Proof by contradiction in "A Concise introduction to Pure mathematics" by M.Liebeck I am going through the first chapter of "A concise Introduction to Pure Mathematics" but can't get my head around how Liebeck explains the proof by contradiction. He starts with:

Suppose we wish to prove the truth of a statement P. A proof by
contradiction would proceed by first assuming that P is false - in
other words, assuming $\bar{P}$. We would try to deduce from this a statement
Q that is palpably false (0=1, for example).

Up until this point, my knowledge matches, but then he goes on:

Having done this, we have shown $ \bar{P} \implies Q$. Hence also $ \bar{Q} \implies P$. Since we know Q is false, $\bar{Q}$ is true, and hence so is P, so we have proved P,
as desired.

I understand why $ \bar{Q} \implies \bar{P} $ is equivalent to $P \implies Q$ (thanks to this question) but why by proving  $ \bar{P} \implies Q$  , then it follows P is true?. Didn't we assume  $ \bar{P}$ was true so the statement $ \bar{P} \implies Q$  is false (because of $T \implies F$)?
 A: Let's distinguish between:

*

*which statements are really true or false

*which statements we could deduce to be true or false if we assumed $\bar P$.

Crucially, in the second sense, it is possible – desired, even – to conclude from the assumption that something is both true and false. That's how we determine that $\bar P$ is a bad assumption, and thus that $P$ must be true.
So, yes, when you assume $\bar P$, then $\bar P \implies Q$ is indeed false, by your argument. Your error was in believing that this means that it can't also be true under that assumption :)
A: No, you do not show that $\bar{P} \implies Q$ is False.
When you ‘assume $\bar{P}$’, you are not declaring $\bar{P}$ to be true. Rather, you are saying ‘what would be the case if $\bar{P}$ were true’. And apparently, if $\bar{P}$ is true, then $Q$ is true. And thus you have shown that $\bar{P} \implies Q$ is True … rather than False.
Now, you can say that $\bar{P} \implies Q$ is False if $\bar {P}$ is True … but from that you can only conclude something like $\bar{P} \implies \neg (\bar{P} \implies Q)$, rather than $\neg (\bar{P} \implies Q)$ by itself. In fact, since you show that $\bar{P} \implies Q$, that would be another way of establishing the contradiction you are looking for, since (as you yourself point out) $\bar{P} \implies  Q$ would be False  if $P$ would be False. So, the assumption that $P$ is False leads to a contradiction (since then $\bar{P} \implies Q$ would be both True and False) and therefore $P$ must in fact be True
