# Formal Logic - Natural deduction: Problem with assumptions about exists-negation

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:

So far what I've accomplished:

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:

• 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! – Juho Mar 10 at 14:29

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 $$P$$.