# Does $NP^{SAT}=NP^{NP}$?

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?

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, $$L\in\mathsf{NP}^\mathsf{SAT}$$.