A complete three-vertex skeleton has several equivalent DAGs

Each graph has all three possible edges, so there is no triple with nonadjacent endpoints and in particular no unshielded collider. Their skeletons and unshielded colliders agree; [F2] proves Markov equivalence. Every distinct pair has a direct active path under every allowed conditioning set by [F1], so neither DAG has a d-separation between distinct vertices. Faithfulness in [F6] makes the oracle dependent on every pair and set.

In the two DAGs, the orientations of $A-B$, $A-C$, and $B-C$ are respectively $A\to B$ versus $B\to A$, $A\to C$ versus $C\to A$, and $B\to C$ versus $C\to B$. Thus each edge is oriented both ways in the equivalence class, and [F3] makes every edge undirected in the CPDAG.

The perfect oracle never reports independence for any pair by step 1.1, so the algorithm deletes no edge. With three vertices, after the level-$0$ tests the complete triangle permits level $1$; at level $1$ each pair has only one possible conditioning vertex at each endpoint, and the oracle again returns dependence by step 1.1. The stopping condition then fails because an endpoint has at most one neighbor other than the other endpoint, fewer than the required two. Thus the skeleton search terminates. There is no unshielded triple to orient, and none of R1–R3 can fire because each requires at least one directed edge; the finite orientation scan therefore terminates as well. PC returns the undirected triangle, which [F5] identifies with the CPDAG found in step 1.2. The diagram is only a rendering of these verified adjacencies and orientations.