Linguistics 280: Language and Formal Reasoning Proofs: finding solutions Robert Levine Autumn Quarter, 2010 Strategies for devising proofs: direct proofs Once you understand how to apply inference rules to formulæ, the next step in modeling reasoning as a series of deductions is to work out just how the rules can get you from a given statement to another statement which is a necessary consequence of the first. Take something very basic: the claim that p ∧ q ⊢ p ∨ q. This is a logically valid statement—it holds every time. When p ∧ q has the value 1, p ∨ q has the value 1, without exception, no matter what sentences of English p and q represent. The truth table for the two formulæ makes this crystal-clear: p q p ∧ q p ∨ q 11 1 1 10 0 1 01 0 1 00 0 0 What the table tells you is that the entailment only goes one way: whenever p ∧ q has the truth value 1, p ∨ q also has the value one, which is what p ∧ q ⊢ p ∨ q means—if we have a proof of p ∧ q, then p ∨ q must be true—but the reverse doesn’t hold: there are two cases in which the disjunction is true, but the conjunction is false. But given that p ∧ q ⊢ p ∨ q does hold, it should be possible—if the Gentzen rules really are the deductive counterpart to the truth tables—to prove this entailment. The trick is figure out what the reasoning steps are. Proofs don’t solve themselves, and staring at the problem for long enough isn’t going to make the answer occur to you. Formal proofs are the purest form of logic puzzle. There is, as you may be aware, an explosion in publishing based on logic puzzles. Amazon.com lists more than 1600 titles in this area; everyone from USA today to Johns Hopkins University Press has published collections, and Soduku—which is a purely deductive number-based logic puzzle genre has probably surpassed crossword puzzles in terms of numbers of ‘players’. But a lot of the logic puzzles that you find in these books involve not just deductive tasks, but misleading or tricky formulations or contexts which lead you to jump to false conclusions, from which it’s impossible to reason to the solution. Formal proofs don’t have this element—everything is laid out in front of you, with no way to deceive you or guide you down a false path. It’s just you against the proof. But without a strategy for solving the proof in front of you, you aren’t going to succeed. And to develop such a strategy you need to think very carefully about what it is that you’re given, and what it is you have to end up with. In this case, you start with a conjunction. What does that tell you? Given a conjunction, we know from the Gentzen rules (themselves based on the truth tables) that in p ∧ q, both p and q must be true. That’s what we’re allowed to deduce from p ∧ q, and when you start a proof such as this one, you must review exactly what it is that you can legitimately infer from the statement you’re asked to assume at the beginning. You then have to go from that knowledge to the conclusion. In the very simple case we’re now looking at, that’s eay. We know that p is true, and we have to get from p to p ∨ q. Are we allowed to do that? Is there an entailment of the form p ⊢ p ∨ q anywhere in our rule system? Look at the rule for ∨ Intro, and you’ll see that that’s exactly what the rule allows: ∨ Intro 1: Γ ⊢ ϕ Γ ⊢ ϕ ∨ ψ where ϕ and ψ range over all possible logical statement. What this means is that we have a green light in our logical setup for the conclusion we want, based on what that setup tells us about conjunctions. All we have to do is set up the proof correctly, as a way of ensuring that it’s really valid. Begin at the beginning, with the startup axiom. We need to get the original premise, p∧q, into play, and the line underneath is a reminder that we’re not just staring at the formula, but are about to make an inference from it: p ∧ q ⊢ p ∧ q | {z } |{z} |{z} Γ ϕ ψ The rule for making inferences from conjunctions is ∧ Elim 1: Γ ⊢ ϕ ∧ ψ Γ ⊢ ϕ According to the rule above, with Γ = {p ∧ q}, ϕ = p, ψ = q, we are allowed to infer p ∧ q ⊢ p: p ∧ q ⊢ p ∧ q | {z } |{z} |{z} Γ ϕ ψ ∧E1 p ∧ q ⊢ p | {z } |{z} Γ ϕ And at this point, the labeling that got us through the inference step from p ∧ q ⊢ p ∧ q to p ∧ q ⊢ p is no longer relevant. As explained in the preceding handout on proofs, one we’ve made an inference from a given assumption set, we need to match the formula we’ve inferred to the rule we want to use. And that may be possible only if we interpret symbols such as Γ, ∆ and others that show in the rule as covering different parts of the formula we wish to apply the next rule to than they did in the previous rule. We therefore need to decide how to treat the next reasoning step so that the result of the previous step can undergo the next rule we wish to apply. We have, so far, p ∧ q ⊢ p ∧ q ∧E1 p ∧ q ⊢ p Based on our earlier informal storyline for the proof, we need to apply the particular ∨ Intro rule given earlier. We take Γ to be {p ∧ q}, ϕ = p, and ψ = q (which happen to be the previous values for these; more about that shortly), and get 2 p ∧ q ⊢ p | {z } |{z} Γ ϕ ∨I1 p ∧ q ⊢ p ∨ q | {z } |{z} |{z} Γ ϕ ψ Suppressing the (now irrelevant) labelling, we have p ∧ q ⊢ p ∨I1 p ∧ q ⊢ p ∨ q By combining the first inference step together with this one, we can display the whole history of the proof, which establishes irrefutably that every step of the proof is licensed by the proof system we’ve adopted: p ∧ q ⊢ p ∧ q ∧E1 p ∧ q ⊢ p ∨I1 p ∧ q ⊢ p ∨ q where the black box notation indicates that the proof is now formally terminated successfully. Trickery: proof by contradiction The strategy followed in the foregoing proof was quite simple, because nothing more indirect was necessary. The rule for ∧ Elimination ‘feeds’ the rule for ∨ Introduction. Such moves are quite common in proofs, but it’s unusual for a proof to consist only of such moves. Sometimes something much more devious is called for—even when the theorem itself seems to be nothing but the essence of common sense. For example, the logical statement ¬p ⊢ ¬(p ∧ q) seems so self-evident as to be trivial: if something is false, the conjunction of it and something else cannot be true. And, unsurprisingly, the truth table show that the statement we’ve been asked to prove is indeed true: p q ¬p p ∧ q ¬(p ∧ q) 110 1 0 100 0 1 011 0 1 001 0 1 When ¬p has 1 as its truth value, so does ¬(p ∧ q), just as we expect. Surely nothing very much is required here by way of proof—one or two lines should do it. But when you first encounter this theorem, it’s far from obvious how to translate the admittedly obvious truth of this statement into a proof. The problem is that you cannot directly relate the negation in ¬p to ¬(p ∧ q). A few moments’ thought should make it clear why: ¬p is definitive—p cannot be true—but ¬(p ∧ q) is not; in principle, the latter can be true even if ¬p itself is not true (as, indeed, the truth table makes evidence). So we cannot proceed directly from the premise ¬p to the conclusion ¬(p ∧ q). How then can the proof go forward? Here’s a general rule of thumb: whenever you see negation on the right side of the turnstile—particularly when that negation applies to a complex formula, as in the present case—it’s going to be difficult to establish 3 the negation directly. But you can almost always achieve such proofs by establishing a contradiction (⊥) and then resolving the contradiction by shifting one of the premises which creates the contradiction over to the right if the turnstile, in negated form. This move is exactly what is mandated by the ¬ Intro rule—we’ve already seen evidence of its effectiveness in proving the theorem p ⊃ q ⊢ ¬q ⊃ ¬p. A very useful tactic in the case of proof by contradiction is to work backwards from the last step to the contradiction which would have preceded it and enabled that last step. It’s clear from the Gentzen rule for ¬ Intro—namely ¬ Intro: Γ, ϕ ⊢⊥ Γ ⊢ ¬ϕ —that if we had a stage in the derivation ¬p,p ∧ q ⊢ ⊥, we obtain the proof we seek directly frome the ¬ Intro rule. In other words, using proof by contraction changes our target: if we can prove ¬p,p ∧ q ⊢ ⊥, then the rest of the proof is trivial. But why should proving this statement be any easier than proving the original statement? Once you’ve worked out a few such proofs, the answer to this question will be clear to you in the sharpest way: with proof by contradiction, you’ve moved the thing you have to prove into the assumption set on the left of the turnstile.