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?**

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