How to show this statement?
I know that if P were to be a strict subset of RP then P not equal NP, but I can't find any other relation or reason why P would be different than NP?
If P is a strict subset of BPP then P not equal NP
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computer-science
computational-complexity
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0Have you considered the contrapositive? – 2017-02-06
1 Answers
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Assume P=NP, then NP=coNP and the poly time hierarchy collapses. Sipser–Lautemann theorem states that $BPP\subseteq \Sigma_2$, therefore BPP=P, a contradition.