Proposition: No deterministic decoder of the witness pair

No fixed quantum instrument and classical decoder perfectly distinguish $\tau (0)$ from $\tau (c_{\epsilon })$ in one execution.

Proof. Let $E$ be the total effect of outcomes decoded as the second input. Perfect decoding would require

Positivity implies $E|e(0)⟩=0$ and $E|e(c_{\epsilon })⟩=|e(c_{\epsilon })⟩$. Hermiticity would make these vectors orthogonal. Their components in (eq:encoding) are strictly positive, so their inner product is positive.