# Is P^SAT with only one query equal to the union of NP and coNP?

I have a following problem:

Let $$P^{SAT[1]}$$ be a class of problems decidable by a deterministic polynomial Turing Machines with SAT oracle. (only one question to oracle).

Assume that: $$\mathrm{co}NP \neq NP \neq P$$

Decide if $$coNP \cup NP \neq P^{SAT[1]}$$ or $$\mathrm{co}NP \cup NP = P^{SAT[1]}$$.

I know how to show that $$\mathrm{co}NP \cup NP \subseteq P^{SAT[1]}$$.

But I have no idea what about inverse direction. Any tips?

Under the assumption that $$NP\neq \mathrm{co}NP$$, we have that:

$$P^{SAT[1]}=NP\cup \mathrm{co}NP\implies P^{NP[1]}\neq P^{NP[2]}$$

Proof: Since $$D^p\subseteq P^{NP[2]}$$, we deduce that $$(SAT\dot{\land} UNSAT)\in P^{NP[2]}$$.

Meanwhile, $$(SAT\dot{\land}UNSAT)\notin NP$$ due to $$UNSAT\notin NP$$ and similarly, $$(SAT\dot{\land}UNSAT)\notin \mathrm{co}NP$$ due to $$SAT\notin \mathrm{co}NP$$. So, $$(SAT\dot{\land}UNSAT)\notin NP\cup \mathrm{co}NP=P^{NP[1]}$$.

We conclude that $$P^{NP[1]}\neq P^{NP[2]}$$.$$\blacksquare$$

In the above, $$(SAT\dot{\land}UNSAT)=\{(\phi_1,\phi_2)\mid\phi_1\in SAT \land \phi_2\in UNSAT\}$$.

So, though it might be tempted to push $$P^{NP[1]}$$ downto $$NP\cup \mathrm{co}NP$$, this result (if true) could tell us more about where the polynomial-time hierarchy $$PH$$ would actually collapse to. And, in this case, your proposition would save $$PH$$ from collapsing to $$P^{NP[1]}$$.

Hint (noting the homework tag on the other post): $P^{SAT}$, by the Cook-Levin theorem, is the same as an $NP$ oracle. Given that $\text{co-}NP \neq NP$, what does this imply?