# Proving the validity of a sequent using Modus Tollens

Problem: Prove $$p \rightarrow (q \vee r), \neg q, \neg r \vdash \neg p$$ using Modus Tollens.

I need to prove the validity of the above sequent by using natural deduction. Initially, I didn't read the entire problem, and went the long way by doing an implication-elimination, two negation-eliminations, an or-elimination, and finally a negation-introduction. Then I read the problem and realized I had to use MT, and now I'm stuck. I don't have the greatest grasp on propositional logic and natural deduction. Am I allowed to do the following?

Since I have $$\neg q$$ and $$\neg r$$ as my premises, can I use an and-introduction and do $$\neg q \wedge \neg r$$, which would allow me to immediately use MT to deduce $$\neg p$$? If not, could someone point me in the correct direction? Thanks.

Given $$A \to B$$ and $$\lnot B$$, deduce $$\lnot A$$.
In your case, you are given $$p \to (q \lor r)$$, and so $$A = p$$ and $$B = q \lor r$$. Therefore you need to prove $$\lnot B = \lnot (q \lor r)$$. This is not the same as $$\lnot q \land \lnot r$$, though the two are logically equivalent. What you have to do is deduce $$\lnot (q \lor r)$$ from the premises $$\lnot q$$ and $$\lnot r$$.