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Holo
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Your axiom schema is precisely the statement "all sets are of finite rank", that isequivalent to being an $V=V_\omega$$\omega$-model.

Let $X$ be any set and let $\phi(x)$ beWorking inside an $x=x$$\omega$-model, byany counter example to your schema will be counter example of the axiom there exists someof foundation of ZFC, as you can use $m\in\omega$ such that there is no$\le$ and the given $m$$\phi$ of the counter-descendingexample to construct the counter-example sequence starting from $X$, so there is no $m-1$-descensingthis sequence starting from any $y\in x$is of infinite length externally, and so on and so forthbut the model has the same finite numbers, so you get that the rank of $X$it is less than $m$ ($X\in V_m$)infinite internally as well.

Now ifassume the model is not an $X\in V_\omega$ we know that$\omega$-model, let $X\in V_m$ for some$\phi(x)$ be $m\in\omega$, so your schema$x=x$ and $X$ be a $k$-nested singleton where $k$ is obviously correct.

(Note that I didn't usesome non standard natural, then $X, \phi$ will be a counter example to the well ordering onschema as the universe)sequence defined as $x_{i+1}=\min_\le\{x\in x_i\}, x_0=X$ won't stop in any standard step.

Your axiom schema is precisely the statement "all sets are of finite rank", that is $V=V_\omega$.

Let $X$ be any set and let $\phi(x)$ be $x=x$, by your axiom there exists some $m\in\omega$ such that there is no $m$-descending sequence starting from $X$, so there is no $m-1$-descensing sequence starting from any $y\in x$, and so on and so forth, so you get that the rank of $X$ is less than $m$ ($X\in V_m$).

Now if $X\in V_\omega$ we know that $X\in V_m$ for some $m\in\omega$, so your schema is obviously correct.

(Note that I didn't use the well ordering on the universe)

Your axiom schema is equivalent to being an $\omega$-model.

Working inside an $\omega$-model, any counter example to your schema will be counter example of the axiom of foundation of ZFC, as you can use $\le$ and the given $\phi$ of the counter-example to construct the counter-example sequence, this sequence is of infinite length externally, but the model has the same finite numbers, so it is infinite internally as well.

Now assume the model is not an $\omega$-model, let $\phi(x)$ be $x=x$ and $X$ be a $k$-nested singleton where $k$ is some non standard natural, then $X, \phi$ will be a counter example to the schema as the sequence defined as $x_{i+1}=\min_\le\{x\in x_i\}, x_0=X$ won't stop in any standard step.

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Holo
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Your axiom schema is precisely the statement "all sets are of finite rank", that is $V=V_\omega$.

Let $X$ be any set and let $\phi(x)$ be $x=x$, by your axiom there exists some $m\in\omega$ such that there is no $m$-descending sequence starting from $X$, so there is no $m-1$-descensing sequence starting from any $y\in x$, and so on and so forth, so you get that the rank of $X$ is less than $m$ ($X\in V_m$).

Now if $X\in V_\omega$ we know that $X\in V_m$ for some $m\in\omega$, so your schema is obviously correct.

(Note that I didn't use the well ordering on the universe)