In the first-order context, "reflection" of a formula $\varphi(x)$ below $\kappa$ refers to the the following situation: 



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> There are many ordinals $\alpha<\kappa$ such that for all $a \in V\_\alpha$, $V\_\alpha \models \varphi(a)$ iff $V\_\kappa \models \varphi(a)$.
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If $\kappa$ is inaccessible, then this holds for any expansion of $V\_\kappa$ in a countable language. "Many" can be taken to mean on a club set.


When we move to higher-order logic, talk of reflection usually shifts to talk of indescribability. A cardinal is $\Pi^m\_n$ indescribable if for any $A \subseteq V\_\kappa$ and any $\Pi\_n$ sentence $\sigma$ in $(m+1)$-order logic with a predicate for $A$, if $(V\_\kappa, \in, A) \models \sigma$, then there is $\alpha<\kappa$ such that $(V\_\alpha,\in,A\cap V\_\alpha) \models \sigma$. It is a standard fact that if $\kappa$ is measurable, then there is a measure-one set of $\alpha< \kappa$ that are $\Pi^m\_n$-indescribable for every $m,n$. One can also show something stronger: If $\kappa$ is measurable, there is a measure-one set of $\alpha < \kappa$ such that if $A \subseteq V\_\alpha$, then there is $\beta < \alpha$ such that $(V\_\alpha,\in,A)$ and $(V\_\beta,\in,A\cap V\_\beta)$ have the same $\omega$-order theory.


Now this is not completely analogous to reflection because we're no longer talking about elementary substructures, but just elementarily equivalent structures, albeit with a common interpretation of a particular predicate. So my question is, what kind of large cardinal $\kappa$ is needed to get the following statement?



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> For any $A \subseteq V\_\kappa$ and any $n \in \omega$, there are many ordinals $\alpha < \kappa$ such that $(V\_\alpha,\in,A \cap V\_\alpha) \prec^n (V\_\kappa,\in,A)$, where $\prec^n$ means elementary in $(n+1)$-order logic.
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It happens at an $\omega$-strong cardinal, but this is clearly not optimal.



higher-order reflection
It depends on how we define elementarity for the $(n+1)^{th}$-order language of set theory, $\mathcal L^{n}\_\in$. For $X\subseteq V\_\kappa$, two salient definitions are:


$V\_\alpha \prec^n\_X V\_\kappa$ iff $\forall \vec{x}\in V\_\alpha(V\_\alpha\vDash \phi(X\cap V\_\alpha,\vec{x}) \leftrightarrow V\_\kappa\vDash \phi(X,\vec{x}))$, for all $\phi\in \mathcal L^{n}\_\in$ with parameters among $X, \vec{x}$. 


$V\_\alpha \prec^{\*n}\_X V\_\kappa$ iff $\forall \vec{x}\in V\_\alpha\forall \vec{Y}\subseteq V\_\alpha(V\_\alpha\vDash \phi(X\cap V\_\alpha, \vec{Y}, \vec{x}) \leftrightarrow V\_\kappa\vDash \phi(X,\vec{Y},\vec{x}))$, for all $\phi\in \mathcal L^{n}\_\in$ with parameters among $X,\vec{Y}, \vec{x}$. 


${\bf Theorem}$ ${\bf 1:}$ There are no $\alpha<\kappa$ such that $V\_\alpha\prec^{\*n} V\_\kappa$ for $n\geq 1$. 


${\it Proof.}$ If $Y = V\_\alpha$, $V\_\alpha\vDash \forall x(x\in Y)$ but not $V\_\kappa\vDash \forall x(x\in Y)$. 


${\bf Theorem}$ ${\bf 2:}$ If $\kappa$ is $\Pi^{n}\_1$-indescribable for $n\geq 1$, then for any $X\subseteq V\_\kappa$ there are unbounded $\alpha<\kappa$ such that $V\_\alpha\prec^{n-1}\_X V\_\kappa$.


${\it Proof.}$ In $n+1^{th}$-order Zermelo set theory we can define a $\Delta^{n}\_1$ satisfaction predicate for $\mathcal L^{n-1}\_\in$. Denote it $Sat$. Then we can define $Y\subseteq V\_\kappa$ such that $V\_\kappa$ satisfies:


(\*) $\langle \vec{x},\phi\rangle\in Y \leftrightarrow Sat(\phi, X, \vec{x})$, for all $\phi\in \mathcal L^{n-1}\_\in$ with parameters among $X, \vec{x}$. 


Clearly, (\*) is $\Delta^n\_1$ and since $\kappa$ is $\Pi^{n}\_1$-indescribable, it follows that there are unbounded $\alpha<\kappa$ which satisfy:


(\*\*) $\langle \vec{x},\phi\rangle\in Y\cap V\_\alpha \leftrightarrow Sat(\phi, X\cap V\_\alpha, \vec{x})$, for all $\phi\in \mathcal L^{n-1}\_\in$ with parameters among $X, \vec{x}$. 


For any such $\alpha$ it is easy to see that $V\_\alpha\prec^{n-1}\_X V\_\kappa$, as required. 


(Note that we can easily strengthen "unbounded" to "stationary" but not "club" - any $\alpha$ for which $V\_\alpha\prec^1 V\_\kappa$ will be inaccessible.)


${\bf Corollary:}$ $\kappa$ is totally indescribable just in case for any $X\subseteq V\_\kappa$ and $n<\omega$ there are unbounded $\alpha$ such that $V\_\alpha\prec^n\_X V\_\kappa$. 


