# If $\overline{3SAT}\in BP\cdot NP$ then $PH=\Sigma_3^P$

I have the problem

If $$\overline{3SAT}\in BP\cdot NP$$ then $$PH=\Sigma_3^P$$

To solve this I am using a result $$BP\cdot NP\subset NP/poly$$ which I can prove (not doing here). I have two solutions but in both case I am failing to give a final statement which kind of goes like this $$x\in L \iff$$ this, this, this is true ('this' means some conditions here)

To Show $$PH=\Sigma_3^P$$ it is enough to show $$\Pi_3^P\subseteq \Sigma_3^P$$

Proof-try 1: Take $$\Pi_3^P-3SAT$$. Hence $$\phi\in\Pi_3^P\iff \forall x\exists y\forall z\ \phi(x,y,z)=1$$ Now consider the language $$L'=\left\{ \langle \phi,y_2\rangle\mid \forall \langle x,z\rangle\ \neg \phi(x,y,z)=0 \right\}$$Since $$L'\in coNP$$ we have a polynomial time turning machine $$M$$ such that $$M(\langle \phi,y\rangle)=\psi$$ such that $$\psi\in \overline{3SAT}$$. Now since $$\overline{3SAT}\in BP\cdot NP\subseteq NP/poly$$ there exists a polynomial-sized circuit $$C$$ such that $$\psi\in \overline{3SAT}\iff \exists s \ C(\psi,s)=1$$Now $$M(\langle \phi,y\rangle)\in\overline{3SAT}\iff \exists C\exists s \ C(M(\langle \phi,y\rangle),s)=1$$ Now as I said I am stuck what should be the final statement at the right side of the if and only if symbol in $$\phi \in \Pi_3^P-3SAT\iff$$

Proof-try 2: Similarly take $$\Pi_3^P-3SAT$$. Hence $$\phi\in\Pi_3^P\iff \forall x\exists y\forall z\ \phi(x,y,z)=1$$Now fix $$y$$ we get $$\psi(\cdot, \cdot)=\phi(\cdot, y,\cdot)$$ Now $$\neg \psi(\langle x,z\rangle)=0\ \forall \langle x,z\rangle$$Hence $$\neg \psi\in\overline{3SAT}$$. Similarly since $$\overline{3SAT}\in BP\cdot NP\subseteq NP/poly$$ there exists polynomial size nondeterministic circuit $$C$$. Hence $$\neg \psi \in \overline{3SAT}\iff \exists s\ C(\neg \psi, s)=1$$Here again I am having trouble forming the last final statement

What should I do in this case