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If $H=G_0\times G_1$ is generic for product forcing over $M$, then $G_1$ is generic over $M[G_0]$. For a name $\dot{D}$ for a dense subset of the second factor, take $p_0\in G_0$ forcing density. Ground-model pairs with first coordinate incompatible with $p_0$, or with first coordinate below $p_0$ forcing the second into $\dot{D}$, form a dense product set. The existential clause of syntactic forcing and the membership clause for a canonical forcing name give the density witnesses. The product filter meets this set, cannot use the incompatible case, and hence its second projection meets ${\dot{D}}^{G_0}$.

Use standard notation for forcing: $q\le p$ means that $q$ is stronger. A generic filter is nonempty, upward closed and downward directed, and meets every dense subset of a forcing order in the ground model. Names below are forcing names in $M$, with pairs ordered as $(\text{name},\text{condition})$.

The semantic forcing relation isHere ${\tau }^K=\mathrm{val}(\tau ,K)$ is the evaluation of a forcing name. Countability of $M$ supplies such generic filters through every condition by the Rasiowa–Sikorski lemma.

Define the syntactic forcing relation by mutual well-founded recursion on the forcing names for the atomic cases, then induction on the first-order formula. Its membership clause isFor equality, first abbreviateand set $p{⊩}^{\ast }\sigma =\tau$ exactly when both inclusions hold. Each recursive call lowers the rank of at least one forcing name without raising the other. This makes the mutual recursion well-founded.

For a basis of connectives consisting of logical conjunction, negation and existential quantification, the remaining clauses areThe last line is the existential clause of syntactic forcing; witnesses need only occur densely, rather than be forced by $p$ itself with one preselected name. Other connectives are defined by logical abbreviations. The recursion gives a definable relation inside $M$ for each fixed first-order formula, with quantification over the class of its names. If $\mathbb{P}$ has no greatest element, use canonical forcing names $\check{a}=\{(\check{b},p):b\in a,p\in \mathbb{P}\}$, which still evaluate to $a$. It also proves monotonicity: strengthening a condition preserves what it forces. The forcing theorem identifies the two relations and supplies the truth lemma.

We prove mutual genericity for product forcing. Let $D\in M[G_0]$ be a dense subset of $\mathbb{Q}$, and take a forcing name $\dot{D}\in M$ evaluating to $D$. By the forcing theorem, some $p_0\in G_0$ forces that $\dot{D}$ is a dense subset of the canonical forcing name for $\mathbb{Q}$.

In $M$ defineThis set is dense in the product forcing order. Given $(p,q)$, the incompatible case is immediate. Otherwise first strengthen $p$ below $p_0$. The forced density assertion and the existential clause of syntactic forcing supply a further $r\le p,p_0$ and a ground-model $q^{\prime }\le q$ with $r⊩{\check{q}}^{\prime }\in \dot{D}$. To justify choosing a ground-model $q^{\prime }$, a name forced to lie in $\check{\mathbb{Q}}$ can densely be made equal to some ${\check{q}}^{\prime }$ by the atomic membership clause; strengthen to that equality and use the forced order comparison. Thus $(r,q^{\prime })\in E$ lies below $(p,q)$.

The generic filter $H$ meets $E$. It cannot meet the first part, because its first projection contains $p_0$ and is directed. Hence there is $(p,q)\in H$ with $p\in G_0$ and $p⊩\check{q}\in \dot{D}$. Soundness of the forcing theorem gives $q\in D$, and the projection gives $q\in G_1$. Since every such $D$ is met,This establishes the stronger property, rather than merely meeting dense ground-model subsets.

Use standard notation for forcing, so $q\le p$ means that $q$ is stronger. Let $\check{p}$ be the canonical forcing name for the ground-model condition $p$. DefineHere $q\perp p$ means that they are incompatible forcing conditions. The checks and their collection are formed by recursion and Axiom schema of replacement in $M$, and the displayed set is selected by axiom schema of separation, so this is a forcing name in $M$.

By evaluation of a forcing name,If $p\in G$, directedness of the generic filter gives a common stronger condition for $p$ and each $q\in G$, so $p$ is absent from this value.

Conversely, for fixed $p\in \mathbb{P}$ the setis a dense subset of a forcing order belonging to $M$. A condition incompatible with $p$ is already in it, and one compatible with $p$ has a common extension below $p$. Genericity supplies $q\in G\cap D_p$. If $p\notin G$, upward closure of $G$ rules out $q\le p$, hence $q\perp p$. ThereforeThis forcing name for the complement of a generic filter works without a separativity assumption on the order.