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$\sf ZF + Def$ is the theory that extends $\mathcal L(=,\in)_{\omega_1,\omega}$ with axioms of $\sf ZF$ (written in $\mathcal L(=,\in)_{\omega, \omega}$) and the axiom of definability:-

$\textbf{Define: } Dx \iff \bigvee x= \{ y \mid \Phi \}$

where $\Phi$ range over all formulas in $\mathcal L(=,\in)_{\omega, \omega}$ in which only the symbol "$y$" occurs free, and the symbol "$y$" never occurs bound.

Axiom of definability: $\forall x Dx$

This theory has its models being exactly the pointwise-definable models of $\sf ZF$ [Hamkins]

Now, if one wants to confine matters to $\mathcal L(=,\in)_{\omega,\omega}$, i.e. the usual $\textbf{FOL}(=,\in)$, then one can add this rule to $\sf ZFC$:

$\textbf{Definability: }$ if $\phi_1,\phi_2, \phi_3,...$ are all formulas in which only symbol "$y$" occurs free, and "$y$" never occur bound, and that doesn't use the symbol "$x$", and $\psi$ is a formula in which only symbol "$x$" occurs free, and "$x$" never occur bound; then:

$\underline {[i=1,2,3,...; \\ \forall x \, (x=\{y \mid \phi_i\} \to \psi)]} \\ \forall x: \psi$

In English: if a parameter free formula holds for all parameter free definable sets, then it holds for all sets.

This was proved by Hamkins to be equivalent over $\sf ZFC$ to the set theoretic axiom $\sf V=HOD$.

The finitary version of definability doesn't manage to confine all of its models to be the pointwise-definable models of $\sf ZFC$. [see here]. But, it can be argued to be the finitary parallel of the infinitary version.

My question here is:

Are there finitary sentences (i.e. in $\mathcal L(=,\in)_{\omega,\omega}$) that are theorems of $\sf ZF + Def$ yet not provable in $\sf ZFC + \text { Definability rule}$ (or equivalently in $\sf ZFC+[V=HOD])$? Or is the former a conservative extension of the latter?

If there are such sentences, are there clear examples?

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Any model of ${\sf ZFC}+V=\sf HOD$ has an elementary equivalent pointwise definable model.

If $M$ models $V=\sf HOD$, it has a parameter free definable well ordering, for each formula $φ$ consider the Skolem function that gives the least witness using that well ordering, the class $M_0$ of definable (without parameters) elements of $M$ will be closed under those Skolem functions, so $M\succ M_0$, because $M_0$ is elementary substructure of $M$, the definable elements of $M_0$ are exactly equal to those of $M$, so by construction $M_0$ is pointwise definable model that is elementary equivalent to $M$.

In other words, $\sf ZFC$+"being pointwise definable" does not prove anything that ${\sf ZFC}+V=\sf HOD$ does not prove in the language of FOST

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  • $\begingroup$ so $\sf ZF + Def$ is a conservative extension of $\sf ZFC + Definability \ rule$, that's what you are saying, right? I don't know why I feel that this is not the case. $\endgroup$ Jun 27, 2023 at 20:45
  • $\begingroup$ I'm getting the impression that what you did is just proving that the Definability rule (which is an omega rule in finitary FOST) and V=HOD are equivalent, which is already mentioned in the posting. I don't think you proved that the infinitary expression of Definability is equivalent to the finitary version over the language of FOST. Not sure though? $\endgroup$ Jun 27, 2023 at 23:09
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    $\begingroup$ @ZuhairAl-Johar I have shown that $ZFC+Def$ is conservative to $ZFC+Definability\ rule$, if the former proves $\phi$ a statement in FOST and the latter does not, let $M$ be a witness (a model of the latter + $\lnot\phi$), then the $M_0$ I defined will be a model of $ZFC+Def+\lnot\phi$, which is a contradiction $\endgroup$
    – Holo
    Jun 28, 2023 at 7:25

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