np complete – 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