Proposition: Causal separability does not ensure an acyclic signalling graph

There is a causally separable bipartite process with both $A{R}_{\inf }B$ and $B{R}_{\inf }A$. Thus its aggregate pairwise signalling relation is not a strict partial order.

Proof. Choose $0<\lambda <1$. With probability $\lambda$, initialize a qubit in $|0⟩$, send it through $A$, then through $B$, and discard the final output. With probability $1-\lambda$, use the reversed circuit. The classical choice is independent of the laboratory settings. This is a convex mixture of definite-order processes, hence a causally separable process in the bipartite sense OCB.

At both laboratories use measure-and-prepare instruments:

Let $x,y$ be the settings at $A,B$, respectively. The resulting complete probability table is

Consequently,

Changing $x$ changes $B$'s unconditional record, and changing $y$ changes $A$'s unconditional record. Both influences are nonzero. The two edges describe alternative order branches; they do not constitute a physical closed loop within either branch.