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.


1 Answer 1


Modus Tollens is the following rule:

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 $\lnot r$.


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