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If $G$ is a generic filter over a countable transitive model $M$, the generic extension $M[G]$ consists of the interpretations by $G$ of all forcing names in $M$.

LetFor each natural number $m$, conditions whose stem has length at least $m$ form a dense subset of a forcing order, so the generic filter meets all of them and $d\in {\omega }^{\omega }$.

Fix $h\in {\omega }^{\omega }\cap M$. The setis dense: from $(s,f)$ replace $f$ by $n\mapsto \max \{f(n),h(n)+1\}$. Choose $(s,f)\in G\cap D_h$. Every stronger condition must put each newly added stem value $d(n)$ above $f(n)$, sofor every $n\ge |s|$. Thus $d$ is a dominating real over $M$.

Solved by gpt-5.6-sol high.

With the convention that $p\le q$ means that $p$ is stronger, a set $G\subseteq \mathbb{P}$ is a generic filter over $M$ when it is a filter and meets every dense set $D\subseteq \mathbb{P}$ with $D\in M$. Explicitly, $G$ is upward closed toward weaker conditions, every two members have a common stronger member in $G$, and $G\cap D\ne \varnothing$ for every such $D$.

Solved by gpt-5.6-sol high.