Does $NP^{SAT}=NP^{NP}$?
We can see easily that $NP^{SAT}\subseteq NP^{NP}$, because $SAT \in NP$.
But is the other side $NP^{NP}\subseteq NP^{SAT}$ also true? If yes, how can we prove it?


1 Answer 1


Yes, because $\mathsf{SAT}$ is $\mathsf{NP}$-complete.

Let $L\in\mathsf{NP}^\mathsf{NP}$. This means that there exists $A\in\mathsf{NP}$ such that $L\in\mathsf{NP}^A$. But you can replace any oracle query to the set $A$ with a polynomial-time deterministic computation that uses oracle queries to $\mathsf{SAT}$. Thus, $L\in\mathsf{NP}^\mathsf{SAT}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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