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(I know nothing about this topic beyond the popular level, so apologies if this question is not well-posed.)

Every time I've seen this problem discussed, it's always implicit that P must either equal NP or not equal NP. Is it possible that P is related to NP by something other than equal or not equal? For example, could it be that P is related to NP in such a way that they are not exactly equal, but they are also not exactly unequal? (e.g. how time is related to space in general relativity)

Somewhat related: How do mathematicians know the problems they're trying to solve are not undecidable? suggests that this question could be undecidable, which would answer this question with "no, it does not have to be equal or not equal" - but that's an outcome that I never see mentioned either.

Allure
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    (Posting this as a comment instead of an answer, since I don't know if it's actually an answer): shortly after writing this question, I found https://www.scottaaronson.com/papers/pnp.pdf which says (page 4) that P is a subset of NP, so it's immediately obvious that either the containment is proper, or it is not -- which suggests that the answer to the title is "yes". – Allure Sep 11 '24 at 08:09
  • That contradicts the linked question though. No idea how to resolve that. – Allure Sep 11 '24 at 08:09
  • How would your first comment contradict the linked question? – Aphelli Sep 11 '24 at 08:14
  • @Aphelli if it's undecidable, then it's neither equal nor not equal, no? – Allure Sep 11 '24 at 08:19
  • Not quite. “Undecidable” means that there are “worlds” in which the axioms hold and $P=NP$, and that there are “worlds” in which the axioms hold and $P \neq NP$. But, in any given world, one either has $P=NP$ or $P \neq NP$. – Aphelli Sep 11 '24 at 08:40
  • @Aphelli I suggest writing that as an answer, then! – Allure Sep 11 '24 at 08:44
  • The P vs NP problem reduces to checking if only one NP-complete problem is or isn't in P. There's no in-between state for any single NP-complete problem, so there's no in-between state for P vs NP. – Randy Marsh Sep 11 '24 at 09:20

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I think the question relies on a misunderstanding of what “undecidable” means (and hopefully my understanding is not too naive here).

First, undecidability is always defined with respect to (a language and) a collection of axioms. A proposition is then undecidable if these axioms cannot prove or disprove the result.

A way to view undecidability is to consider that the objects we are familiar with (integers, sets, ordinals, cardinals), satisfying the axioms could exist in many possible “mathematical worlds” – so that we cannot tell them apart using the axioms.

In any given “mathematical world”, a given statement is either true or false. If I really “have” a real line, either it contains subsets of smaller cardinality than itself that are not countable, or it doesn’t.

In any given world, $P$ and $NP$ are sets with $P \subset NP$. So one either has $P=NP$, or $P \neq NP$.

If $P=NP$ is undecidable (with respect to, say, ZFC), it means that there are mathematical worlds in which the ZFC axioms hold and where $P=NP$ and others where $P \neq NP$ and the ZFC axioms also hold.

[Note that a statement that is undecidable for a certain collection of axioms can be decidable by another. You can of course trivially add the statement or its negation as an axiom, but this can be more subtle, as in the following MO question about the Riemann hypothesis: https://mathoverflow.net/questions/79685/can-the-riemann-hypothesis-be-undecidable].

Aphelli
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  • +1,would you check my this question, https://math.stackexchange.com/questions/4976044/the-language-is-in-np-setminus-p, –  Sep 29 '24 at 12:24