Let $M$ denote a well-founded set-sized model of ZFC. The *descent value* of $M$ will be defined as the value of $n$ returned by the following process.


**Initialization.** Let $n$ equal $0$ and $X$ equal $M$.


**Step.** If $L\_{\omega\_1^X}^X$ doesn't satisfy ZFC according to $M$, halt and output $n$. Otherwise, increment $n$, let $X$ equal $L\_{\omega\_1^X}^X$, and repeat.



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> **Question.** Assuming sufficiently powerful large cardinal axioms, is it true that for every natural $n\geq 0,$ there exists a well-founded model $M$ of ZFC whose descent value is $n$?
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Do models of ZFC have arbitrarily large descent values?
No, the descent value is always at most $1$ in any model of ZFC, whether it is well-founded or not. To see this, observe that if the descent value isn't $0$, then on the next step you have $X=L\_{\omega\_1^M}^M$, and so $X$ satisfies $V=L$, and so $\omega\_1^X$ is the $\omega\_1^L$ inside $M$, and $L\_{\omega\_1^L}$ never satisifes ZFC, as it thinks that every ordinal is countable. 


