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 dense embeddings of $\mathcal B(\Bbb P)\restriction p$ into $\mathcal B(\Bbb P)\restriction q$

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 which 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?