Lemma: Exact payload transport

The successive adjacent exchanges transport the original unknown payload to $A_n$, return all preceding payload registers to blank, and preserve all payload–reference operators exactly.

Proof. One exchange swaps complete tensor factors and their correlations. Induction over the chain moves the original factor forward and leaves behind the next original blank. The resulting tensor-factor permutation gives the identity channel from the initial to the final payload interface, including arbitrary references.