Let $B$ be a complete Boolean algebra. Jech defines a Boolean-valued model $\mathfrak{A}$ of the language of set theory to consist of a Boolean universe $A$ and functions of two variables with values in $B$, $\qquad \qquad \| x=y \|, \qquad \| x \in y \|$

that safisfy the following:

$ (i)\ \ \ \| x=x \| = 1 \\ (ii) \ \ \| x=y \| = \|y = x \| \\ (iii)\ \| x=y \| \cdot \| y=z \| \leq \| x=z \| \\ (iv) \ \ \| x\in y \| \cdot \|v=x \| \cdot \|w=y \| \leq \|v \in w \| $

For every formula $\phi(a_1,\ldots,a_n)$, we define the Boolean value $\| \phi(a_1,\ldots,a_n) \|$ of $\phi$ as follows:

$(a)$ For atomic formulas, we have the functions $\| x=y \|, \| x \in y \| $

$(b)$ If $\phi$ is negation, disjunction, $\exists x \, \psi$,

$\qquad \| \lnot \phi(a_1,\ldots,a_n) \| = - \| \phi(a_1,\ldots,a_n) \| $

$\qquad \| (\phi \vee \psi)(a_1,\ldots,a_n) \| = \| \phi(a_1,\ldots,a_n) \| + \| \psi (a_1,\ldots,a_n) \|$

$\qquad \| \exists x \, \psi(x,a_1,\ldots,a_n) \| = \underset{a\in A}{\sum} \| \psi (a,a_1,\ldots,a_n)\|$

Next, Jech says that it's easy to prove

$\qquad \qquad \qquad \| x=y \| \cdot \|\phi(x)\| \leq \|\phi(y)\|$

However, I'm not seeing it. It seems clear that it should be a proof by induction on the complexity of $\phi$, and the inductive step is easy in the case of disjunction and existential, but why does it hold for negation? That is, why is it that if $ \| x=y \| \cdot \|\phi(x)\| \leq \|\phi(y)\|$, then $\| x=y \| \cdot \|\lnot \phi(x)\| \leq \| \lnot \phi(y)\|$?

Thank you!

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