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.




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**Summary**  Harry Altman [cogently commented](https://mathoverflow.net/questions/108433/for-which-millennium-problems-does-undecidable-true#comment279309_108433):



> 
> 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](https://mathoverflow.net/questions/108433/for-which-millennium-problems-does-undecidable-true#comment279309_108433))
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](https://mathoverflow.net/a/163127/11394));
3. "$\mathsf{P}\ne\mathsf{NP}$" … *no obvious $\Pi^0\_1$ counterexample*  
(per [Bjørn Kjos-Hanssen's answer](https://mathoverflow.net/a/163124/11394));
4. "Navier–Stokes is globally regular" … *no obvious $\Pi^0\_1$ counterexample*  
(per [Terry Tao's answer](https://mathoverflow.net/a/163160/11394));
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*](http://en.wikipedia.org/wiki/Arithmetical_hierarchy#The_arithmetical_hierarchy_of_formulas) explains the notation of Harry Altman's answer.


**What "No Obvious $\Pi^0\_1$ Counterexample" Means**   As was [noted](http://rjlipton.wordpress.com/2014/04/10/multiple-credit-tests/#comment-45395) on Dick Lipton and Ken Regan's weblog *Gödel's Lost Letter and P=NP*, the authority of the [Scientific Advisory Board](http://www.claymath.org/millennium-problems/rules-millennium-prizes) (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](https://www.youtube.com/watch?v=rAaWvVFERVA), the logical possibility of amending a Millennium Prize question to accommodate $\Pi^0\_1$ counterexamples — via ingenious "[burning arrows](http://media-cache-ak0.pinimg.com/736x/2c/a4/67/2ca467dbb02e1b3eb5d57c467334fc05.jpg)," to adopt [Dick Lipton and Ken Regan's phrase](http://rjlipton.wordpress.com/2009/09/13/are-impossibility-proofs-possible/) — cannot be formally excluded.



For which Millennium Problems does undecidable -> true?
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?](https://mathoverflow.net/questions/111075/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.) 


