P≠NP, yet NP=co-NP

Question

I know , if P = NP then it can easily be proven that P = NP = Co-NP. But I was wondering, if we assume P≠ NP, also then can it be proven that, NP =co-NP??

Answer 1

We don't know anything either way: it is consistent with our knowledge so far that $\mathsf{P\not=NP}$ and $\mathsf{NP=coNP}$, and it is consistent with our knowledge so far that $\mathsf{P\not= NP}$ and $\mathsf{NP\not=coNP}$.

Of course in some sense that's a purely negative statement since it's just about the limits of our current knowledge, not really about the mathematics itself. Here's an actual theorem which says, in a precise sense, that the "standard" tools of complexity theory cannot show anything about $\mathsf{NP}$ vs $\mathsf{coNP}$ even assuming $\mathsf{P\not=NP}$:

There are oracles $A$ such that $\mathsf{P}^A\not=\mathsf{NP}^A$ and $\mathsf{P}^B\not=\mathsf{NP}^B$ but $\mathsf{NP}^A=\mathsf{coNP}^A$ and $\mathsf{NP}^B\not=\mathsf{coNP}^B$.

This should recall the Baker-Gill-Solovay theorem that $\mathsf{P}$ vs. $\mathsf{NP}$ cannot be decided via relativizing methods. The proof is similarly a forcing argument, but it's a bit more intricate. I'll add a reference once I can track it down.

(I do not know whether we similarly have a "natural proofs" barrier here. I suspect we do, but that argument is more involved.)