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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 $T \implies F$)?

juanjo12x
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  • But $Q$ is known to be false (e.g. $0=1$). Thus, if the inference from $\lnot P$ to $Q$ is logically correct (valid) we have that $\lnot P \to Q$ must be True and thus we have that $\lnot P \to \text {False}$ is True and this is just in case that $\lnot P$ is False. – Mauro ALLEGRANZA Sep 12 '21 at 12:31
  • The gist is: having derived a contradiction from the assumption that $P$ is false, we have to conclude that the assumption is untenable, and thus $P$ must be true. – Mauro ALLEGRANZA Sep 12 '21 at 12:32
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    See also Proof by contradiction – Mauro ALLEGRANZA Sep 12 '21 at 12:34
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    The author is convoluting their explanation by unnecessarily invoking Contrapositive to explain Contradiction. In case you'd like another Answer to your question, read my Remarks here and my Point 3 here. The short of it is that in a proof by contradiction, we first make an assumption, then derive some absurdity, then observe that since the assumption was the only possible "leak" (source of the absurdity), it must have been falsely made. – ryang Sep 12 '21 at 13:39
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    Thank you very much! Mauro's references made me understand what a contradiction means while Ryan's remark enlightened me on why the contrapositive seemed out of place. – juanjo12x Sep 12 '21 at 15:49
  • See fifth century Neo-platonic philosopher Proclus: "Every reduction to impossibility takes the contradictory of what it intends to prove and from this as a hypothesis proceeds until it encounters something admitted to be absurd and, by thus destroying its hypothesis, confirms the proposition it set out to establish. (Commentary on Euclid’s Elements)" – Mauro ALLEGRANZA Sep 13 '21 at 08:46

2 Answers2

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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 :)

Ben Millwood
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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

Bram28
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