$P\ne BPP$ implies $P\ne NP$ Prove: $P\ne BPP$ implies $P\ne NP$.


Obviously $P\subseteq BPP$ so the assumption is really $BPP\subsetneq P$. I'm trying to show the implication by contradiction. Let's assume that $P=NP$ and let's look at some language $L\in BPP$. Then, it has a proper probabilistic TM which decides it. 

At first, I thought that I could just show some $NP$ machine which uses the $BPP$ machine but a relation between the two classes it not yet known, so it's probably not what I'm supposed to do.

We did learn Adleman's theorem and I think it could be used here (though we haven't define $P/poly$). Adelman's theorem claims that for every $x$ there's a $y$ s.t.

$$x\in A \implies M(x,y) = 1 \\ x\not\in A \implies M(x,y) = 0$$

so it boils down to the problem of finding such $y$, but I don't think this problem (finding $y$) is in $NP$ (or even possible)

Also, I understood that one can use the polynomial hierarchy theorem but we were not taught it.
Use the fact that $\mathsf{BPP}\subseteq\Sigma_2$. Also note that if $P=NP$ then the entire polynomial hierarchy collapses to $P$.