5

Suppose someone proved that the $P = NP$ question is undecidable from $ZFC$. Would that imply it is true? Would that imply it is false? I know that there are certain mathematical statements which, if they were proved undecidable, it would automatically imply that they are true.

user107952
  • 23,768

1 Answers1

5

As recently described here, statements of the form $\forall n : P(n)$ where $P(n)$ is a statement about a natural number $n$ that can be verified by a finite computation, have the property that if they are unprovable then they are true, since if they were false they would have a counterexample which would constitute a disproof. These statements are referred to as $\Pi_1^0$ statements.

However, P = NP is not such a statement: it has the form "there exists an algorithm such that all on all instances of [some NP-complete problem, e.g. 3SAT]..." so the quantifier structure looks like $\exists n \forall m : P(n, m)$. This means it's what is called a $\Sigma_2^0$ statement, and as far as I know such statements can be true and unprovable or false and unprovable.

Thinking in terms of nonstandard models,

  • if P = NP were true but unprovable, this would mean that there exists a polynomial-time algorithm for 3SAT which works on all "actual" instances of 3SAT, but that some nonstandard model of arithmetic contains a "nonstandard instance" of 3SAT on which it fails, and
  • if P = NP were false but unprovable, this would mean that there does not exist a polynomial-time algorithm for 3SAT, but that some nonstandard model of arithmetic contains a "nonstandard algorithm" (really a nonstandard natural number which PA interprets as coding such an algorithm) which PA interprets as a polynomial-time algorithm for 3SAT.

P = NP could turn out to be equivalent to a $\Pi_1^0$ statement but I'm not aware of any such equivalence.

Qiaochu Yuan
  • 468,795