Names determined by a condition, 2 Given a c.t.m. $M$, a preorder $\mathbb{P}\in M$, a condition $p\in\mathbb{P}$, and a $\mathbb{P}$-name $\tau$, let us say that "$p$ determines $\tau$" if $\tau_G$ is the same set for all $\mathbb{P}$-generic filters $G$ that contain $p$. By other words, for all $\mathbb{P}$-generic filters $G$, $H$ such that $p\in G\cap H$ we have $\tau_G=\tau_H$.
If $p$ determines $\tau$ and $G$ is any $\mathbb{P}$-generic filter such that $p\in G$ then $\tau_G\in M$. Indeed, it is possible to find another $\mathbb{P}$-generic filter $H$ such that $p\in H$ and $G,H$ are mutually generic (that is, $G\times H$ is $(\mathbb{P}\times\mathbb{P})$-generic). Then we have $\tau_G=\tau_H\in M[G]\cap M[H]$, and, by this answer, $M[G]\cap M[H]=M$. (See also this question.)
It follows that the relation "$p$ determines $\tau$" is definable in $M$, for example, by the formula "$(\exists x)\ p\Vdash\tau=\check x$".
It could be interesting to find some recursive formula that directly defines this relation, similar to the definition of $p\Vdash\sigma\in\tau$ and $p\Vdash\sigma=\tau$.
We can also do the following. Given a $\mathbb{P}$-name $\tau$, we can define $(\mathbb{P}\times\mathbb{P})$-names $l(\tau)$, $r(\tau)$ recursively by $l(\tau)=\{(l(\sigma),(p,1))\!:(\sigma,p)\in\tau\}$ and $r(\tau)=\{(r(\sigma),(1,p))\!:(\sigma,p)\in\tau\}$. Then for all $\mathbb{P}$-generic filters $G,H$ we have $l(\tau)_{G\times H}=\tau_G$ and $r(\tau)_{G\times H}=\tau_H$. Let us note that $G\times H$ need not be $(\mathbb{P}\times\mathbb{P})$-generic.
We have the following implications/equivalences:
"$p$ determines $\tau$"
$\Leftrightarrow$
"for all $\mathbb{P}$-generic filters $G,H$, if $p\in G\cap H$ then $\tau_G=\tau_H$"
$\Leftrightarrow$
"for all $\mathbb{P}$-generic filters $G,H$, if $p\in G\cap H$ then $l(\tau)_{G\times H}=r(\tau)_{G\times H}$"
$\Rightarrow$
"for all $(\mathbb{P}\times\mathbb{P})$-generic filters $G$, if $(p,p)\in G$ then $l(\tau)_G=r(\tau)_G$"
$\Leftrightarrow$ "$(p,p)\Vdash l(\tau)=r(\tau)$".
Question: Is the relation "$p$ determines $\tau$" equivalent to "$(p,p)\Vdash l(\tau)=r(\tau)$"?
 A: Yes. For any $G, H$ $\mathbb{P}$-generic over $M$ and any condition $p\in\mathbb{P}$, there is a $J\ni p$ which is $\mathbb{P}$-generic over $M$ such that $G\times J$ and $H\times J$ are both $\mathbb{P}^2$-generic over $M$. Now if $G, H\ni p$, we just argue (assuming $(p, p)\Vdash l(\tau)=r(\tau)$) that:


*

*$\tau_G=\tau_J$, and

*$\tau_J=\tau_H$,
so we get $\tau_G=\tau_H$.

Why does such a $J$ always exist?
Fix $G$ and $H$ appropriate. Since $M$ is countable, let's pick (externally) lists of the conditions and of the dense open subsets of $\mathbb{P}^2$ in $M$: $$\mathbb{P}=\{p_i: i\in\omega\},\quad\mathcal{P}_{denseopen}(\mathbb{P}^2)=\{D_i: i\in\omega\}.$$
We now build a sequence of conditions $j_i$ ($i\in\omega$) by stages, the key lemma being the following:

Suppose $I$ is $\mathbb{P}$-generic over $M$, $q\in \mathbb{P}$, and $D\subseteq\mathbb{P}^2$ is dense open. Then there is some $\hat{q}\le q$ and $r\in I$ such that $(r, \hat{q})\in D$.

Proof: Otherwise for some $p\in I$ we must have no condition below $(p, q)$ in $D$, contradicting the density of $D$. $\quad\Box$
Now we simply iterate the lemma, building a sequence of conditions $q_i$ ($i\in\omega$) such that for each $i$ we have $$q_0\le p,\quad q_{i+1}\le q_i\quad\mbox{ and }\quad \exists p_1\in G, p_2\in H((p_1, q_i), (p_2, q_i)\in D).$$ Let $J$ be any $\mathbb{P}$-generic-over-$M$ filter containing all of the $q_i$s. By construction, $G\times J$ and $H\times J$ are each 
