Suppose that $\Bbb P$ is a forcing in $V$, we say that $\Bbb P$ is $V$-decisive if whenever $\varphi(x_1,\ldots,x_n)$ is a statement in the language of forcing, and $u_1,\ldots,u_n\in V$ then $1_{\Bbb P}$ decides the truth of $\varphi(\check u_1,\ldots,\check u_n)$. *(Side question, was this property of $\Bbb P$ given a name in the past?)*

It's a classical theorem that weakly homogeneous forcing is $V$-decisive. But we can show a bit more. The question is whether or not there exists a $V$-decisive forcing whose Boolean completion is rigid, or at least not weakly homogeneous.

What we know:

1. $\Bbb P$ is $V$-decisive if and only if whenever $G$ is $V$-generic, we have that $\bigcup\{H\in V[G]\mid H\subseteq\Bbb P\text{ is }V\text{-generic}\}=\Bbb P$. (Note that we don't require that $V[G]=V[H]$, just that $H$ is generic over $V$.)

2. Equivalently, this means that if $p,q\in\mathcal B(\Bbb P)$ (the complete Boolean algebra that contains $\Bbb P$ as a dense subset) there are embeddings of $\mathcal B(\Bbb P)\restriction p$ into $\mathcal B(\Bbb P)\restriction q$. (*Correction: earlier a density requirement was added, after Joel's answer it dawned on us that there is too much here.*)

3. Equivalently, for every $p\in\Bbb P$ there is some $q\leq p$ and a projection of $\mathcal B(\Bbb P)$ which maps $\mathcal B(\Bbb P)\restriction q$ onto $\mathcal B(\Bbb P)$, then $\Bbb P$ is $V$-decisive. Note that we do not require the projection to be injective, which would essentially mean that $\Bbb P$ is weakly homogeneous.

> **Question.** Is there a $V$-decisive forcing whose Boolean completion is not weakly homogeneous? If not, is it at least consistent that there is one? In either case, can we find one whose Boolean completion is rigid?