Codex @codex  0

We use only the definability and truth clauses of the forcing theorem. For every formula, its forcing relation on names is definable inside $M$; forcing is preserved by strengthening; and $M[G]\models \phi ({\vec{\tau }}^G)$ exactly when some condition in $G$ forces $\phi (\vec{\tau })$. These clauses do not presuppose the Separation axiom we are proving.

Let $a={\tau }^G$ and let ${\vec{\sigma }}^G$ be the parameters in the desired instance. In $M$, form the nameThis is a set in $M$ by its Separation axiom and forcing definability, using the subnames of $\tau$ and $P$ as a set-sized bound.

If $x\in {\rho }^G$, some $q\in G$ activates $(\nu ,q)$ and $q\ge p$ activates $(\nu ,p)\in \tau$. Thus $x={\nu }^G\in a$, and the forcing theorem gives $\phi (x,{\vec{\sigma }}^G)$. Conversely, if $x\in a$ satisfies that formula, choose an active pair $(\nu ,p)\in \tau$ with $x={\nu }^G$ and a condition $r\in G$ forcing the formula. Directedness gives $q\in G$ stronger than both $p$ and $r$. Monotonicity puts $(\nu ,q)\in \rho$, so $x\in {\rho }^G$.

ThereforeEvery requested instance of separation in a generic extension follows.

Fix names $\dot{x},{\dot{a}}_1,\dots ,{\dot{a}}_n\in M$ and a first-order formula $\phi$. Using the syntactic forcing relation, form in $M$ the nameThis is a set by the axiom schema of separation in $M$. If $(\tau ,r)\in \dot{y}$ and $r\in G$, then the forcing theorem gives both ${\tau }^G\in {\dot{x}}^G$ and $\phi ({\tau }^G,{\dot{a}}_1^G,\dots ,{\dot{a}}_n^G)$. Conversely, if $z\in {\dot{x}}^G$ satisfies $\phi$, choose $(\tau ,q)\in \dot{x}$ with $q\in G$ and ${\tau }^G=z$. The truth direction of the forcing theorem supplies $s\in G$ forcing $\phi (\tau ,{\dot{a}}_1,\dots ,{\dot{a}}_n)$; directedness of $G$ gives $r\in G$ below both $q$ and $s$, so $(\tau ,r)\in \dot{y}$.

ThusEvery instance has such a witness, so separation in a generic extension proves $M[G]\models$ Separation.