One definition of $\mathsf{PH}$ uses Oracles and in this definition both $\mathsf{NP}$ and $\mathsf{coNP}$ are contained in P^NP which equals $\mathsf{P^{coNP}}$. It is believed that $\mathsf{NP}$ does not equal $\mathsf{coNP}$, in other words they are not many-one reducible to each other. If indeed this is proved, then doesn't it also hold that $\mathsf{P^{NP}}$ doesn't equal $\mathsf{P^{coNP}}$ since many-one reducibility imply Turing reducibility?
Many-One Reductions vs Turing-Reductions and PH
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computational-complexity