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These Proofs Giving Me Trouble
Can only use the following rules of inference...

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These Proofs Giving Me Trouble

Postmodern Beatnik
Or Is It?

Usergroup: Administrators
Joined: Nov 18, 2005
Total Topics: 16
Total Posts: 1464

Posted 08/25/08 - 11:42 AM:	#11
Here are the best proofs I could come up with: 1. ~(P -> Q) : P & ~Q (Supposed to be able to do this in 12 lines) 1. ~(P -> Q) [Premise] \| 2. ~P [Assumption for RAA] \| \| 3. P [Assumption for CP] \| \| 4. P v Q [3, vI] \| \| 5. Q [2, 4, vE] \| 6. P -> Q [3-5, CP) 7. P [1, 6, RAA] \| 8. ~Q [Assumption for CP] \| 9. ~Q v ~P [vI] \| 10. ~P [vE] 11. ~Q -> ~P [9-11, CP] 12. ~Q [7, 13, MT] 13. P & ~Q [7,14, &I] Obviously, DeMorgan's laws would make this a lot easier -- the more dependent you are on the primitive rules, the harder you have to work. But it's good to know how to work a proof without the derived rules. This one's a bit tricky, though, so don't let it get you down! 2. ~P : P -> Q (supposed to be able to do this in 9 lines) 1. ~P [Premise] \| 2. P [for CP] \| 3. P v Q [2, vI] \| 4. Q [1, 3, vE] 5. P -> Q [2-4, CP] Eventually, the rule of conditional exchange ( P -> Q :: ~P v Q) will make this kind of problem irrelevant. But in the absence of such shortcuts, the quickest route to proving a conditional will almost always be conditional proof. While the step-by-step deduction may not be immediately obvious, however, what you are being asked to prove should be rather intuitive. Think truth tables: to get a true conditional, either the antecedent must be false or the consequence must be true. (I'm not sure why this is supposed to take 9 lines.) I hope this is helpful.
"It is not a mistake to have strong views. The mistake is to have nothing else." --Anthony Weston "If you can't imagine how anyone could hold the view you are attacking, you just don't understand it yet." --Anthony Weston

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