[proofplan] The result is exactly the membership form of the recursive definition of the valuation map $\sigma \mapsto \sigma_G$. We prove both implications by unpacking the set-builder defining $\tau_G$. The forward direction extracts the witnessing pair that produced $x$ as an element of $\tau_G$, and the reverse direction inserts such a witnessing pair back into the same definition. [/proofplan] [step:Extract a witnessing name and condition from membership in $\tau_G$] Assume that $x \in \tau_G$. By the defining equation for the valuation of the $\mathbb{P}$-name $\tau$ by the filter $G$, \begin{align*} \tau_G = \{ \sigma_G : \text{there exists } p \in G \text{ such that } (\sigma,p) \in \tau \}. \end{align*} Membership of $x$ in this set means precisely that there is an object of the displayed form equal to $x$. Hence there exist a $\mathbb{P}$-name $\sigma$ and a condition $p \in G$ such that $(\sigma,p) \in \tau$ and $x = \sigma_G$. [guided] Assume that $x \in \tau_G$. The only information we use is the recursive definition of valuation. Applied to the particular name $\tau$, that definition says \begin{align*} \tau_G = \{ \sigma_G : \text{there exists } p \in G \text{ such that } (\sigma,p) \in \tau \}. \end{align*} This is a set-builder description: every element of $\tau_G$ is obtained by first choosing a pair $(\sigma,p)$ from the name $\tau$ whose second coordinate $p$ lies in the filter $G$, and then evaluating the first-coordinate name $\sigma$ by $G$. Since $x \in \tau_G$, the set-builder condition supplies witnesses. Therefore there exist a $\mathbb{P}$-name $\sigma$ and a condition $p \in G$ such that $(\sigma,p) \in \tau$ and the element produced from this witness is exactly $x$, namely \begin{align*} x = \sigma_G. \end{align*} This gives the required witnesses for the forward implication. [/guided] [/step] [step:Insert a witnessing name and condition into the valuation definition] Conversely, assume that there exist a $\mathbb{P}$-name $\sigma$ and a condition $p \in G$ such that $(\sigma,p) \in \tau$ and $x = \sigma_G$. Since $p \in G$ and $(\sigma,p) \in \tau$, the defining equation \begin{align*} \tau_G = \{ \rho_G : \text{there exists } q \in G \text{ such that } (\rho,q) \in \tau \} \end{align*} implies that $\sigma_G \in \tau_G$. Because $x = \sigma_G$, substitution gives $x \in \tau_G$. [/step] [step:Conclude the equivalence] The first step proves that $x \in \tau_G$ implies the existence of a $\mathbb{P}$-name $\sigma$ and a condition $p \in G$ with $(\sigma,p) \in \tau$ and $x = \sigma_G$. The second step proves the converse implication from the same witnesses. Therefore, for every set $x$, \begin{align*} x \in \tau_G \iff \text{there exist a } \mathbb{P}\text{-name } \sigma \text{ and a condition } p \in G \text{ such that } (\sigma,p) \in \tau \text{ and } x = \sigma_G. \end{align*} This is the desired characterization of membership in the evaluated name $\tau_G$. [/step]