0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Why not just a truth table? $\endgroup$
    – quasi
    May 1, 2017 at 6:59
  • $\begingroup$ Professor will not allow TautCon or the use of truth tables. $\endgroup$
    – Matt
    May 1, 2017 at 7:00
  • $\begingroup$ Are you using a textbook? If so, which one? $\endgroup$
    – quasi
    May 1, 2017 at 7:34
  • $\begingroup$ Also duplicate of proving $\lnot (A \rightarrow B) \vdash A \land \lnot B$ $\endgroup$ May 2, 2017 at 6:43

1 Answer 1

0
$\begingroup$

enter image description here

No words needed. A picture speaks a thousand words!

$\endgroup$
2

Not the answer you're looking for? Browse other questions tagged .