Let $A, B, C..X, Y$ range over potentially proper classes in NBG set theory and $x,y,z$ range over sets.

Since I don't know the proper symbols for formula classes that have $n$ proper class quantifiers I'll just defined by analogy with number/set quantifiers:

• $P(C)$ is a $\Sigma^2_0=\Pi^1_0$ class predicate if $P(C) \iff \phi(C)$ where $\phi(C)$ only contains set quantifiers.
• $P(C)$ is a $\Sigma^2_{n+1}$ ($\Pi^2_{n+1}$) class predicate if $P(C)$ has the form $(\exists X) Q(C,X)$ ($(\forall X) Q(C,X)$ ) where $Q$ is a $\Sigma^2_{n}$ ($\Pi^2_{n}$) class predicate.

Suppose $P(C)$ is a non-empty $\Sigma^2_0$ class predicate, i.e., $P(C)$ is equivalent to a formula quantifying only over sets. Must there be a definable class $B$ satisfying $P(B)$ (where definable means definable via quantification over sets no proper classes...so same thing as in ZFC)? What if $P(C)$ is instead $\Sigma^2_n/\Pi^2_n$?

I'm tempted to think the answer is yes (at least for $\Sigma^2_0$). However, it seems to me that you can think of this as asking whether there is a (full height of $V$) path through a certain tree of classes and what is going on seems very similar to the question of whether every nonempty arithmetic tree has a $\Delta^1_1$ path where the answer is no. Indeed there is a $\Pi^0_2$ class with no $\Delta^1_1$ path and as that is exactly where things started to fail for me with the class question it seems the evidence says no even in that case.

Also I'm probably just failing to think of something simple since my set theory is rusty.