I'm a little confused about the difference between these two types of proof. As I have been taught them, it seems like proofs by contrapositive are just a subset of proofs by contradiction.

Say we want to prove $P \implies Q$. Here is how I've been taught to use both proofs:

Contrapositive: We assume $\lnot Q$ and from this we try to conclude $\lnot P$.

Contradiction: We assume $P$ and $\lnot Q$ and from this we try to deduce a contradiction.

My question is, let say we proved $P \implies Q$ by contrapositive, so we proved $\lnot Q \implies \lnot P$. Wouldn't this be equivalent to assuming $P$ and $\lnot Q$ and then use $\lnot Q$ to prove $\lnot P$, so we would deduce the contradiction $P \land \lnot P$?

• They're very similar. When contraposition works, a proof by contradiction feels unneccesarily clunky (you've just added a few lines to the top and bottom for me to read and process with no actual value), while some times a proof by contradiction (or, equivalently, by contraposition and then splitting into the two cases $P$ and $\lnot P$) is necessary. – Arthur Aug 6 '18 at 11:35
• @Arthur sorry for wasting your time sir. Feel free to edit the question so no one else has to process useless information. – Yagger Aug 6 '18 at 11:42
• I'm sorry. That's not what I meant. Your question is fine. I meant that when you make such a proof (i.e. a proof by contradiction which is basically a proof by contraposition with an added assumption of $P$ at the top), there are lines in that proof with no value. Specifically, the lines "assume $P$" at the top of the proof and "Thus $P$ and $\lnot P$, which is a contradiction" at the bottom. – Arthur Aug 6 '18 at 11:43

Every proof by contraposition can be reformulated as a proof by contradiction, as you correctly noticed. Indeed, suppose you have a proof $\pi$ of $\lnot Q \implies \lnot P$; then you can prove $P \implies Q$ by contradiction, by assuming $P$ and $\lnot Q$, which yields a proof of $\lnot P$ (via modus ponens between the assumption $\lnot Q$ and the conclusion $\lnot Q \implies \lnot P$ of $\pi$) and then you get the contradiction $P \land \lnot P$.
But not every proof by contradiction can be reformulated as a proof by contraposition, as well explained in the accepted answer of this question. Indeed, proofs by contradiction are "more general" (i.e. they can be applied to a wider set of propositions to prove) because you can stop when you find any contradiction, not only a contradiction directly involving the hypotheses. More precisely, in a proof by contradiction of $P \implies Q$, we assume $P$ and $\lnot Q$ and we deduce a contradiction $R \land \lnot R$; in the particular case where $R = P$ then (usually) you can reformulate your proof as a proof by contraposition, otherwise you cannot.