It is known that:

  • $\Sigma_2^P \subseteq \Delta_3^P$ i.e. $NP^{NP} \subseteq \Delta_3^P$.
  • $BPP^{NP} \subseteq \Delta_3^P$

If $\Sigma_2^P = PSPACE => BPP^{NP} \subseteq NP^{NP}$

Doesn't that imply: $\Sigma_2^P = PSPACE => BPP \subseteq NP$?

Am I missing something?


I have not verified your claims, but I will still try to answer.

In the general case, given a language $A$ and two complexity classes $\mathcal{C}_1$ and $\mathcal{C}_2$, the equality $\mathcal{C}_1^A = \mathcal{C}_2^A$ does not imply $\mathcal{C}_1 = \mathcal{C}_2$ (and same with inclusion).

In particular, Baker, Gill and Solovay proved in 1975 that there exists two languages $A$ and $B$ such that $\mathsf{P}^A=\mathsf{NP}^A$ and $\mathsf{P}^B\neq \mathsf{NP}^B$ (but it is still not known whether $\mathsf{P} =\mathsf{NP}$ or not). The idea is that giving and oracle to a non-determinist class let that class use this oracle non-deterministicaly, meaning potentially using it independently on every computing path, resulting in virtually calling the oracle an exponential number of times.

That means that $\mathsf{NP}^A \subseteq \mathsf{P}^A$ but it is not necessarily true that $\mathsf{NP}\subseteq \mathsf{P}$.

  • $\begingroup$ thanks a lot! I understand the general case. But if I understand correctly to show we can't have a proof using an oracle we have to show both cases i.e ($X^a \subseteq Y^a$) and ($X^b \nsubseteq Y^b$). Are we aware of an oracle result that shows: $BPP^a \nsubseteq NP^a$? $\endgroup$ Nov 4 at 11:47
  • $\begingroup$ I don't know about this particular case, sorry. $\endgroup$
    – Nathaniel
    Nov 4 at 11:59

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