If P is a strict subset of BPP then P not equal NP

507 Views Asked by At

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?

1

There are 1 best solutions below

0
On

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.