# 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 $$\mathsf{NP} = \mathsf{coNP}$$.