Introduction The first two parts of the course concern propositional logic. This is, roughly, the logic of combining statements. Here, the defining property of a statement is that it is either true or false. We'll use statement and proposition interchangeably. So, for instance, `7 is a prime' is a statement. So is `the moon is made of cheese'. But `is the moon made of cheese?' and `add 8 and 7' are not: the first is a question, the second an instruction. What about `x is prime'? This isn't a statement, unless we also specify x, in a way that makes sense. (E.g. x = 7, but perhaps not x = the moon.) In propositional logic, we will have propositional variables to represent statements, these might be called say p or q. We combine our propositional variables to form more complicated statements, called propositional terms. We combine them using connectives: `and' (^), `or' (_), `not' (:), `implies' (!), and `if and only if' $. When we do this at first it will be purely syntactic (that is, just formal manipulation of strings of symbols, with nothing to give meaning). When we do add meaning (using valuations), the crucial thing to note is that the truth or falsity of a combined statement, say `7 is prime and the moon is made of cheese' depends only on the truth or falsity of the two statements that are combined. It doesn't depend in any way on what those two statements actually say. In the next section, we start making this all precise, with the definition of proposi- tional terms. 1. Propositional terms Definition 1.1. A propositional language is a nonempty set L of symbols, called propo- sitional variables, usually denoted p; r; q; : : : ; p1; p2;:::. We abbreviate propositional variable by p.v.. We define the set of propositional terms (or propositional formulas) of L, which we write SL, by induction: • S0L = L, • Sn+1L = SnL [ f(s ^ t); (s _ t); (:s); (s ! t); (s $ t): s; t 2 SnLg. And then we define S • SL = n≥0 SnL. We will always assume that L \ f^; _; !; $; :; (; )g = ;. Let's consider the simplest case L = fpg. Then we have S0L = fpg. And S1L = fp; (p^p); (p_p); (p ! p); (p $ p); (:p)g. And, for instance ((p^p)^(p^p)) 2 S2L. And so on. And for instance, with L = fp; q; rg, we have (p ^ q); (:r) 2 S1L (among other things) and ((p ^ q) _ (:r)) 2 S2L and so on. For the time being, this is completely formal. We will read (p ^ q) as `p and q', and so on, but so far we haven't done anything to give these formal strings meaning. Note that the definition is inductive. And correspondingly, it is often helpful to prove results about terms by induction (which we call induction on complexity of terms). 1 2 There are many brackets. This is to avoid any ambiguity (for instance how should we read p _ q ^ r?). But there is still some possible ambiguity. In simple terms such as (p^q) or ((p^q)_r) we can see which connective is introduced last. This is called the principal connective. And we have a clear sense of how these terms are built up from more complicated terms. We can draw trees to represent this. For (p ^ q) we have: (p ^ q) p q And for ((p ^ q) _ r) we have: ((p ^ q) _ r) (p ^ q) r p q For more complicated terms, it isn't immediately clear that we can tell which con- nective is the principal connective. For instance, consider the term (((:p) ! ((p ! (:q)) ! (:r))) ! (p ! r)): How can we tell which is the principal connective? We can count brackets! So for the example above we have: (1(2(3:p)2 ! (3(4p ! (5:q)4)3 ! (4:r)3)2)1!(p ! r)) (1(2:p)1!((p ! (:q)) ! (:r))) (p ! r) (:p) (1(2p ! (3:q)2)1!(:r)) p r p (p ! (:q)) (:r) p (:q) r q Here, we keep count of the number of open left brackets, until we get to the con- nective after we have one open bracket. This will be the principal connective (this will fall out of our proof below). In the tree, the principal connective is underlined. Here is the example from class: 3 (1(2(3(4:p)3 ^ q)2 ^ (3(4:r)3 _ (4r ^ p)3)2)1_((p ^ r) _ (:q))) (1(2(3:p)2 ^ q)1^((:r) _ (r ^ p))) (1(2p ^ r)1_(:q)) (1(2:p)1^q) (1(2:r)1_(r ^ p)) (p ^ r) (:q) (:p) q (:r) (r ^ p) p r q p r r p We now prove that there is no ambiguity. (Note, for those who used truth tables in 10101 or 10111, you've been implicitly assuming the following theorem!) Theorem 1.2 (Unique readability). Let s 2 SL. Then exactly one of the following holds. (a) s 2 L (that is, s is a p.v.); (b) s is (:t) for some t 2 SL; (c) s is (t ^ u) for some t; u 2 SL; (d) s is (t _ u) for some t; u 2 SL; (e) s is (t ! u) for some t; u 2 SL; (f) s is (t $ u) for some t; u 2 SL. Moreover, in (b), the term t is unique and in (c)-(f) both t and u are unique. Proof. To begin, we prove that every term can be written in at least one of these forms. We do this by induction on complexity of terms. First, note that, by definition, every term in S0L has one of the required forms (as they are all p.v's). This gives the base case of our induction. For the inductive step, suppose that n ≥ 0 and that every s 2 SnL has one of the above forms. Let s 2 Sn+1L. Then either s 2 SnL, in which case we're done by our inductive hypothesis, or s has one of the required forms by the definition of Sn+1L. This completes the induction and shows that every term is in one of these forms. The uniqueness step is more difficult. To prove it, we introduce some notation. Suppose that s 2 SL. We let l(s) = the number of left brackets ( in s; and r(s) = the number of right brackets ) in s: We need the following. Lemma 1.3. Suppose that s 2 SL. Then l(s) = r(s): Proof. This is again by induction on complexity. First suppose that s 2 S0L. Then s is a p.v. and so l(s) = 0 and r(s) = 0. So the result holds in this case. Suppose that n ≥ 0 and that for all s 2 SnL we have l(s) = r(s). Suppose that s 2 Sn+1L n SnL. Then s has one of the forms (b)-(f). If s is (:t) for some t 2 SnL, then l(s) = l(t) + 1 and r(s) = r(t) + 1. And by our inductive hypothesis, l(t) = r(t), so we have l(s) = r(s). 4 If s is (t ^ u), for some t; u 2 SnL then we have l(s) = l(t) + l(u) + 1 and r(s) = r(t) + r(u) + 1. And again by the inductive hypothesis, we have l(t) = r(t) and l(u) = r(u). So l(s) = r(s). And then there are similar cases for _; ! and $. Before continuing with the proof of the unique readability theorem, we need some further terminology. Every propositional term is a string of symbols, each of which is either in L, or in f^; _; $; !; :g or ( or ). The terms have a certain particular form, and we refer to a general finite string of these symbols as a word. And if x; y; z are words then xyz is the word obtained by first writing x then y and then z. If a word x has the form yz then we say that y is a left subword of x, and we say y is a proper left subword of x if z is nonempty. Similarly, z is a right subword of x and is said to be proper if y is nonempty. Lemma 1.4. Suppose that s is a propositional term and that x is a proper left subword of s. Then either x is empty or l(x) > r(x): In particular, x is not a propositional term. Proof. This is again by induction on the complexity of terms. Suppose that s 2 S0L. Then s is a p.v. so if x is a proper left subword of s then x must be empty. So the lemma holds for all s in S0L. Suppose that the lemma holds for all s 2 SnL. Let s 2 Sn+1L n SnL. If s is (:t) for some t 2 SnL and x is a proper left subword of s then either x is empty, x is (, x is (:, or x is (:y, for a nonempty left subword y of t. We're clearly done in the first three of these cases, so suppose that x is (:y, with y a nonempty left subword of t. By 1.3 and the inductive hypothesis, we have l(y) ≥ r(y): So l(x) = l(y) + 1 ≥ r(y) + 1 > r(y) = r(x) as required. Next suppose that s is (t∗u) where ∗ 2 f^; _; !; $g and t; u 2 SnL.