1
$\begingroup$

I'm stuck on how to progress with this proof, despite I have tried, I cannot see the next move.

Given this proof without predicate:

Proof

So far what I've accomplished:

My approach

My idea is, as I can't see any other option using (-(Sv(P->Q)) as the first assumption in order to introduce a conditional, so the assumption must end in P ^ -Q ^ -S. As you can see I have obatined -Q and -S but, how do I proof P?

SOLUTION:

sol

$\endgroup$
1
  • $\begingroup$ Note that there's no need to add "SOLVED" or anything like that to the title/question, once you mark an answer as accepted, it's done :) Welcome to the site! $\endgroup$ – Juho Mar 10 at 14:29
2
$\begingroup$

It is a bit hard to say exactly which introductions you should perform since we don't have your rule set, but I think the over all strategy should be as follows:

From $\neg (S \lor (P \to Q))$ you get $\neg (P \to Q)$. Since you have $\neg Q$, you should try to introduce $\neg P$ and thus obtain $P \to Q$. That should be a contradiction and you can conclude $P$.

$\endgroup$
2
  • 1
    $\begingroup$ Damn now I see it: $\endgroup$ – Jesús Díaz Castro Mar 10 at 14:14
  • $\begingroup$ Good! Nicely done! $\endgroup$ – Pål GD Mar 10 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.