Joint Forcing Extension Property

Definition 1: A class $\mathcal{K}$ of countable transitive models of $\text{ZF}$ has strong "joint forcing extension property" (JFEP) iff for all $M,N\in \mathcal{K}$ there are forcing notions $\langle\mathbb{P}, <_{\mathbb{P}}\rangle\in M,\langle\mathbb{Q}, <_{\mathbb{Q}}\rangle\in N$ and $\mathbb{P}$-generic filter $G$ on $M$ and $\mathbb{Q}$-generic filter $H$ on $N$ such that $M[G]=N[H]$.

Also we say $\mathcal{K}$ has "weak JFEP" if $M[G]\equiv N[H]$.

Definition 2: A expansion $T$ of $\text{ZF}$ has strong/weak JFEP if the collection of its countable transitive models has strong/weak JFEP.

Question: Does $\text{ZF}$ have strong/weak JFEP? If not, is there any expansion of $\text{ZF}$ with strong/weak JFEP? In the other words, can one add strong/weak JFEP to $\text{ZF}$ by adding some new axioms like large cardianl assumptions or axiom of constructibility?

First, I note that the strong JFEP is the same as the medium JFEP, since two transitive sets such as $M[G]$ and $N[H]$ are isomorphic if and only if they are equal.

Next, note that the strong property is just too much.

Theorem. No consistent extension of ZF, which allows a model to be a forcing extension by adding a Cohen real, has the strong JFEP or the medium JFEP.

Proof. For any countable transitive model $M$ of ZF, there are $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common forcing extension. One can build $c$ and $d$ so that each of them separately is $M$-generic, but the combination $c\oplus d$ codes a real that collapses all of $M$. This is explained in my answer to this question. QED

This argument can be generalized beyond the forcing to add a Cohen real to any forcing that allows that coding argument to go through, and I believe this might include all forcing.

Theorem. There is a theory $T$ extending ZFC with the strong JFEP.

Proof. Assume without loss that there are transitive models of ZFC. Thus, there is a minimal transitive model. Let $T$ be the complete theory of the minimal transitive model $L_\alpha\models ZFC$. Now, the point is that if $M$ and $N$ are models of $T$, then actually $M=N$, since both must have the form $L_\beta$ for some $\beta$ but it cannot be that $\beta\gt\alpha$ and so both are equal to the minimal transitive model itself. Since $M=N$, of course it follows that they have a common forcing extension. QED

This idea seems to generalize to many other theories.

Meanwhile, it is easy to get the weak JFEP.

Theorem. Every complete theory trivially has the weak JFEP.

Proof. If $T$ is complete and $M,N\models T$, then they are already elementarily equivalent and they are forcing extensions of themselves by trivial forcing. QED

• Thank you Joel. I will edit the triviality on Medium JFEP. My emphasis in this question is on searching for some expansion of $\text{ZF}$ with strong/weak JFEP. – user42090 Nov 25 '13 at 18:32

Note that it is consistent that $\sf ZFC$ doesn't have weak JFEP either. Suppose that there are exactly two different height of transitive models, that is some $\alpha<\beta$ such that $L_\alpha$ and $L_\beta$ are both countable transitive models of $\sf ZFC$.

In $L_\beta$ we have that $L_\alpha$ is a countable transitive model, so in $L_\beta$ there exists a transitive countable model of $\sf ZFC$, whereas in $L_\alpha$ there are no transitive models of $\sf ZFC$ at all. Since the models agree on $\omega$, they also agree on the theory they think is $\sf ZFC$.

Since we cannot add a countable transitive model of $L_\alpha$ by forcing, and we cannot destroy the fact that $L_\alpha\in L_\beta$ is a countable transitive model, it is impossible for any $L_\alpha[G]$ and $L_\beta[H]$ to be elementary equivalent.

Note that it is consistent that $\sf ZFC$ doesn't have transitive models at all, in which case it has strong JFEP trivially.