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i.e. $A\leq_pB\:\wedge\:B\in\text{co-NP}\rightarrow A\in\text{co-NP}$ ?

I feel like it's the case but I can't think of a straightforward proof.

Clarification: I am talking about polynomial-time Turing reductions.

nc404
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  • Do you know how to solve this with coNP replaced by NP? If so, try connecting the two. – Yuval Filmus Apr 19 '21 at 05:47
  • @YuvalFilmus It seems not. I know there are a few questions about this here but none of the answers make sense to me. For example, https://cs.stackexchange.com/questions/128187/if-x-is-polynomial-reduction-to-y-and-y-is-in-np-then-x-is-in-np here there only seems to be a question of a 1-to-1 reduction. But a polynomial-time reduction can have multiple calls to problem B, not only one. – nc404 Apr 19 '21 at 14:52
  • Are you sure the reduction is not many-one? – Yuval Filmus Apr 19 '21 at 15:30
  • @YuvalFilmus What do you mean? In that problem? It says the reduction is "x in X iff phi(x) in Y", that's a single call to Y, isn't it? – nc404 Apr 19 '21 at 15:39
  • So the reduction in your case is many-one. You mentioned that "a polynomial-time reduction can have multiple calls to problem B", but this is not the case here. – Yuval Filmus Apr 19 '21 at 15:41
  • @YuvalFilmus Is that what many-one means? I'm not sure about this. If you solve problem A with 10 calls to problem B for example, is that many-one? – nc404 Apr 19 '21 at 15:52
  • No. A many-one reduction is a function $\phi$ as in your preceding comment. – Yuval Filmus Apr 19 '21 at 15:53
  • @YuvalFilmus Ok. So I am talking about a polynomial-time Turing reduction. How do you prove that if a problem A reduces to a problem in NP, then that problem A is in NP? I can't see a solution because of the non-determinism of turing machines which solve NP problems. – nc404 Apr 19 '21 at 16:02
  • Your reduction is not a Turing reduction. It’s a many-one reduction. – Yuval Filmus Apr 19 '21 at 16:13
  • @YuvalFilmus But I meant Turing reductions. That is what my question was about. I edited the question to clarify. – nc404 Apr 19 '21 at 16:19
  • Any problem Turing-reduces to its complement, so your question is essentially equivalent to NP=coNP. – Yuval Filmus Apr 19 '21 at 16:35
  • @YuvalFilmus which is an open question. Interesting. Thank you. – nc404 Apr 19 '21 at 16:44
  • @YuvalFilmus But how do you prove that statement but with NP instead of coNP? It that feasible? If a problem A poly-time turing reduces to a problem B in NP, is A in NP? – nc404 Apr 19 '21 at 17:57
  • The statements for NP and coNP are equivalent. Either both hold or both don't. – Yuval Filmus Apr 19 '21 at 19:37
  • @YuvalFilmus thank you. – nc404 Apr 19 '21 at 21:35

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