Given a (set-sized) logic $\mathcal{L}$, let $\mathcal{L}'$ be gotten from $\mathcal{L}$ by adding the ability to quantify over $\mathcal{L}$-definable relations. *(I'm being a bit vague here, since e.g. we would want to be able to say "Every $\{+\}$-definable set is periodic" in $(\mathbb{N};+,\times)$ for monotonicity reasons; see [this earlier post of mine](https://mathoverflow.net/questions/438181/what-is-the-iterated-definability-limit-of-first-order-logic)) for the detailed definition.)* For $\alpha$ an ordinal, let $\mathcal{L}^{(\alpha)}$ be gotten by iterating the $'$-operation along $\alpha$, taking unions at limit ordinals. Note that $\mathcal{L}^{(\alpha)}$ can quantify over $\mathcal{L}^{(\beta)}$-definable relations for each fixed $\beta<\alpha$. A priori we have that the class-sized logic $\mathcal{L}^{(\infty)}=\bigcup_{\alpha\in\mathsf{Ord}}\mathcal{L}^{(\alpha)}$ is contained in the closure of $\mathcal{L}$ under all set-sized Boolean combinations, call this $\mathbb{I}_{\infty,\omega}(\mathcal{L})$.

My question is the following:

> Is it consistent with $\mathsf{ZFC}$ (in particular, does it follow from large cardinals) that $\mathcal{L}^{(\infty)}<\mathbb{I}_{\infty,\omega}(\mathcal{L})$ for every set-sized logic $\mathcal{L}$ extending first-order logic?

I know that this does hold for a wide variety of logics. For example:

 - If $\mathcal{L}$ has a downward Lowenheim-Skolem number (= some $\kappa$ such that for all finite languages $\Sigma$, every $\Sigma$-structure has an $\mathcal{L}$-elementary substructure of size $\le\kappa$), then the result holds. 

 - If $\mathcal{L}$ is at least as strong as $\mathsf{SOL}$, then the result holds. 

However, this still leaves a large gap. The only "big-hammer" theorem I can think of that feels relevant is Makowsky's result that $\mathsf{VP}\implies$ every logic has a strong compactness cardinal, but I don't really see how to apply that here.

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Here are sketches of the proofs of the above results.

 - The first is basically just a lift of the proof that "gap ordinals" exist in the $L$-hierarchy. Let $\kappa$ be the downward Lowenheim-Skolem number of $\mathcal{L}$ (and assume $\mathcal{L}$ has $<\kappa$-many formulas in a finite language), and consider the $\mathcal{L}$-analogue of the $L$-hierarchy-with-urelements built on the structure $(\kappa;<)$. Applying dLS to an appropriate level of this hierarchy, we get that there is some ordinal $\lambda<\kappa^+$ such that no new subsets of $\kappa$ get added at stage $\lambda+1$. This gives that every $\mathcal{L}^{(\infty)}$-definable subset of $\kappa$ appears in this hierarchy by level $\lambda$, but this means we miss lots of subsets of $\kappa$ (and every subset of $\kappa$ is $\mathbb{I}_{\infty,\omega}(\mathsf{FOL})$-definable).

 - For the second fact, the point is that *on pure sets* if $\mathcal{L}\ge\mathsf{SOL}$ then $\mathcal{L}\equiv\mathcal{L}'$ and so $\mathcal{L}\equiv\mathcal{L}^{(\infty)}$. On the other hand, every pure set is characterized up to isomorphism by an $\mathbb{I}_{\infty,\omega}(\mathsf{SOL})$-sentence.