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$\text{Ordinal Replacement:}$ if $\phi(x,y)$ is a formula in two free variables $x,y$, then:

$\forall x \ [ordinal(x) \to \exists! y \ (ordinal (y) \wedge \phi(x,y)) ] \to \forall A \ (\forall x \in A (ordinal(x)) \to \exists B \ \forall y \ (y \in B \leftrightarrow \exists x \in A \ \phi(x,y))) $

is an axiom.

In English for every set $A$ of von Neumann ordinals and a function $F$ from von Neumann ordinals to von Neumann ordinals, then we can get a set $B$ whose elements are all the $F$-images of elements of $A$.

i.e. we can replace elements of a set of ordinal by ordinals.

Now is full Replacement provable in $\text{Z + Ordinal Replacement}$?

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No. Consider the model $\langle V_{\omega_1},\in\rangle$. This is a model of Zermelo's theory, but all ordinals in it are countable. Thus, it satisfies the ordinal-replacement axiom, since the image of a countable ordinal under any function will be countable, and $V_{\omega_1}$ contains all its countable subsets, since $\omega_1$ is regular.

In fact, in this case, we needn't restrict $B$ to consist of ordinals. Rather, we get the general version of replacement for functions whose domain is a set contained in the ordinals.

Since this model does not satisfy full replacement, it shows that full replacement is not provable from ordinal replacement over Z.

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