A partially ordered set $\mathfrak{A}$ with a least element $0$ has the disjunction property of Wallman if for every pair $(a,b)$ of elements of the poset, either $b\leq a$ or there exists an element $c\leq b$ such that $c\ne 0$ and $c$ has no nontrivial common predecessor with $a$ . That is, in the latter case, the only $x$ with $x\leq a$ and $x\leq c$ is $x=0$ .

For the case if the poset $\mathfrak{A}$ is a $\cap$ -semilattice disjunction property of Wallman is equivalent to every of the following three formulas:

1. $\forall a,b\in\mathfrak{A}:(\{c\in\mathfrak{A}|c\cap a\ne 0\} = \{c\in\mathfrak{A}|c\cap b\ne 0\} \Rightarrow a = b)$ ;
2. $\forall a,b\in\mathfrak{A}:(\{c\in\mathfrak{A}|c\cap a\ne 0\} \subseteq \{c\in\mathfrak{A}|c\cap b\ne 0\} \Rightarrow a \subseteq b)$ ;
3. $\forall a,b\in\mathfrak{A}:(a\subset b \Rightarrow \{c\in\mathfrak{A}|c\cap a\ne 0\} \subset \{c\in\mathfrak{A}|c\cap b\ne 0\})$ .

The proof of this equivalence can be found in this online article.