TITLE: If two posets have same dense open sets, are they equivalent as notions of forcing?
QUESTION [5 upvotes]: Suppose that $\mathbb{P}_0=(P,\leq_0)$ and $\mathbb{P}_1=(P,\leq_1)$ are partial orderings (in the weak sense, i.e., reflexive and transitive relations) on the same underlying set $P$, and such that $\leq_0$ is stronger than $\leq_1$, i.e., $p\leq_0q$ implies $p\leq_1 q$. Note that this implies that any $\leq_0$-dense subset $D$ of $P$ is also $\leq_1$-dense.
Suppose, moreover, that whenever $D$ is a $\leq_1$-dense open subset of $P$, it is also $\leq_0$-dense. Note that then, any $M$-generic filter $G$ for $\mathbb{P}_0$ generates an $M$-generic filter for $\mathbb{P}_1$.
My question is, are $\mathbb{P}_0$ and $\mathbb{P}_1$ equivalent as notions of forcing? The key issue seems to be that incompatibility need not be preserved in the passage from $\leq_0$ to $\leq_1$.

REPLY [4 votes]: No - let $\mathbb{P}=2^{<\omega}$ with the usual ordering, $\mathbb{P}_0=\mathbb{P}\setminus\{\emptyset\}$, and $\mathbb{Q}$ be the lottery sum of $\mathbb{P}_0$ and the one-point forcing. Then - identifying the "dull point" of $\mathbb{Q}$ with the empty string - $\mathbb{Q}$ and $\mathbb{P}$ have the same underlying set, and the same dense open subsets$^*$, but are clearly not equivalent. 
I suspect this type of counterexample can be modified to get something less trivial.

$^*$Specifically, if $D\subseteq\mathbb{P}$ is dense and open, then $\emptyset\in D$ and $D\cap \mathbb{P}_0$ is dense open in $\mathbb{P}_0$. Meanwhile, a dense open in $\mathbb{Q}$ is precisely the "dull point" together with a dense open subset of $\mathbb{P}_0$.