Cores in Boolean valued models

Question

The following is taken from Bell's Set Theory: Boolean-Valued Models and Independence Proofs. $V^{(B)}$ is a Boolean valued model, where $B$ is some (complete) Boolean algebra.

I'm stuck in verifying the second part of the definition of a core. Namely, let $y\in V^{(B)}$ with $\|y\in u\|=1$. Pick $z\in w$ with $f_z=f_y$. This gives, by definition of $\|\cdot \in\cdot\|$,

$$1=\|y\in u\|=\bigvee \text{ran}(f_y)=\bigvee\text{ran}(f_z)=\|z\in u\|$$

Thus $\|z=x\|=1$ for some (unique) $x\in v$. Ideally, I'd like to conclude that $\|x=y\|=1$, but I don't see why that holds. Expanding the definition of $\|x=y\|$ does not seem very promising.

Answer 1

For posterity, here's one way of doing it:

Lemma: Say $x,y\in V^{(B)}$, with $f_x=f_y$. If $\|y\in u\|=1$, then $\|x=y\|=1$.

Proof: Fix $z\in\text{dom}(u)$.The assumption $f_x=f_y$, says that $u(z)\wedge \|x=z\|=u(z)\wedge \|z=y\|$. Then $$ \|x=y\|\ge \|x=z\|\wedge \|z=y\| \ge u(z)\wedge \|z=y\| $$ Taking the supremum over $z\in\text{dom}(u)$, we get that $$ \|x=y\|\ge \bigvee_{z\in\text{dom}(u)}u(z)\wedge \|z=y\|=\|y\in u\|=1 $$ Therefore, $\|x=y\|=1$. $\square$.

Now assume $y\in V^{(B)}$ with $\|y\in u\|=1$. Pick $z\in w$ with $f_y=f_z$. We have $\|z\in u\|=\|y\in u\|=1$, so there exists (a unique) $x\in v$ with $\|z=x\|=1$. Note that $$ \|x=y\|\ge \|x=z\|\wedge\|z=y\|=1\wedge 1=1 $$ by the lemma, so $\|x=y\|=1$, as desired.