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For forcing names $\sigma ,\tau$ and a generic filter $G$, ${\sigma }^G\in {\tau }^G$ exactly when some $p\in G$ syntactically forces $\sigma \in \tau$. Assuming the equality truth lemma, interpreted membership supplies an active pair $(\rho ,r)\in \tau$ and a condition in $G$ forcing $\sigma =\rho$; a common strengthening forces membership. Conversely the defining equality witnesses occur dense below a forcing condition forcing membership. The dense-below generic meeting lemma supplies one in $G$, and the equality truth lemma recovers interpreted membership.

If a forcing is $\kappa$-closed in the ground model, then it adds no functions from a ground ordinal $\alpha <\kappa$ into a ground set $B$. Recursively decide each value inside the ground model and take common stronger bounds at limit stages and after the final step. Conditions deciding a whole ground function are dense below any condition asserting this type of function. The dense-below generic meeting lemma ensures the generic filter meets that dense subset of a forcing order. One arbitrary bound need not be in the filter in an ordered set.

If a ground-model order is $\lambda$-closed, it adds no ordinal-valued functions with domain $\mu <\lambda$. Recursively decide each name value inside the ground model, using closure at intermediate limit stages and once more at the end. Below each starting condition, conditions deciding the entire function are dense. A generic filter meets them by the dense-below generic meeting lemma. The recursion must be internal to the ground model, since its closure property only covers its own sequences.

Closure preserves the ground-model cardinal $\lambda$. We use the stronger fact that closed forcing adds no short ordinal sequences. Fix an ordinal $\mu <\lambda$ in $M$ and a forcing name $\dot{h}$ that a condition $p$ forces to be an ordinal-valued function with domain $\mu$.

Inside $M$, recursively strengthen $p$ to decide each value $\dot{h}(\xi )$, for $\xi <\mu$. At a limit stage take a lower bound of the previously chosen descending sequence; its length is below $\lambda$, so $\lambda$-closure supplies one. After all $\mu$ values have been decided, closure supplies a common lower bound $q$. The decided values form a function $h\in M$ by the Axiom schema of replacement, andThe recursion is carried out within $M$ using the definable forcing theorem and a ground-model choice of deciding extensions, so every sequence to which closure is applied belongs to $M$. Repeating the construction below any stronger condition shows that such full-decision conditions are dense below a forcing condition $p$. The dense-below generic meeting lemma ensures that a generic filter containing $p$ meets them. Thus the interpreted function is in $M$.

If $\lambda$ ceased to be a cardinal number in $M[G]$, there would be a surjection $\mu \to \lambda$ for some ordinal $\mu <\lambda$. The forcing theorem gives a name and a condition in $G$ forcing this. The argument above makes its interpretation a ground-model function, but $M$ regards $\lambda$ as a cardinal number and admits no such surjection. ThereforeNo regularity of $\lambda$ is required; every recursion used has length strictly below $\lambda$. Finite cardinals are preserved automatically. This is cardinal preservation by closed forcing.

Inside $M$, the finite-function order has the countable chain condition for finite-function forcing. In an uncountable family of binary conditions there are uncountably many distinct finite domains, since each domain supports only finitely many conditions. These domains would have an uncountable Delta-system subfamily by the Delta-system lemma. There are only finitely many assignments to the common finite root, so thin further to an uncountable family agreeing there. Any two such conditions have a union that is again a function and is a common stronger extension. Thus no uncountable antichain in a forcing order exists.

Here is the relevant preservation argument, rather than an appeal to the condition alone. Fix $p\in G$ forcing that a forcing name $\dot{h}$ is a function $\omega \to {\alpha }_1$. For each $n$, choose inside $M$ a maximal antichain in a forcing order below $p$ deciding $\dot{h}(n)$. By the countable chain condition for forcing, this antichain is countable in $M$, so the possible values form a countable $B_n\subseteq {\alpha }_1$. The union $B={\bigcup }_n{B}_n$ is countable in $M$, hence bounded below its regular ${\omega }_1={\alpha }_1$. The interpreted $h$ has range contained in $B$, by maximality and the dense-below generic meeting lemma. It therefore cannot be a surjection onto ${\alpha }_1$.

If ${\alpha }_1$ were not a cardinal number in $M[G]$, it would be equinumerous with a smaller ordinal, which was countable already in $M$; composing with that ground-model enumeration would give a surjection $\omega \to {\alpha }_1$. This is impossible. ThereforeThis is the possible-values lemma for chain-condition forcing specialized to ${\omega }_1$. Countability and regularity in this proof are computed inside $M$, not inferred from the external countability of $M$.

Conversely, suppose $p\in G$ and $p{⊩}^{\ast }\sigma \in \tau$. The syntactic forcing relation says thatis dense below a forcing condition $p$. This is a set in $M$ by definability of forcing and axiom schema of separation. The dense-below generic meeting lemma gives $s\in G\cap D$: augment $D$ by all incompatible forcing conditions with $p$ to obtain a globally dense ground-model set, and use $p\in G$ to exclude the incompatible part.

Choose the witnessing $(\rho ,r)\in \tau$. Since $s\le r$ and $s\in G$, upward closure gives $r\in G$. The assumed equality truth lemma gives ${\sigma }^G={\rho }^G$, while $r\in G$ gives ${\rho }^G\in {\tau }^G$. Combining both directions,This completes the atomic membership truth lemma for forcing; no separate assumption of the membership truth lemma was made.