ClassicalPropNotes.pdf

7/30/2019 ClassicalPropNotes.pdf

1/18

Notes on Classical Propositional Logic

Melvin FittingLehman College, CUNY, 250 Bedford Park Boulevard West, Bronx, NY 10468-1589

CUNY Graduate Center, 365 Fifth Avenue, New York, NY 10016

February 1, 2008, revised August 31, 2010

Contents

1 Introduction 1

2 The Language 23 The Truth Table Method 2

4 Axiom Systems 4

5 The Goal and General Outline 5

6 Lindenbaums Theorem 7

7 Implication and the Deduction Theorem 9

8 Negation 14

9 Implication 15

10 Conjunction 15

11 Disjunction 16

12 Completeness At Last 16

13 Summary of Axiom System 18

1 Introduction

Classical propositional logic is the simplest and most nicely behaved of any logic (whatever thatmeans). In a course discussing a wide variety of logics, this is a natural place to start. Many differentproof procedures have been developed for it: axiom systems, tree (tableau) systems, sequent calculi,natural deduction, resolution, and more. Some of these carry over to some other logics, and somedo not. That is the way of things.

There are many, conceptually different, completeness proofs for axiomatic formulations of clas-sical propositional logic. Often, for a particular way of proving completeness, one set of axioms is

1


2/18

Notes on Propositional Logic 2

easier to work with than another. Then, one way of developing an axiomatic system is to use acompleteness proof as a guide to determining the axioms needed. In these notes we present a stan-dard completeness argument that uses maximal consistent sets. This way of doing things is due toLindenbaum, though it is often described as Henkin-style (incorrectlyhe extended Lindenbaumsideas to incorporate quantication). We use this argument to motivate our choice of axioms.

Our style here is to begin as broadly as possible, then narrow things down as required. So werst talk about axiom systems in general, and the notion of an axiomatic proof. Even withoutspecifying particular axioms or rules, there are still results of interest and use that can be proved.In fact, we go a long way before we need to get specic.

If you have comments or suggestions on these notes, please e-mail them to me.

2 The Language

Before we get to our subject of primary interest, there are some preliminary items we must get outof the way. This is the rst of two sections in which we do this.

Formulas are built up from a countable list of propositional letters , P 1 , P 2 , .... Commonly a

unary operator of negation is allowed. Here it is more convenient to take negation as dened andto assume we have a falsehood connective instead; X abbreviates X . We write the falsehoodconstant as . Then propositional atoms are the propositional letters, together with .

Formulas are built up using various binary connectives . Ternary connectives, quatenary (isthere such a word) connectives, and so on, could be allowed, but it is not necessary since withenough other machinery they can all be dened. We will take as binary connectives: implication,; disjunction, ; and conjunction, . Fewer are possible since some can be dened from others.More are also possible. This is our basic set.

Given all this, here is the denition of the set of propositional formulas. It is a recursivedenition, and says how formulas are generated.

Denition 2.1 (Formula) The set of formulas is the smallest set meeting the following condi-tions:

1. Every propositional atom is a formula;

2. If X and Y are formulas, and is a binary connective, then ( X Y ) is a formula.

Using this denition it is easy to show that, say, (( X Y ) (X Y )) is a formula. It is less easyto show that (( X Y (X Y )) is not a formula. Devising tests for formula-hood is somethingthat is done in automata theory. Here well assume you can recognize formulas when you see them.Further, we will often be somewhat informal and omit outer parentheses when convenient, writing(X Y ) (X Y ) for ((X Y ) (X Y )), for instance.

Exercise 2.1 Find a way of proving that ((X Y (X Y )) is not a formula according to the denition above.

3 The Truth Table Method

Formulas are pure syntax. They have to be assigned some meaning, and this is the job of semantics.For classical propositional logic, the meaning of a formula is simply truth or falsehood, and onediscusses meaning in a particular context. Truth tables are the standard tool for thiseach line


3/18


represents a context. I assume you all have seen truth tables, and I wont go into their details.What I will do is extract their mathematical essence, because it will be convenient later on.

Let us assume we have two truth values , true and false . (Exactly what these are is notimportant, only that they are different from each other.) We assume there are various operationsdened on the set {true , false } of truth values, corresponding to our choice of connectives. For

convenience, we use as both a connective and as the name of an operation on truth values. Theseare really quite different things, but you can tell from context which is meant. It is analogous to thesituation one encounters in mathematics, where one sees similar ambiguity. On the one hand, wemight talk about the leftmost occurrence of the symbol + in the equatiion ( x + y) = ( y + x). Herewe are using syntax. Or we might say 3 + 5 is the number 8. Here we are applying an operation.While these are quite different, there is usually no problem keeping the differences sorted out.

Now, here are our denitions of three operations on the set {true , false } of truth values.

true true true true truetrue false false false truefalse true true false truefalse false true false false

Now to connect these operations with formulas, we have the following.

Denition 3.1 (Boolean Valuation) A boolean valuation is a mapping v from the set of formu-las to the set of truth values that meets the following conditions:

1. v() = false

2. v(X Y ) = v(X ) v(Y )

3. v(X Y ) = v(X ) v(Y )

4. v(X Y ) = v(X ) v(Y )

Take a typical line of this denition, say v(X Y ) = v(X ) v(Y ). The occurrence of on theleft is syntacticalit is part of the formula ( X Y ). The occurrence on the right is the operationfrom the table above.

It can be shown (by induction on formula complexity) that any two boolean valuations thatagree on propositional letters will agree on all formulas. It can also be shown that a booleanvaluation is completely specied by giving its values on propositional letters. And nally, it canbe shown that the value of a boolean valuation v on a formula X is not affected by changing v onpropositional letters that dont occur in X . Well assume all this here.

As an example, suppose v(P 1 ) = true and v(P 2 ) = false . We compute v(P 1 (P 2 )).

v(P 1 (P 2 )) = v(P 1 ) v(P 2 )

= v(P 1 ) (v(P 2 ) v())= true (false false )= true false= false

You should recognize in the calculation above all the steps involved in lling in a truth tableline for v(P 1 (P 2 )) where the line is the one that assigns P 1 the value true and P 2 the valuefalse .

Now we use the semantics just dened to characterize the fundamental notions we study.


4/18


Denition 3.2 (Tautology) A formula X is a tautology if v(X ) = true for every boolean valu-ation v.

Informally, this amounts to saying X is a tautology if every line of a truth table for X assignsto X the value true . The utility of boolean valuations over truth tables is that they make it easierto give mathematical arguments concerning the notions of logic.

Being a tautology has a generalization to being a consequence of a set of formulas.

Denition 3.3 (Semantic Consequence) Let S be a set of formulas, and let X be a singleformula. We say X is a semantic consequence of S just in case every boolean valuation v that mapsevery member of S to true also maps X to true .

Loosely, X is a semantic consequence of S if X must be true whenever all members of S aretrue. Notice that saying X is a tautology is equivalent to saying X is a semantic consequence of .Also note that if X is a tautology, it is a semantic consequence of every set. (Why?)

Exercise 3.1 Let w be a boolean valuation such that w(P 1 ) = w(P 3 ) = true and w(P 2 ) = false .Evaluate w((P 1 P 3 ) (P 2 )) .

4 Axiom Systems

Constructing a truth table for a formula with many propositional letters is a very large task. Formore complex logics, the semantics may not even provide an effective method for determiningvalidity. What is often used instead is some notion of formal proof. Axiom systems provide a verycommon such methodologyan axiom system proof can be a relatively small object compared toa truth table (though a proof may be hard to nd). In this section we set up the basics of theaxiomatic approach, then in subsequent sections we work towards soundness and completeness (andexplain what these terms mean).

To have an axiom system two things are needed: we need some class of formulas called axioms,whose truth is simply assumed; and we need rules of derivation, for producing new truths from old,so to speak. A word about each of these.

Any set of formulas could be taken as a set of axioms, but practicality issues come into it. If wetake the set of all tautologies as axioms, all axiomatic proofs would be trivial, and nothing wouldbe gained over truth tables. Or, we could take as a set of axioms some innite set of formulas whoseGodel numbers (if you know what they are) form a non-recursive set. In this case we would haveno general way of telling what is and what is not an axiom. Reasonable restrictions are needed.

It is very common to specify a set of axioms by giving a set of axiom schemes any formulaof such-and-such form is an axiom. For example, we might say that any formula of the form(X (Y X )) is an axiom. If we did this, among the axioms would be ( P 1 (P 2 P 1 )) andalso ((P 1 P 2 ) (P 3 (P 1 P 2 ))). In this approach ( X (Y X )) is an axiom scheme , and we

have given examples of specic axioms that are instances of this scheme. This is the way we willdo things herewe will specify axiom systems by giving a nite number of axiom schemes.

Rules of derivation must be nitary and effective. Finitary means we need a nite number of formulas to trigger a rule application. Effective means we can tell when a rule can be applied. Inpractice, our rules will be specied by writing something like the following:

X 1 , X 2 , . . . , X nY


5/18


where this means that if we have specic formulas that match the pattern above the line, we canconclude the formula that results, below the line. A familiar example is modus ponens .

X, X Y Y

Now, to get a little more specic (but not very much so yet).Suppose A is a nite set of axiom schemes. We will say a formula is an axiom of A if it is an

instance of one of the schemes in A. And suppose R is a nite set of rules of derivation. These twotogether determine the notions of a derivation, and a proof, as follows.

Denition 4.1 (Derivation, Proof) Let S be a set of formulas (not schemes, and not necessarilynite). By a derivation from S , in the system with axiom schemes A and rules R , we mean anynite sequence of formulas

X 1X 2...

X n

in which each formula is either: an axiom from A, a member of the set S , or follows from earlierformulas using a rule from R . The last formula, X n , is the formula that is derived.

If the set S is empty, we say the derivation is a proof , and the last formula is the formula thathas been proved.

If X has a derivation from S , we symbolize this with S X . If X is provable, that is, if X ,we just write X .

There are some easy items about these notions that can now be established.

Proposition 4.2 (Monotonicity) If S X and S S , then S X .

Proof If S S then any derivation from S is also a derivation from S .

Proposition 4.3 (Compactness) If S X then there is some nite subset S of S such that S X .

Proof Suppose S X , and so we have a derivation of X from S . That derivation has only nitelymany lines, and so only uses a nite number of members of S . Let S be the set of members of S actually used in the derivation. Then S is nite, and the same derivation of X from S is also aderivation of X from S .

5 The Goal and General OutlineWe want to nd an axiom system with a nite number of axiom schemes and rules that can proveexactly the tautologies. More specically, we want an axiom system with a nite number of axiomschemes and rules for which we can prove soundness and completeness , where these terms aredened as follows.

Denition 5.1 (Soundness and Completeness) An axiom system is sound if it only provestautologies. An axiom system is complete if it proves every tautology.


6/18


Each of these separately is easy to manage. An axiom system with no axioms at all cant proveanything, hence is sound (nd me something it proves that is not a tautology). It is not very usefulsince it is badly incomplete. An axiom system that proves every formula is obviously completeitproves everything hence it proves all tautologies. It too is not very useful since it is badly unsound.We need the two conditions together. In fact, soundness is rather easy to manage.

Proposition 5.2 (Soundness) Suppose we have an axiom system that meets the following twoconditions:

1. Every axiom is a tautology.

2. Every rule is sound , which means that any boolean valuation that maps all the premises of a rule application to true must also map the conclusion of the rule to true .

Then the axiom system is sound; it only proves tautologies.

Proof It is easy to see that if the conditions are met, every line of a proof must be a tautology,hence the last line, which is what the proof proves.

In fact a stronger result can be proved, which we leave to you as Exercise 5.1.So, we will make sure to choose axiom schemes whose instances are tautologies, and rules of

derivation that are sound. Completeness is much harder, however. This will occupy the next severalsections. Since it is more complex, here is something of a guide to it.

Suppose we have an axiom system, and we wish to prove it is complete. We must show thatevery tautology is provable. Actually, we will show the converse: if X is not provable, then X isnot a tautology. To show X is not a tautology, we must nd a boolean valuation v that maps X to false . Let us take a look, again, at how boolean valuations behave.

Let v be a boolean valuation, and consider the set of all formulas that v maps to true ; whatis such a set like? Well, the following should be clear. Such a set must contain exactly one of Z

or Z for each formula Z . Such a set must contain Z W exactly when it contains both Z andW . Such a set must contain Z W exactly when it contains at least one of Z or W . And so on.Suppose we could show that if a formula X is not provable in our axiom system, it must fail tobe a member of a set meeting conditions like this. Then we could use the set to create a booleanvaluation mapping X to false , and we would have completeness.

In the next section we will introduce enough machinery so that we can dene a notion of axiomatic consistency . We will show that there are consistent sets that cannot be enlarged withoutbecoming inconsistentthese are called maximally consistent sets. Then we will take as our guidingprinciple: choose our axioms so that maximally consistent sets must have all the properties sketchedabove: exactly one of Z or Z belongs, Z W belongs exactly when both Z and W belong, andso on. Then these maximally consistent sets will serve as the appropriate tools to let us provecompleteness.

In the next section consistency and maximal consistency are dened. In the sections that followwe choose axioms schemes that guarantee maximal consistent sets have the properties we want.And nally, we use them to prove completeness of the axiom system we have built up.

Exercise 5.1 Suppose we have an axiom system for which the two conditions of Proposition 5.2 are met. Show that if S X , then X is a semantic consequence of S (Denition 3.3 ).


7/18


6 Lindenbaums Theorem

We have not yet made any use of the falsehood constant, , that is part of our language. Now it istime. As you read, please notice that we dont really need much about it. We dont actually makeuse of the fact that it represents falsehood, only that it is some particular formula.

Denition 6.1 (Consistency) A set S of formulas is called inconsistent if S . The set S isconsistent if it is not inconsistent, in other words, if S .

There are some easy results about consistency that we can now prove.

Proposition 6.2 Suppose S S . If S is inconsistent, so is S . If S is consistent, so is S .

Proof This follows immediately from Proposition 4.2, taking X to be .

Proposition 6.3 A set S of formulas is consistent if and only if every nite subset of S is con-sistent.

Proof The assertion is equivalent to: S is inconsistent if and only if S has some inconsistent nite

subset.Suppose S is inconsistent. Then S , so by Proposition 4.3, S for some nite S S ,and hence S has an inconsistent nite subset.

Conversely, suppose S has an inconsistent nite subset, S . Then S , and since S S , wehave S by Proposition 4.2, and hence S is inconsistent.

It is common in mathematics to call something maximal if it cannot be made bigger. Thisapplies to the notion of consistency too.

Denition 6.4 (Maximal Consistent) A set S of formulas is maximally consistent if: 1) it isconsistent, but 2) no proper enlargement is consistent.

Another, very useful, way of saying part 2) of the denition above is this. Suppose that,whenever S {X } is consistent (that is, the result of adding X to S is consistent), it turns outthat X was already in S . Then any enlargement of S that is consistent turns out not to be reallyan enlargement, so S has no proper enlargements that are consistent. This says S is maximal. So,our test for maximality of a consistent set is:

if S {X } is consistent, then X S

Maximally consistent sets have many nice and useful properties. For instance, they are closedunder derivation.

Proposition 6.5 Suppose S is a maximally consistent set. If S X , then X S .

Proof Assume S X , and S is maximally consistent. If we can show that S {X } is consistent, itfollows that X S , so this is our goal. We do this by contradiction. Let us suppose that S {X }is not consistent, and derive a contradiction.

Suppose S {X } . (Recall, we are trying for a contradiction.) Then we have a derivationof from S {X }. Say it is

Z 1Z 2...

Z n


8/18


where Z n is . Each line is an axiom, comes from earlier lines by a rule of derivation, or is amember of S , or is X (since this is a derivation from S {X }). Suppose some line, say Z i , isactually the formula X . Since X itself has a derivation from S , simply insert between lines Z i 1and Z i all the steps of a derivation of X from S . Do this for each occurrence of X . This convertsthe derivation from S {X } into one from S alone. Since the derivation ends with , we have

shown that S is inconsistent, contrary to our assumption that S was maximally consistent, andhence consistent.

We have, indeed, established several nice properties but this is not enough, because it is notobvious that there are any maximally consistent sets at all. If there were none, they would all haveevery nice property you might dream of, and this would be of no use at all. Fortunately a famousresult due to Lindenbaum says there are lots of maximally consistent sets. Before we get to this, itwill be useful to have the following availableit plays a central role in establishing Lindenbaumsresult.

Proposition 6.6 (Chain Limit) Suppose we have a sequence of sets of formulas, S 1 , S 2 , S 3 , . . . ,so that S 1 S 2 S 3 . . . . (Such things are called chains .) Let S = S 1 S 2 S 3 . . . . Then

S is a kind of limit of the chain, and we have S 1 S 2 S 3 . . . S . We claim that, for any formula X :

S X if and only if S n X for some positive integer n.

Proof The proof has two parts.Part 1: Suppose S n X for some positive integer n . Since S n S we have S X by

Proposition 4.2.Part 2: Suppose S X . By Proposition 4.3. S X for some nite S S .We rst show that S S n for some positive integer n. To keep the clutter down, lets say

that S has three membersthe general case follows the same idea. Say S = {Z 1 , Z 2 , Z 3 }. SinceS S , Z 1 S , so by the denition of S , Z 1 S a for some integer a. Similarly Z 2 S b and

Z 3 S c for some integers b and c. Let n be whichever of a, b, c is largest. Since we have a chainstructure, all of Z 1 , Z 2 , and Z 3 will be in S n , and thus S S n .

We now have that S S n for some integer n. Since S X we also have S n X byProposition 4.2.

Now for the main event.

Theorem 6.7 (Lindenbaums Theorem) Let S be any consistent set. Then S S for some maximally consistent set S .

This is often stated informally as every consistent set can be extended to a maximally consistent one.

Proof We have in our language a countable list of propositional letters. There is a simple resultof set theory, that would take us too far out of our way to prove here, that says we have thefollowing consequence: the set of all formulas of our language is countable. This means it ispossible (constructively, in fact), to set up an innite list, X 1 , X 2 , X 3 , . . . , in such a way that everyformula appears somewhere in the list. Let us assume this has been done.

We now dene a sequence of sets, S 1 , S 2 , S 3 , ..., as follows.To start things off, S 1 = S . Since S is consistent by assumption, the sequence of sets starts

with a consistent set.


9/18


Next, suppose S n has been dened, and is consistent. We dene the next set as follows.

S n +1 =S n {X n } if S n {X n } is consistentS n if S n {X n } is not consistent

Notice that, in either case, S n +1 must be consistent. Notice also that S n S n +1 . We thus have a

chain structure, as in Proposition 6.6. Let S = S 1 S 2 S 3 . . .. Then we have:S 1 S 2 S 3 . . . S

It is obvious that S extends S = S 1 . We claim that S is consistent, and maximal, and sotaking S to be S will nish the proof.

First, S is consistent because, if it were not, S , and hence S n for some integer n, byProposition 6.6, which would mean that S n is not consistent. But by construction, S n is consistentfor every integer n .

Next we show that S is maximal. Suppose S {Z } is consistent; we must show that Z S .Well, since X 1 , X 2 , X 3 , ...is a listing of all formulas, Z is somewhere in the list; say Z is formulaX k . Then S k {X k } must be consistent, for if it were not, S {X k } would not be consistent

either, by Proposition 4.2, but this is S {Z }, and our assumption is that this is consistent. SinceS k {X k } is consistent, it is this set that constitutes S k +1 , hence Z = X k S k +1 S .

7 Implication and the Deduction Theorem

This section is a bit of a detour on the route to choosing axioms to make maximally consistentsets behave well. We will take as a kind of basic connective, and for this we have a fundamentalresult due independently to Tarski and Herbrand. The result says that if we pick the right axiomsand rules for implication, we can prove an extremely useful result called the Deduction Theorem.The Theorem tells us that the notion of derivation, from Denition 4.1, and implication are closelyconnected. But rst, what axioms and what rules must we adopt? We begin with the following

very common rule of derivation.

Rule of Derivation, Modus Ponens

X X Y Y

And next, we adopt two axiom schemes for the implication connective.

1 Axiom Scheme X (Y X )

2 Axiom Scheme (X (Y Z )) ((X Y ) (X Z ))

Let us give an (important) example of a proof in the axiom system we have thus far. We willshow A A is provable, for any formula A. We do not claim a proof is easy to discover, but hereis a proof. Line numbers are so that we can give explanations.

1. A ((W A) A)2. (A ((W A) A)) ((A (W A)) (A A))3. (A (W A)) (A A)4. A (W A)5. A A


10/18


In this line 1 is an axiom; in 1 take X to be A and Y to be W A, where W is any formula youmay choose; its exact value wont matter. Line 2 is an instance of 2; take X to be A, Y to beW A and Z to be A. Line 3 follows from lines 1 and 2 by modus ponens . Line 4 is an instance of 1, take X to be A and Y to be W . Finally, line 5 follows from lines 3 and 4 by modus ponens .

As we noted above, this proof is not easy to discover. In fact, ( A B ) ((B C ) (A C ))

has a proof too, and you might convince yourselves that it is really difficult to nd (Ill bet youdont nd it, in fact). The power of the Deduction Theorem is that it gives us an easier way of discovering such proofs. Recall that in Denition 4.1 we had a notion of derivation, as well as anotion of proof. It is the notion of derivation, for the axiom system given so far, that we use now.

Theorem 7.1 (Deduction) Let S be a set of formulas, and let X and Y be single formulas. In the axiom system given so far, and in every extension of it that results by adding further axiom schemes, if S {X } Y , then S (X Y ). Further, there is an algorithm for converting a derivation showing S {X } Y into one showing S (X Y ).

Proof Let us assume we have a derivation of Y from S {X }, say it looks like this:

Z 1Z 2Z 3...

Z n 1Z n

In this derivation the last line is Y , that is, Z n = Y . And otherwise, since it is a derivation fromS {X }, each line is either an axiom, or a member of S {X }, or comes from earlier lines by modus ponens . Members of S {X } are members of S , and X itself. Our goal is to turn this derivationinto one of X Y , that is, X Z n , from S alone.

As a rst step, append X to the beginning of each line, geting the following sequence.

X Z 1X Z 2X Z 3

...X Z n 1X Z n

This has the last line we want, but it is very unlikely that it is a legal derivation. Now we add somelines to turn it into one.

Consider one of the lines above, say X Z k . We are going to insert some lines immediatelybefore this. What lines depends on the justication for Z k being in the original derivation, and sothere are three different cases.

Case 1: Z k is an axiom. Then expand the line X Z k in the second sequence to the following:

Z kZ k (X Z k )

X Z k

We can now justify the presence of each of these. The rst line is an axiom, because that is thecase we are in. The second line is an axiom, an instance of scheme 1. The third follows from


11/18


the rst two by modus ponens . All this is allowed in a derivation from S , indeed, in a derivationfrom anything.

Case 2: Z k is a member of S . This is handled exactly as in Case 1, we expand the line in thesame way. The only difference is that now the presence of Z k is justied because it is a member of S , and so is allowed in a derivation from S .

Case 3: Z k is X . In this case the formula Z k X is just X X . We know it has an axiomaticproof; we showed this earlier. Insert the steps of that proof just above the line Z k X .

Case 4: Z k comes from earlier lines by modus ponens . That is, there are lines Z i and Z jwhere i, j < k , and where Z j = ( Z i Z k ). Then, in the second list we must have X Z i andX (Z i Z k ) somewhere before the line X Z k . This time, expand the line X Z k in thesecond list to the following.

(X (Z i Z k )) ((X Z i ) (X Z k ))(X Z i ) (X Z k )

X Z k

The rst of these formulas is an axiom, an instance of scheme 2. Since X (Z i Z k ) occurssomewhere earlier, the second line follows by modus ponens . And then, since X Z i also occurssomewhere earlier, the third line also follows by modus ponens .

By adding these extra lines, we have converted the sequence of formulas into a proper derivation,and only members of S are now used as assumptions.

Here is an example to show how useful this theorem can be. Earlier we said that ( A B ) ((B C ) (A C )) was provable, but it was hard to nd a proof. Heres how to nd one. Firstwe show there is a derivation to justify the following.

{A B, B C, A} C

Here is the derivation, with line numbers added for convenience.1. A2. A B3. B4. B C 5. C

In this, line 1 is one of the premises. So is line 2. Line 3 follows from 1 and 2 by modus ponens .Line 4 is a premise, and line 5 follows from 3 and 4 by modus ponens .

Now, we convert this derivation into one showing

{A B, B C } A C

by following the algorithm given in the proof of Theorem 7.1. First, we append A to each line,getting the following.

1. A A2. A (A B )3. A B4. A (B C )5. A C


12/18


Next, we insert some lines to turn this into a legal derivation from {A B, B C }.Line 1 is provablewe insert the lines of the proof, as given earlier. This is case 3 in the proof

of the deduction theorem.

A ((W A) A)(A ((W A) A)) ((A (W A)) (A A))(A (W A)) (A A)A (W A)

1. A A2. A (A B )3. A B4. A (B C )5. A C

Line 2 in the original derivation is a premise other than A, so we are in case 2 of the deductiontheorem, and we insert lines as follows.


1. A AA B(A B ) (A (A B ))

2. A (A B )3. A B4. A (B C )5. A C

The rst of the new lines is a member of {A B, B C }, the second is an axiom, and the linenumbered 2 follows by modus ponens .Next, line 3 of the original derivation, B , comes from lines 1 and 2 by modus ponens . We must

justify the presence of line number 3 in the new derivation, A B . As it happens, this is a memberof {A B, B C } so we could just assume it, or we could follow the steps of the algorithmexactly, adding some more lines. To keep things (relatively) simple, well just accept it as one of the premises.


13/18


Line 4 in the original derivation, B C , is a premise, and so case 2 of the algorithm has usinserting extra lines into the new derivation, as follows.


1. A AA B(A B ) (A (A B ))

2. A (A B )3. A B

B C (B C ) (A (B C ))

4. A (B C )5. A C

The rst of these new lines is a member of {A B, B C }. The second is an axiom, an instanceof 1. Then line number 4 follows from these by modus ponens .

Finally, in the original derivation line 5 followed from lines 3 and 4 by modus ponens . This putsus in case 4 of the algorithm, and in the new derivation we insert extra lines, as follows.


1. A AA B(A B ) (A (A B ))

2. A (A B )3. A BB C (B C ) (A (B C ))

4. A (B C )(A (B C )) ((A B ) (A C ))(A B ) (A C )

5. A C

The rst of these new lines is an instance of axiom scheme 2. The second follows from this andline 4 by modus ponens . Then line 5 follows from the second of the new lines and line 3, by modus ponens again.

We now have a properly constructed derivation showing the following{A B, B C } A C

Now, use the Deduction Theorem again, to produce a derivation showing

{A B } (B C ) (A C )

and then once more to get a derivation showing

(A B ) ((B C ) (A C ))


14/18


and since this is a derivation from the empty set, we nally have created a proof of ( A B ) ((B C ) (A C )), as promised. And, as promised, it is long and complex, which is why wehavent actually given the complete proof.

Remark Before closing this section, we should note that our two axiom schemes and the ruleof modus ponens dont quite give us everything needed for classical negation. In fact, what isaxiomatized so far is the implication of intuitionistic logic . We dont prove that here, or evenexplain what it means. Were just warning you.

Exercise 7.1 Show how the Deduction Theorem can be used to establish the provability of (A (B C )) (B (A C )) .

8 Negation

We are now starting the business of adding axioms so that maximally consistent sets have the rightproperties to correspond to boolean valuations. We begin with negation, which is actually a denedoperatorrecall that a falsehood constant, , is part of our language, though we have made nospecial axiomatic assumptions about it yet. We wont until the next section, in fact.

Denition 8.1 (Negation) The expression X is taken as an abbreviation for X .

Please note, we could have taken negation as basicit is commonly done. Our way makescertain things easier to establish; thats all.

A boolean valuation maps exactly one of X or X to true . If we want maximally consistent setsof formulas to correspond to boolean valuationsthey should be the sets that boolean valuationsmap to true we should have the following. We have called it a Proposition, because in fact wedo have it.

Proposition 8.2 Let M be a maximally consistent set of formulas. For each formula X , exactly one of X or X is in M .

Proof First we show we cant have both X and X in M . Well, suppose X M and ( X ) M .Then a simple application of modus ponens tells us M , and so M is inconsistent, contrary tothe assumption that it is a maximally consistent set.

Next we show we must have at least one of X or X in M . If X M , we are done. Nowsuppose X M . If M {X } were consistent, X M since M is maximally consistent, but thisis not the case. So, M {X } is not consistent, that is, M {X } . Then, using the DeductionTheorem, 7.1, M (X ). Since M is maximally consistent, ( X ) M by Proposition 6.5,that is, X M .

Remark The negation we have axiomatized so far is not yet the one of classical propositional logic.It is not even the one of intuitionistic logic (see the Remark at the end of the previous section).It is the negation of a relatively unknown logic called minimal propositional logic , introduced byJohansson in 1937. It becomes the negation of intuitionistic logic by adding the axiom scheme X (roughly, anything follows from a falsehood). It becomes classical negation if we also add X X . Eventually we will be forced to add both of these, the rst in the next section and thesecond toward the end.


15/18


9 Implication

We continue toward our goal of making maximally consistent sets correspond to boolean valuations.In this section we t implication into the pattern and we will see that this forces us to add onemore axiom scheme.

A boolean valuation maps X Y to true exactly when it either does not map X to trueor does map Y to true . If maximally consistent sets are to correspond to boolean valuations weshould have, for each such set M : (X Y ) M if and only if X M or Y M . We do have partof this.

Proposition 9.1 Let M be a maximally consistent set of formulas. If (X Y ) M then either X M or Y M .

Proof Suppose ( X Y ) M . If X M we are done. If X M then it follows that M Y , andhence Y M because M is maximally consistent, Proposition 6.5.

We dont quite have the converse, however. Certainly, if Y M then ( X Y ) M because

maximally consistent sets are closed under derivability, and Y (X Y ) is an axiom. But if X M we cant conclude ( X Y ) M . Heres the best we can do. If X M , by Proposition 8.2we must have ( X ) M . Thats not quite what we want, but if we had Y as an axiomwe would be able to conclude ( X Y ) M . In fact, all formulas of this form are tautologies, soadding them as axioms leaves us with a sound system. We now make this official.

Axiom Scheme Z

We can now state the followingwe have already given the proof. Please note that from thispoint on the axiom scheme above is part of our system.

Proposition 9.2 Let M be a maximally consistent set of formulas. Then (X Y ) M if and only if X M or Y M .

10 Conjunction

We continue adding axioms to make maximally consistent sets correspond to boolean valuations.Each boolean valuation maps X Y to true exactly when it maps X to true and also Y to true .We must add enough axioms so that if M is a maximally consistent set, ( X Y ) M if and onlyif X M and Y M . And of course our axioms should be tautologies, so that we still have asound axiomatic system.

Making sure that ( X Y ) M implies X M and Y M is simple. We take the followingtwo axiom schemes.

1 Axiom Scheme (X Y ) X

2 Axiom Scheme (X Y ) Y

Making sure that X M and Y M implies (X Y ) M is almost equally simple. We adoptthe following axiom scheme.

3 Axiom Scheme X (Y (X Y ))


16/18


We now have the following.

Proposition 10.1 Assume the three axioms for conjunction given above have been added to the axiom system. Then, if M is any maximally consistent set, (X Y ) M if and only if X M and Y M .

Exercise 10.1 Supply a proof for Proposition 10.1.

11 Disjunction

Disjunction is the last of the basic binary connectives we will consider. Each boolean valuationmaps X Y to true exactly when it maps at least one of X or Y to true . We must add axioms toensure that maximally consistent sets mimic this behavior: ( X Y ) M if and only if X M orY M . One direction is simple; the one from right to left. We adopt the following axiom schemes(which are tautologies).

1 Axiom Scheme X (X Y )

2 Axiom Scheme Y (X Y )

We still need to guarantee the other direction holds: if ( X Y ) M then X M or Y M .To achieve this, we reason as follows. What we desire, written in the contrapositive form, is: if X /M and Y /M then ( X Y ) /M . And recall, from Proposition 8.2, for any formula Z ,Z /M exactly when ( Z ) M . So what we desire can be restated as: if ( X ) M and(Y ) M then (( X Y ) ) M . This leads us to the following axiom scheme.

3 Axiom Scheme (X ) ((Y ) ((X Y ) )

This does the job.

Proposition 11.1 Assume the three axioms for disjunction given above have been added to the axiom system. Then if M is any maximally consistent set, (X Y ) M if and only if X M or Y M .

Remark Axiom scheme 3 could have been stated using our dened negation symbol. It becomesX (Y (X Y )), which is part of one of deMorgans laws. We could also have adoptedan axiom scheme that looks more general: ( X Z ) ((Y Z ) ((X Y ) Z ). This is alsoa tautology, so it is safe to adopt, and it has 3 as a special case. As it happens, the narrowerversion we did take is sufficient to get us a complete axiom system.

Exercise 11.1 Supply a proof for Proposition 11.1.

12 Completeness At Last

We have (almost) everything we need to prove completeness for our axiom system. It turns outone axiom scheme is still missing. We will let it arise in its natural place, rather than introducingit up front.

We would like to show that every tautology has a proof in our axiom system. We go aboutthings in the contrapositive direction: we show that if X does not have a proof, then X is not atautology.


17/18


Assume X does not have a proof. We have arranged things so that maximally consistent setsand boolean valuations correspond closely. Suppose we could nd a maximally consistent set M that did not contain X ; then we could easily show X was not a tautology as follows. Dene amapping v by setting v(Z ) = true if Z M and v(Z ) = false if Z /M , for every formula Z . Wehave adopted enough axiom schemes to guarantee that v will be a boolean valuation. Since X /M

it must be that v(X ) = false , and so X is not a tautology.So, we must establish that if X does not have a proof, there is some maximally consistent setM that does not contain X . By Proposition 8.2, M will omit X just in case M contains X . Sowe must nd a maximally consistent set containing X . Lindenbaums Theorem, 6.7, providesus with a way of creating maximally consistent sets. According to it, if {X }were consistent,it could be extended to a maximally consistent set M , and we would be done.

So nally, we must establish that if X is not provable, then {X } is consistent. And it ishere that we are forced to add our nal axiom. Let us try proving that {X } is consistent,where X itself is not provable. We will attempt to do this by contradiction; we begin by supposingthat {X } is not consistent. Well, if {X }werent consistent, {X } . Then by theDeduction Theorem, 7.1, (X ) would be a theorem. We know X is not a theorem. Tomake a connection between these, we add our nal axiom scheme.

Double Negation Axiom Scheme ((X ) ) X , or equivalently, X X

With this scheme added, we are nished. All the parts of a completeness proof are now present.To bring it all together, we restate the work above, but this time as a proper proof rather than asan exploration of the landscape, so to speak.

Theorem 12.1 (Completeness) Assume all the axiom schemes given so far, and the modusponens rule. If X is a tautology, then X has an axiomatic proof.

Proof We actually show that if X does not have an axiomatic proof then X is not a tautology.

So, assume X is not provable. Then the set {X } is consistent, for otherwise ( X ) would be provable, using the Deduction Theorem, and then X would be provable, using the DoubleNegation Axiom Scheme. Since {X } is consistent, there is some maximally consistent set M such that {X } M , by Lindenbaums Theorem. Now dene a mapping v by setting

v(Z ) = true if Z M false if Z /M

Over the last several sections we have veried that such a mapping v meets all the conditions of a boolean valuation. Since ( X ) M , then X /M , so v(X ) = false , and thus X is not atautology.

In fact, a stronger version of completeness can also be shown.Theorem 12.2 (Strong Completeness) Assume all the axiom schemes given so far, and the modus ponens rule. Let S be a set of formulas and X be a single formula. If X is a semantic consequence of S (Denition 3.3 ), then X has an axiomatic derivation from S , that is, S X .

Exercise 12.1 Give a proof for Theorem 12.2 .


18/18


13 Summary of Axiom System

Our axioms have been presented over several sections. Here is the whole system in one place. First,our rule of derivation.

Rule of Derivation, Modus Ponens

X X Y Y

And then our axiom schemes.

1 Axiom Scheme X (Y X )

2 Axiom Scheme (X (Y Z )) ((X Y ) (X Z ))

Axiom Scheme Z

1 Axiom Scheme (X Y ) X

2 Axiom Scheme (X Y ) Y

3 Axiom Scheme X (Y (X Y ))

1 Axiom Scheme X (X Y )

2 Axiom Scheme Y (X Y )

3 Axiom Scheme (X ) ((Y ) ((X Y ) )

Double Negation Axiom Scheme ((X ) ) X , or equivalently, X X

Exercise 13.1 We did not consider equivalence, , as a basic connective, but we could have. We could also have considered the Sheffer stroke connective (also known as NAND), , or the joint denial connective (also known as NOR), . Here is how they behave semanticallythat is, here are the boolean valuation conditions for them.

true true true true falsetrue false false true falsefalse true false true falsefalse false true false true

1. Invent appropriate axiom schemes for and show the completeness proof extends to incor-porate them. That is, one can prove completeness where formulas are al lowed to contain as a connective.

2. Do the same for .

3. Do the same for .

Documents

ClassicalPropNotes.pdf