Intermediate submodels without Boolean algebras

Question

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:

1. 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 $RO(\mathbb{R})$.
2. 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.)
3. 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.

Answer 1

The "mere poset" approach has something in it. Of course a BA is a partial case of a poset. However the cba-hull of a poset may lack properties which the poset has.

Take eg the Cohen forcing $\mathbb C$. It is countable while its CBA-hull is not. Now, by the Solovay $\Sigma$-method, if $a$ is Cohen-generic over $V$ and $b$ a real in $V[a]$ then $V[a]$ is a $\Sigma$-generic extension of $V[b]$, where $\Sigma\subseteq\mathbb C$, hence, either still a Cohen extension of $V[b]$ or else $V[b]=V[a]$ -- just because any subset $\Sigma$ of $\mathbb C$ is countable and hence is either trivial of Cohen itself. Try to prove this old fact on the cba base and you will see the difference.

Further, $\Sigma$ itself is a $\mathbb Q$ generic set over $V$, where $\mathbb Q$ is a strengthening of $\mathbb C$ (the same domain but a $\subseteq$-larger order relation). Once again this implies that $V[b]=V[\Sigma]$ is a Cohen extension of $V$ (or else $V[b]=V[\Sigma]=V$). Again try to prove this on the cba base.

Answer 2

One possible way to prove the theorem without using the Boolean valued techniques is to use Bokovsky's theorem given in Characterization of generic extensions of models of set theory. I may mention there are several proofs of this fact without refering to Boolean algebras, see for example Schindler's paper The long extender algebra or the paper On the set-generic multiverse by Friedman-Fuchino-Sakai.

Note that the theorem is an immediate corollary of Bukovsky's theorem.