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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

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  • $\begingroup$ 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, 2021 at 14:29

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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 $P$.

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    $\begingroup$ Damn now I see it: $\endgroup$
    – Jesús Díaz Castro
    Mar 10, 2021 at 14:14
  • $\begingroup$ Good! Nicely done! $\endgroup$
    – Pål GD
    Mar 10, 2021 at 14:26

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