# Many-One Reductions vs Turing-Reductions and PH

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?

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## migrated from cstheory.stackexchange.comNov 25 '11 at 15:24

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Whomever voted to close it can you please tell me why. Thanks –  Tayfun Pay Nov 25 '11 at 1:52
I am guessing that the reason for close votes is that they feel that the question is really not research level, and is probably more suitable for Mathematics. Based on the answer to the question I have to agree with them. –  Kaveh Nov 25 '11 at 5:34
You may delete it –  Tayfun Pay Nov 25 '11 at 12:07
Since there is an answer, it wont let me delete it. You can moce it to math as well. It is okay! :-) –  Tayfun Pay Nov 25 '11 at 15:17