Three good answers were received — by Alex Gavrilov, Bjørn Kjos-Hanssen, and Terry Tao — and the bounty has been awarded (somewhat arbitrarily) to Alex Gavrilov.

The answers are summarized below; because they are open-ended and technically subtle, the question has been flagged for conversion to community Wiki.

Thanks are extended to all who contributed.

Summary  Harry Altman cogently commented:

This is essentially asking which of these statements are equivalent to a $\Pi^0_1$ statement, right?

We embed this insight into a better version of the question:

Which of the Millennium Prize problems can be stated as a postulate that can be falsified by a $\Pi^0_1$ counterexample?

to which the answers given (as I understand them) amount to:

  1. "The Riemann Hypothesis is true" …a $\Pi^0_1$ counterexample could exist;
    (per Noam Elkies' comment)
  2. "The Birch and Swinnerton-Dyer Conjecture is true" … conceivably a $\Pi^0_1$ counterexample could be constructed, but not with present knowledge (per Alex Gavrilov's answer);
  3. "$\mathsf{P}\ne\mathsf{NP}$" … no obvious $\Pi^0_1$ counterexample
    (per Bjørn Kjos-Hanssen's answer);
  4. "Navier–Stokes is globally regular" … no obvious $\Pi^0_1$ counterexample
    (per Terry Tao's answer);
  5. "Yang–Mills has a mass gap" … no obvious $\Pi^0_1$ counterexample (?);
  6. "The Hodge Conjecture is true" … no obvious $\Pi^0_1$ counterexample (?);

Resource  Wikipedia's article Arithmetical Hierarchy explains the notation of Harry Altman's answer.

What "No Obvious $\Pi^0_1$ Counterexample" Means   As was noted on Dick Lipton and Ken Regan's weblog Gödel's Lost Letter and P=NP, the authority of the Scientific Advisory Board (SAB) of the Clay Mathematics Institute (CMI) extends to:

"The SAB may consider recommending the award of the prize to an individual who has published work that, in the judgement of the SAB, fully resolves the questions raised by one of the Millennium Prize Problems even if it does not exactly meet the wording in the official problem description.”

In view of the CMI/SAB's supreme executive authority, the logical possibility of amending a Millennium Prize question to accommodate $\Pi^0_1$ counterexamples — via ingenious "burning arrows," to adopt Dick Lipton and Ken Regan's phrase — cannot be formally excluded.

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    $\begingroup$ @darij: This is the point of mathoverflow.net/questions/31846/… $\endgroup$ – Andrés E. Caicedo Sep 30 '12 at 2:51
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    $\begingroup$ The point is that if there's a counterexample then one can prove it by a contour integral: no need to actually locate the zero, only to prove there's one in a circle disjoint from the critical line. $\endgroup$ – Noam D. Elkies Sep 30 '12 at 3:03
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    $\begingroup$ @zeb: "True" means "true in the standard model". $\endgroup$ – Andrés E. Caicedo Sep 30 '12 at 3:19
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    $\begingroup$ So if I understand correctly, this is essentially asking which of these statements are equivalent to a $\Pi_1^0$ statement, right? $\endgroup$ – Harry Altman Sep 30 '12 at 4:01
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    $\begingroup$ Will - yes, but the undecidable bit is to prove that a CW-complex is not simply connected. There is a partial algorithm that terminates if and only if an input CW-complex is simply connected or, equivalently, an group presentation presents the trivial group, and this is enough for these purposes. The partial algorithm simply tries to write each generator as a product of conjugates of relators. $\endgroup$ – HJRW Sep 30 '12 at 6:54

As Harry Altman pointed out, for a conjecture undecidable -> true means that it can be formulated as a $\Pi_1^0$ statement. To put it simply, if the conjecture is false, one can prove this by an explicit (finite) calculation. I would leave Yang–Mills and Navier–Stokes to someone more familiar with mathematical physics then I am. The other three conjectures, apparently, aren't $\Pi_1^0$ statements for now.

By the way, the Poincare conjecture wasn't a $\Pi_1^0$ statement before Rubinstein's algorithm was discovered (which determines whether or not a 3-manifold is the 3-sphere, see the comment by HJRW). (Expert guys, fix me here if I am wrong). So, whether or not a particular conjecture is in this calss depends on the awailable knowledge. After all, once a conjecture is proven, it is in $\Pi_1^0$ by definition.

Let me begin with Birch and Swinnerton-Dyer Conjecture. Technically, what you need to get \$1M is to prove or disprove the following (http://www.claymath.org/millenium-problems/birch-and-swinnerton-dyer-conjecture). If $E$ is an elliptic curve over $\mathbb{Q}$, then $r = rank(E(Q))$, called arithmetic rank, is equal to the order $r_*$ of zero of $L(E, s)$ at $s=1$, called analytic rank. This conjecture actually consists of two rather different parts, namely $r\le r_*$ and $r\ge r_*$. The first one is a $\Pi_1^0$ statement. If $r>r_*$ for a particular curve $E$, you can prove it by a direct computation (with some luck). The other half of the conjecture is more tricky, as Will Sawin pointed out. The problem is, there is no known algorithm to compute the group $E(Q)$. (Do not take me wrong: there are some algorithms, and they apparently work. We just don't have a proof that they work.)
Theoretically, it is possible that while $r<r_*$ for some curve $E$, you will never know it because you won't be sure if there are some more generators of $E(Q)$ which you did not find yet. In fact, Manin used this very argument in the opposite direction, and proposed an algorithm of computing the Mordell-Weil group assuming the Birch and Swinnerton-Dyer Conjecture to be true. (See Hindry, Silverman "Diophantine Geometry" for details. To be pedantic, you need a bit more then $r=r_*$ for this.)
So, the inequality $r\ge r_*$ is not a $\Pi_1^0$ statement yet, for the best of my knowledge.

I can't say anything sensible about the Hodge conjecture, except that it definitely does not look like $\Pi_1^0$. In order to disprove it by an explicit example, you need to prove that on a particular variety there is no algebraic cycles with a given cohomology class. Maybe experts know better, but I have never heard about any algorithm for a job like this.

Bjørn Kjos-Hanssen already gave an answer about $P\neq NP$, and I don't have much to add. Except for, as I pointed out in a separate question, How to prove a $\Pi_2$ statement properly?, there is a technical possibility that that problem is decidable, but the "decision" is wrong. (Do not take me seriously, I don't really believe in this.)

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    $\begingroup$ What do you mean by, "After all, once a conjecture is proven, it is in $\Pi^0_1$ by definition"? Logically speaking, a sentence $\sigma$ is not in general equivalent to "there is a proof of $\sigma$," as Goedel shows. $\endgroup$ – Monroe Eskew Apr 11 '14 at 17:54
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    $\begingroup$ Yes, but I mean it is proven and assumed to be true. $\endgroup$ – Alex Gavrilov Apr 11 '14 at 17:58
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    $\begingroup$ I don't see how assuming a statement to be true makes it $\Pi^0_1$. $\endgroup$ – Monroe Eskew Apr 11 '14 at 19:10
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    $\begingroup$ Once it's proven, the sentence is proven to be equivalent to a $\Pi^0_1$ statement, namely $0=0$. When we speak of "a sentence" we probably mean "up to provable equivalence" since the Millennium Problems are only defined up to provable equivalence... $\endgroup$ – Bjørn Kjos-Hanssen Apr 11 '14 at 20:48
  • $\begingroup$ Alex Gavrilov, you have received the bounty. Thank you for taking the time to give such a well-considered answer. $\endgroup$ – John Sidles Apr 18 '14 at 0:24

Global regularity for Navier-Stokes on the torus is (logically equivalent to) a $\Pi_2^0$ statement; this is essentially an unpublished observation of Bourgain (who made it in the more general context of supercritical equations), which I sketched out in this paper http://arxiv.org/abs/0710.1604 . Basically, Navier-Stokes is equivalent to the assertion that for all $T, E > 0$, there exists an $M$ such that all initial data with $H^1$ norm at most $E$, there exists a solution up to time $T$ whose $H^1$ norm is always bounded by $M$. Now if such a claim is the case, it can be verified (using rigorous perturbation theory) by constructing a sufficient number of approximate solutions to a sufficient number of choices of initial data, and verifying that all of these approximate solutions have norm at most $M/2$ (say) up to time $T$, where "sufficient number" is something explicit that depends on $T, E, M$, and the nature of the approximation can be discretised at an explicit scale that also depends on $T,E,M$. So Navier-Stokes is equivalent to a statement of the form $\forall T,E \exists M: P(T,E,M)$ where $P(T,E,M)$ is something that can be verified in finite time for each $T,E,M$ (which can be taken to be rational numbers). (In fact, for each $E$ there is an explicit time $T = T(E)$ for which one needs to verify $P(T,E,M)$, and beyond which global regularity is easily obtained from the energy dissipation, so the claim even simplifies a little to $\forall E \exists M: P( T(E), E, M)$.)

It looks very difficult to me, however, to make Navier-Stokes equivalent to a $\Pi^0_1$ statement. This would basically amount to being able to describe all possible "blowup scenarios" by a countable set, and to be able to determine whether each such blowup scenario can actually happen in finite time. While some blowup scenarios (particularly "stable", "approximately self-similar" ones) can be described and verified in such a manner, it's not clear whether all of them can.

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    $\begingroup$ Are there examples of $ E $ where $ M $ can be calculated? $\endgroup$ – Bjørn Kjos-Hanssen Apr 12 '14 at 17:34
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    $\begingroup$ For sufficiently small E, perturbation theory allows one to conclude that M depends linearly on E in the limit as E goes to zero. $\endgroup$ – Terry Tao Apr 13 '14 at 21:13
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    $\begingroup$ Terry Tao, you deserve a bounty, for taking the time to give such a well-considered answer (from which I have quoted). I have flagged the amended question/answer for conversion to Community Wiki; please do not hesitate to amend it as you see fit. And, thank you. $\endgroup$ – John Sidles Apr 18 '14 at 0:26
  • $\begingroup$ Here is a version of the Navier Stokes problem due to the physicist Giovanni Gallavotti. Given a divergence free vector field on periodic three space defined by a finite fourier series , at a fixed point in space give an algorithm which describes the solution vector field one second later at that point to any degree of precision with an estimated time for the computation at that precision. Is this version of the problem decidable? Dennis Sullivan $\endgroup$ – Dennis Sullivan Jan 25 at 14:03

$P\ne NP$ is a $\Pi^0_2$ statement:

for each polynomial $p$ and Turing machine $M$ implementing an algorithm attempting to decide SAT, there is a formula $\phi_M$ such that if we look at the computation of $M$ on input $\phi_M$ after $p(|\phi_M|)$ computation steps, $M$ has either not halted or it has answered incorrectly.

(Edit: added more detail to the $\Pi^0_2$ statement above.)

If we place a computable bound on the size of $\phi_M$ as a function of $M$ then we can improve this to a $\Pi^0_1$ statement, and hence will have the property $\text{undecidable}\rightarrow\text{true}$, since true $\Sigma^0_1$ statements are provable (in Peano Arithmetic). Of course in a formal sense, any provable statement is equivalent to $0=0$ which is $\Pi^0_1$.

Is it plausible that that $P\ne NP$, but there is no computable bound on $|\phi_M|$? This would mean that there are algorithms that do "arbitrarily well" at solving SAT in polynomial time. That is, relative to their description, they are correct for an Ackermann-function or busy-beaver-function level many $\phi$'s.

But intuitively, there is not much structure in SAT to exploit, and so once you have so many variables $x_1,\ldots,x_v$ that a random assignment of truth values has higher Kolmogorov complexity than $M$: $$K(M)\le v + O(1) $$ then there ought to be a $\phi(x_1,\ldots,x_m)$ on which $M$ fails. Since $K(M)$ has a computable bound as a function of $M$, it would follow that $M\mapsto |\phi_M|$ is computably bounded. This is probably off by a "factor" somewhere; maybe a better way is to say that $P^A\ne NP^A$ for a random oracle $A$, and in that case $\phi_M$ is computably bounded.

So we can make a plausible "computably bounded $P\ne NP$ conjecture" that $M\mapsto \phi_M$ is computably bounded (with a specific bound being incorporated into the conjecture), which does have the property $\text{undecidable}\rightarrow\text{true}$.

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    $\begingroup$ Bjorn, your bound-restriction nicely addressed Scott Aaronson's weblog remark "Independence from ZFC is not a property of the language L itself, but only a property of how I described L to you! … something like 'the subclass of P consisting of all languages that ZFC can prove are in P' doesn’t even make sense" (an objection that a TCS StackExchange wiki addressed too). Restrictions can be tricky, and so I'll reflect further on your fine answer in the next few days. $\endgroup$ – John Sidles Apr 12 '14 at 9:31
  • $\begingroup$ Bjorn, in regard to "true $\Sigma^0_1$ statements are provable in PA", and specifically in regard to the exemplar "$M\in\text{P}$ and fails for no formula length-bounded by $\phi_M$", is the latter really in $\Sigma^0_1$? Isn't "$M\in\text{P}$" itself undecidable in PA (and even in all formal systems)? Your answer's overall strategy of "improving" $\text{P}\ne\text{NP}$ to a $\Pi^0_1$ statement is promising, but perhaps the "$M\in\text{P}$" element of it requires adjustment? $\endgroup$ – John Sidles Apr 14 '14 at 9:07
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    $\begingroup$ There is no need to decide whether $M$ runs in polynomial time (which I assume is what you mean by "$M\in P$"); I have put the added detail into an edited version of my answer $\endgroup$ – Bjørn Kjos-Hanssen Apr 14 '14 at 18:19
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    $\begingroup$ Thank you Bjorn. Your construction (as it seems to me) definitely establishes $\text{P}\ne\text{NP}$ as a $\Pi^0_2$ statement. Per recent "burning arrow" comments on Gödel’s Lost Letter, and along the lines of the second half of your answer, it is natural to wonder whether (in your view) there exists any restriction of $\text{P}\ne\text{NP}$ that reduces it to $\Pi^0_1$, while respecting sufficiently the spirit of the problem to "reasonably resolve it" (per the concluding phrase of the Millenium Prize rules)? $\endgroup$ – John Sidles Apr 14 '14 at 20:08
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    $\begingroup$ Well, if there was a short and simple algorithm running very fast, that provably was correct on all $\phi$ of length at most $2^{2^{2^{1000000}}}$, I guess they would consider awarding the prize... $\endgroup$ – Bjørn Kjos-Hanssen Apr 14 '14 at 20:25

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