Exercises for Propositional Logic II

Reasoning with Inference Rules For proofs on this homework, remember that each step must be justified by one of the following: a premise, a WaterWorld axioms, a listed inference rule with the referenced line numbers (and, if ambiguous, substitutions for the inference rule's meta-variables), or a subproof shown inline, or equivalently, a theorem/lemma shown previously. Except where otherwise directed, you may use any theorem shown in the text or by a previous exercise, even if that exercise was not assigned. Fill in the blank reasons in the following proof that ∨ commutes, that is, χ υ ⊢ υ χ. 1χ υ Premise 2subproof:χ ⊢ υ χ 2.a χ Premise for subproof 2.b υ χ ∨Intro, line 2.a 3subproof:υ ⊢ υ χ 3.a υ Premise for subproof 3.b υ χ ____________________ 4υ χ ____________________

Show that φ ψ, φ θ, ψ δ ⊢ θ δ. It should take around 8 steps. Show what is often called the implication chain rule: φ ψ, ψ θ ⊢ φ θ. [Practice problem—solution provided.] Show what is often called negated-or-elimination (left): φ ψ ⊢ φ. Think backwards. How can we end with φ? One way is to end with RAA, under the premise φ. Using that premise φ and the starting premise φ ψ can you derive the contradiction? 1φ ψ Premise 2subproof:φ ⊢ 2.a φ Premise for subproof 2.b φ ψ ∨Intro, line 2a 2.c Intro, lines 1,2b 3φ RAA, line 2

Using the inference rule RAA, prove φ ⊢ φ ψ. Show that W-safe Y-unsafe ⊢ W-unsafe Y-safe. The proof is a bit longer than you might expect. Use the ∨Elim inference rule to get the final result. In our inference rules, unlike our equivalences, we chose to not include any corresponding to distributivity. Prove a left-hand version of one direction of distributivity: φ ψθ ⊢ φ ψ φ θ . Use the previous part's result, plus ∧'s commutativity to prove the corresponding right-hand version: ψθ φ ⊢ ψ φ θ φ . In our inference rules, unlike our equivalences, we chose to not include any corresponding to DeMorgan's Law. Show that each of the following versions is still provable. φ ψ ⊢ φ ψ φ ψ ⊢ φ ψ φ ψ ⊢ φ ψ φ ψ ⊢ φ ψ The above exercise suggests that it would be useful to have an inference rule or theorem that says given θ ⊢ δ, then θ ⊢ δ. Or, equivalently, because of ⇒Intro and ⇒Elim, θ δ ⊢ θ δ. Why don't we? In our inference rules, unlike our equivalences, we have nothing that directly equates φ ψ and φ ψ. Prove each of the following. φ ψ ⊢ φ ψ φ ψ ⊢ φ ψ Prove the following: φ ψ, ψ φ ⊢ φ ψ φ ψ Prove what is commonly called the Law of Excluded Middle: ⊢ χ χ. Give a short proof citing our previous proof of ⊢ χ χ and the relevant version of DeMorgan's Law from above. Give a direct version without using previous theorems. Use RAA two or three times. Prove the missing steps and reasons in the following WaterWorld proof of X-has-1 ⊢ W-unsafe Y-unsafe. 1X-has-1 ____________________ 2____________________ WaterWorld axiom 3____________________ ⇒Elim, lines 1,2 4subproof: W-safe Y-unsafe ⊢ W-unsafe Y-unsafe 4.a W-safe Y-unsafe Premise for subproof 4.b Y-unsafe ____________________ 4.c W-unsafe Y-unsafe ____________________ 5subproof: W-safe Y-unsafe ⊢ W-unsafe Y-unsafe 5.a W-safe Y-unsafe Premise for subproof 5.b W-unsafe Y-safe CaseElim (left), lines ____________________ where φ ____________________, and ψ ____________________ 5.c ____________________ ____________________ 5.d W-unsafe Y-unsafe ____________________ 6 W-safe Y-unsafe W-safe Y-unsafe Theorem: Excluded Middle, where χ ____________________ 7W-unsafeY-unsafe ____________________

[Practice problem—solution provided.]

Given the above figure, and using any of the immediately obvious facts as premises, prove that location P is safe by using our proof system and the WaterWorld axioms. While this proof is longer (over two dozen steps), it's not too bad when sub-proofs are used appropriately. To make life easier, you may use the following theorem: Q-has-1 P-safe R-safe P-safe W-safe R-safe W-safe , along with any proven previously. When looking at the given board, you can use premises like Y-safe as well as Y-unsafe. 1Q-has-1 Premise 2X-has-1 Premise 3Y-unsafe Premise 4W-unsafe Y-unsafe Theorem: above problem, line 2 5Y-unsafe W-unsafe Theorem: ∨ commutes, line 4 6W-unsafe CaseElim, lines 3,5 7subproof:P-safe W-safe ⊢ 7.a P-safe W-safe Premise for subproof 7.b P-safe W-safe ¬Elim, line 7.a 7.c W-safe ∧Elim, line 7.b 7.d W-safe W-unsafe WaterWorld axiom 7.e W-unsafe ⇒Elim, lines 7.c,7.d 7.f Intro, lines 6,7.e 8P-safe W-safe RAA, line 7 9subproof:R-safe W-safe ⊢ 9.a R-safe W-safe Premise for subproof 9.b R-safe W-safe ¬Elim, line 9.a 9.c W-safe ∧Elim, line 9.b 9.d W-safe W-unsafe WaterWorld axiom 9.e W-unsafe ⇒Elim, lines 9.c,9.d 9.f Intro, lines 6,9.e 10R-safe W-safe RAA, line 9 11 Q-has-1 P-safe R-safe P-safe W-safe R-safe W-safe Theorem: Allowed by problem statement 12 P-safe R-safe P-safe W-safe R-safe W-safe ⇒Elim, lines 1,11 13 R-safe W-safe P-safe R-safe P-safe W-safe Theorem: ∨ commutes, line 12 14 P-safe R-safe P-safe W-safe CaseElim, lines 8,13 15 P-safe W-safe P-safe R-safe Theorem: ∨ commutes, line 14 16P-safe R-safe CaseElim, lines 10,15 17P-safe ∧Elim, line 16

Alternatively, the subproofs could easily have been pulled out into lemmas. Just like using subroutines in a program, that would make the proof somewhat clearer, even though in this case each lemma would be used only once. Observe how the two subproofs have some identical lines (7.c-7.f and 9.c-9.f). It would be incorrect to replace those lines in the second subproof with a citation of the results of the first subproof. First, because the previous subproof had been completed, and moreover, the two subproofs have different premises. This is analogous to two subroutines that happen to have some identical code lines, even through they are called separately and have different parameters. Interestingly, we didn't need to use R-safe as a premise. (In fact, we nearly proved that R-safe would have been inconsistent with the other premises.) Starting from the WaterWorld axiom Q-has-1 P-safe R-safe W-unsafe P-safe R-unsafe W-safe P-unsafe R-safe W-safe , we could prove the following theorem cited in the previous problem: Q-has-1 P-safe R-safe P-safe W-safe R-safe W-safe . Prove the following theorem which is slightly simpler: φ ψ θ δ ε ⊢ φ ψ δ . If you have trouble, first prove an even simpler version: φ ψ θ ⊢ φ ψ. [Practice problem—solution provided.] Show that the ¬Elim inference rule is redundant in our system. In other words, without using ¬Elim, prove that φ ⊢ φ. 1φ Premise 2subproof:φ ⊢ 2.a φ Premise for subproof 2.b Intro, lines 1,2.a 3φ RAA, line 2

Show that the ¬Intro inference rule is redundant in our system. In other words, without using ¬Intro, prove that φ ⊢ φ. To make sure that you're not hiding any uses of ¬Intro, also do not use any previous theorems. Show that the CaseElim inference rule is redundant in our system. For brevity, we'll just consider the left-hand version. In other words, without using CaseElim, prove that φ ψ, φ ⊢ ψ. To make sure that you're not hiding any uses of CaseElim, also do not use any previous theorems. State where on a board pirates could be positioned, so that: P-has-1 U-has-1 W-has-1, but X-safe. Compare this with a previous theorem, B-has-1 G-has-1 J-has-1 K-unsafe , the same idea shifted down a couple of rows. Suppose we try to translate this theorem's proof so as to conclude X-safe (clearly untrue, by the above). What is the first step of the modified proof which doesn't hold when B,G,J,K are mindlessly replaced with P,U,W,X, respectively? (Just give a line number; no explanation needed. Your answer will be of the form Lemma A line 1 or main proof line 2.) We've just seen that the mindless changing of location-names introduces false steps. But we can be a little smarter, and modify the false step to get a formula which is true, and is also still in the spirit of the original proof. We can thus patch the problem from the previous part, and continue on modifying the original proof for several more steps. But clearly we can't translate the entire original proof; we eventually hit a more fundamental snag: a formula which isn't true, yet can't be patched up, either. What is the first line that can't be patched? (Again, just give a line number; no explanation needed. Your answer will be of the form Lemma A line 1 or main proof line 2.) Which is worse, having an unsound (but complete) inference system or an incomplete (but sound) one? Why?