Let $M=M_0$ be a ctm of ZF.
If $(M_0, \ldots, M_n)$ is a sequence of ctms of ZF, where for all $i,$ either $M_{i+1}$ is an inner model of $M_i$ or a generic extension of $M_i,$ then call $M_n$ an $n$-model of $M.$
Call $M_n$ an $n'$-model of $M$ if we can further demand that $M_1$ is an inner model of $M_0.$
Call $M_n$ an $n''$-model if we can instead demand that $M_1$ is a generic extension of $M_0.$
Let $C_n^M, C_n^{M'}, C_n^{M''}$ respectively denote the sets of $n$-models, $n'$-models, and $n''$-models of $M.$
Let $C_{\infty}^M=\bigcup_{n<\omega} C_n^M.$
My intention is that $C_{\infty}^M$ represents the set of $o(M)$-height models which $M$ is capable of "reasoning" about.
My main question is whether there is some $n$ such that $C_{\infty}^M=C_n^M.$
It's not hard to cook up $M$ such that $C_2^{M'} \not \subset C_2^{M''}$ or $C_2^{M''} \not \subset C_2^{M'},$ so this hierarchy doesn't seem to collapse immediately (if at all), at least in general. It's a priori possible that such $n$ exists for some $M$ and not others; I'm not sure how find an example either way.
If such an $n$ cannot exist, what if we instead ask for something weaker, say $n$ such that $\{\varphi \in \mathcal{L}_{\{\in\}}: \exists N \in C_{\infty}^M(N \models \varphi)\}=\{\varphi \in \mathcal{L}_{\{\in\}}: \exists N \in C_n^M(N \models \varphi)\}?$ It seems like some strengthening of the Inner Model Hypothesis could justify the existence of such an $n.$
