Proof Worksheet 2, Math 187 Fall 2017 (with solutions) Dr. Holmes October 17, 2017 The instructions are the same as on the first worksheet, except you can use all the rules in the strategies handout. We now have rules involving disjunction (or) and rules combining disjunction and negation (or and not). You are most welcome to come to office hours to ask me about these. Due date extended to Friday the 6th (really I ought to have extended it to the 4th, but I said Friday and I ll leave it at Friday). 1. Prove (P Q) (Q P ). This looks like one of the problems on the previous set, but it doesn t go quite the same way; and notice that I m only having you prove the implication in one direction, so you don t have to repeat yourself. Remember the use of the rule of addition to make your conclusions look right in each case [that s a hint!]. Assume (1): P Q Comments: Prove this by cases using (1). Case 1: Assume (1a): P (2a): Q P addition 1a Case 2: Assume (1b): Q (2b): Q P addition 1b (3): Q P proof by cases 1, 1a-2a, 1b-2b. (4): main theorem, deduction 1-3 1
It s not the only way to do it. Assume (1): P Q Comments: To prove this, assume Q and deduce P (or vice versa). Assume(2): Q Goal: P (3): P disjunctive syllogism 2,1 (4): Q P, alternative elimination 2-3 (5): main theorem deduction 1-4 2
2. Prove ((P Q) (Q R)) (P R). Hint: set everything up using the rules of implication and conjunction, and you will have a goal P R to prove. Use the alternative elimination strategy: assume P and prove R as a new goal, or vice versa. Disjunctive syllogism can be your friend here. Comments: I break up the hypothesis into two. Assume (1): P Q Assume(2): Q R Goal: P R Assume (3): P Goal: R (4): Q disjunctive syllogism 3,1 (5): R disjunctive syllogism 2,4 (6): P R alternative elimination 3-5 (7): main theorem deduction 1-6 3
3. Verify the classical rule of constructive dilemma, which takes the form (typo fixed above) P Q P R Q S R S We set up the first few lines (just to show you how to verify a rule as opposed to a theorem: we give the premises as initial lines and argue to the conclusion as a goal): [I ll complete the solution starting with these lines] (1): P Q premise (2): P R premise (3): Q S premise Goal: R S Comments: I m using the alternative elimination approach: assume R and show that S has to follow. Assume (4): R Goal: S (5): P modus tollens 4,2 (6): Q disjunctive syllogism 5,1 (7): S modus ponens 6,3 (8): R S alternative elimination 4-7, which completes the verification of the rule. (the presentation just above was not afflicted with a typo: if you used it you were all right) Hint: there are two very different ways to do this. One works by proof by cases on line 1 and goes directly in each case (with use of the rule of addition to get the conclusions of the cases to line up). The other approach involves setting up the proof of R S by alternative elimination (assume R and argue to S as a goal, or vice versa), followed by applications of modus tollens and disjunctive syllogism. 4
If you can write both proofs, I might give a little extra credit. 4. Prove (A B) ( A B) (or you may prove the other de Morgan law if you prefer, or both for additional credit). Please note that you cannot use the de Morgan laws in this proof! I stole the solution from another of my texts which presented the problem with different letters. Theorem: (P Q) ( P Q) Proof: Part 1: Assume (1): (P Q) Goal: P Q We use the alternative elimination strategy: assume the negation of one alternative and show that the other alternative follows. Assume (2): P Goal: Q Assume (3): Q Goal: (a contradiction) Goal: P Q (in order to get a contradiction with line 1) 4 P double negation, line 3 5 P Q (lines 3 and 4) 6 1,5 contradiction. This resolves the goal at line 3. 7 Q lines 3-6 negation introduction. This resolves the goal at line 2. 8 P Q 2-7 disjunction introduction. This resolves the goal at line 1. 5
Part 2: Assume (9): P Q Goal: (P Q) Assume (10): P Q Goal: (a contradiction) We use the strategy of proof by cases on line 9. Case 1 (9a): P Goal: 11 : P from line 10 12 : 9a, 11 contradiction (this resolves the goal after 9a) Case 2 (9b): Q Goal: 13 Q from line 10 14 9b, 13 contradiction (this resolves the goal after 9b) 15 9, 9a-14 proof by cases. 16 (P Q) 9-15 negation introduction. This resolves the goal at line 9. 6