Lemma: Recursive recovery normal form

There is an ancilla-free unitary $U_h$ on $A_h$ such that

Every exact recovery channel on $A_h$ is therefore the same channel

whose Choi rank is $2^{n_h-1}$.

Proof. At $h=1$, use (eq:normalform). For the induction, apply the three child unitaries without yet measuring their syndrome factors. Each child outputs its intermediate logical qubit tensored with a maximally mixed syndrome independent of that qubit and every reference. The three intermediate qubits have the root accessible channel ${\mathcal{N}}_A(\rho )$. Apply $U$ to them. The root contributes two further maximally mixed syndrome qubits. All syndrome factors are independent, giving (eq:treenormal).

The faithful-syndrome lemma proves uniqueness. Partial trace over $n_h-1$ qubits has exactly $2^{n_h-1}$ independent Kraus maps, and unitary precomposition does not change this rank.