Let $\Delta$ be a finite set of formulas. I define $R_\Delta(\Sigma)$ for a partial type $\Sigma$ to be the Cantor-Bendixson rank of $Y := \{q \in S_\Delta(\mathbb M) \mid \text{$q$ is consistent with $\Sigma$}\}$ in $S_\Delta(\mathbb M)$, where $\mathbb M$ is the monster model. That is, $R_\Delta(p)$ is the least ordinal $\alpha$ such that $Y^{(\alpha + 1)} = 0$. (Is that called the local $\Delta$-rank?)
I'm trying to understand the proof that $R_\Delta(q) < R_\Delta(p)$ iff $q$ is a forking extension of $p$, where $p \in S_\Delta(A)$, $p \subseteq q \in S_\Delta(B)$, $A \subseteq B$, and the ranks are defined for $p$ and $q$.
The proof I'm looking at goes like the following: Assume $q$ is nonforkng. First, there is $\phi \in q$ such that $R_\Delta(\phi) = R_\Delta(q)$ by compactness. Then it claims the existence of $\psi \in p$ which is a positive Boolean combination of $A$-conjugate of $\phi$. Then, by following easy facts about the rank, we have $R_\Delta(\psi) \le R_\Delta(\psi)$ and thus $R_\Delta(p) \le R_\Delta(q)$, whence the two ranks are the same.
I don't see why there is such a $\psi$. The proof says it follows from the proof of the open mapping theorem, but I don't understand the argument.
Why is there such $\psi$?