Theorem: Uniform secret-bit obstruction

Every normalized output with a transcript-coherence certificate obeys

for every normalized comparison law $\sigma$.

Consequently, if $R\in \mathcal{C}$, the same bound holds for every $n\ge 1$ and every completed protocol on $R^{\otimes n}$ satisfying Specification spec:protocol. It holds for every initial purification retained by Eve, and $K_D(R)=0$ for this model.

Proof. Use a common dominating measure for the actual transcript law and the ideal comparison law. This retains their singular parts. Homogeneity preserves (eq:TC).

Set

The classical bit blocks give

The total trace deficit in the two matching blocks is $e$; hence $a_0+a_1\ge e$, so

Lemma lem:fidelity also gives

or

Cauchy–Schwarz and the arithmetic–geometric mean inequality simplify (eq:TC) to

If $e\ge 1/10$, (eq:eta2e) proves (eq:gap). If $0\le e\le 1/10$, then

Equation (eq:etaf) again proves (eq:gap).

For $R^{\otimes n}$, tensor closure and Lemma lem:Ccertificate supply the same certificate. For an arbitrary purification, an Eve-side channel recovers the canonical purification from it: invert its purifying isometry on its range and complete the map by fixed-state preparation on the orthogonal complement. This channel fixes the bits and public transcript, maps ideal keys to ideal keys, and contracts trace norm.

Finally, retaining the first bit locally from any nonempty ideal uniform key gives an ideal bit and cannot increase distance. A secure sequence therefore cannot have ${\ell }_n\ge 1$ for infinitely many $n$. Its lengths are eventually zero, and its rate is zero. The empty key attains zero rate.