# Is every Boolean algebra a separative partial order?

A partially ordered set $\langle P,\leq\rangle$ is separative iff it satisfies the following condition: $\neg x\leq y\Rightarrow\exists z(z\leq x\wedge z\bot y)$ where: $x\bot y\iff\neg\exists z(z\leq x\wedge z\leq y).$ If $\langle B,\leq\rangle$ is a complete Boolean lattice (every subset of $B$ has supremum), then $B^+=B\setminus\{0\}$ is separative.

My question is: what if $B$ is not complete? Is it separative or does not have to be separative?

Recall that $\lnot x\leq y$ is the same as saying that $x\cdot y\neq x$. Let $z=x\cdot\bar y$ (where $\bar y$ is the complement of $y$).
Then $z\leq x$, and of course $z\leq\bar y$. But also $z\cdot y=x\cdot\bar y\cdot y=x\cdot 0=0$. Therefore $z\perp y$ as wanted.