DAG adjacency is equivalent to having no separating set

If $i$ and $j$ are adjacent, the one-edge path between them has no internal vertices, so it is active given every $S\subseteq V\setminus \{i,j\}$ by [F1]. Hence no such $S$ d-separates them.

Suppose $i$ and $j$ are nonadjacent. Acyclicity implies that at least one of $i,j$ is not a descendant of the other; after interchanging labels, assume $j$ is not a descendant of $i$. Since $i,j$ are nonadjacent, $j$ is a non-descendant non-parent of $i$. Put $S={\mathrm{pa}}_G(i)$. Then $S$ excludes both endpoints. Consider any path $i=v_0,v_1,\dots ,v_r=j$. If its first arrow points into $i$, then $v_1\in S$ and the occurrence of $v_1$ is a noncollider, because the arrow $v_1\to i$ has its tail at $v_1$; the path is blocked. If the first arrow points out of $i$, follow arrows that point forward along the path until the first arrow that points backward. Such a reversal must occur because $j$ is not a descendant of $i$. At the reversal vertex $v_t$, both incident arrows point into $v_t$, so it is a collider and the initial segment is a directed path $i\to \cdots \to v_t$. If $v_t$ had a descendant $p\in S$, then $i\to \cdots \to v_t\to \cdots \to p\to i$ would be a directed cycle. Thus this collider has no descendant in $S$, and the path is blocked by [F1]. Every path is blocked, so $S$ d-separates $i$ and $j$.

The adjacent and nonadjacent cases prove both implications.