= Solution
A <forcing name> is built by well-founded recursion: every member of a $\mathbb P$-name $\tau$ is a pair $(\sigma,p)$ with $p\in\mathbb P$ and $\sigma$ itself a $\mathbb P$-name of lower <forcing name rank>. In the ground model all such <forcing names> form the recursively defined class $V^{\mathbb P}$. Its evaluation by a <generic filter> is
$$
\boxed{\tau_G=\{\sigma_G:\exists p\in G\ ((\sigma,p)\in\tau)\}.}
$$
The recursion is on <forcing name rank>, not on the <forcing> order. <Forcing names> need not have a unique evaluation across different generics, and many different <forcing names> may have the same evaluation.
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