Can I use argument by cases while proving a contradiciton?

Question

Can I use proof by cases (argument by cases) while proving contradiction ? So in my case it is like that. I need to prove $\lnot (P \lor Q)$. So I need to derive contradiction from $P \lor Q$. And I can prove contradiciton from $P$ and $Q$ (by "and" I don`t mean conjunction, I mean I can prove contradiction from $P$ and also I can do it from $Q$), so have I proved $\lnot (P \lor Q)$ ?

Answer 1

Yes. Technically, you would be doing a proof by Cases within a Proof by Contradiction.

Here is what the basic template looks like formally:

Line 9 is where you complete the Proof by Cases, and line 10 is where you complete the Proof by Contradiction.

Answer 2

Sure.

If $P$ leads to a contradiction, and $Q$ leads to a different contradiction than the neither $P$ nor $Q$ can be correct.

It may be helpful to note $\lnot(P\lor Q)\equiv (\lnot P)\land (\lnot Q)$. So it's okay to prove the two statements 1) $\lnot P$ and 2) $\lnot Q$.