Logic 101 (#43): Proof Practice #1

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To see how one solves proofs in practice, I'm going through a problem set for the first time to solve questions I have never seen before. Today, we have a candidate for proof by contradiction...

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You said not P in the beginning and threw me off so hard. I thought I wasn't cut out for this logic thing. Then you all of the sudden say not M. Geez you playing with my heart there. I thought my life was over.

appleman

Proof practice? More like "Perfect, that is!" These videos fit together very nicely, and I can't wait to watch more!

PunmasterSTP

Mine solution:
1 | ~M => (N n O)
2 | N => P
3 | O => ~P | … M
4 | ~M | Assumption for PBC
5 | N n O | 1, 4 Modus ponens
6 | N | 5 Simplification
7 | P | 2, 6 Modus ponens
8 | O | 5 Simplification
9 | ~P | 3, 8 Modus ponens
10 | P n ~P | 7, 9 Conjunction introduction
11 | M | 4-10 PBC

Happy that I'm correct. Thx for the videos.

no_name_

Hi William, can I work like this?

4. ~M (Assumption)
5. N^O (1, 4 Modus Ponens)
6. N (5 Simplification)
7. O (5 Simplification)
8. (P^~P) (2, 3, 5 Destructive Dilemma) Contractdiction!

Thanks so much!

lokkkan

i solved this through making the first line into a disjunction with material implication instead which took one more line to come to a conclusion but i guess i feel more comfortable with ands/ors instead of implications and stuff.

Nyllet

Is this method okay?
Since N implies P, and O implies not P, therefore, in the first line, not M implies N and O, which means not P and P with happen. But since Not P and P at the same time is not possible, to maintain the equality, not M will also be impossible. hence, M is proved.

Silver

4. - -P => -O; 3 Contraposition
5. P => -O; 4 Double Negation
6. N => -O; 2, 5 Hypothetical Syllogism
7. -N v -O; 6 Material Implication
8. -(N ^ O); 7 DeMorgan
9. - - M; 1, 7 Modus Tollens
10. M; Double Negation

The way i did it without PBC

jackharper

I did it by proof by contradiction. I assume that -M is true. Then I apply modus ponens to first line to get (N and O). Then I apply simplification to get N. Then I proceed to apply modus ponens to second line to get P. Then apply simplification to the first line to get O. Then apply modus ponens to the third line to get: -P. Finally I introduce a conjunction to arrive (P and -P). Because that is a contradiction it must follow that the assumed premise -M has to be false. If negation of M is false, thus M is proved as true.

lea

Could I say premises are falses ? I thought a premise is always given as true. Thanks for your great work.

eugenepoubelle

Could it be done by direct method? And if it can't, why not?. Please could you help me out here, cause im trying it and i'm stock. Thank you so much

christianenriquez

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