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Is proof by contradiction always a sufficient proof technique ?

A proof by contradiction has the form:

Let $P$ and $Q$ be statements. If $ P \rightarrow Q \land \lnot Q $ then you can conclude $\lnot P$.

However just because $P$ lead to a contradiction, how can one be sure that $\lnot P$ is true ? What if the axiomatic system allows both $P$ and $\lnot P$ to be false in different context? I mean you could prove $P$ is true by showing $\lnot P$ leads to a contradiction, but what if a different context you can show $\lnot P$ is true by showing $P$ leads to a contradiction?

Is the theory of mathematics always constructed so $P$ and $\lnot P$ cannot both be false in different context?

Asaf Karagila
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Shuzheng
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  • See also: http://math.stackexchange.com/questions/240/are-the-proofs-by-contradiction-weaker-than-other-proofs and http://math.stackexchange.com/questions/243770/can-every-proof-by-contradiction-also-be-shown-without-contradiction and http://math.stackexchange.com/questions/50764/are-proofs-by-contradiction-just-more-powerful-or-just-more-convenient-than-othe and http://math.stackexchange.com/questions/51767/why-some-people-dont-like-proofs-by-contradiction and probably other links as well... – Asaf Karagila Sep 12 '13 at 12:05

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What $P\to Q\land \neg Q$ means is that in every context where $P$ is true, $Q$ and $\neg Q$ will both be true. Since there is no situation in which $Q$ and $\neg Q$ are both true, there cannot be any situation where $P$ is true either. So instead $\neg P$ must be true always.

Sure, one can imagine a $Q$ that is sometimes true and sometimes false. But then $Q\land \neg Q$ will still always be false, because there's no case in which $Q$ and $\neg Q$ is true simultaneously. And for such a $Q$ you shouldn't be able to prove $P\to Q\land \neg Q$, unless $P$ itself is never true ...

hmakholm left over Monica
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