Bernays' Reflection Principle Holding in Ranks? I wanted to see for which ranks Bernays' Reflection Principle holds; that is, for every class and every property (allowing quantification over all classes) which is true about that class, there is a transitive set $u$ for which $\mathcal{P}(u)$ satisfies that property about that class's intersection with $u$. Like Vopenka did with Vopenka cardinals, I allow "arbitrary $A\subseteq V_\kappa$" to be classes in $V_\kappa$.
It turns out that once analyzed, these ranks bear a striking similarity to those ranks of Indescribable cardinals. I formalized this definition in ZFC as follows:
Let $\kappa$ be $\Pi_m$-Bernays when for every first-order $\Pi_m$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is an unary predicate symbol:
$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists u\in V_\kappa(u\;\mathrm{is}\;\mathrm{transitive}\land\langle \mathcal{P}(u);\in,A\cap u\rangle\models\phi))$$
A similar definition in ZFC would be that for every first-order $\Pi_m$ unary formula $\phi$ in the language $\mathcal{L}_\in$:
$$\forall A\subseteq V_\kappa(\phi^{V_{\kappa+1}}(A)\rightarrow\exists u\in V_\kappa(u\;\mathrm{is}\;\mathrm{transitive}\land\phi^{\mathcal{P}(u)}(A\cap u)))$$
Of course, the existence of a $\Pi_{<\omega}$-Bernays cardinal is equivalent to the existence of a cardinal rank which satisfies Bernays' Reflection Principle. Thus, $\Pi_{<\omega}$-Bernays cardinals, in consistency strength, are somewhere above Bernays' Reflection Principle, which oddly enough implies the existence of an Inaccessible.
These cardinals are therefore consistency-wise stronger than that of Inaccessible cardinals.
Every cardinal is $\Pi_0$-Bernays. This is pretty simple once one considers that all of the unary first-order formulas in the language $\mathcal{L}_\in$ are either true for every set or true for no set.
This brings me to my question:
Are these cardinals inconsistent with ZFC? Are they any weaker or stronger than Indescribability?
 A: Every $\Pi_n^1$-Indescribable cardinal is $\Pi_n$-Bernays, and thus the $\Pi_{n+1}^1$-Indescribable cardinals are strictly of greater consistency strength than that of the $\Pi_n$-Bernay cardinals, and also the $1$-Indescribable cardinals are of strictly greater consistency strength than that of Bernay's principle itself.
Here's the proof:
A little Lemma before the proof is to show that if $\exists\alpha<\kappa(\langle V_{\alpha+1};\in,A\cap V_\alpha\rangle\models\phi)$ then $\exists u\in V_\kappa(\langle\mathcal{P}(u);\in,A\cap u\rangle\models\phi)$. Specifically, this becomes quite obvious when one realizes that $V_\alpha$ is such a $u$ described when the first condition holds for $\alpha$.
A cardinal $\kappa$ is $\Pi_n^1$-Indescribable iff for every first-order $\Pi_n$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is an unary predicate symbol:
$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists\alpha<\kappa(\langle V_{\alpha+1};\in,A\cap V_\alpha\rangle\models\phi))$$
By the previous lemma, this implies that for every first-order $\Pi_n$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is a unary predicate symbol:
$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists u\in V_\kappa(\langle\mathcal{P}(u);\in,A\cap u\rangle\models\phi))$$
Of course, this is the first definition I gave for a $\Pi_n$-Bernays cardinal. Thus, every $\Pi_n^1$-Indescribable cardinal is $\Pi_n$-Bernays.
A: I would like to add that, for $n>1$, the $\Pi_n-$Bernays cardinals are precisely the $\Pi_n^1-$indescribable cardinals. We can do this really simply by giving a $\Pi_2$ definition of the Axiom of limitation of size:
$$ALZ\leftrightarrow\forall x(\exists W\ni x\leftrightarrow\exists F,y(F\text{ is a bijection } F:x\rightarrow y\land\exists U\ni y))$$
Then $2^u\vDash ALZ$ precisely when $2^u$ satisfies limitation of size. Let $AU$ be the axiom of union (Which is also $\Pi_2$). Then whenever $u$ is infinite, transitive, and $2^u\vDash ALZ\land AU$, then $u$ is a Grothendrick universe and therefore of the form $V_\lambda$ for some inaccessible $\lambda$.
Then, for any $\Pi_2^1$ formula $\phi$ in the language of $\{\in,S\}$ such that $(V_{\kappa+1},\in,S)\vDash\phi$, to get some  $(V_{\lambda+1},\in,S\cap V_\lambda)\vDash\phi$, let $\psi$ be the formula $ALZ\land AU\land\phi$, and take some transitive $u$ such that $(2^u,\in,S\cap u)\vDash\psi$. It is easy to see that $\kappa$ is inaccessible.
