I want to show that for atomic $\varphi$,
$$p\Vdash^\ast \varphi \iff \{p_1\le p:p_1\Vdash^\ast \varphi\} \text{ is dense below $p$}$$
The pertinent definitions are as follows:
For any forcing poset $\mathbb{P}$ and names $\tau,\vartheta\in V^\mathbb{P}:$
$p\Vdash^\ast \tau=\vartheta\ \text{ iff }\ \forall\sigma \in\text{dom}(\tau)\cup\text{dom}(\vartheta)\forall q\le p [q\Vdash^\ast\sigma\in\tau\leftrightarrow q \Vdash^\ast \sigma\in\vartheta].$
$p\Vdash^\ast\pi \in\tau\ \text{ iff }\ \{q\le p:\exists\langle \sigma,r\rangle\in\tau[q\le r \wedge q\Vdash^\ast \pi=\sigma]\} \text{ is dense below $p$}.$
The case where $\varphi$ is $\pi\in\tau$ is already done, and the hint for when $\varphi$ is of the form $\tau=\vartheta$ is to use the $\in$ case for $\sigma\in\tau$ and $\sigma\in\vartheta$. My attempted solution:
Suppose $\{p_1\le p:p_1\Vdash^\ast \tau=\vartheta\}$ is dense below $p$. We need to show $(1.)$ holds, so fix $\sigma\in\text{dom}(\tau)\cup\text{dom} (\vartheta)$ and $q\le p$. We need to check
$$q\Vdash^\ast \sigma\in \tau \leftrightarrow q\Vdash^\ast \sigma\in\vartheta.$$
By the $\in$ case, it suffices to see
$$\{q_1\le q:q_1\Vdash^\ast \sigma\in\tau\} \text{ is dense below $q$}\leftrightarrow \{q_1\le q:q_1\Vdash^\ast \sigma\in\vartheta\} \text{ is dense below $q$}$$
Assume the left side holds, and fix $q_2\le q$. There is some $q_1\le q_2$ such that $q_1 \Vdash^\ast \sigma\in\tau$. Since $q_1\le q_2\le q\le p$, by our original hypothesis $\exists p_1\le q_1(p_1\Vdash^\ast \tau=\vartheta)$. In particular, by (1.):
$$p_1\Vdash ^\ast \sigma\in\tau \leftrightarrow p_1\Vdash^\ast \sigma\in\vartheta$$
Now, since $p_1\le q_1\Vdash \sigma \in \tau$, it follows (by a previously proven result) that $p_1\Vdash ^\ast \sigma\in\tau$. But also $p_1\le q_1\le q_2$, and since $q_2\le q$ was arbitrary, $\{q_3\le q: q_3\Vdash^\ast \sigma \in\vartheta\}$ is dense below $q$.
The other implication is analogous.
Is this correct? In any case, it seems rather long, is there a shorter way to do it?