But is it a proof that neither is provable? Or just a statement of giving up?
Assuming it's a proof, there's two cases:
1) it's really P=NP, but we can't prove it
2) it's really P!=NP, but we can't prove it
If it's case 1, that's impossible. If P=NP, then there exist polytime algorithms for NP-Complete problems, and any of those algorithms are a proof that P=NP. So that's a contradiction.
So the _only_ possibility given a proof that it's unprovable would be case 2, thus proving P!=NP and contradicting itself.
Am I missing something? Even assuming I'm right, this isn't a proof that one of P=NP or P!=NP are provable, just that it's impossible to prove that neither can be proven.
Math proofs start given a set of axioms. The modern set of Axioms are ZFC. If you can find a new axiom A such that ZFC + A is a consistent system[1] and P=NP, and also find a new axiom B such that ZFC + B is a consistent system[1] and P!=NP, then logically, P?=NP cannot have a solution either way in just ZFC.
Are Turing machines representable in ZFC (pretty sure they are)? Seems like it'd be at least _really_ difficult to do that trick if so.
That would mean that in ZFC+A there exist algorithms that can't exist in ZFC+B? Hard for me to imagine what those would even be, since they'd have to be representable in ZFC itself anyway, right, since that's all that an algorithm would be this model, a particular Turing machine?
Assuming it's a proof, there's two cases: 1) it's really P=NP, but we can't prove it 2) it's really P!=NP, but we can't prove it
If it's case 1, that's impossible. If P=NP, then there exist polytime algorithms for NP-Complete problems, and any of those algorithms are a proof that P=NP. So that's a contradiction.
So the _only_ possibility given a proof that it's unprovable would be case 2, thus proving P!=NP and contradicting itself.
Am I missing something? Even assuming I'm right, this isn't a proof that one of P=NP or P!=NP are provable, just that it's impossible to prove that neither can be proven.