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?**



Bernays' Reflection Principle Holding in Ranks?
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.


