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What's the consistency strength of adding the following single sentence replacement like statement to $\sf Z + \forall x \exists \alpha: x \in V_\alpha$ ?

$$\forall \varphi \forall A \ [\forall x \in A \exists \alpha \forall \theta > \alpha: V_\theta \models \exists! y \varphi(x,y) \to \\ \exists \beta: \forall x \in A \exists y \in V_\beta: V_\beta \models \varphi(x,y)]$$

The idea is that $\sf ZF$ is not aximatizable by adding a single sentence to axioms of Zermelo set theory, on the other hand the above statement seems to be a theorem of $\sf ZFC$, so this means that the above theory must be strictly weaker than $\sf ZFC$.

Which of the known fragments of $\sf ZFC$ this theory would be equivalent to?

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  • $\begingroup$ Are you assuming that each $V_\alpha$ is a set? $\endgroup$
    – Hanul Jeon
    Commented May 29, 2021 at 11:24
  • $\begingroup$ @HanulJeon, yes, I think this is a theorem of $\sf Z + \forall x \exists \alpha: x \in V_\alpha$, since the indices are von Neumann ordinals as usual. $\endgroup$
    – Zuhair Al-Johar
    Commented May 29, 2021 at 11:36
  • $\begingroup$ Do we need replacement to ensure $V_\alpha$ is a set? $\mathsf{Z}$ lacks replacement (especially, $\Pi_1$-replacement), so I am skeptical about $\mathsf{Z}$ proves $V_\alpha$ exists for all $\alpha$. $\endgroup$
    – Hanul Jeon
    Commented May 29, 2021 at 11:38
  • $\begingroup$ @HanulJeon, $\sf Z$ alone doesn't prove that of course, I mean $\sf Z + \forall x \exists \alpha: x \in V_\alpha$ can prove it $\endgroup$
    – Zuhair Al-Johar
    Commented May 29, 2021 at 11:40

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