Lemma: Exact scheduled locality

The terminal reduced channel factors through the initial snapshot on $N_s(Y)$, and

If two runs have the same schedule, equal initial snapshots on $N_s(Y)$, and equal channels on the blocks entering this backward recursion, their terminal snapshots agree.

All statements are reference preserving when the corresponding joint initial reductions agree.

Proof. An output observable on $Y$ is the identity on blocks disjoint from $Y$. Their channel adjoints are unital, so they contribute only identities. The remaining adjoints act on the union of blocks meeting $Y$. Iterating proves the support inclusion and the dependence only on the indicated channels. Each backward layer adds distance at most one. Lemma lem:comparison gives factorization and the reference statement.