Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 / 5 / i / a / 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 isThe 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.