<p><span class="math-container">$BPP\subset PH$</span> does NOT imply that <span class="math-container">$BPP^{NP}\subset PH$</span>. You have to go through the proof of <span class="math-container">$BPP\subset PH$</span> and notice that it relativizes, i.e., also holds if the Turing-machines have access to an oracle.</p>

CC BY-SA 4.0
2019-05-07T19:25:32.083Z
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
CC BY-SA 4.0
2019-05-08T07:17:21.067Z
These are not really research level questions, so maybe you should rather ask them at https://cs.stackexchange.com/.
domotorp
CC BY-SA 4.0
2019-05-07T11:32:52.660Z
2019-05-07T11:32:52.660Z
domotorp
<p>Why is <span class="math-container">$BPP^{NP}$</span> in the polynomial hierarchy?
I know that <span class="math-container">$BPP$</span> is contained in <span class="math-container">$NP^{NP}$</span>, so <span class="math-container">$BPP$</span> is inside <span class="math-container">$PH$</span>. However, how does that imply <span class="math-container">$BPP^{NP}$</span> is inside the polynomial hierarchy?</p>

<cc.complexity-theory><complexity-classes><time-complexity>
Why is $BPP^{NP}$ in polynomial hierarchy?
Herman Chu
# Why is $BPP^{NP}$ in polynomial hierarchy?

(Asked by: username_0 on 2019-05-07)

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?

**Tags:** cc.complexity-theory, complexity-classes, time-complexity


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## 1 answers

### Answer by username_1 on 2019-05-07. Score: 1

$BPP\subset PH$ 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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: These are not really research level questions, so maybe you should rather ask them at https://cs.stackexchange.com/.  

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