Absoluteness of $\Delta_0$ formulas for Boolean-Valued models.

Question

I am stuck on the following lemma from Jech's forcing chapter and need a little bit of help.

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Lemma. If $\varphi(x_1, \dots, x_n)$ is a $\Delta_0$ formula, then $$\varphi(x_1, \dots, x_n) \hspace{0.5cm} \leftrightarrow \hspace{0.5cm} \|\varphi(\check{x}_1, \dots, \check{x}_n)\| = 1.$$

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Now i have proved this for atomic formulas by a simultanious induction on both $x \in y$ and $x = y$. The connectives are easy. But i'm stuck on the existential quantifier. The $(\exists y \in x)\varphi(y) \rightarrow \|(\exists y \in \check{x})\varphi(y)\| = 1$ part is relatively easy. But the converse seems a little bit tricky. I have tried to use the fact that $V^B$ is full but that doesn't guarantee that the element that we get by eliminating the quantifier is a set in $x$.
Any hints or answers would be very helpful. Thanks for your patience.

Answer 1

but that doesn't guarantee that the element that we get by eliminating the quantifier is a set in $x$.

It's easier to approach this from a slightly different angle: $$ \begin{align*} 1 & = \| \exists y \in \check{x} \phi(y, \ldots) \| \\ &= \sum_{y \in \mathrm{dom}(\check{x})} \check{x}(y) \cdot \| \phi(y, \ldots) \| \end{align*} $$

Now recall that

1. $\check{x}(y) \in \{0,1\}$ for all $y \in \mathrm{dom}(\check{x})$ and
2. $\mathrm{dom}(\check{x}) = \{ \check{y} \mid y \in x \}$.

Puzzle this together to conclude that there is some $y \in x$ such that $$ 1 = \| \phi(\check{y}, \ldots) \|, $$ as desired.