in $A$. So we must show that ${\bm{\langle} x ]} \cap C_y$ and ${\bm{\langle} x ]}$ are adjacent. As $D' \sim D$ and ${\bm{\langle} x ]} \sim {\bm{\langle} x ]}$ in $G$ we have that $$({\bm{\langle} x ]} \cap D') \sim ({\bm{\langle} x ]} \cap D) = {\bm{\langle} x ]}.$$ But as $y \not\in D'$ we have that $D' \leq C_y$, and so we also have $${\bm{\langle} x ]}\cap D' \leq {\bm{\langle} x ]} \cap C_y  \leq {\bm{\langle} x ]};$$ and so by we get ${\bm{\langle} x ]} \sim ({\bm{\langle} x ]} \cap C_y)$, as needed.

On the other hand, let $D$ and $D'$ be non-adjacent downsets of $J_L$. Then we must show that there is some arc $(y,x) \in J_L$ for $x,y \in (D {-}D')$ or $x,y \in (D' {-}D)$ that is not in $A$. Assume that all such arcs with $x,y \in (D {-}D')$ are in $A$. Then for each, we saw above that ${\bm{\langle} x ]} \sim ({\bm{\langle} x ]} \cap C_y)$. Fixing $x$ and taking the intersection over all $y \leq x$ in $D {-}D'$, we get $${\bm{\langle} x ]} = \bigcap {\bm{\langle} x ]} \sim \bigcap ({\bm{\langle} x ]} \cap C_y) = \bigcap ({\bm{\langle} x 