Suppose $T$ is a strongly minimal theory. If $A \subset M$ and $\bar{a} \in M$, then $\operatorname{RM}(\bar{a}/A) = \operatorname{dim}(\bar{a}/A)$.

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To be more specific, I'm asking about a detail in the proof of this theorem (6.2.19 on Marker's Introduction to Model Theory). The proof is very straightforward, by induction: we show that if $a_1,\dots,a_k$ are independent over $A$, then $\operatorname{RM}(\bar{a}/A)=k$. In the inductive step, they show first that $\operatorname{RM}(\bar{a}/A) \geq k$, which I can understand. However, for the other inequality, they take again a minimal formula $\phi(v)$ witnessing the type's rank, and the following argument is carried out:

"Suppose that $\psi(v)$ is a formula such that $\psi(M) \subset \phi(M)$ and $\neg\psi(a)$, then it suffices to show that $\operatorname{RM}(\psi)<k$."

Why does it suffice? I would use that $\phi = (\phi \land \psi) \lor (\phi \land \neg \psi)$ to get some kind of contradiction if $\operatorname{RM}(\psi)\geq k $, but this doesn't seem to bound anything from above. My intuition is that, if removing the tuple $\bar{a}$ from $\phi(M)$, makes the rank drop then the rank originally could not be higher than $k$, but it is not clear to me just from the statement made in the book. Any help would be appreciated. The rest of the proof is very much understandable.

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If the rank of $\phi$ is $(k+1)$ or greater, then this is witnessed by infinitely many disjoint definable subsets of rank $k$ or greater. But every definable subset of rank $k$ or greater contains $a$. So there are not even two disjoint definable subsets of rank $k$ or greater.