Proving $\|x=y\|\cdot \|\phi(x)\|\le\|\phi(y)\|$ in Boolean valued models

Question

This question relates to the Boolean algebra approach to forcing.

Fix a complete Boolean algebra $B$. I'm writing $\|\sigma\|$ for the Boolean value of $\sigma$, where $\sigma$ is a sentence of the forcing language.

I'm trying to show the inequality $\|x=y\|\cdot \|\phi(x)\|\le\|\phi(y)\|$ (which is stated without proof in Jech) by induction on the complexity of $\phi$. I'm stuck on the negation case. Namely, how do we go from

$$ \|x=y\|\cdot \|\phi(x)\|\le \|\phi(y)\| $$ to $$ \|x=y\|\cdot \|\neg\phi(x)\|\le \|\neg \phi(y)\| $$ (recall that $\|\neg\psi\|:=-\|\psi\|$).

I've tried a few manipulations, but the negation switches the inequality, which is proving to be quite annoying.

Answer 1

Assuming that symmetry of equality has already been proved, you have, from the induction hypothesis about $\phi$, that $\Vert x=y\Vert\land\Vert\phi(y)\Vert\leq \Vert\phi(x)\Vert$. Take the Boolean negation of both sides, remembering that, as you said, the direction of the inequality gets reversed, you get $\neg\Vert\phi(x)\Vert\leq(\neg\Vert x=y\Vert)\lor(\neg\Vert\phi(y)\Vert)$. Now take the Boolean meet of both sides with $\Vert x=y\Vert$ and apply the distributive law to the right side. Also remember that Boolean negation gives truth values of negated formulas. You get $$ \Vert x=y\Vert\land\Vert\neg\phi(x)\Vert\leq(\Vert x=y\Vert\land \neg\Vert x=y\Vert)\lor(\Vert x=y\Vert\land\Vert\neg\phi(y)\Vert). $$ On the right side, the first disjunct is $0$ and so the formula simplifies to $$ \Vert x=y\Vert\land\Vert\neg\phi(x)\Vert\leq\Vert x=y\Vert\land\Vert\neg\phi(y)\Vert \leq\Vert\neg\phi(y)\Vert, $$ as required.