TITLE: Proof by Contradiction, Circular Reasoning?
QUESTION [16 upvotes]: Suppose we wish to prove $P$ implies $Q$. We assume $P$. Proof by contradiction begins by assuming not $Q$, and from these two assumptions, a "contradiction" is derived. Now, sometimes that contradiction is that $Q$ is true. I can't help but to feel the argument is circular, yet I can't make precise what it is that bothers me. We're saying not $Q$ implies $Q$, hence $Q$ must be true to begin with. Why are proofs by contradiction of these kind rigorous?

REPLY [4 votes]: I think I can see your problem. Ignoring $P$ (which is not really relevant for the confusion), we're proving $Q$ by the following technique:

Assume $\neg Q$. Then (bla bla bla bla) and therefore $Q$. But this contradicts our assumption that $\neg Q$, so we conclude $Q$.

On the face of it, this does indeed look circular -- it seems as if we're allowed never to justify our assumption of $\neg Q$ simply because it happens to contradict our eventual conclusion, in the sense that the conclusion and the assumption can't both be true at the same time.
Indeed, that is not a sound reasoning principle in general:

Assume $0=1$. Then, multiplying both sides by $2$ we get $0=2$. But this contradicts $0=1$ (because $1\ne 2$, so $0$ cannot equal both of them at the same time), so we don't need to justify our assumption that $0=1$. Thus, we have proved that $0=2$.

Luckily this is not really what happens in a proof by contradiction. One way to look at a general scheme is

Assume $\neg Q$. Then (bla bla bla bla) and therefore $R$. But $\neg Q$ and $R$ contradict each other (because bla bla bla), so we conclude $Q$.

The important point is that it is $Q$ rather than $R$ that is being concluded. The first argument above is then just an instance of this when $R=Q$ -- in which case it is so easy to see that $\neg Q$ and $R$ cannot both be true that it deserves no argument at all.
A different way of looking at the argument is to consider it an abbreviated form of this:

We know at least that either $Q$ or $\neg Q$ is true. Proceed by case analysis:

Assume $Q$. Then clearly $Q$ is true.

Assume $\neg Q$. Then (bla bla bla) and therefore $Q$.


Since both branches of the case analysis reached the same conclusion $Q$ (and there must always be one of the branches that holds), we can confidently conclude that $Q$ is always true.