why is $\Pi_2$ smaller than $NP\cap coNP$ Consider the language $A=\{(\phi_1, \phi_2) | \phi_1 \in SAT, \phi_2\in \overline{SAT}  \}$. What is the smallest class that $A$ is known to belong to?
Apparently, the answer is $\Pi_2$, although I thought it was $NP \cap coNP$, since if we have a non-deterministic machine $M_1$ that decides $SAT$ and another non-deterministic machine $M_2$ that decides $\overline {SAT}$, then we can define a machine $M_3$ that on input $(\phi_1, \phi_2)$, runs $M_1$ on $\phi_1$ and $M_2$ on $\phi_2$, and only accepts if both machines return True. If the input does belong to $A$, then for some set of nondeterministic choices, we know $M_3$ will accept. If the input doesn't belong to $A$, then either $M_1$ or $M_2$ always return False, and so $M_3$ always rejects.
My first question is whether my explanation correctly shows $A \in NP \cap coNP$. More importantly though, why is $\Pi_2$ smaller than $NP\cap coNP$!?
You seem to be using the following definition of $\mathsf{NP} \cap \mathsf{coNP}$: it consists of all languages of the form $L_1 \cap L_2$, where $L_1 \in \mathsf{NP}$ and $L_2 \in \mathsf{coNP}$. However, in reality $\mathsf{NP} \cap \mathsf{coNP}$ consists of all languages $L$ such that $L \in \mathsf{NP}$ and $L \in \mathsf{coNP}$.