]} {-}{[ y \bm{\rangle}})
       =:T_x$$ where $T_x = {\bm{\langle} x ]} {-}\bigcup {[ y \bm{\rangle}}$ is contained in ${\bm{\langle} x ]} {-}D'$ as the union is over all $y \leq x$ that are in $D {-}D'$. Now taking the union over all $x \in D {-}D'$ we get that $$D = \bigcup {\bm{\langle} x ]} \sim \bigcup T_x \subseteq \bigcup ({\bm{\langle} x ]} {-}D') \subseteq (D {-}D').$$ By we get that $D \sim D {-}D'$.

Similarily we get that $D' \sim (D \cap D')$. But as $D \cap D' \leq D, D'$ we get from that $D \sim D'$, a contradiction.

Now, Lemma \[lem:extension1\] allows us to assume that $A$ satisfies . The uniqueness of $A$ then follows by observing that $G(J_L,A')$ would be different for any other sub-digraph $A'$ of $J_L$ satisfying : this is simple let $(y,x)$ be an arc of $A'$ but not $A$. Then the edge ${\bm{\langle} x ]} \sim {\bm{\langle} y ]}$ is in $G(J_L,A')$ but not in $G(J_L,A)$.

Corollary \[cor:char2\] is immediate from Theorem \[thm:char2\] and the following lemma.

\[lem:pographiscompat\] If $P$ is a poset and $A \subseteq P$, then $\cD(P)$ is compatible with $G = G(P,A)$.

We use Lemma \[lem:altGPA\] for the definition 