Logic: Proving that ~(P->Q) and P^~Q are equivalent

Question

Informally, I realize that ~(P->Q) is the same as P^~Q, since the only way that the original conditional P->Q would be false is if we had both P^~Q.

I think that given ~(P->Q), the next step would be to assume ~(P^~Q) and try to obtain P^~Q via proof by contradiction, but after this step, I am not sure how to proceed.

The system I am using is Fitch. Unfortunately, I cannot use AnaCon or TautCon, nor can I use truth tables.

Professor will not allow TautCon or the use of truth tables. — Matt, May 01 '17 at 07:00
Also duplicate of proving $\lnot (A \rightarrow B) \vdash A \land \lnot B$ — Mauro ALLEGRANZA, May 02 '17 at 06:43

score 0 · Answer 1 · answered May 01 '17 at 07:05

0

No words needed. A picture speaks a thousand words!

answered May 01 '17 at 07:05

Itsnhantransitive

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Wish it were the case, unfortunately my professor will not allow truth tables, nor the use of TautCon. – Matt May 01 '17 at 07:09
https://math.stackexchange.com/questions/1726708/how-to-solve-p-%E2%86%92-q-p-q-by-natural-deduction?noredirect=1&lq=1 – Itsnhantransitive May 01 '17 at 07:34

Logic: Proving that ~(P->Q) and P^~Q are equivalent

1 Answers1