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I have been doing some research on the P versus NP problem. On multiple occasions, I have seen people say that the problem itself is an NP problem. I have been curious about how we know this. If we know that the problem is NP, then has anyone come up with an algorithm that could be run on a nondeterministic Turing machine to solve the problem in polynomial time? Or is there some other reason that we know that the problem is NP?

Thanks for any replies in advance

Bobjoesmith
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Determining for any statement if there is a proof with $n$ symbols or less is an $NP$ problem (i.e. the proof can be checked in polynomial time with respect to the length of the proof and the statement), that's probably the sense in which they meant that "P versus NP is itself NP". However, it does not really make sense to assign a complexity class to proving any particular statement (such as $P\neq NP$), as that technically takes constant time.

sbares
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    OK, but of course it's entirely possible that there is no proof (in your favourite formal system) of whichever of $P=NP$ and $P \ne NP$ happens to be true. If $P=NP$, there is a polynomial-time algorithm for your favourite NP-complete problem, but there is no guarantee that we could prove that the algorithm is correct or that the algorithm is polynomial-time. – Robert Israel Jun 17 '20 at 17:51
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    @RobertIsrael That is true, but if we put a bound on the size of the proof, then whether it exists or not becomes decidable (indeed, it becomes $NP$). – sbares Jun 17 '20 at 17:54
  • @RobertIsrael: If $P = NP$ but you can't prove it, then our axioms are far weaker than we currently believe they are. If you really have a P-time algorithm that decides SAT, it ought to be straightforward to prove that it is P-time, and not too hard to prove that it decides SAT. I could just about believe the reverse ("$P \ne NP$ but we can't prove it"), but even that seems unlikely to me. – Kevin Jun 18 '20 at 21:59
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    Not really. For example, Miller's test for primality is a polynomial-time algorithm which the generalized Riemann hypothesis (GRH) implies is correct. It is quite reasonable to suppose that the GRH is true but unprovable (after all, people have tried very hard to prove it and haven't done so yet). So this is an example of a polynomial-time algorithm (for a problem which we happen to know, by other means, is in P) that is so far not proven to be correct, and as far as we know its correctness may be unprovable). – Robert Israel Jun 19 '20 at 02:24
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The "P versus NP" problem is a single yes-no question: is $NP = P$? The correct answer is either "yes" or "no", we just don't know which. But the complexity of the answer is $1$.

Robert Israel
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    What is the correct answer if it is undecidable in a given formal system? (As a slight constructivist, I object to the premise of yes-no questions being either yes or no :-)) – ComFreek Jun 18 '20 at 08:32
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    As I commented to SBareS's answer, either is possible, since there's no guarantee we could prove it either way. – Robert Israel Jun 18 '20 at 13:10
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While naïvely this statement makes little sense, there might be more to it than it seems.

This specific statement seems to be linked to the natural proofs barrier, one of several formal results (the so called barriers) on why it is so hard to proof $\smash{\mathrm P\not= \mathrm{NP}}$. Here is a great expository article on this:

It starts as follows:

Have you ever wondered whether the reason there is (apparently) no simple proof that $\mathrm P\not= \mathrm{NP}$ is that $\mathrm P\not= \mathrm{NP}$? Or to turn it around, that an easy proof that $\mathrm P = \mathrm{NP}$ would somehow solve a problem that is hard not only in the Millennium Prize sense but also in the computational-complexity sense?

Stated this naïvely, the above idea does not quite make sense, but in a paper that won them the 2007 Gödel Prize, Alexander Razborov and Steven Rudich [3] proved a result that showed that there is something to this intuition after all. [...]

M. Winter
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