Proof by contradiction of $P\implies Q$?

Question

If I want to proof the following by contradiction: $$P\implies Q.$$ Say $R$ is a contradiction$(R\equiv\bot)$. By constructing $$\begin{array}{l} (P\land\lnot Q)\implies R.\\ \lnot R.\\ \hline\therefore\lnot(P\land\lnot Q).\end{array}$$ Since $$\lnot(P\land\lnot Q)\iff\lnot P\lor Q,$$ which implies $$P\implies Q.$$

Is the above correct? Is the above relate to the Law of Excluded Middle?

Answer 1

Yes, your argument looks correct.