For a logic $\mathcal{L}$, say that a cardinal $\kappa$ is $\mathcal{L}$-correct iff every satisfiable $\mathcal{L}$-theory of size $<\kappa$ has a model of size $<\kappa$. First-order correctness is of course boring, but quite quickly we enter the realm of strong large cardinal properties (see e.g. [here](https://mathoverflow.net/questions/211855/what-sort-of-cardinal-number-is-the-l%C3%B6wenheim-skolem-number-for-second-order-log)).


I'm interested in a kind of "iterated correctness:" repeatedly add to a given logic the ability to quantify over cardinals which are correct for it (or rather, for the previous iteration of this process). This doesn't make obvious sense in general, but for logics with reasonably nice syntax things are better. In particular, I'm interested in what happens when we iteratively "add correctness quantifiers" to second-order logic, as follows:


Let $\mathcal{L}^2\_0$ be usual second-order logic, and let $\mathcal{L}^2\_{n+1}$ be $\mathcal{L}\_n^2$ augmented with a quantifier $$\mathsf{C}\_nx\varphi(x)\equiv\mbox{$\vert\{x: \varphi(x)\}\vert$ is $\mathcal{L}^2\_n$-correct.}$$ So $\mathcal{L}^2\_k$, in addition to the usual logical symbols, has $k+2$ different types of quantifier: the first-order quantifiers, the second-order quantifiers, and $k$-many correctness quantifiers. These quantifiers can alternate however is desired.


My question is:



> 
> Do any of the "usual" large cardinal properties imply $\mathcal{L}^2\_n$-correctness for every $n$?
> 
> 
> 


This is not at all obvious to me. On the other hand, I can't pin down a concrete obstacle to e.g. supercompactness having the above property.



Lowenheim-Skolem numbers for SOL + correctness quantifiers
Assume ZFC. Let $n$ be a meta-integer. Then $\kappa$ is $\mathcal{L}^2\_n$-correct iff $\kappa$ is $\Sigma\_{n+2}$-reflecting, i.e. $V\_\kappa\preccurlyeq\_{n+2}V$.


Proof: For $n=0$, i.e. 2nd order logic, we have: If $V\_\kappa\preccurlyeq\_2 V$ then easily $\kappa$ is $\mathcal{L}^2\_0$-correct. Suppose now that $\kappa$ is $\mathcal{L}^2\_0$-correct. Now for each $\alpha<\kappa$, if $|V\_\alpha|<\kappa$ then $|V\_{\alpha+1}|<\kappa$, as is easily seen using $\mathcal{L}^2\_0$-correctness. So there is a limit ordinal $\eta\leq\kappa$ such that $|V\_\eta|=\kappa$. Now observe that $V\_\eta\preccurlyeq\_2 V$ by using $\mathcal{L}^2\_0$-correctness. It follows that $\eta=\kappa$ (if $\eta<\kappa$, take $\alpha<\eta$ such that $\eta\leq\beta=|V\_\alpha|<\kappa$, and using $V\_\eta\preccurlyeq\_2 V$, show $\beta<\eta$ for a contradiction). So $V\_\kappa\preccurlyeq\_2 V$ as desired.


For $n=1$: Suppose $V\_\kappa\preccurlyeq\_3V$. Let $T\in V\_\kappa$ be a satisfiable $\mathcal{L}^2\_1$-theory. Then there is an ordinal $\eta$ such that $V\_\eta\preccurlyeq\_2 V$ and $M\in V\_\eta$ such that $V\_\eta\models$"$M\models T$", where the latter statement uses $V\_\eta$'s own notions to compute $\Sigma\_2$-elementarity, and hence the $C\_0$ quantifier. But this is a $\Sigma\_3$ statement, so $V\_\kappa$ models it, and note that this really gives a true model of $T\in V\_\kappa$.


Conversely, suppose $\kappa$ is $\mathcal{L}^2\_1$-correct. So in particular it is $\mathcal{L}^2\_0$-correct, and hence $V\_\kappa\preccurlyeq\_2V$. We want to see $V\_\kappa\preccurlyeq\_3V$. Let $x\in V\_\kappa$ and suppose $V\models\exists w\varphi(x,w)$ where $\varphi$ is $\Pi\_2$. Let $\alpha<\kappa$ with $x\in V\_\alpha$. Consider the $\mathscr{L}^2\_0$-theory theory $T$ in parameters in $V\_\alpha$, which describes a rank segment $V\_\eta$ of $V$ such that $\eta$ is a cardinal and $V\_\eta\preccurlyeq\_2V$ (by using the $C\_0$-quantifier to require $\eta$ to be $\mathcal{L}^2\_0$-correct), and $V\_\eta\models\exists w\varphi(w,x)$. Note this is satisfiable, and hence satisfiable in cardinality ${<\kappa}$, and hence $V\_\kappa\models\exists w\varphi(w,x)$.


Now proceed in this manner.


So in particular, least $\mathcal{L}^2\_0$-correct cardinal is $<$ least strong
and least supercompact, these are $<$
least $\mathcal{L}^2\_1$-correct $<$ least extendible
< least $\mathcal{L}^2\_2$-correct.


