of adjacency in $G$. Assume that $D \sim D'$ and $E \sim E'$. We must show that $(D \cup E) \sim (D' \cup E')$ and $(D \cap E) \sim (D' \cap E')$. For the former, let $x \in D \cup E$ and $(x, y)$ be in ${{A^{\rm c}}}$. Then $x \in D$ or $E$, so as $D \sim D'$ and $E \sim E'$, we have that $y \in D'$ or $E'$. Thus $y \in D' \cup E'$. That $x \in D' \cup E'$ implies $y \in D \cup E$ is the same, so $(D \cup E) \sim (D' \cup E')$. The proof of the latter is similar.

Now, consider $G(C,A)$ where $C$ is a chain. All downsets are of the form ${\bm{\langle} c ]}$ for some $c \in C$, or $\emptyset=:{\bm{\langle} -1 ]}$. As we may assume that $A$ satisfies , two downsets ${\bm{\langle} x ]}$ and ${\bm{\langle} y ]}$, for $y \leq x$ are adjacent if and only if $(y+1, x)$ is an arc of $A$. The following is then clear, and is the starting point of our next characterisation of reflexive $\operatorname{DL}$-graphs.

\[fact:PI\] For a chain $C$ and a sub-digraph $A$, $G(C,A)$ satisfies , so is a proper 