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Today at lunch, I brought up this issue with my colleagues, and to my surprise, Jeff E.'s argument that the problem is decidable did not convince them (here's a closely related post on mathoverflow). A problem statement that is easier to explain ("is P = NP?") is also decidable: either yes or no, and so one of the two TMs that always output those answers decides the problem. Formally, we can decide the set $S :=\{|\{P, NP\}|\}$: either the machine that outputs $1$ only for input $1$ and otherwise $0$ decides it, or the machine that does so for input $2$.
One of them boiled it down to basically this objection: if that's how weak the criterion of decidability is - which implies that every question which we can formalize as a language that we can show to be finite is decidable - then we should formalize a criterion that doesn't render any problem with finitely many possible answers that's formalizable in this way decidable. While the following is possibly a stronger criterion, I suggested that maybe this could be made precise by requiring that decidability should depend on being able to show a TM, basically proposing an intuitionist view of the matter (which I don't incline towards - nor do any of my colleagues, all of them accept the law of excluded middle).
Have people formalized and possibly studied a constructive theory of decidability?
My suspicion is that constructive computability would be quite boring. (I find their objection weaker than the definition they complain about.) – Raphael – 2016-03-22T15:18:03.443
2How about a machine that searches in parallel for proofs of $\mathsf{P} = \mathsf{NP}$ and of $\mathsf{P} \neq \mathsf{NP}$ and acts accordingly? Assuming that the question is decidable, the machine will always halt and accepts the language. Do you allow that? – Yuval Filmus – 2016-03-22T15:49:52.537
@YuvalFilmus Hm, it's circular to show I can decide a problem under the assumption that I can decide it - maybe I misunderstood you? – G. Bach – 2016-03-22T16:08:17.257
@G.Bach The difference is that my machine is explicit, only the proof that it halts is non-constructive. – Yuval Filmus – 2016-03-22T16:55:21.950
@YuvalFilmus The problem I'm having is that I don't see that a proof for either P = NP or P $\neq$ NP must exist; if I knew that, then what you suggest would work assuming we can enumerate proofs (e.g. via some form of Gödelization). Could you elaborate a bit? – G. Bach – 2016-03-22T19:53:30.620
1@G.Bach You don't see it because we don't know that it exists. But if you assume that $\mathsf{P} = \mathsf{NP}$ is not independent, then the program works. If it is independent, then your language itself is model-dependent, which is somewhat strange. – Yuval Filmus – 2016-03-22T22:33:37.070
there seem to be closely related questions of P vs NP that are not obviously decidable such as (say P≠NP) "what is the smallest $c$ such that SAT is solvable in $O(n^c)$ time." scott aaronson has a survey on P vs NP decidable question. but the short story is that there does not seem to be any compelling reason to expect undecidability problems with complexity theoretic questions, "yet". the Blum axioms (speedup/ gap etc) are worth some study as a case where there are some paradoxical results. also oracle results such as Baker Gill Solovay 1975 might be relevant. – vzn – 2016-03-23T02:06:58.150
@vzn That survey by Aaronson, do you mean "Is P v NP formally independent?"? Having read the first page of it, that seems like a worthwhile read. If you'd like credit for it, please post it as an answer. – G. Bach – 2016-03-23T09:04:49.023
@YuvalFilmus I thought about it some more, and the connection between decidability and independence is more relevant here than I would've thought. If you'd like credit for it, please post an answer - and if you have any additional remarks, that's even better. – G. Bach – 2016-03-25T21:40:02.953