In Bell's Boolean-valued Models and Independence Proofs (third edition), the author describes the "forcing over the universe" approach in the first two chapters and how to get corresponding ctms in chapter 4. However, instead of the usual definition $M[G]=\{\sigma_G:\sigma\in M^{(B)}\}$, where $B$ is a complete boolean algebra in a ctm $M$, $M^{(B)}$ is the collection of all $B$-names in $M$ and $\sigma_G$ is the evaluation of $\sigma$ w.r.t. the generic filter $G$, the author defines an equivalence relation $\sigma\sim_G\tau$ as $\big\|\sigma=\tau\big\|\in G$ and forms the quotient $M^{(B)}/G$, and then spends several pages showing that this model is well-founded. Does the usual definition $M[G]$ work here (of course it does, but I am wondering if there is a quick way to transfer previous results about $M^{(B)}$ to $M[G]$)?
I believe $M[G]$ works so I attempted to show that for any $B$-sentence $\phi$, $M[G]\models\phi\Leftrightarrow\big\|\phi\big\|\in G$ by induction on $\phi$. Here is the quantifier step (although similar thing appears in atomic case):
$M[G]\models\forall x\phi(x)\Leftrightarrow\forall\sigma\in M^{(B)}, M[G]\models\phi(\sigma)\Leftrightarrow\forall\sigma\in M^{(B)}, \big\|\phi(\sigma)\big\|\in G$,
where the second equivalence is by induction hypothesis. Now I want to conclude that $G\ni\big\|\forall x\phi(x)\big\|=\bigwedge_{\sigma\in M^{(B)}}\big\|\phi(\sigma)\big\|$. Let $U=\{\big\|\phi(\sigma)\big\|:\sigma\in M^{(B)}\}$ and $D=\{x\in B\setminus\{0\}:(\forall a\in U,x\leq a)\lor(\exists a\in U,x\leq a^*)\}$ ($a^*$ is the complement of $a$). To me it seems like $D$ is a dense set that belongs to $M$ (since $U\in M$). Then there is some $x\in G\cap D$, and because each $a$ belongs to $G$ it cannot be $x\leq a^*$, so it can only be $\forall x\in U, x\leq a$, i.e., $x\leq\bigwedge_{\sigma\in M^{(B)}}\big\|\phi(\sigma)\big\|$.
Is this argument correct? I thought it was but then I realized the same argument shows that if $D\subseteq G$ and $D\in M$ then $D$ has a (nonzero) lower bound in $G$. This seems too strong and probably wrong, plus I haven't found it in any book.