similar to Subcase 4a; I leave thedetails as an exercise.

4c: neither B nor C contains a disjunction. Then A (B ^ C ),which is already in DNF and is exactly as long as A itself.

This completes the induction.

Again, this Lemma implicitly provides an algorithm; this time, for pullingdisjunctions outside the scope of any conjunction, by repeatedly applying theDistribution Laws:

((A _ B) ^ C ) ((A ^ C ) _ (B ^ C ))(A ^ (B _ C )) ((A ^ B) _ (A ^ C ))

Applying these to our earlier example, we get, successively:((A ^ (B _ :C)) _ ((:D _ :E) ^ (D _ E)))

(((A ^B) _ (A ^ :C)) _ ((:D _ :E) ^ (D _ E)))(((A ^B) _ (A ^ :C)) _ (((:D _ :E) ^D) _ ((:D _ :E) ^ E)))(((A ^B) _ (A ^ :C)) _ (((:D ^D) _ (:E ^D)) _ ((:D ^ E) _ (:E ^ E))))

And this is in DNF, as may become clearer if we temporarily allow ourselvesto remove brackets, following the notational conventions of fx C10.3:

(A ^B) _ (A ^ :C) _ (:D ^D) _ (:E ^D) _ (:D ^ E) _ (:E ^ E)Together, then, these Lemmas implicitly provide us with an algorithm for com-puting an equivalent sentence, in DNF. Moreover, this algorithm althoughlengthy was much more ecient at generating a DNF sentence than thetruth-table method would have been. (The truth table for our example wouldhave 32 lines, 22 of which are true.)

It remains to assemble these Lemmas, into a proof of the DNF Theorem:Proof of DNF Theorem via substitution. Let A be any TFL sentence.

Lemma 3.2 tells us that there is a sentence, B , satisfying (dnf1), such thatA B .

Lemma 3.3 tells us that there is a sentence, C , satisfying (dnf1) and (dnf2),such that B C .

Lemma 3.4 tells us that there is a sentence, D, in DNF, such that C D.And, by the transitivity of entailment, A D.

3. Normal forms 27

3.4 Cleaning up DNF sentencesI have oered two dierent proofs of the DNF Theorem. They also describedtwo dierent algorithms for putting sentences into DNF. But these algorithmsdo not always leave you with the most elegant output. Indeed, just as we canclean up the output of the algorithm for calculating interpolants (see 2.2), wecan clean up the output of our algorithms for calculating DNF-equivalents.Here are the clean-up rules; I leave it as an exercise to justify them. First,we rewrite our DNF sentence using the relaxed notational conventions of fx C.Now:

Rule 1: remove repetitious conjuncts. If any disjunct contains the same con-junct more than once, remove all but one instance of that conjunct (respectingbracketing conventions as appropriate).

Rule 2: remove overly-specic disjuncts. Delete any disjunct which entails anyother disjunct. (If two or more disjuncts entail each other, delete all but therst of those disjuncts.)Example: the output of our example in 3.3 was:

(A ^B) _ (A ^ :C) _ (:D ^D) _ (:E ^D) _ (:D ^ E) _ (:E ^ E)Noting that a contradiction entails everything, we should remove both contra-dictory disjuncts, obtaining (still with relaxed conventions):

(A ^B) _ (A ^ :C) _ (:E ^D) _ (:D ^ E)Rule 3: invoke excluded middle. Suppose there is a disjunct whose conjunctsare just A1; : : : ;Am and B (in any order), and another disjunct whose conjunctsare just A1; : : : ;Am and :B (in any order), so that the two disjuncts disagreeonly on whether whether B should be prexed with a negation. In that case,delete both disjuncts and replace them with the simpler disjunct (A1^: : :^Am).(However, if we have just B and :B as disjuncts, then delete the entire sentenceand leave the tautology (B _ :B).)Example: the output of the example in 3.2 was:

(A ^B ^ C) _ (A ^ :B ^ C) _ (:A ^ :B ^ C) _ (:A ^ :B ^ :C)which can be simplied to:

(A ^ C) _ (:A ^ :B)

3.5 Conjunctive Normal FormSo far in this chapter, I have discussed disjunctive normal form. Given theduality of disjunction and conjunction (see 2.3), it may not come as a surpriseto hear that there is also such a thing as conjunctive normal form (CNF).

The denition of CNF is exactly analogous to the denition of DNF. So, asentence is in CNF i it meets all of the following conditions:(cnf1) No connectives occur in the sentence other than negations, conjunc-

tions and disjunctions;

3. Normal forms 28

(cnf2) Every occurrence of negation has minimal scope;(cnf3) No conjunction occurs within the scope of any disjunction.Generally, then, a sentence in CNF looks like this

(A1 _ : : : _ Ai) ^ (Ai+1 _ : : : _ Aj) ^ : : : ^ (Am+1 _ : : : _ An)

where each Ak is an atomic sentence.Since : is its own dual, and _ and ^ are the duals of each other, it

is immediate clear that if a sentence is in DNF, then its dual is in CNF; andvice versa. Armed with this insight, we can immediately prove another normalform theorem:

Theorem 3.5. Conjunctive Normal Form Theorem. For any sentence,there is a tautologically equivalent sentence in conjunctive normal form.

Proof. Let S be any sentence. Applying the DNF Theorem to S , there is aDNF sentence, D, such that S D. So, by Theorem 2.7, S D. Whence,by Lemma 2.5, S D. But, since D is in DNF, D is in CNF.

This slick proof is a further illustration of the power of duality. However, itmight suggest that the DNF Theorem enjoys some kind of precedence overthe CNF Theorem. That would be misleading. We can easily prove the CNFTheorem directly, using either of the two proof techniques that we used toprove the DNF Theorem (whereupon the DNF Theorem could be proved as aconsequence of the CNF Theorem and duality). I shall sketch the main ideasof the direct proofs of the CNF Theorem; I leave it as an exercise to make theseideas more precise.

Proof sketch of CNF Theorem via truth tables. Given a TFL sentence, S , webegin by writing down the complete truth table for S .

If S is true on every line of the truth table, then S (A1 _ :A1).If S is false on at least one line of the truth table then, for every line on

the truth table where S is false, write down a disjunction (A1 _ : : : _ An)which is false on (and only on) that line. Let C be the conjunction of all ofthese disjuncts; by construction, C is in CNF and S C .

Proof sketch of CNF Theorem via substitution. Using Lemmas 3.23.3, we canobtain a sentence satisfying (cnf1) and (cnf2). To turn this into a sentencein CNF, we apply a modied version of Lemma 3.4, obtained by invoking theduals (what else!) of the Distribution Laws invoked there; i.e.:

((A ^ B) _ C ) ((A _ C ) ^ (B _ C ))(A _ (B ^ C )) ((A _ B) ^ (A _ C ))

This allows us to pull conjunctions outside the scope of any disjunction.

Of course, after obtaining a sentence in CNF however we chose to do it wecan clean-up the resulting sentence using rules similar to those of 3.4. I leaveit as an exercise, to determine what the appropriate rules should be.

3. Normal forms 29

Practice exercisesA. Consider the following sentences:

1. (A! :B)2. :(A$ B)3. (:A _ :(A ^B))4. (:(A! B) ^ (A! C))5. (:(A _B)$ ((:C ^ :A)! :B))6. ((:(A ^ :B)! C) ^ :(A ^D))

For each sentence: use both algorithms (by truth table, and by substitution) to write down

sentences in DNF that are tautologically equivalent to these sentences. use both algorithms to write down sentences in CNF that are tautologi-

cally equivalent to these sentences.B. Oer proofs of the two trivial inductions mentioned in the proof of Theo-rem 3.1 via truth tables.

C. Justify the Principle of Strong Induction on Disjunctive-Length, appealedto in Lemma 3.4.

D. Prove that each of the rules mentioned in 3.4 is admissible, i.e. thatapplying them preserves tautological equivalence.

E. Fill out the details of the two proof sketches of the CNF Theorem. Deter-mine what the appropriate clean-up rules should be, for making CNF sentencestidy.

Expressive adequacy 4

In chapter 3, we established two normal form theorems. In this chapter, weshall use them to demonstrate the expressive power of TFL.

4.1 The expressive adequacy of TFLIn discussing duality (2.3), I introduced the general idea of an n-place connec-tive. We might, for example, dene a three-place connective, ~, into existence,by stipulating that it is to have the following characteristic truth table:

A B C ~(A ;B ;C )T T T FT T F TT F T TT F F FF T T FF T F TF F T FF F F F

Probably this new connective would not correspond with any natural Englishexpression (in the way that ^ corresponds with and). But a question arises:if we wanted to employ a connective with this characteristic truth table, mustwe add a new connective to TFL? Or can we get by with the connectives wealready have?

Let us make this question more precise. Say that some connectives arejointly expressively adequate i, for any possible truth function, there isa scheme containing only those connectives which expresses that truth function.Since we can represent truth functions using characteristic truth tables, wecould equivalently say the following: some connectives are jointly expressivelyadequate i, for any possible truth table, there is a scheme containing onlythose connectives with that truth table.

I say scheme rather than sentence, because we are not concerned withsomething as specic as a sentence (it might help to reread fx C7 and 2.1). Tosee why, consider the characteristic truth table for conjunction; this schemat-ically encodes the information that a conjunction (A ^ B) is true i both Aand B are true (whatever A and B might be). When we discuss expressiveadequacy, we are considering something at the same level of generality.

30

4. Expressive adequacy 31

The general point is, when we are armed with some jointly expressivelyadequate connectives, no truth function lies beyond our grasp. And in fact, weare in luck.1

Theorem 4.1. Expressive Adequacy Theorem. The connectives of TFLare jointly expressively adequate. Indeed, the following pairs of connectives arejointly expressively adequate:

1. : and _2. : and ^3. : and !

Proof. Given any truth table, we can use the method of proving the DNFTheorem (or the CNF Theorem) via truth tables, to write down a scheme whichhas the same truth table. For example, employing the truth table method forproving the DNF Theorem, I can tell you that the following scheme has thesame characteristic truth table as ~(A ;B ;C ), above:

(A ^ B ^ :C ) _ (A ^ :B ^ C ) _ (:A ^ B ^ :C )

It follows that the connectives of TFL are jointly expressively adequate. I nowprove each of the subsidiary results.

Subsidiary Result 1: expressive adequacy of : and _. Observe that thescheme that we generate, using the truth table method of proving the DNFTheorem, will only contain the connectives :, ^ and _. So it suces toshow that there is an equivalent scheme which contains only : and _. Toshow do this, we simply consider the following equivalence:

(A ^ B) :(:A _ :B)

and reason as in Lemma 3.2 (I leave the details as an exercise).Subsidiary Result 2: expressive adequacy of : and ^. Exactly as in

Subsidiary Result 1, making use of this equivalence instead:

(A _ B) :(:A ^ :B)

Subsidiary Result 3: expressive adequacy of : and !. Exactly as inSubsidiary Result 1, making use of these equivalences instead:

(A _ B) (:A ! B)(A ^ B) :(A ! :B)

Alternatively, we could simply rely upon one of the other two subsidiary results,and (repeatedly) invoke only one of these two equivalences.

In short, there is never any need to add new connectives to TFL. Indeed, thereis already some redundancy among the connectives we have: we could havemade do with just two connectives, if we had been feeling really austere.

1This, and Theorem 4.5, follow from a more general result, due to Emil Post in 1941.For a modern presentation of Posts result, see Pelletier and Martin Posts Functional Com-pleteness Theorem (1990, Notre Dame Journal of Formal Logic 31.2).

4. Expressive adequacy 32

4.2 Individually expressively adequate connectivesIn fact, some two-place connectives are individually expressively adequate.These connectives are not standardly included in TFL, since they are rathercumbersome to use. But their existence shows that, if we had wanted to, wecould have dened a truth-functional language that was expressively adequate,which contained only a single primitive connective.

The rst such connective we shall consider is ", which has the followingcharacteristic truth table.

A B A " BT T FT F TF T TF F T

This is often called the Sheer stroke, after Harry Sheer, who used it to showhow to reduce the number of logical connectives in Russell and WhiteheadsPrincipia Mathematica.2 (In fact, Charles Sanders Peirce had anticipated Shef-fer by about 30 years, but never published his results.)3 It is quite common,as well, to call it nand, since its characteristic truth table is the negation ofthe truth table for ^.

Proposition 4.2. " is expressively adequate all by itself.Proof. Theorem 4.1 tells us that : and _ are jointly expressively adequate.So it suces to show that, given any scheme which contains only those twoconnectives, we can rewrite it as a tautologically equivalent scheme which con-tains only ". As in the proof of the subsidiary cases of Theorem 4.1, then, wesimply apply the following equivalences:

:A (A " A)(A _ B) ((A " A) " (B " B))

to Subsidiary Result 1 of Theorem 4.1.

Similarly, we can consider the connective #:A B A # BT T FT F FF T FF F T

This is sometimes called the Peirce arrow (Peirce himself called it ampheck).More often, though, it is called nor, since its characteristic truth table is thenegation of _.

Proposition 4.3. # is expressively adequate all by itself.2Sheer, A Set of Five Independent Postulates for Boolean Algebras, with application

to logical constants, (1913, Transactions of the American Mathematical Society 14.4)3See Peirce, A Boolian Algebra with One Constant, which dates to c.1880; and Peirces

Collected Papers, 4.2645.

4. Expressive adequacy 33

Proof. As in Proposition 4.2, although invoking the dual equivalences:

:A (A # A)(A ^ B) ((A # A) # (B # B))

and Subsidiary Result 2 of Theorem 4.1.

4.3 Failures of expressive adequacyIn fact, the only two-place connectives which are individually expressively ad-equate are " and #. But how would we show this? More generally, how canwe show that some connectives are not jointly expressively adequate?

The obvious thing to do is to try to nd some truth table which we cannotexpress, using just the given connectives. But there is a bit of an art to this.Moreover, in the end, we shall have to rely upon induction; for we shall needto show that no scheme no matter how long is capable of expressing thetarget truth table.

To make this concrete, lets consider the question of whether _ is expres-sively adequate all by itself. After a little reection, it should be clear that itis not. In particular, it should be clear that any scheme which only containsdisjunctions cannot have the same truth table as negation, i.e.:

A :AT FF T

The intuitive reason, why this should be so, is simple: the top line of the desiredtruth table needs to have the value False; but the top line of any truth tablefor a scheme which only contains disjunctions will always be True. But so far,this is just hand-waving. To make it rigorous, we need to reach for induction.Here, then, is our rigorous proof.

Proposition 4.4. _ is not expressively adequate by itself.Proof. Let A by any scheme containing no connective other than disjunctions.Suppose, for induction on length, that every shorter scheme containing onlydisjunctions is true whenever all its atomic constituents are true. There aretwo cases to consider:

Case 1: A is atomic. Then there is nothing to prove.Case 2: A is (B _ C ), for some schemes B and C containing only disjunc-tions. Then, since B and C are both shorter than A , by the inductionhypothesis they are both true when all their atomic constituents are true.Now the atomic constituents of A are just the constituents of both B andC , and A is true whenever B and C . So A is true when all of its atomicconstituents are true.

It now follows, by induction on length, that any scheme containing no connec-tive other than disjunctions is true whenever all of its atomic constituents aretrue. Consequently, no scheme containing only disjunctions has the same truthtable as that of negation. Hence _ is not expressively adequate by itself.

4. Expressive adequacy 34

In fact, we can generalise Proposition 4.4:

Theorem 4.5. The only two-place connectives that are expressively adequateby themselves are " and #.Proof. There are sixteen distinct two-place connectives. We shall run throughthem all, considering whether or not they are individually expressively ade-quate, in four groups.

Group 1: the top line of the truth table is True. Consider those connectiveswhere the top line of the truth table is True. There are eight of these, including^, _, ! and $, but also the following:

A B A 1 B A 2 B A 3 B A 4 BT T T T T TT F T T T FF T T F F TF F T T F F

(obviously the names for these connectives were chosen arbitrarily). But, ex-actly as in Proposition 4.4, none of these connectives can express the truthtable for negation. So there is a connective whose truth table they cannotexpress. So none of them is individually expressively adequate.

Group 2: the bottom line of the truth table is False. Having eliminated eightconnectives, eight remain. Of these, four are false on the bottom line of theirtruth table, namely:

A B A 5 B A 6 B A 7 B A 8 BT T F F F FT F T T F FF T T F T FF F F F F F

As above, though, none of these connectives can express the truth table fornegation. To show this we prove that any scheme whose only connective is oneof these (perhaps several times) is false whenever all of its atomic constituentsare false. We can show this by induction, exactly as in Proposition 4.4 (I leavethe details as an exercise).

Group 3: connectives with redundant positions. Consider two of the re-maining four connectives:

A B A 9 B A 10 BT T F FT F F TF T T FF F T T

These connectives have redundant positions, in the sense that the truth valueof the overarching scheme only depends upon the truth value of one of theatomic constituents. More precisely:

A 9 B :AA 10 B :B

4. Expressive adequacy 35

Consequently, there are many truth functions that they cannot express. Inparticular, they cannot express either the tautologous truth function (givenby 1), or the contradictory truth function (given by 8). To show this, itsuces to prove that any scheme whose only connective is either 9 or 10(perhaps several times) is contingent, i.e. it is true on at least one line and falseon at least one other line. I leave the details of this proof as an exercise.

Group 4. Only two connectives now remain, namely " and #, and Propo-sitions 4.2 and 4.3 show that both are individually expressively adequate.

The above inductions on complexity were fairly routine. To close this chapter, Ishall oer some similar results about expressive limitation that are only slightlymore involved.

Proposition 4.6. Exactly four 2-place truth functions can be expressed usingschemes whose only connectives are biconditionals.Proof. I claim that the four 2-place truth functions that can be so expressedare expressed by:

A $ AAB

A $ BIt is clear that we can express all four, and that they are distinct.

To see that these are the only functions which can be expressed, I shallperform an induction on the length of our scheme. Let S be any schemecontaining only (arbitrarily many) biconditionals and the atomic constituentsA and B (arbitrarily many times). Suppose that any such shorter scheme isequivalent to one of the four schemes listed above. There are two cases toconsider:

Case 1: S is atomic. Then certainly it is equivalent to either A or to B .Case 2: S is (C $ D), for some schemes C and D. Since C and D areboth shorter than S , by supposition they are equivalent to one of thefour mentioned schemes. And it is now easy to check, case-by-case, that,whichever of the four schemes each is equivalent to, (C $ D) is alsoequivalent to one of those four schemes. (I leave the details as an exercise.)

This completes the proof of my claim, by induction on length of scheme.

Proposition 4.7. Exactly eight 2-place truth functions can be expressed usingschemes whose only connectives are negations and biconditionals.Proof. Observe the following equivalences:

::A A(:A $ B) :(A $ B)(A $ :B) :(A $ B)

We can use these tautological equivalences along the lines of Lemmas 3.33.4 to show the following: for any scheme containing only biconditionals and

4. Expressive adequacy 36

negations, there is a tautologically equivalent scheme where no negation isinside the scope of any biconditional. (The details of this are left as an exercise.)

So, given any scheme containing only the atomic constituents A and B(arbitrarily many times), and whose only connectives are biconditionals andnegations, we can produce a tautologically equivalent scheme which containsat most one negation, and that at the start of the scheme. By Proposition 4.6,the subscheme following the negation (if there is a negation) must be equivalentto one of only four schemes. Hence the original scheme must be equivalent toone of only eight schemes.

Practice exercisesA.Where 7 has the characteristic truth table dened in the proof of Theorem4.5, show that the following are jointly expressively adequate:

1. 7 and :.2. 7 and !.3. 7 and $.

B. Show that the connectives 7, ^ and _ are not jointly expressivelyadequate.

C. Complete the exercises left in each of the proofs of this chapter.

Soundness 5

So far, I have discussed only the syntax and semantics of TFL. In this chapter,and the next, I shall relate TFLs semantics to its proof system (as dened infx C). I shall prove that the semantics and the formal proof system mirror areperfectly suited to one another, in a sense to be made more precise.

5.1 Soundness dened, and setting up the proofIntuitively, a formal proof system is sound i it does not allow you to proveany invalid arguments. This is obviously a highly desirable property. It tellsus that our proof system will never lead us astray. Indeed, if our proof systemwere not sound, then we would not be able to trust our proofs. The aim of thischapter is to prove that our proof system is sound.

Let me make the idea more precise. A formal proof system is sound (rela-tive to a given semantics) i, if there is a formal proof of C from assumptionsamong , then genuinely entails C (given that semantics). Otherwise put,to prove that TFLs proof system is sound, I need to prove the following

Theorem 5.1. Soundness. For any sentences and C : if ` C , then C

To prove this, we shall check each of the rules of TFLs proof system individu-ally. We want to show that no application of those rules ever leads us astray.Since a proof just involves repeated application of those rules, this will showthat no proof ever leads us astray. Or at least, that is the general idea.

To begin with, we must make the idea of leading us astray more precise.Say that a line of a proof is shiny i the assumptions on which that linedepends tautologically entail the sentence on that line.1 To illustrate the idea,consider the following TFL-proof:

1 F ! (G ^H)2 F

3 G ^H !E 1, 24 G ^E 35 F ! G !I 24

Line 1 is shiny i F ! (G^H) F ! (G^H). You should be easily convincedthat line 1 is, indeed, shiny! Similarly, line 4 is shiny i F ! (G ^H); F G.

1The word shiny is not standard among logicians.

37

5. Soundness 38

Again, it is easy to check that line 4 is shiny. As is every line in this TFL-proof.I want to show that this is no coincidence. That is, I want to prove:

Lemma 5.2. Every line of every TFL-proof is shiny.

Then we will know that we have never gone astray, on any line of a proof.Indeed, given Lemma 5.2, it will be easy to prove the Soundness Theorem:

Proof of Soundness Theorem, given Lemma 5.2. Suppose ` C . Then thereis a TFL-proof, with C appearing on its last line, whose only undischargedassumptions are among . Lemma 5.2 tells us that every line on every TFL-proof is shiny. So this last line is shiny, i.e. C .

It remains to prove Lemma 5.2.To do this, we observe that every line of any TFL-proof is obtained by

applying some rule. So what I want to show is that no application of a ruleof TFLs proof system will lead us astray. More precisely, say that a rule ofinference is rule-sound i for all TFL-proofs, if we obtain a line on a TFL-proof by applying that rule, and every earlier line in the TFL-proof is shiny,then our new line is also shiny. What I need to show is that every rule in TFLsproof system is rule-sound.

I shall do this in the next section. But having demonstrated the rule-soundness of every rule, Lemma 5.2 will follow immediately:

Proof of Lemma 5.2, given that every rule is rule-sound. Fix any line, line n,on any TFL-proof. The sentence written on line n must be obtained using aformal inference rule which is rule-sound. This is to say that, if every earlierline is shiny, then line n itself is shiny. Hence, by strong induction on the lengthof TFL-proofs, every line of every TFL-proof is shiny.

Note that this proof appeals to a principle of strong induction on the length ofTFL-proofs. This is the rst time we have seen that principle, and you shouldpause to conrm that it is, indeed, justied. (Hint: compare the justicationfor the Principle of Strong Induction on Length of sentences from 1.4.)

5.2 Checking each ruleIt remains to show that every rule is rule-sound. This is actually much easierthan many of the proofs in earlier chapters. But it is also time-consuming, sincewe need to check each rule individually, and TFLs proof system has plentyof rules! To speed up the process marginally, I shall introduce a convenientabbreviation: i will abbreviate the assumptions (if any) on which line idepends in our TFL-proof (context will indicate which TFL-proof I have inmind).

Lemma 5.3. Introducing an assumption is rule-sound.

Proof. If A is introduced as an assumption on line n, then A is among n,and so n A .

Lemma 5.4. ^I is rule-sound.

5. Soundness 39

Proof. Consider any application of ^I in any TFL-proof, i.e. s