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Here is a well-known example of this kind of proof.

Proposition: Show that there are irrational numbers $a,b$ such that $a^b$ is rational.

Proof: Consider the statement "$\sqrt{2}^{\sqrt{2}}$ is irrational."

If the statement is true, then $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}=2$ and we win. If the statement is false, meaing $\sqrt{2}^{\sqrt{2}}$ is rational and we win as well. $\square$

Note that this proof uses the statement but require neither of the statement nor its negation to be proved.

In another more advanced example, the generalized Riemann hypothesis could be used as the statement to show that Gauss's list of imaginary quadratic fields with class number 1 is complete. The proof scheme is similar: both the statement and the negation of the statement implies the conclusion.

I am looking for clever proofs of such scheme, since the rational $a^b$ problem is the only one I am familiar with.

JonathanZ
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William Sun
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    It's fair from clear what you are asking. The proof method you are describing is just proof by cases. What do you mean by "clever proofs of such scheme"? Are you asking for more interesting examples of the technique? Or What? – Rob Arthan Oct 22 '23 at 20:13
  • If you assume that something is false, then that something is not an hypothesis – Exodd Oct 22 '23 at 20:14
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    @RobArthan Yes this is indeed a way to prove by cases, but the interesting and clever part is actually a right choice of the “hypothesis”. – William Sun Oct 22 '23 at 20:29
  • So you are asking for interesting examples of the technique. You should make that clearer in your question. – Rob Arthan Oct 22 '23 at 20:32
  • Your use of the word "hypothesis" is incorrect. That is making your question not clear. – jjagmath Oct 22 '23 at 20:32
  • In Mathematics, if you want to prove a theorem or a proposition you always takes the hypothesis to be true. – jjagmath Oct 22 '23 at 20:34
  • @jjagmath in that case I would call it a “claim” instead of a “hypothesis.” Is there a word you would suggest instead of “hypothesis”? – William Sun Oct 22 '23 at 20:36
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    It could be "statement". And I wouldn't write that statement before the proposition. Considering the statement is part of the proof. – jjagmath Oct 22 '23 at 20:41
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    @WilliamSun - I rewrote your proof because I don't think the previous structure was actually valid, and it was definitely confusing to many of the commenters. Please feel free to rollback my edit if you think it didn't improve things. – JonathanZ Oct 22 '23 at 21:19
  • So you're looking for a proof of $q$ that shows $(p\to q)\land(\neg p\to q)$? – J.G. Oct 22 '23 at 22:10
  • In classical logic, a proof of $$\lnot(P\to Q)\vdash(Q\to R)$$ doesn't rely on $P$ or $R$: $$\begin{array}{c|c|c}1. & \lnot(P\to Q) & Premiss \ 2. & \lnot(Q\to R) & Assump. \ 3. & \lnot Q & 1\ 4. & Q & 2\end{array}$$ – Shaun Oct 31 '23 at 21:57

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