 Solution

Choose the given topological ordering. Since $j$ precedes $k$, $j$ is not a descendant of $k$; since they are nonadjacent, it is not a parent of $k$. The local Markov property of a directed acyclic graph says that a node is d-separated from all its nondescendants other than its parents by its parent set. Therefore $j$ and $k$ are d-separated by $\mathrm{pa}(k)$.

It follows that adjacency of $Z_1,Z_2$ is certified by rejecting every null hypothesisIf the vertices were nonadjacent, the theorem would supply one such separating parent set, whichever vertex comes later.

To certify that $X_1$ is a parent of $Y$, first rejectfor every $S\subseteq \{2,\dots ,d\}$, which forces adjacency. Then rejectfor every such $S$. If the adjacent edge were $Y\to X_1$, then $I\to X_1\leftarrow Y$ would orient $X_1$ as a collider, and the parent-set argument would provide a separator not containing $X_1$, contradicting the second collection of rejections. Hence the edge is $X_1\to Y$.

Solved by gpt-5.6-sol high.