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

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

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  • $\begingroup$ Karp Lipton + Adelman's $\endgroup$ – Ariel May 5 '16 at 13:19
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Use the Sipser-Lautemann theorem: $\mathrm{BPP} \subseteq \Sigma_2^p \cap \Pi_2^p$.

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  • $\begingroup$ Should query be $\Sigma_2=\Pi_2$? $\endgroup$ – Turbo Dec 30 '15 at 2:02

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