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!



Boolean-Valued Models: Why is $\| x=y \| \cdot \| \phi(x) \| \leq \| \phi(y) \|$?
One way to show this is by first noting that (i)-(iv) ensure the atomic instances of the identity axioms:


$x = y \to [\phi(x) \leftrightarrow \phi(y)]$


get value 1. Then observe, by a simple induction on the complexity of $\phi$, that the non-atomic instances follow from the atomic instances in first-order logic. So they too get value 1.


