TITLE: Cantor-Bernstein for notions of forcing
QUESTION [13 upvotes]: For forcing notions $\mathbb{P}$ and $\mathbb{Q}$ let us write $\mathbb{P}\triangleleft\mathbb{Q}$ if forcing with $\mathbb{Q}$ always adds a $\mathbb{P}$-generic filter over $V$. In other words, $\mathbb{P}\triangleleft\mathbb{Q}$ holds if there is a $\mathbb{Q}$-name $\tau$ such that $\tau[H]$ is a $\mathbb{P}$-generic filter over $V$ whenever $H$ is a $\mathbb{Q}$-generic filter over $V$. If $\mathbb{P}\triangleleft\mathbb{Q}$ and $\mathbb{Q}\triangleleft\mathbb{P}$ need $\mathbb{P}$ and $\mathbb{Q}$ be forcing isomorphic, ie must they produce the same generic extensions? I expect not, but haven't been able to think of a counterexample.
I believe this question can be phrased in terms of Boolean algebras; ie if two complete Boolean algebras are each isomorphic to a complete Boolean subalgebra of the other, need they be forcing isomorphic?
(Incidentally, if the generics the forcings add are done in a 'reversible' way, ie if $\tau,\sigma$ are the $\mathbb{Q},\mathbb{P}$ names for the added generics and always $\tau[\sigma[G]]=G$ and $\sigma[\tau[H]]=H$ then the forcing notions will be equivalent. This is mentioned in the beginning of Shelah's text Proper Forcing).

REPLY [6 votes]: I realize I'm a bit late to the party, but I wanted to add that counterexamples can be easily constructed out of almost any type of forcing.  I'll give a particular example, but the method is flexible.  All we need are three forcings, $\mathbb{R}_1$, $\mathbb{R}_2$ and $\mathbb{R}_3$, pairwise incomparable by the $\lhd$ relation described in the question.  For concreteness, let's take $\mathbb{R}_1 = Add(\omega,1)$,   $\mathbb{R}_2 = Add(\omega_1,1)$ and  $\mathbb{R}_3 = Add(\omega_2,1)$.  
Let $\mathbb{P}$ be the lottery sum of $\mathbb{R}_1$ and  $\mathbb{R}_1 \times \mathbb{R}_2$ (so $\mathbb{P}$ first chooses generically whether to force with just $\mathbb{R}_1$  or with the product $\mathbb{R}_1  \times \mathbb{R}_2$, and then forces accordingly).  Similarly, let $\mathbb{Q}$ be the lottery sum of $\mathbb{R}_1$ and  $\mathbb{R}_1 \times \mathbb{R}_3$.
$\mathbb{P}$ and $\mathbb{Q}$ are clearly not forcing equivalent, since the first has the option of adding a generic for $\mathbb{R}_2$, which the second cannot (and vice versa, adding a generic to $\mathbb{R}_3$).  However, forcing with one will always add a generic for the other, since either one will add a generic for $\mathbb{R}_1$, which is one option in the lottery in each case.
This does not answer the second question, about Boolean algebras, and I think it demonstrates that the order relation on Boolean algebras given by "embedding into a subalgebra" is really distinct from the $\lhd$ order -- here we have two forcings satisfying $\mathbb{P} \lhd \mathbb{Q}$ and  $\mathbb{Q} \lhd \mathbb{P}$, but neither Boolean algebra embeds into a subalgebra of the other. 
This suggests a natural additional condition, namely that $H$ give the entire extension $V[G]$.  That is, in every extension $V[G]$ by  $\mathbb{P}$ there is $H$ generic for $\mathbb{Q}$ such that $V[G]=V[H]$.  I believe this guarantees forcing equivalence in the case where $\mathbb{P} \lhd \mathbb{Q}$ and  $\mathbb{Q} \lhd \mathbb{P}$, since every extension by $\mathbb{P}$ equals and extension by $\mathbb{Q}$ and vice versa.