Following Jech's Set Theory, fix a complete Boolean algebra $B$ and form the Boolean-valued model $V^B$ of ZFC. We then have the name $\check{B}\in V^B$. Apparently, it is the case that $\|\check{B}\text{ is a Boolean algebra}\|=1$ How come? Certainly we can define $\check{p}\wedge\check{q}=(p\wedge q)\check{\phantom{p}}$ for $p,q\in B$ etc., but is this enough? Or do we have to specify $B$-names for functions, which will represent the Boolean operations on $\check{B}$?
This question is motivated by the later discussion of the name of a generic object $\dot{G}$, defined by $\text{dom}(G)=\{\check{p};p\in B\}$ and $\dot{G}(\check{p})=p$. Various sources state that $\|\dot{G} \text{ is a generic ultrafilter on }\check{B}\|=1$ but for this to hold, of course, $\check{B}$ has to be a Boolean algebra (in $V^B$), and this is what confuses me. The (outline of a) proof I've seen of this, seems only to consider canonical names of elements of $B$ when proving, for example, upward closedness, and I'm not sure why this should work.