How can I prove that if P=NP then for each non-trivial language $L,L'\in NP$ there exists a polynomial reduction $L\leq L'$?
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1I don't really know much about complexity but this sounds wrong to me. Take L to be a polynomial recognizable language and take L' to be an RE-hard language. I don't think, even if P=NP, that there is a polytime reduction from L' to L. – Jake Jun 09 '15 at 15:22
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The correct version adds the requirement that $L,L'$ belong to $\mathsf{NP}$. – Yuval Filmus Jun 09 '15 at 15:55
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you are right. I edited my question. – TT8 Jun 09 '15 at 16:09
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Think: what can a polynomial-time reduction do? Also, you have to require that $L'\notin {\Sigma^*,\emptyset}$, otherwise this is false. – Shaull Jun 09 '15 at 16:18
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You can define a reduction to take every x that belongs to L to y that belongs to L' and every x' that does not belong to L to y' that does not belong to L'? @Shaull – TT8 Jun 09 '15 at 16:30
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Yes, that's the idea. You should first fix x' and y', of course, otherwise the reduction might not be polynomial (you can't compute them mid-reduction). – Shaull Jun 09 '15 at 16:33
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This is pretty much by definition. – Raphael Jun 09 '15 at 18:02