Boolean-valued models have no essentially new ordinals 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.
 A: Let us run through the proof of the lemma (Jech's Set Theory, 3rd Millennium Edition, Lemma 14.23):
$\renewcommand{\Ord}{\mathrm{Ord}}
\renewcommand{\Dom}{\operatorname{Dom}}$
Lemma 14.23: For every $x\in V^B$, $$\|x\text{ is an ordinal}\|=\sum_{\alpha\in\Ord}\|x=\check\alpha\|$$
For shortness sake, I'll denote $\phi(x)=x\text{ is an ordinal}$.
Proof: By a previous lemma, $\phi(x)$ is a $\Delta_0$ formula and thus absolute between $V$ and $V^B$. This means that $\|\phi(\check\alpha)\|=1$ for every $\alpha\in\Ord$, and clearly $\|x=\check\alpha\rightarrow\phi(x)\|=1$, therefore $\|x=\check\alpha\|\le\|\phi(x)\|$, for every $\alpha$. Therefore, $\sum\|x=\check\alpha\|\le\|\phi(x)\|$.
On the other hand, let $\|\phi(x)\|=u$, then for every ordinal $\gamma$ we have:
$$\|\phi(x)\land x\in\check\gamma\|\le\sum_{\alpha\in\gamma}\|x=\check\alpha\|\tag 1$$
This is because $\check\gamma(t)=1$ for every $t\in\Dom(\check\gamma)$ and $\|x\in\check\gamma\|=\sum_{t\in\Dom(\check\gamma)}\|x=t\|\cdot\check\gamma(t)$, inductively we can show that $t\in\Dom(\check\gamma)\iff t=\check\alpha$ for $\alpha\in\gamma$.
Since ordinals are comparable then for every $\alpha$ we have:
$$u=\|\phi(x)\|\le\|x\in\check\alpha\|+\|x=\check\alpha\|+\|\check\alpha\in x\|$$
This statement is equivalent to saying that $\phi(x)$ implies that $x$ and $\alpha$ are comparable, which is what we wanted to say.
Now since there is only set many $\alpha$ such that $x(\check\alpha)\neq 0$, we have that there is some $\gamma$ such that $u\le\|x\subseteq\check\gamma\|$, thus $\|\phi(x)\|\le\|x\subseteq\check\gamma\|$, and it means that $\|\phi(x)\|\le\|\phi(x)\land x\in\check\gamma\|$. Using $(1)$ we yield:
$$u=\|\phi(x)\|\le\sum_{\alpha\in\gamma}\|x=\check\alpha\|\le\sum_{\alpha\in\Ord}\|x=\check\alpha\|\le u$$
A: Actually, it seems everything is all right. If I haven't made a mistake, all you need to prove that ordinals are comparable is Extensionality, and Jech proves that Extensionality is valid in $V^B$ in Lemma 14.17. So the result does need some validity of set theory in the Boolean-valued model
