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REPORTS > KEYWORD > P^{NP[1]}=P^{NP[2]}:
Reports tagged with P^{NP[1]}=P^{NP[2]}:
TR13-144 | 8th October 2013
VyasRam Selvam

The two queries assumption and Arthur-Merlin classes

We explore the implications of the two queries assumption, $P^{NP[1]}=P_{||}^{NP[2]}$, with respect to the polynomial hierarchy and the classes $AM$ and $MA$.
We prove the following results:

1. $P^{NP[1]}=P_{||}^{NP[2]}$ $\implies$ $AM = MA$
2. $P^{NP[1]}=P_{||}^{NP[2]}$ $\implies$ $PH \subset MA_{/1}$
3. $\exists\;B\;P^{NP[1]^B}=P^{NP[2]^B}$ and $NP^B \not\subseteq coMA^B$.
4. $P^{NP[1]}=P_{||}^{NP[2]}$ $\implies$ $PH ... more >>>

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