Lemma: Bell-projection pair-distance bound

Let $\mathcal{N}$ have a qubit input and complement $\mathcal{C}$. For any orthonormal input pair $|0⟩,|1⟩$, put

Every deterministic decoder satisfies

Proof. Write $\epsilon =d_{\diamond }(\mathcal{DN},{\mathrm{id}}_M)$. Dilate the decoder by an isometry $T:A\to M^{\prime }\otimes G$ and apply the combined isometry to the Bell input, obtaining a pure state $|\Omega {⟩}_{Q{M}^{\prime }GE}$. Its Bell overlap $f$ satisfies $f\ge 1-\epsilon$ by the effect bound.

If $f=0$, then $\epsilon =1$, which already proves the claim. Otherwise project $\Omega$ onto $|\Phi {⟩}_{Q{M}^{\prime }}$. After normalization this gives $|\Phi ⟩\otimes |\chi {⟩}_{GE}$, whose squared overlap with $\Omega$ is $f$. The pure-state distance and partial-trace contraction give

Local decoding leaves the original $QE$ reduction unchanged.

Pinching $Q$ in its input-pair basis yields

The triangle inequality makes the left side at least $\ell /2$, giving (eq:pairBound).