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Suppose a machine $T$, and oracles $A$ and $B$ solve all problems in the complexity classes $\mathcal C_T$, $\mathcal C_A$ and $\mathcal C_B$ respectively.

Let $T^{\{A,B\}}$ denote a machine that is equivalent to a machine $T$ that has access to oracles $A$ and $B$. What would be the complexity class of all problems solvable by such a machine? Is it $\mathcal C_T^{\mathcal C_A\cup\mathcal C_B}$, $\mathcal C_T^{\mathcal C_A}\cup\mathcal C_T^{\mathcal C_B}$ or something else entirely? E.g. could it be that $\mathsf{P^{\{NP, coNP\}}=PSPACE}$?

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The definition of an oracular complexity class is key to answering this question, it is often defined as: $$ \mathsf{C^{A}=\bigcup_{L_A\in A}C^{L_A}} $$

But how do we define two oracles? Well the union definition ($\mathsf{C^{A,B}}=\mathsf{C^{A\cup B}}$) would imply that a Turing machine has access to only one language from either $\mathsf{A}$ or $\mathsf{B}$, which is not quite what we want. We would like that it has a language from both:

$$ \mathsf{A^{B,C} = \bigcup_{L_B\in B,\hspace{0.05cm}L_C\in C} A^{L_B, L_C}}. $$

Choosing this definition would imply that $\mathsf{A^{B,C}\not = \mathsf{C^A \cup C^B}}$ either. For counter-example assume there exists some $\mathsf{L_A\in C^A\setminus C^B}$ and some $\mathsf{L_B\in C^B\setminus C^A}$. Deciding the language: $$ \mathsf{L}=\{x\#y: x\in L_A, y\in L_B \} $$ is in $\mathsf{C^{A,B}}$ under our definition (assuming C is capable of Turing reductions), but it is in neither $\mathsf{C^A}$ or $\mathsf{C^B}$, so it can not be in $\mathsf{C^A \cup C^B}$.

So what is $\mathsf{C^{A,B}}$? Well, it seems to be it is its own object, without a simple equality to some other understood concept. Although, it often can be simplified when $\mathsf{A}$, $\mathsf{B}$ or $\mathsf{C}$ are known. For example, in the case you state there is a simple reduction.

$$ \mathsf{P^{NP,CoNP}=P^{NP} }. $$

This is because $\mathsf{P^{NP}=P^{CoNP}}$ and polynomial time TMs can simulate other polynomial time TMs.

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  • $\begingroup$ So, it's not possible to simulate alternations with having an access to both $\mathsf{NP}$ and $\mathsf{coNP}$ oracles. And I suppose that would hold even if you replace oracles with their functional counterparts - $\mathsf{FNP}$ and $\mathsf{coFNP}$. $\endgroup$
    – rus9384
    Nov 23, 2023 at 11:03

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