# MATH20302 Propositional Logic. Mike Prest School of Mathematics Alan Turing Building Room

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1 MATH20302 Propositional Logic Mike Prest School of Mathematics Alan Turing Building Room April 10, 2015

2 Contents I Propositional Logic 3 1 Propositional languages Propositional terms Valuations Beth trees Normal forms Adequate sets of connectives Interpolation Deductive systems A Hilbert-style system for propositional logic Soundness Completeness A natural deduction system for propositional logic II Predicate Logic 32 3 A brief introduction to predicate logic: languages and structures Predicate languages The basic language Enriching the language L-structures Some basic examples Definable Sets If you come across any typos or errors, here or in the examples/solutions, please let me know of them. 1

3 Introduction: The domain of logic By logic I mean either propositional logic (the logic of combining statements) or first-order predicate logic (a logic which can be used for constructing statements). This course is mostly about the former; we will, however, spend some time on predicate logic in the later part of the course. In any case, propositional logic is a part of predicate logic so we must begin with it. Predicate Logic is dealt with thoroughly in the 3rd/4th-year course by that title; other natural follow-on courses from this one are Model Theory and Non-Standard Logics. Propositional logic can be seen as expressing the most basic laws of thought which are used not just in mathematics but also in everyday discourse. Predicate logic, which can also be thought of as the logic of quantifiers, is strong enough to express essentially all formal mathematical argument. Most of the examples that we will use are taken from mathematics but we do use natural language examples to illustrate some of the basic ideas. The natural language examples will be rather bare, reflecting the fact that these formal languages can capture only a small part of the meanings and nuances of ordinary language. There are logics which capture more of natural language (modality, uncertainty, etc.) though these have had little impact within mathematics itself (as opposed to within philosophy and computer science), because predicate logic is already enough for expressing the results of mathematical thinking. 1 1 One should be clear on the distinction between the formal expression of mathematics (which is as precise and as formal as one wishes it to be) and the process of mathematical thinking and informal communication of mathematics (which uses mental imagery and all the usual devices of human communication). 2

4 Part I Propositional Logic 3

5 Chapter 1 Propositional languages 1.1 Propositional terms Propositional logic is the logic of combining already formed statements. It begins with careful and completely unambiguous descriptions of how to use the propositional connectives which are and, or, not, implies. But first we should be clear on what is meant by a statement (the words assertion and proposition will be used interchangably with statement ). The distinguishing feature of a statement is that it is either true or false. The moon is made of cheese is a (false) statement and = 2 is a (true, essentially by definition) statement. Fortunately, in order to deal with the logic of statements, we do not need to know whether a given statement is true or false: it might not be immediately obvious whether = is true or false but certainly it is a statement. A more interesting example is There are infinitely many prime pairs. where by a prime pair we mean a pair, p, p + 2, of numbers, two apart, where both are prime (for instance 3 and 5 form a prime pair, as do 17 and 19 but not 19 and 21). It is a remarkable fact that, to date, no-one has been able to determine whether this statement is true or false. Yet it is surely 1 either false (after some large enough number there are no more prime pairs) or true (given any prime pair there is always a larger prime pair somewhere out there). On the other hand, the following are not statements. Is 7 a prime number? Add 1 and 1. The first is a question, the second a command. What about x is a prime number. : is this a statement? The answer is, It depends. : if the context is such that x already has been given a value then it will be a statement (since then either x is a prime number or it is not) but otherwise, if no value (or other sufficient information) has been assigned to x then it is not a statement. Here s a silly example (where we can t tell whether something is a statement or not). Set x = 7 if there are infinitely many prime pairs but leave the value of 1 There are some issues there but they are more philosophical than mathematical. 4

8 are using then we will let L ( L for language ) denote the set of propositional variables. We also introduce notation for the set of propositional terms built up from these, namely, set S 0 L = L and, having inductively (on n) defined the set S n L we define S n+1 L to be the set of all propositional terms which may be built from S n L using a single propositional connective and, so as to make this process cumulative, we also include S n L in S n+1 L. More formally: S n+1 L = S n L {(s t), (s t), ( s), (s t), (s t) : s, t S n L}. We also set SL = n 0 S nl to be the union of all these - the set of all propositional terms (sometime called sentences, hence the S in SL ) which can be built up from the chosen base set, L = S 0 L, of propositional variables. 3 Notice that we place parentheses around all the propositional terms we build; we discussed already that leaving these out could give rise to ambiguity in reading them: was s t u - a term in S 2 L - built up by applying to s, t u S 1 L or by applying to s t, u S 1 L, that is, should it be read as s (t u) or as (s t) u? In practice we can omit some pairs of parentheses without losing unique readability but, formally, those pairs are there. In fact, although intuitively it might at first seem obvious that if we look at a propositional term in SL, then we can figure out how it was constructed - that is, there is a unique way of reading it - a bit more thought reveals that there is an issue: how do we detect the last connective in its construction? Clearly, if we can do that then we can proceed inductively to reconstruct its construction tree. We have been precise in setting things up so we should be able to prove unique readability (if it is true - which it is, as we show now). Theorem Let s SL be any propositional term. Then exactly one of the following is the case: (a) s is a propositional variable; (b) s has the form (t u) for some t, u SL; (c) s has the form (t u) for some t, u SL; (d) s has the form ( t) for some t SL; (e) s has the form (t u) for some t, u SL; (f) s has the form (t u) for some t, u SL. Proof. Every propositional term s does have at least one of the listed forms: because s SL it must be that s S n L for some n and then, just by the definitions of S 0 L and S n+1 L, s does have such a form. We have to show that it has a unique such form. For this we introduce two lemmas and the following definitions: if s SL then by l(s) we denote the number of left parentheses, (, occurring in s and by r(s) we denote the number of right parentheses, ), occurring in s (for purposes of this definition we count all the parentheses that should be there). Lemma For every propositional term s we have l(s) = r(s). Proof. This is an example of a proof by induction on complexity/construction of terms. 3 A word about notation: I will tend to use p, q, r for propositional variables, s, t, u for propositional terms (which might or might not be propositional variables) and v for valuations (see later). That rather squeezes that part of the alphabet so I will sometimes use other parts and/or the Greek alphabet for propositional variables and terms. 7

9 If s S 0 L then l(s) = 0 = r(s) so the result is true if s S 0 L. For the induction step, suppose that for every s S n L we have l(s) = r(s). Let s S n+1 L; then either there is t S n L such that s = ( t) or there are t, u S n L such that s = (t u) or (t u) or (t u) or (t u). Since t, u S n L, we have l(t) = r(t) and l(u) = r(u) by the inductive assumption. In the first case, s = ( t), it follows that l(s) = 1+l(t) = 1+r(t) = r(s), as required. In the second case, s = (t u), we have, on counting parentheses, l(s) = 1 + l(t) + l(u) and r(s) = r(t) + r(u) + 1, and so l(s) = r(s), as required. The other three cases are similar and so we see that in all cases, l(s) = r(s). Thus the inductive step is proved and so is the lemma. Digression on proof by induction on complexity. At the start of the proof of above I said that the proof would be by induction on complexity of terms but you might have felt that the proof was shaped as a proof by induction on the natural numbers N = {0, 1, 2,... }. That s true; we used the sets S n L to structure the proof, and the proof by induction on complexity of terms was reflected in the various subcases that were considered when going from S n L to S n+1 L. But the proof could have been given without reference to the sets S n L. The argument - the various subcases - would be essentially the same; the hitherto missing ingredient is the statement of the appropriate Principle of Induction. Recall that, for N that takes the form Given a statement P (n), depending on n N, if P (0) is true and if from P (n) we can prove P (n + 1), then P (n) is true for every n N. 4 The corresponding statement for our construction tree for propositional terms is: Given a statement P (s), depending on s SL, if P (p) is true for every propositional variable p and if, whenever P (s) and P (t) are true so are P (s t), P (s t), P ( s), P (s t) and P (s t), then P (s) is true for every s SL. Before the next lemma, notice that every propositional term can be thought of simply as a string of symbols which, individually, are either: propositional variables (p, q etc.), connectives (,,,, ), or parentheses (left, right). Then the statement that s, as a string, is, for instance, xyz will mean that x, y, z are strings and, if we place them next to each other in the given order, then we get s. For instance if s is (s (t u)) then we could write s as xyz where x, y, z are the strings x =, y = (s, z = (t u)); we could even write s = xyzw with x, y, z as before and w the empty string (which we allow). We define the length, lng(x), of any string x to be the number of occurrences of symbols in it. We extend the notations l(x) and r(x) to count the numbers of left parentheses, right parentheses in any string x. If the string x has the form yz then we say that y is a left subword of x, a proper subword if z ; similarly z is a right subword of x, proper if y is not the empty string. (We will use the terms string and word interchangably.) Proposition For every propositional term s, either s is a propositional variable or there is just one way of writing s in either of the forms s = ( t) for some propositional term t or s = (t u) for some propositional terms t, u where is one of the binary propositional connectives. Proof. We can suppose that s is not a propositional variable. Note that if s has the form ( t) then the leftmost symbols of s are (, whereas if s has the 4 I follow the convention that 0 is a natural number; not followed by everyone but standard in mathematical logic. 8

10 form (t u) then its leftmost symbols are (( or (p where p is a propositional variable, so we can treat these two cases entirely separately. In the first case, s = ( t), this is the only possible way of writing it in this form because t is determined by s. Therefore, since, as we observed above, it cannot be written in the form (t u), there is no other way of writing s as a propositional term. In the second case, we argue by contradiction and suppose that we can write s = (t u) = (t u ) where t, u, t, u are propositional terms and, are propositional connectives and, for the contradiction, that these are not identical ways of writing s, hence that either t is a proper left subword of t or t is a proper left subword of t. A contradiction will follow immediately once we have proved the following lemma. Lemma If s is a propositional term and if s is a proper left subword of s then either s = or l(s ) r(s ) > 0; in particular s is not a propositional term. Similarly, if s is a proper right subword of s then either s = or r(s ) l(s ) > 0, and s is not a propositional term. Proof. We know that s has the form ( t) or (t u). In the first case, s has one of the forms, ( or ( t where t is a left subword (possibly empty) of t. By induction on lengths of propositional terms we can assume that t = or l(t ) r(t ) 0 ( > if t is a proper left subword of t, = by in the case t = t) and so, in each case, it follows that l(s ) r(s ) > 0. In the second case, s has one of the forms, (, (t where t is a left subword of t, (t u where u is a left subword of u. Again by induction on lengths of propositional terms we can assume that l(t ) r(t ) 0 and l(u ) r(u ) 0; checking each case, it follows that l(s ) r(s ) > 0. By we deduce that s is not a propositional term. Similarly for the assertion about right subwords. 1.2 Valuations Now for the key idea of a (truth) valuation. Fix some set L = S 0 L of propositional variables, and hence the corresponding set SL of propositional terms. A valuation on the set of propositional terms is a function v : SL {T, F} to the 2-element set 5 {T, F} which satisfies the following conditions. 6 For all propositional terms s, t we have v(s t) = T iff v(s) = T and v(t) = T; v(s t) = T iff v(s) = T or v(t) = T; v( s) = T iff v(s) = F; v(s t) = T iff v(s) = F or v(t) = T; v(s t) = T iff the values of v(s) and v(t) are the same: v(s) = v(t). 5 really, the two-element boolean algebra 6 Of course, T represents true and F false. Often the 2-element set {1, 0} is used instead, normally with 1 representing true and 0 representing false. 9

11 There s quite a lot to say about this definition. We start with a key point. Namely, because all propositional terms are built up from the propositional variables using the propositional connectives, any valuation is completely determined by its values on the propositional variables (this, see 1.2.1((b) below, is the formal statement of the point we made (the crucial observation ) when discussing mice, cheese and homology groups). For instance if v(p) = v(q) = T and v(r) = F then we have, since v is a valuation, v(p r) = T and hence v( (p r)) = F. Similarly, for any propositional term, t, built from p, q and r, the value v(t) is determined by the above choices of v(p), v(q), v(r). That does actually need proof. There is the, rather obvious and easily proved by induction, point that this process works (in the sense that it gives a value), but there s also the more subtle point that if there were more than one way of building up a propositional term then, conceivably, one construction route might lead to the valuation T and the other to F. But we have seen already in that this does not, in fact, happen: every propositional term has a unique construction tree. Therefore if v 0 is an function from the set, S 0 L, of propositional variables to the set {F, T} then this extends to a unique valuation v on SL. In particular, if there are n propositional variables there will be 2 n valuations on the propositional terms built from them. We state this formally. Proposition Let L be a set of propositional variables. (a) If v 0 : L {F, T} is any function then there is a valuation v : SL {F, T} on propositional terms in L such that v(p) = v 0 (p) for every p L. (b) If v and w are valuations on SL and if v(p) = w(p) for all p L then v = w (so the valuation in part (a) is unique). (c) If t is a propositional term and if v and w are valuations which agree on all propositional variables occurring in t then v(t) = w(t). The proof of part (c), which is a slight strengthening of (b), is left as an exercise. In order to prove it we could prove the following statement first (by induction on complexity of terms): if L L are sets of propositional variables then for every n, SL n S n L; furthermore, if v is a valuation on SL and v is a valuation on SL such that v(p) = v (p) for every p L then v(t) = v (t) for every t SL. From that, part (c) follows easily (take L to be the set of propositional terms actually occurring in t). (You might have noticed that I didn t actually define what I mean by a propositional variable occurring in a propositional term; I hope the meaning is clear but it is easy to give a definition by, what else, induction on complexity of terms.) Truth tables are tables showing evaluation of valuations on propositional terms. They can also be used to show the effect of the propositional connectives on truth values. Note that or is used in the inclusive sense ( one or the other or both ) rather than the exclusive sense ( one or the other but not both ). p q p q T T T T F F F T F F F F p q p q T T T T F T F T T F F F 10

12 p q p q T T T T F F F T T F F T p p T F F T p q p q T T T T F F F T F F F T You might feel that the truth table for does not capture what you consider to be the meaning of implies but, if we are to regard it as a function on truth values (whatever the material connection or lack thereof between its input propositions) then the definition given is surely the right one. Or just regard p q as an abbreviation for p q (not-p) or q, since they have the same truth tables. The following example might make the reading of p q as meaning p q reasonable: let p be n = 1 and let q be (n 1)(n 2) = 0, so p q reads n = 1 implies (n 1)(n 2) = 0 or If n = 1 then (n 1)(n 2) = 0 and then consider setting n = 1, 2, 3,... in turn and think about the truth values of p, q and p q. You will have seen examples of truth tables in the first year Sets, Numbers and Functions course. Recall that they can be used to determine whether a propositional term t is a tautology, meaning that v(t) = T for every valuation v. The opposite notion is: if v(t) = F for every valuation v; then we say that t is unsatisfiable (also called a contradiction though that s not good terminology to use when we ll be drawing the distinction between syntax and semantics). Notice that the use of truth tables implicitly assumes part (c) of We say that two propositional terms, s and t, are logically equivalent, and write s t, if v(s) = v(t) for every valuation v. It is equivalent to say that s t is a tautology. Let s prove that. Suppose s t so, if v is any valuation, then v(s) = v(t) so, from the definition of valuation, v(s t) = T. This is so for every valuation so, by the definition of tautology, s t is a tautology. For the converse, suppose that s t is a tautology and let v be any valuation. Then v(s t) = T and so (again, by the definition of valuation) v(s) = v(t). Thus, by definition of equivalence, s and t are logically equivalent. We see that the proof was just an easy exercise from the definitions. Now for the semantic notion of entailment; we contrast semantics ( meaning or, at least, notions of being true and false) with syntax (construction and manipulation of strings of symbols). If S is a set of propositional terms and t is a propositional term then we write S = t if for every valuation v with v(s) = T, by which we mean v(s) = T for every s S, we have v(t) = T: whenever S is true so is t. Extending the above notions we say that a set S of propositional terms is tautologous if v(s) = T for every valuation v and S is unsatisfiable if for every valuation v there is some s S with v(s) = F - in other words, if no valuation makes all the terms in S true. We also say that S is satisfiable if there is at least one valuation v with v(s) = T. So note: tautologous means every valuation makes all terms in S true; satisfiable means that some valuation makes all terms in S true; unsatisfiable means that no valuation makes all terms in S true. Lemma Let S be a set of propositional terms and let t, t, u be proposi- 11

13 tional terms. (a) S = t iff S { t} is unsatisfiable (b) S {t} = u iff S = t u (c) S {t, t } = u iff S {t t } = u Proof. These are all simple consequences of the definitions. Before we begin, we introduce a standard and slightly shorter notation: instead of writing S {t 1,..., t k } = u we write S, t 1,..., t k = u. (a) S { t} is unsatisfiable iff for all valuations v, we have v(s) = F for some s S or v( t) = F iff for all valuations v, if v(s) = T for all s S then v( t) = F iff for all valuations v, if v(s) = T for all s S then v(t) = T iff S = t. (b) S {t} = u iff for every valuation v, if v(s) = T for all s S and v(t) = T then v(u) = T iff for every valuation v with v(s) = T for all s S then, if v(t) = T then v(u) = T iff for every valuation v with v(s) = T for all s S then v(t u) = T (by the truth table for ) iff S = t u. (c) S {t t } = u iff for every valuation v with v(s) = T for all s S and v(t t ) = T we have v(u) = T iff (by the truth table for ) for every valuation v with v(s) = T for all s S and v(t) = T and v(t ) = T, we have v(u) = T iff S {t, t } = u. We can use truth tables to determine whether or not S = u (assuming S is a finite (and, in practice, not very large) set) but this can take a long time: if there are n propositional variables appearing then we need to compute a truth table with 2 n rows. The next section describes a method which sometimes is more efficient. 1.3 Beth trees Beth trees provide a method, often more efficient than and perhaps more interesting than, truth tables, of testing whether a collection of propositional terms is satisfiable or not (and, if it is satisfiable, of giving a valuation demonstrating this). Note that this includes testing whether a propositional term is a tautology, whether one term implies another, whether S = t, et cetera. The input to the method consists of two sets S, T of propositional terms; to distinguish between these we will write the typical input as S T. The output will, if we carry the method to its conclusion (which for some purposes will be more than we need to do), be all valuations with v(s) = T and v(t ) = F. So if the output is nonempty then we know that S { t : t T } is satisfiable. For instance, t is a tautology if the output from the pair {t} is empty (which often will be easier than checking whether the output of {t} is all valuations). The actual computation has the form of a tree (as usual in mathematics, trees grow downwards) and, at each node of the tree, there will be a pair of the 12

14 form S T. A node (of a fully or partially-computed Beth tree) is terminal if it has no node beneath it. A node is a leaf if all the propositional terms at it are propositional variables. Directly underneath each non-terminal node is either a branch segment with another node at its end, or two branch segments with a node at the end of each. A key feature of the tree is that if a node lies under another then the lower one contains fewer propositional connectives. That means that if the initial data contains k propositional connectives then no branch can contain more than k+1 nodes. And that means that the computation of the tree will terminate. Before we describe how to compute such trees here, in order to anchor ideas, is an example. Example We determine whether or not p, (p q) r = r (q p). We will build a tree beginning with the input p, (p q) r r (q p) since there will be a valuation satisfying this condition exactly if p, (p q) r = r (q p) does not hold. p, (p q) r r (q p) (p q) r p, r (q p) r, (p q) r p, q p r, q, (p q) r p, p q, (p q) r r, p q r, p, p q q, r r, p q q, r, p q r, p The property ( ) below implies that a valuation v satisfies the input conditions (making both p and (p q) r true but making r (q p) false) iff it satisfies at least one of the leaves. But we can see immediately that the only leaf satisfied by any valuation is q r, p, which is satisfied by the valuation v with v(q) = T, v(r) = F, v(p) = F. So there is a valuation making both p and (p q) r true but making r (q p) false. That is r (q p) does not follow from p and (p q) r. We will list the allowable rules for generating the nodes directly under a given node. To make sense of these, we first explain the idea. The property that we want is the following: ( ) If, at any stage of the construction of the tree with initial node S T, the currently terminal nodes are S 1 T 1,...,S k T k then, for every valuation v, we have v(s) = T and v(t ) = F iff v(s i ) = T and v(t i ) = F for some i. 13

15 For this section, when I write v(t ) = F I mean v(t) = F for every t T. This is a convenient, but bad (because easily misinterpreted), notation. In order for this property to hold it is enough to have the following two: ( 1 ) if a node S T is immediately followed by a single node S 1 T 1 then, for every valuation v we have v(s ) = T and v(t ) = F iff v(s 1 ) = T and v(t 1 ) = F; ( 2 ) if a node S T is immediately followed by the nodes S 1 T 1 and S 2 T 2 then, for every valuation v we have: v(s ) = T and v(t ) = F iff [v(s 1 ) = T and v(t 1 ) = F] or [v(s 2 ) = T and v(t 2 ) = F]. (The fact that these are enough can be proved by an inductive argument.) In the pair S T you can think of the left hand side as the positive statements and those on the right as the negative ones. Each rule involves either moving one term between the positive and negative sides (with appropriate change to the term) or splitting one pair into two. Here are the allowable rules. S, t T S, t, T S t, T S, t T S, s t T S, s, t T S, s t, T S s, T S t, T S, s t T S, s T S, t T S s t, T S s, t, T S, s t T S s, T S, t T S s t, T S, s t, T In lectures we will explain a few of these but you should think through why each one is valid (that is, satisfies ( 1 ) or ( 2 ), as appropriate). You should also note that they cover all the cases - together they allow a single pair to be input and will output a tree where every terminal node is a leaf. When constructing a Beth tree there may well be some nodes where there is a choice as to which rule to apply but no choice of applicable rule is wrong (though some choices might lead to a shorter computation). Example We use Beth trees to show that p q p is a tautology. We already suggested that it might be easier to do the equivalent thing of showing that (p q p) is unsatisfiable; here s the computation for that. 14

16 p q p p q p p, q p - and clearly no valuation can make both p, q true but make p false; we conclude that p q p is a tautology. For comparison here is the direct check that p q p is a tautology. p q p p q p p q Now note that every valuation satisfies the condition expressed by at least one of the leaves, so p q p is indeed a tautology. 1.4 Normal forms First, we look at some more basic properties of logical equivalence where, recall, two propositional terms s, t are said to be logically equivalent, s t, if v(s) = v(t) for every valuation v (and by 1.2.1(c) it is enough to check for valuations on just the propositional variables actually occurring in s or t). Lemma If s, t are propositional terms then: (i) s t iff (ii) s = t and t = s iff (iii) = s t iff (iv) s t is a tautology. Proof. All this is immediate from the definitions. For instance, to prove (iv) (i) let v be any valuation; then, assuming (iv), v(s t) = T and by definition of valuation, we see this can happen only if v(s) = v(t), as required. Note that this is an equivalence relation on SL; that is, it is reflexive (s s), symmetric (s t implies t s) and transitive (s t and t u together imply s u). Here are some, easily checked, basic logical equivalences. For any propositional terms s, t, u: s t t s; s t t s; (s t) s t; (s t) s t; s s; 15

17 s t s t; (s t) u s (t u), so we can write s t u without ambiguity; (s t) u s (t u), so we can write s t u without ambiguity; (s t) u (s u) (t u); (s t) u (s u) (t u); s s s; s s s. Proposition Suppose that s s and t t are propositional terms. Then: (i) s s ; (ii) s t s t ; (iii) s t s t ; (iv) s t s t. Proof. To prove (ii): suppose v(s t) = T. Then by the truth table for, both v(s) = T and v(t) = T; so v(s ) = T and v(t ) = T and hence v(s t ) = T. The other parts are equally easy. We introduce notations for multiple conjunctions and disjunctions; they are completely analogous to the use of for repeated +. Given propositional terms s 1,..., s n we define n i=1 s i by induction: 1 i=1 = s 1, k+1 i=1 s i = k i=1 s i s k+1. Similarly we define n i=1 s i. Because of associativity and commutativity of, respectively of, if we permute the terms in such a repeated conjunction or disjunction, then we obtain an equivalent propositional term. Indeed, we have the following (the proofs of which are left as exercises). Proposition If s 1,..., s n are propositional terms and v is a valuation then: (i) v( n i=1 s i) = T iff v(s i ) = T for all i = 1,..., n; (ii) v( n (iii) n i=1 s i) = T iff v(s i ) = T for some i {1,..., n}; i=1 s i n i=1 s i; (iv) n i=1 s i n i=1 s i; Proposition Suppose that s 1,..., s n and t 1,..., t m are sequences of propositional terms such that {s 1,..., s n } = {t 1,..., t m } (thus the sequences differ only in the order of their terms and possible repetitions). Then n i=1 s i = m j=1 t j and n i=1 s i = m j=1 t j. If S = {s 1,..., s n } is a finite set of propositional terms then we write S for n i=1 s i and S for n i=1 s i. What if S =? Since, roughly, the more conjuncts there are in S the harder it is to be true, it makes some sense to define to be any tautology (i.e. always true). Dually we define to be any unsatisfiable term (so false under every valuation). (Because we are only interested in the truth values of and it doesn t matter which tautology and which contradiction are chosen.) A little more terminology: given a set L of propositional variables, we refer to any propositional variable p, or any negation, p, of a propositional variable as a literal. We are going to show that every propositional term is equivalent to one which is in a special form (indeed, there are two special forms: disjunctive and conjunctive). 16

18 A propositional term is in disjunctive normal form if it has the form n i=1 mi j=1 g ij where each g ij is a literal. Proposition (Disjunctive Normal Form Theorem) If t SL then there is a propositional term s SL which is in disjunctive normal form and such that s t. If {p 1,..., p k } are the propositional variables appearing in t then we may suppose that s has the form n i=1 n 2 k. mi j=1 g ij with each m i k and with Proof. Let v 1,..., v n be the distinct valuations v on {p { 1,..., p k } such that pj if v v(t) = T. For each i = 1,..., n and j = 1,..., k, set g ij = i (p j ) = T p j if v i (p j ) = F. Note that v i ( k j=1 g ij) = T and that if v v i is any other valuation on {p 1,..., p k } then v ( k j=1 g ij) = F. It follows that if w is any valuation on {p 1,..., p k } then w( n k i=1 j=1 g ij) = T iff w is one of v 1,..., v n. Therefore for any valuation v, v( n k i=1 j=1 g ij) = v(t), so t and n mi i=1 j=1 g ij are logically equivalent, as required. For the final statement, note that there are 2 k distinct valuations on {p 1,..., p k }. The proof shows how to go about actually constructing an equivalent propositional term in disjunctive normal form, using either truth tables or, the proof slightly modified, Beth trees. Example Consider the propositional term t = (p q) ( p r). If we construct its truth table then we find 7 rows/valuations on {p, q, r} which make it true. For each of these we form the corresponding g ij. For instance, the valuation v 1 (p) = v 1 (q) = v 1 (r) = T is one of those making t true and the corresponding term is p q r. Another row where t is true is that where p is true and q and r are false, so the corresponding term is p q r. Et cetera, giving the disjunctive normal form term (p q r) (p q r) (p q r) ( p q r) ( p q r) ( p q r) ( p q r) equivalent to t. Normal forms are, however, not unique and you might note that, for example, the last four disjuncts can be replaced by the logically equivalent term p. From this point of view, Beth trees are more efficient, as we can illustrate with this example. If we construct a Beth tree starting with p q p r then very quickly we reach the single leaf p, q r, which corresponds to the single valuation making (p q) ( p r) false. That corresponds to the term p q r, so t is equivalent to the negation of this, namely (p q r), which is equivalent to p q r - a much simpler disjunctive normal form. You might instead construct a Beth tree starting from (p q) ( p r). Taking an obvious sequence of steps leads to a completed tree with the leaves p, q and r. These correspond to the (conjunctions of) literals: p, q, r respectively. Therefore this also leads to the form p q r. The dual form is as follows: A propositional term is in conjunctive normal form if it has the form n mi i=1 j=1 g ij where each g ij is a literal. 17

19 Proposition If t SL then there is a propositional term s SL which is in conjunctive normal form and such that s t. If {p 1,..., p k } are the propositional variables appearing in s then we may suppose that s has the form n mi i=1 j=1 g ij with each m i k and with n 2 k. Proof. The term t is logically equivalent to t and, by 1.4.5, t is equivalent to some term of the form n mi i=1 j=1 g ij. So t is equivalent to n mi i=1 j=1 g ij which, using DeMorgan s laws (the third and fourth on the list of identities after 1.4.1), is in turn equivalent to n mi i=1 j=1 g ij. Since each g ij is a literal (at least once we cancel double negations), the result follows. 1.5 Adequate sets of connectives The proof of actually shows that every truth table on a set, p 1,..., p k, of propositional variables can be generated from them by using the propositional connectives,,. More precisely, every propositional term t in p 1,..., p k defines a function, evaluation-at-t, from the set Val p1,...,p k of valuations v on p 1,..., p k, to {T, F}. Conversely, given any function e : Val p1,...,p k {T, F}, one may construct, using, and, a propositional term t such that e is just evaluation at t. If we change the set of propositional connectives that we are allowed to use then we can ask the same question. For instance, using just and can we construct every truth table/build a term inducing any given evaluation e? What if we use and? And other such questions (the five we have introduced are not the only possible connectives, indeed not even the only ones which occur in nature, or at least in Computer Science, where one also sees NAND=Sheffer stroke, NOR, XOR). 7 We say that a set, S, of propositional connectives is adequate if for every propositional term t (in any number of propositional variables) there is a term t constructed using just the connectives in S such that t and t are logically equivalent. 8 By logically equivalent we mean that they have the same truth tables or, a bit more precisely, they define the same function from Val p1,...,p n to {T, F}. Example The set {, } is adequate. We have already commented that {,, } is adequate so we need only note that s t ( s t). Example The NAND gate/operator or Sheffer stroke is a binary (i.e. has two inputs) propositional connective whose effect is as shown in the truth table below. p q p q T T F T F T F T T F F T 7 We won t formulate the general question because then we would have to give a general definition of (n-ary) propositional connective and would be hard-pressed to distinguish these from propositional terms. 8 Notice that if S includes some new propositional connectives then we have to extend our definitions of propositional term etc. to allow these. That s why I used quotation marks just then. 18

20 You can see from this that p q is logically equivalent to (p q), hence the name NAND. If we take our set of connectives to be just S = { } then we have to re-define propositional term by saying: every propositional variable is a propositional term; if s, t are propositional terms then so is s t. We can refer to these as (propositional) terms build using (only) and can write S (L) for the set of such terms. It is easy to show that { } is adequate. All we have to do is to show that, given propositional variables p, q we can find terms using only which are equivalent to p and p q - because we know that {, } is an adequate set of connectives. Indeed, it is easy to check see that p p is equivalent to p and hence that (p q) (p q) is equivalent to p q. Showing that a given set S of connectives is not adequate can take more thought: how can one show that some propositional terms do not have equivalents built only using connectives from S? Example One might feel that, intuitively, and together are not adequate since they are both positive. How can one turn that intuition into a proof? One would like to show, for instance, that no term built only using and can be logically equivalent to p but, even using only the single propositional variable p, there are infinitely many propositional terms to check. That might suggest trying some sort of inductive proof. But a proof of what statement? What we can do is to prove by induction on complexity/length of a term that: if t is any term built only from and then for every valuation v such that all the propositional variables appearing in t are assigned the value T by v, we also have v(t) = T. Once that is done, we can deduce, in particular, that no term built only using and can be equivalent to p. 1.6 Interpolation Suppose that s and t are propositional terms and that s = t, equivalently s t is a tautology. This could be for the trivial reasons that either s is always false (unsatisfiable) or that t is always true (a tautology). But if that s not the case then the interpolation theorem guarantees that there is some propositional term u which involves only the propositional variables appearing in both s and t such that s = u and u = t. Such a u is referred to as an interpolant between s and t. Theorem (Interpolation Theorem) Suppose that s S(L 1 ) and t S(L 2 ) are such that s = t. Then either s is unsatisfiable or t is a tautology or there is u S(L 3 ), where L 3 = L 1 L 2, such that s = u and u = t. Proof. We suppose that s is satisfiable and that t is not a tautology; we must produce a suitable u. Since s is satisfiable there is some valuation v 1 on L 1 such that v 1 (s) = T and since t is not a tautology there is some valuation v 2 on L 2 such that v 2 (t) = F. First we show that L 3. If this were not so, that is, if L 1 and L 2 had no propositional variables in common, then we could { define a valuation v 3 on v1 (p) if p L S(L 1 L 2 ) by setting, for p L 1 L 2, v 3 (p) = 1. Then, v 2 (p) if p L 2 19

21 by 1.2.1(c), we would have v 3 (s) = v 1 (s) = T and v 3 (t) = v 2 (t) = F, which contradicts the assumption that s = t. Now choose, by 1.4.7, a formula of S(L 1 ) in disjunctive normal form which is equivalent to s, say n i=1 ( l i j=1 g ij m i k=1 h ik) where we have separated out the literals into two groups: the g ij - those belonging to S(L 1 ) \ S(L 3 ); the h ik - those belonging to S(L 3 ). (We allow that some of these conjuncts might be empty.) We can assume that each disjunct is satisfiable (we can drop any which are not). Define u to be n mi i=1 k=1 h ik. Clearly u S(L 3 ) and, if v is a valuation on S(L 1 ) then, if v(s) = T, it must be that v( l i j=1 g ij m i k=1 h ik) = T for some i (by 1.4.3) and hence 9 v( m i k=1 h ik) = T and hence v(u) = T. Thus s = u and we have just seen that u is satisfiable. It remains to prove that u = t. So let v be a valuation on S(L 2 ) such that v(u) = T. Then there must be some i 0 such that v( m i0 k=1 h i 0k) = T. We define a valuation w on S(L 1 L 2 ) by setting, for p L 1 L 2, v(p) if p L 2 T if p = g i0k for some k w(p) = F if p = g i0k for some k T, say if p L 1 \ L 2 and is not already assigned a value, that is, if p does not occur in the i 0 th disjunct. Note that w( l i0 j=1 g i 0j m i0 k=1 h i 0k) = T by construction and hence w(s) = T. But we assumed that s = t and so w(t) = T. But w and v agree on all propositional variables in L 2 ; hence v(t) = T. We conclude that u = t, which was what had remained to be proved. The proof gives an effective procedure for computing interpolants. Example Given that (p ( r s)) (p (r s)) = ((s r) t) ( t (r s)), how do we find an interpolant involving r and s only? (Note that, in the notation of the proof, L 1 = {p, r, s}, L 2 = {r, s, t}, L 3 = {r, s}. We find a term in disjunctive normal form which is logically equivalent to (p ( r s)) (p (r s)); one such is (p s r) ( p r s). Following the procedure in the proof, we obtain the interpolant u which is (s r) (r s). 9 You might reasonably ask what happens if, for this value of i, there are no h ik conjuncts. That could happen but there must be at least one such value of i (that is, with v making the ith conjunct true) such that there is an h ik. Otherwise, arguing as before, we could adjust the valuation v, keeping the same values on propositional variables in S(L 1 )\S(L 3 ) but adjusting it on those belonging to S(L 3 ), so as to make t false, while keeping s true, contradicting that s = t. 20

22 Chapter 2 Deductive systems The design of the following treatise is to investigate the fundamental laws of those operations of the mind by which reasoning is performed; to give expression to them in the symbolical language of a Calculus, and upon this foundation to establish the science of Logic and construct its method; to make that method itself the basis of a general method for the application of the mathematical doctrine of Probabilities; and, finally, to collect from the various elements of truth brought to view in the course of these inquiries some probable intimations concerning the nature and constitution of the human mind. Thus begins Chapter 1 of George Boole s An Investigation of the Laws of Thought (on which are founded the Mathematical Theories of Logic and Probabilities) (1854) We have already seen the symbolical language (though not the way Boole wrote it) and what Boole meant by a Calculus (or Algebra). Now we discuss proof/deductive systems further. Given a propositional term, we may test whether or not it is a tautology by, for example, constructing its truth table. This is regarded as a semantic test because it is in terms of valuations. The test is recursive in the sense that we have a procedure which, after a finite amount of time, is guaranteed to tell us whether or not the term is a tautology. More generally, suppose that S is a finite set of propositional terms and that t is a propositional term. Recall that we write S = t to mean that every valuation which makes everything in S true also makes t true. Checking whether or not this is true also is a recursive procedure. In the case of predicate logic, however, it turns out that there is no corresponding algorithm for determining whether or not a propositional term ( sentence in that context) is a tautology or whether the truth of a finite set of propositions implies the truth of another proposition. 1 The best we can do is to produce a method of generating all tautologies or, more generally of starting with a set, S, of sentences/statements/propositions which we treat as axioms and then generating all consequences of those axioms. Such a method of 1 In fact the set of tautologies of predicate logic is recursively enumerable but not recursive. Saying that the set is recursively enumerable means that there is an algorithm which will output only tautologies and such that any tautology eventually will be output; but we can t predict when. 21

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