Suppose $B$ is a complete Boolean and let $\dot{C}\in V^B$ be such that $$\|\dot{C}\text{is a complete Boolean algebra}\|_B=1.$$
Let us consider all $\dot{c}\in V^B$ such that $\|\dot{c}\in\dot{C}\|=1$ and the equivalence relation $c_1\equiv c_2$ if and only if $\|\dot{c_1}=\dot{c_2}\|=1$. Let $D$ be the set of equivalence classes of this relation. We can make $D$ an equivalence relation as follows: if $\dot{c_1}$ and $\dot{c_2}$ are in $D$, there is a unique $\dot{c}\in D$ such that $\|\dot{c}=\dot{c_1}+_{\dot{C}}\dot{c_2}\|=1$, then we let $\dot{c}=\dot{c_1}+_D\dot{c_2}$, we define similarly $\cdot_D$ and $-_D$.
Furthermore, $D$ is a complete Boolean algebra: If $X\subseteq D$, let $\dot{X}\in V^B$ be given by $dom(\dot{X})=X$ and $\dot{X}(\dot{c})=1$, then by the fullness of $V^B$ there is some $\dot{c}\in D$ such that $\|\dot{c}=\sum_{\dot{C}}\dot{X}\|=1$, then $\dot{c}=\sum_D X$.
In the proof of lemma 16.3 of Jech's Set Theory, the author states the following fact:
Given any $b\in B$, there is a unique $\dot{c}\in D$ such that $\|\dot{c}=1_{\dot{C}}\|=b$ and $\|\dot{c}=0_{\dot{C}}\|=-b$.
Does anyone know why this is true?
Thanks
This is simply definition by cases of names.
$\{b,-b\}$ is a maximal antichain (or a partition of $1$ in terms of Boolean algebras). So we can define a name whose values are determined by that antichain.