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 $P^{NP[1]}$.

| cite | improve this answer | |

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?

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.