A query regarding Complexity class BPP and NP

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?

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 $$\mathsf{NP}\subseteq \mathsf{P}$$.
• 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: $BPP^a \nsubseteq NP^a$? Nov 4 at 11:47