From my understanding,

Proof by contrapositive: Prove $P \implies Q$, by proving that $\neg Q \implies \neg P$ since they are logically equivalent.

Proof by contradiction: Prove $P \implies Q$ by showing that $P \wedge \neg Q$ yields an absurdity and hence false. So $\neg (P \wedge \neg Q)$ is equivalent to $\neg (\neg (P \implies Q))$ and $P \implies Q$ by double negation so showing that $\neg (P \wedge \neg Q)$ proves $P \implies Q$.

If the absurdity derived during the procedure for a proof by contradiction is $P \wedge \neg Q \implies\neg P$, we have essentially already proven $P \implies Q$ by contrapositive since $\neg Q \implies \neg P$ is precisely the required condition for proof by contrapositive. But $(P \wedge \neg Q) \implies \neg P$ is also a contradictory statement which means that $P \implies Q$ must be true.

Now the question is this. Is this proof by contradiction still a valid form of proof even though its a proof by contrapositive in disguise? To me, this proof by contradiction also seems to be a valid proof as it does seem to satisfy the conditions(if they are correct) for proof by contradiction.

Additionally, if you have a contrapositive proof, so you have shown that $\neg Q \implies \neg P$, is it possible to rephrase this in a proof by contradiction by supposing that $P \wedge \neg Q$ instead of just $\neg Q$.

If this is the case, what is the point in distinguishing proof by contradiction from proof by contrapositive?

edit: My thought is that proof by contrapositive is a direct proof while proof by contradiction, in this case, depends on the validity of the double negation law which apparently isn't valid in intuitionistic logic.

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    $\begingroup$ The $(\neg Q\implies\neg P)\implies(P\implies Q)$ direction of the equivalence also requires something like double negation elimination assuming you are starting from a reasonably typical constructive logic. $\endgroup$ Commented Dec 26, 2018 at 0:25
  • $\begingroup$ If every proof by contrapositive can be rephrased into a proof by contradiction, why do so many mathematicians prefer proof by contrapositive when it can be shown that way? $\endgroup$
    – Sei Sakata
    Commented Dec 26, 2018 at 0:29
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    $\begingroup$ @SeiSakata sometimes rephrasing the problem in the contrapositive form makes the proof easier or adds some intuition to the statement. Sometimes it's merely a matter of preference. $\endgroup$ Commented Dec 26, 2018 at 0:33
  • $\begingroup$ @spaceisdarkgreen But what we are discussing here is only in one direction. The statement that every proof by contrapositive can be rephrased into a proof by contradiction is seemingly true but the converse doesn't necessarily hold. I am not sure if you were referring to this in your comment" any proof by contradiction “can be rephrased” as a direct proof". BTW I said seemingly true since the line of reasoning that a proof by contrapositive can be rephrased into a proof by contradiction seems to be general enough to account for all cases. But I might be mistaken and not true at all. $\endgroup$
    – Sei Sakata
    Commented Dec 26, 2018 at 1:33
  • $\begingroup$ Sorry, I didn't read your definition of what you're calling a proof by contradiction carefully... you are right that contrapositive is a special case of contradiction here (coincidentally, I wrote an answer about precisely this a couple days ago math.stackexchange.com/questions/3050738/…). $\endgroup$ Commented Dec 26, 2018 at 7:40

1 Answer 1


Yes it is valid... it doesn't really matter if it's something else 'in disguise', just that is it correct. And deriving $\lnot P$ from $P\land\lnot Q$ is certainly leads to a contradiction that implies $\lnot (P\land \lnot Q)$ is true, which implies that $P\to Q$ is true. One thing to note (that I think you have noticed based on your second question) is that you have an additional assumption of $P$ open and available for use when you derive $\lnot P,$ unlike in the case of just deriving $\lnot Q\to \lnot P$ by assuming $\lnot Q$ and deriving $\lnot P.$

So the second question amounts to whether it is admissible to assume $P$ in a proof of $\lnot Q\to \lnot P.$ It is, and the easiest way to see this is reasoning semantically and using the completeness/soundness theorem. If $P\vdash \lnot Q\to \lnot P$ then every interpretation in which $P$ is true has $Q$ true, which means precisely the same thing as $\vdash P\to Q,$ so we have $\vdash \lnot Q\to \lnot P.$

As a result, proof by contrapositive is essentially a special case of what you're calling proof by contradiction, where the contradiction takes the special form $\lnot P$ contradicting with $P.$

I think the reason you might have seen framing contrapositive proofs as proofs by contradiction discouraged is for style reasons. Unless the outstanding assumption of $P$ is used (unnecessarily), the part of the proof outside the inner proof of $\lnot Q\to \lnot P$ is just boilerplate that can be omitted. It is also more informative to call it a proof by contrapositive since that is a special case of contradiction.

(As Derek indicates in the comments, proof by contrapositive is not intuitionistically valid, so it doesn't have anything to do with this.)


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