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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?

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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 $P^{NP[1]}$.

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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?

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