Since I still have some trouble with transferring theorems of ZFC into the Boolean-valued framework, would someone more at ease with this check the following short calculation?
This is a proposition in Jech's book, stating that $\|z\text{ is an ordinal}\|=\bigvee_{\alpha\in\mathbf{ON}}\|z=\check{\alpha}\|$ I can follow the proof given, but want to confirm the following proof of the inequality $\|z\text{ is an ordinal}\|\leq\|z\in\check{\alpha}\|\vee\|z=\check{\alpha}\|\vee\|\check{\alpha}\in z\|$
Fix a $B$-name $z$ and an ordinal $\alpha$. Classically, if $z$ were an ordinal, it would be comparable with $\alpha$, i.e. either $z\in\alpha,z=\alpha$ or $\alpha\in z$. However, the universal closure of this statement with respect to $z$ is not $\Delta_0$, so we can't directly use absoluteness (some sort of closure is required to avoid producing a check on $z$). Instead, we find $z$'s place in the von Neumann hierarchy, $z\in V_\gamma$. We can then take the universal closure of the above statement, bounded to $V_\gamma$, which is $\Delta_0$, and get $\|\forall y\in\check{V_\gamma}:y\text{ is an ordinal}\Rightarrow y\in\check{\alpha}\vee y=\check{\alpha}\vee\check{\alpha}\in y\|=1$ We can then specify this to $z$ and get the desired result.
I suppose this "trick" of replacing an unbounded quantifier with a bounded one is pretty standard. I realize all of this might be moot since we know that all of the axioms of ZFC are valid in $V^B$, which means all of the theorems are also valid. But since Jech proves the result in question before proving that $V^B$ models ZFC, I thought there must be a simpler way.
EDITED: After further thought, what I wrote above is wrong. We can't specify to $z$, but only to $\check{z}$. I'm at a loss again.