Theorem: Exact complete regular benchmark

For every input operator $X$ on $C\otimes P\otimes Q_n$, the complete output of the benchmark is

where $D_a=⟨a{|}_C\otimes I_{P{Q}_n}$, and ${\omega }_{\mathrm{blank}}$ contains the final blank $Q_0,\dots ,Q_{n-1}$ and scratch $E=0$. The two matter output factors are at $A,B$, respectively. Equation (eq:benchmark-channel) holds with every untouched reference.

On the canonical clock preparation $|01⟩⟨01|$, the matter channel is exactly ${\mathrm{Ad}}_{\mathrm{SWAP}}$. It has

No input outside the physically certified backward region signals to the declared joint output.

Proof. Successive exchanges with blank chain registers move the entire $Q_n$ factor, with all its correlations, into $Q_0$. The other chain registers become blank. The clock identity gives $D_{a}X{D}_a^{†}$ on branch $a$. The nine-gate calculation gives (eq:event-code). The three CNOTs exchange the local matter factors and the record-controlled phase gives $V_a$. Outbound exchanges move the second factor into $Q_n$, again returning all intermediate qubits to their initial blanks. The record is already classical; copying its specified basis word creates precisely the repeated projectors in (eq:benchmark-channel), including their correlations with matter. No unknown quantum state is copied. This proves the formula on all operators and references.

For the witness, fix the clock to $01$, hold the initial $B$ qubit fixed, and replace the $A$ qubit by $|0⟩$ or $|1⟩$. The $B$-outputs are orthogonal, hence their trace distance is one. Theorem thm:physical gives the final no-signalling assertion.