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May be this is trivial but I do not know the answer.

As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.

As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?

If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?

That is is there a natural problem (not amalgamated ones) with a randomized poly algorithm nevertheless does not have a short yes or no certificate? A problem that comes close is PIT which is not known to be in $\mathsf{NP}$ but is in $\mathsf{coNP}$ which is the basis of the concocted problem below by Ricky Demer.

May be this is trivial but I do not know the answer.

As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.

As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?

If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?

That is is there a natural problem (not amalgamated ones) with a randomized poly algorithm nevertheless does not have a short yes or no certificate? A problem that comes close is PIT which is not known to be in $\mathsf{NP}$ but is in $\mathsf{coNP}$ which is the basis of the amalgamated problem (non-natural) below by Ricky Demer.

May be this is trivial but I do not know the answer.

As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.

As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?

If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?

That is is there a natural problem (not amalgamated ones) with a randomized poly algorithm nevertheless does not have a short yes or no certificate? A problem that comes close is PIT which is not known to be in $\mathsf{NP}$ but is in $\mathsf{coNP}$ which is the basis of the concocted problem below.

May be this is trivial but I do not know the answer.

As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.

As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?

If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?

That is is there a natural problem (not amalgamated ones) with a randomized poly algorithm nevertheless does not have a short yes or no certificate? A problem that comes close is PIT which is not known to be in $\mathsf{NP}$ but is in $\mathsf{coNP}$ which is the basis of the concocted problem below by Ricky Demer.

May be this is trivial but I do not know the answer.

As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.

As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?

If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?

Could there be any natural problems and not amalgamated ones?

May be this is trivial but I do not know the answer.

As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.

As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?

If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?

That is is there a natural problem (not amalgamated ones) with a randomized poly algorithm nevertheless does not have a short yes or no certificate? A problem that comes close is PIT which is not known to be in $\mathsf{NP}$ but is in $\mathsf{coNP}$ which is the basis of the concocted problem below.