Why does SAT-UNSAT $\in NP \implies NP = coNP$ I was reading this post about the DP completeness of the problem SAT-UNSAT (both are well defined in this post). The answer added a note at the end that states the class complexity DP differs from NP, unless NP = coNP.
I fail to see why.
I searched and I came across multiples posts such at this one and that one that prove that if SAT-UNSAT is in coNP, then NP = coNP. But unless the fact that SAT-UNSAT $\in NP \implies$ SAT-UNSAT $\in coNP$ (which I do not see), then those proofs are not exactly what would help me. The same goes for this question, I would need SAT-UNSAT $\in coNP$.
Question : Considering the first question (and the answer associated), if the problem SAT-UNSAT $\in NP$, why does NP = coNP.
My take :
Well, I can see that the problem SAT-UNSAT is NP-hard and coNP-hard. If SAT-UNSAT $\in NP$, then SAT-UNSAT is NP-complete. This implies things such that the problem UNSAT (which is coNP-complete) is NP-hard since we can reduce UNSAT to SAT-UNSAT which is NP-complete. That's all I got and that doesn't really help.
I'd appreciate any clarification on the subject. Thanks to you all
Suppose that $X$ is a coNP problem which is NP-hard.
Let $A$ be any problem in NP. Since $X$ is NP-hard, there is a polytime reduction from $A$ to $X$. Since $X$ is in coNP, this shows that $A$ is in coNP.
We have shown that $\mathsf{NP} \subseteq \mathsf{coNP}$. Now suppose $B$ is in coNP. Then $\overline{B}$ is in NP, and so in coNP. Hence $B$ is in NP. This shows that also $\mathsf{coNP} \subseteq \mathsf{NP}$, and so $\mathsf{NP} = \mathsf{coNP}$.