# What if $NP\subseteq BPP$?

I'm new to complexity and came upon the following exercise which I'm unable to solve.

Prove that if $NP\subseteq BPP$ then $\Sigma_2^p=\Pi_4 ^p$.

• Karp Lipton + Adelman's – Ariel May 5 '16 at 13:19

Use the Sipser-Lautemann theorem: $\mathrm{BPP} \subseteq \Sigma_2^p \cap \Pi_2^p$.
• Should query be $\Sigma_2=\Pi_2$? – Turbo Dec 30 '15 at 2:02