$P^{NP}\subseteq BPP^{NP}$ holds. According to current knowledge $BPP$ is in $\Sigma_2^P\cap\Pi_2^P$ holds. So according to current knowledge is following true?

  1. $P^{\Sigma_2^P\cup\Pi_2^P}\subseteq BPP^{NP}\subseteq(\Sigma_2^P\cap\Pi_2^P)^{NP}\subseteq\Sigma_3^P\cap\Pi_3^P$?

  2. Is $P^{\Sigma_2^P\cup\Pi_2^P}$ the largest standard polynomial hierarchy class in $BPP^{NP}$ and is $(\Sigma_2^P\cap\Pi_2^P)^{NP}$ the smallest standard polynomial hierarchy class containing $BPP^{NP}$?

  • $\begingroup$ How is $(\Sigma_2^\mathrm{P} \cap \Pi_2^\mathrm{P})^{\mathbf{NP}}$ defined? Note relativized classes only make sense if you start out with a class which has a machine characterization (and AFAIK $\Sigma_2^\mathrm{P} \cap \Pi_2^\mathrm{P}$ is not one such class). $\endgroup$ – dkaeae May 27 at 7:59
  • $\begingroup$ @dkaeae I thought $NP\cap coNP$ had a machine charaterization and likewise.. no? $\endgroup$ – T.... May 27 at 8:05
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    $\begingroup$ Well, we have $\mathbf{NP} \cap \mathbf{coNP} = \mathbf{P}^{\mathbf{NP} \cap \mathbf{coNP}}$, but I'm unsure how you'd make the oracles "stack". $\endgroup$ – dkaeae May 27 at 8:10
  • $\begingroup$ @dkaeae I see.... $\endgroup$ – T.... May 27 at 8:12
  • $\begingroup$ @dkaee we know $BPP$ is in $\Sigma_2^P\cap\Pi_2^P$. What is the best way to say where $BPP^{NP}$ is properly best contained in (I thought it was $(\Sigma_2^P\cap\Pi_2^P)^{NP}$ and you have shot it down then)? $\endgroup$ – T.... May 27 at 8:17

The NP-machine hypothesis implies that $BPP^{NP}=P^{NP}$.

So combining with the hypothesis that $PH$ does not collapse, this will falsify the proposition that $P^{\Sigma_2^p\cup\Pi_2^p}\subseteq BPP^{NP}$.

Reference for NP-machine hypothesis and its aforementioned consequence: https://www.researchgate.net/publication/225526797_Hardness_Hypotheses_Derandomization_and_Circuit_Complexity


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