# Why is $BPP^{NP}$ in polynomial hierarchy? [closed]

Why is $$BPP^{NP}$$ in the polynomial hierarchy? I know that $$BPP$$ is contained in $$NP^{NP}$$, so $$BPP$$ is inside $$PH$$. However, how does that imply $$BPP^{NP}$$ is inside the polynomial hierarchy?

## closed as off-topic by Emil Jeřábek, Aryeh, Gamow, Hsien-Chih Chang 張顯之, domotorpMay 8 at 7:17

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$$BPP\subset PH$$ does NOT imply that $$BPP^{NP}\subset PH$$. You have to go through the proof of $$BPP\subset PH$$ and notice that it relativizes, i.e., also holds if the Turing-machines have access to an oracle.
• are you saying that by checking the proof, I am able to see that $BPP^{NP}\in(NP^{NP})^{NP}$ something like that? Thanks! $(NP^{NP})^{NP}$ seems to just be $NP^{NP}$ since the last oracle seems to be redundant? – Herman Chu May 7 at 19:25