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Examine the first order theory with the following extra-logical axioms:

Specification:: $$[\forall A \, \exists! X \, \forall Y \, (Y \in X \iff Y \in A \land \phi)]$$; whenever $\phi$ doesn't use the symbol $X$

Reflection: $$\varphi \implies \exists \alpha : \varphi^{V_\alpha}$$

where $\varphi$ is a first order sentence (i.e. a formula with no free variables), not using defined predicate symbols [defined functions allowed]. $\varphi^X$ is the "$\in X$" bounded form of $\varphi$.

Now is this theory equivalent to $\sf ZF$?

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  • $\begingroup$ If you don't allow parameters in your reflection scheme, I don't think this has a hope of working: consider $V_{\kappa+\omega}$ for sufficiently large $\kappa$. $\endgroup$
    – Noah Schweber
    Commented Mar 10, 2022 at 18:19
  • $\begingroup$ How is the formula $U = V_\alpha$ expressed? (Or maybe you just need to express $\exists\alpha\in\mathit{On} : U = V_\alpha$, perhaps that's easier.) I mean, I know what it means in ZFC, but I'm not convinced that there aren't multiple versions which would be equivalent in ZFC but not in the axiom system you're proposing. (Also, it matters to the elegance of the system being proposed: it seems nice, but not if there's a lot of complexity hidden in that $V_\alpha$.) $\endgroup$
    – Gro-Tsen
    Commented Mar 10, 2022 at 19:15
  • $\begingroup$ (Just to belabor the point, if $\psi(\alpha,U)$ is the formula which says “lots of ZFC axioms are true, and also $U = V_\alpha$”, then in ZFC it's equivalent to $U = V_\alpha$, and clearly “$\exists \alpha: \exists U: \psi(\alpha,U) \land \varphi^U$” is going to state much more than you intended.) $\endgroup$
    – Gro-Tsen
    Commented Mar 10, 2022 at 19:18
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    $\begingroup$ @ZuhairAl-Johar Think about what happens if $\kappa$ is large enough that every theory satisfied by some level of the $V$-hierarchy is satisfied by some level of the $V$-hierarchy below $\kappa$. Then any level of the $V$-hierarchy above $\kappa$ will satisfy "parameter-free reflection" for a silly reason. $\endgroup$
    – Noah Schweber
    Commented Mar 10, 2022 at 19:57
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    $\begingroup$ @ZuhairAl-Johar There are only set-many (continuum-many to be precise) theories, without parameters, in the language of set theory. $\endgroup$
    – Noah Schweber
    Commented Mar 10, 2022 at 21:10

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