[proofplan] We construct the evaluation assignment by class recursion on the name-rank of $\mathbb{P}$-names. The filter axioms are not needed for the recursion itself, so we work with an arbitrary subset $G\subseteq\mathbb{P}$. At the stage for a name $\tau$, every name $\sigma$ appearing in a pair $(\sigma,p)\in\tau$ has already been evaluated, so the displayed formula defines an actual set by Replacement and Separation applied to the set $\tau$. Uniqueness is proved by the same rank induction: two class assignments satisfying the recursion agree on all lower-rank names, hence give the same value on $\tau$. [/proofplan]

[step:Construct the value of each name from previously evaluated lower-rank names]Fix a subset $G \subseteq \mathbb{P}$. For each $\mathbb{P}$-name $\tau$, define its name-rank $\operatorname{rank}_{\mathbb{P}}(\tau)$ by recursion as the least ordinal strictly larger than $\operatorname{rank}_{\mathbb{P}}(\sigma)$ for every $\sigma$ such that $(\sigma,p)\in\tau$ for some $p\in\mathbb{P}$. This least ordinal exists because $\tau$ is a set, so the collection \begin{align*} R_\tau:=\{\operatorname{rank}_{\mathbb{P}}(\sigma): \exists p\in\mathbb{P} \text{ such that }(\sigma,p)\in\tau\} \end{align*} is a set of ordinals by Replacement, and one may take $\sup R_\tau+1$. We invoke the class recursion principle on the well-founded class relation $\sigma\prec\tau$ defined by $\sigma\prec\tau$ iff $(\sigma,p)\in\tau$ for some $p\in\mathbb{P}$. This relation is well-founded because $\sigma\prec\tau$ implies \begin{align*} \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau). \end{align*} Thus it is legitimate to define a class assignment $E_G$ on all $\mathbb{P}$-names by recursion on $\operatorname{rank}_{\mathbb{P}}(\tau)$. Suppose that $E_G(\sigma)$ has already been defined for every $\mathbb{P}$-name $\sigma$ with \begin{align*} \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau). \end{align*} For every pair $(\sigma,p)\in\tau$, the definition of name-rank gives \begin{align*} \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau), \end{align*} so $E_G(\sigma)$ is already defined. Define \begin{align*} E_G(\tau):=\{E_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\}. \end{align*} This is a set: first separate from $\tau$ the subset \begin{align*} A_\tau:=\{(\sigma,p)\in\tau:p\in G\}, \end{align*} and then apply Replacement to the map $(\sigma,p)\mapsto E_G(\sigma)$ on $A_\tau$. Thus the recursion defines a value $E_G(\tau)$ for every $\mathbb{P}$-name $\tau$.[/step]

[guided]Fix a subset $G \subseteq \mathbb{P}$. The point of the rank condition on names is that a name is built only from earlier names. More precisely, define $\operatorname{rank}_{\mathbb{P}}(\tau)$ to be the least ordinal strictly larger than $\operatorname{rank}_{\mathbb{P}}(\sigma)$ whenever $(\sigma,p)\in\tau$ for some $p\in\mathbb{P}$. Why does this ordinal exist? Since $\tau$ is a set, Replacement gives the set of predecessor ranks \begin{align*} R_\tau:=\{\operatorname{rank}_{\mathbb{P}}(\sigma): \exists p\in\mathbb{P} \text{ such that }(\sigma,p)\in\tau\}. \end{align*} The ordinal $\sup R_\tau+1$ is strictly larger than every ordinal in $R_\tau$, and the least such ordinal exists by well-ordering of the ordinals. Therefore, if we are evaluating $\tau$, every name $\sigma$ appearing inside $\tau$ has already appeared at a strictly lower rank. We now use class recursion on the well-founded class relation $\sigma\prec\tau$, where $\sigma\prec\tau$ means that $(\sigma,p)\in\tau$ for some $p\in\mathbb{P}$. This relation is well-founded because every predecessor has strictly smaller name-rank: \begin{align*} \sigma\prec\tau \implies \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau). \end{align*} The recursion is a class recursion, not a set recursion, because the collection of all $\mathbb{P}$-names is generally a proper class. The recursion principle permits us to define a class assignment $E_G$ as long as the value at $\tau$ depends only on values at predecessors of $\tau$. Assume inductively that $E_G(\sigma)$ has already been defined for every $\mathbb{P}$-name $\sigma$ satisfying \begin{align*} \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau). \end{align*} Now take any pair $(\sigma,p)\in\tau$. Since $\tau$ is a $\mathbb{P}$-name, $\sigma$ is itself a $\mathbb{P}$-name and $p\in\mathbb{P}$. By the definition of name-rank, \begin{align*} \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau). \end{align*} Hence $E_G(\sigma)$ is already available before we define $E_G(\tau)$. We now define \begin{align*} E_G(\tau):=\{E_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\}. \end{align*} This formula defines a set, not merely a class. Indeed, $\tau$ is a set, so Separation gives the set \begin{align*} A_\tau:=\{(\sigma,p)\in\tau:p\in G\}. \end{align*} On this set, the assignment $(\sigma,p)\mapsto E_G(\sigma)$ is already defined by the induction hypothesis. Replacement therefore gives the image set \begin{align*} \{E_G(\sigma):(\sigma,p)\in A_\tau\}, \end{align*} which is exactly \begin{align*} \{E_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\}. \end{align*} Thus the recursive clause produces a legitimate set at the stage $\tau$.[/guided]

[step:Verify that the recursively constructed map satisfies the displayed formula] By construction, for every $\mathbb{P}$-name $\tau$, \begin{align*} E_G(\tau)=\{E_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\}. \end{align*} Writing $\tau_G:=E_G(\tau)$ for each $\tau\in V^{\mathbb{P}}$, this becomes \begin{align*} \tau_G=\{\sigma_G:\exists p\in G \text{ such that }(\sigma,p)\in\tau\}. \end{align*} Therefore every $\mathbb{P}$-name has a value determined by the displayed recursion. [/step]

[step:Prove uniqueness by induction on name-rank] Let $F_G$ be another class assignment with domain $V^{\mathbb{P}}$ satisfying the same recursion: \begin{align*} F_G(\tau)=\{F_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\} \end{align*} for every $\mathbb{P}$-name $\tau$. We prove by induction on $\operatorname{rank}_{\mathbb{P}}(\tau)$ that $F_G(\tau)=E_G(\tau)$. Suppose $F_G(\sigma)=E_G(\sigma)$ for every $\mathbb{P}$-name $\sigma$ with \begin{align*} \operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau). \end{align*} If $(\sigma,p)\in\tau$, then $\operatorname{rank}_{\mathbb{P}}(\sigma)<\operatorname{rank}_{\mathbb{P}}(\tau)$, so the induction hypothesis gives $F_G(\sigma)=E_G(\sigma)$. Hence \begin{align*} F_G(\tau)=\{F_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\} \end{align*} and \begin{align*} E_G(\tau)=\{E_G(\sigma): \exists p\in G \text{ such that }(\sigma,p)\in\tau\} \end{align*} are the same set. Therefore $F_G(\tau)=E_G(\tau)$. By induction, $F_G=E_G$ on all of $V^{\mathbb{P}}$. Thus the evaluation recursion has a unique solution, and each value $\tau_G$ exists and is uniquely determined. [/step]