Let $\mathfrak{A}$ is a poset. For $a, b \in \mathfrak{A}$ we will denote $a \curlyvee b$ if only if there is a non-least element $c$ such that $c \leqslant a \wedge c \leqslant b$.
I call a poset $\mathfrak{A}$ separable if and only if $\forall x \in \mathfrak{A}: \left( x \curlyvee a \Leftrightarrow x \curlyvee b \right) \Rightarrow a = b$.
Let $\mathfrak{A}$ is a family of posets indexed by some set $n$. We introduce a partial order (called product order) on $\prod \mathfrak{A} = \prod_{i \in n} \mathfrak{A}_i$ by the formula (for every $a, b \in \prod \mathfrak{A}$) $$ a \leqslant b \Leftrightarrow \forall i \in n : a_i \leqslant b_i . $$ It is easy to prove that $$ a \curlyvee b \Leftrightarrow \exists i \in \operatorname{dom}\mathfrak{A}: a_i \curlyvee b_i . $$ Let $\mathfrak{A}$ is an indexed family of separable posets. Can we infer that $\mathfrak{A}$ \prod\mathfrak{A}$ is a separable poset?
I think the answer is ``no'', because I have failed to prove it. But I don't know a counter-example.
