Theorem: Schedule-past factorization

The terminal channel satisfies

Its output reduction depends only on the initial reduction to $N_0(Y)$. Joint output with an untouched reference $E$ depends only on the initial reduction to $N_0(Y)E$. Every scheduled ancestor maps into the terminal metric past.

Proof. On a block disjoint from the current backward support, the observable is the identity. A complete channel adjoint preserves that identity. Other blocks enlarge support only within their union. Backward induction proves the inclusion.

Equality of all pulled-back expectation values proves reduced state dependence; appending the reference identity proves the joint statement. Concatenation of causally embedded edges proves metric containment.