My question is motivated by the following well-known fact regarding intermediate submodels of generic extensions. I would like to know if it can be proven using posets without the need for Boolean algebras.

**Fact:** Let $\mathbb{R}$, $\mathbb{S}$ be partial orders in $V$. Let $J$ be $\mathbb{R}$-generic over $V$ and let $K \in V[J]$ be $\mathbb{S}$-generic over V. Then there is some partial order $\mathbb{T} \in V[K]$ and $H$ that is $\mathbb{T}$-generic over $V[K]$ such that $$V[K][H]=V[J].$$

**Remark:** As I understand it, a standard way to prove this can be sketched as follows:

- By Jech03-Lem15.4, there is some Boolean algebra $\mathbb{B}$ which is a complete subalgebra of $RO(\mathbb{R})$ and such that $V[K]=V[\mathbb{B} \cap J']$, where $J'$ is the obvious counterpart of $J$ in $RO(\mathbb{R})$.
- Let $\mathbb{T} = RO(\mathbb{R})/(\mathbb{B} \cap J')$. (There are a number of ways of doing this but they all come to the same thing. I'm more comfortable treating the Boolean algebras as posets.)
- Let $H = J'/(\mathbb{B} \cap J')$. Then it can be shown that we may define $J'$ from $K$ and $H$ and vice versa; so, $$V[J]=V[J']=V[K][H].$$

Alternatively instead of steps 2. and 3., we might define $\mathbb{T}$ as the $\mathbb{S}$-name, $RO(\mathbb{R})/\mathbb{B}$, and show that $\mathbb{S} \times \mathbb{T}$ is forcing equivalent (or even isomorphic) to $\mathbb{R}$.

**My question:** Can the fact be demonstrated without recourse to Boolean algebras; i.e., with posets alone? Clearly steps, (2) and (3) above (or something like them) will work, but step (1) seems problematic.

If we want to follow the plan above, we want to find some $\mathbb{P}$ which is completely embedded in $\mathbb{R}$ and such that $V[K]=V[\mathbb{P} \cap J]$; thus, avoiding a detour through Boolean algebras.

However, Kunen provides a counter-example in his new Kunen11-ExV.4.50. Since the book is relatively new, I'll write it in:

"**Exercise V.4.50** Assume in $M$ that $\mathbb{Q} = Fn(\omega,\kappa)$ with $\kappa > 2^\omega$, and let $\mathbb{H}$ be $\mathbb{Q}$-generic over $M$. Then there is a real number $A \in M[H]$ such that: $A$ is random over $M$; $A \subseteq I \in M$, where $I$ is the set of rationals; but there is no poset $\mathbb{P}$ such that $\mathbb{P} \subseteq_c \mathbb{Q}$ and $M[A]=M[H \cap \mathbb{P}]$."

This suggests that the path through 1. is blocked. Is there another way through?

In Kanamori, Proposition 10.10, we have the statement of similar fact. Kanamori seems to suggest that it can be established with partial orders.

**A More Precise Problem:** Can the fact above be established when V is a countable transitive model of $ZF-P$?

**Remark:** Without powerset we cannot guarantee the existence of $RO(\mathbb{R})$ so the obvious implementation of the strategy above is blocked.

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