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Information-set rate tends to symmetric private information

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QAlgorithms.Papers.Independent.Intakethird100.P10.wiretap_rate

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let x↦ρxBEx\mapsto\rho_x^{BE}x↦ρxBE​ be a binary cq wiretap channel with finite nonempty registers. Its Bob and Eve marginals are WWW and W∗W^*W∗, one CPTP map degrades WWW to W∗W^*W∗ for both letters, and the two Eve letter states commute. Fix 0<β<1/20<\beta<1/20<β<1/2. At each N=2nN=2^nN=2n, use the prescribed fidelity-defined sets and put AN=PN(W∗,β)∩GN(W,β)\mathcal A_N=\mathcal P_N(W^*,\beta)\cap\mathcal G_N(W,\beta)AN​=PN​(W∗,β)∩GN​(W,β). The polar construction places its information bits on these indices, so its information rate satisfies

lim⁡n→∞∣A2n∣2n=I(W)−I(W∗).\lim_{n\to\infty}\frac{|\mathcal A_{2^n}|}{2^n}=I(W)-I(W^*).n→∞lim​2n∣A2n​∣​=I(W)−I(W∗).

This is the rate of the specified index-set construction. The separate preshared-key consumption is not subtracted, and no unrestricted private-capacity optimization is asserted.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P10
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P10

open scoped BigOperators Matrix ComplexConjugate

open Filter

-- Accepted definitions, without paper-local aliases.

open QAlgorithms.Papers.Independent.Information100.P58 (basisDensity)

open QAlgorithms.Papers.Independent.Information100.P67 (squaredFidelity)

open QAlgorithms.Papers.Independent.Claimed50.P48 (indexedKrausDilation)

/-- PDF pp.5–6, Theorem 5: prescribed information-set rate along N=2^n.
This is |A|/N without subtracting the separate secret-key consumption. -/
theorem wiretap_rate {B E : Type*} [Fintype B] [DecidableEq B] [Nonempty B]
    [Fintype E] [DecidableEq E] [Nonempty E]
    (V : Wiretap B E) (beta : ℝ) (hpos : 0 < beta) (hbeta : beta < 1 / 2) :
    Tendsto (fun n => leafProbability (informationSet V n beta)) atTop
      (nhds (symmetricInformation V.bob - symmetricInformation V.eve)) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P10

end
Source
Polar codes for degradable quantum channels, arXiv:1109.5346v3, PDF p. 5, Theorem 5: quantum wiretap polar-code rate
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let B,EB,EB,E be finite, nonempty sets with decidable equality. Let VVV consist of two positive semidefinite trace-one matrices Ω0,Ω1\Omega_0,\Omega_1Ω0​,Ω1​ indexed by B×EB\times EB×E, together with a map DDD from matrices on BBB to matrices on EEE admitting a finite Kraus representation D(X)=∑ℓ=0m−1KℓXKℓ†D(X)=\sum_{\ell=0}^{m-1}K_\ell XK_\ell^\daggerD(X)=∑ℓ=0m−1​Kℓ​XKℓ†​, where ∑ℓKℓ†Kℓ=IB\sum_\ell K_\ell^\dagger K_\ell=I_B∑ℓ​Kℓ†​Kℓ​=IB​. Define Rx(b,b′)=∑eΩx((b,e),(b′,e))R_x(b,b')=\sum_e\Omega_x((b,e),(b',e))Rx​(b,b′)=∑e​Ωx​((b,e),(b′,e)) and Qx(e,e′)=∑bΩx((b,e),(b,e′))Q_x(e,e')=\sum_b\Omega_x((b,e),(b,e'))Qx​(e,e′)=∑b​Ωx​((b,e),(b,e′)), and assume Qx=D(Rx)Q_x=D(R_x)Qx​=D(Rx​) for both Boolean values xxx, and Q0Q1=Q1Q0Q_0Q_1=Q_1Q_0Q0​Q1​=Q1​Q0​. Let 0<β<120<\beta<\tfrac120<β<21​. For any binary matrix family W=(W0,W1)W=(W_0,W_1)W=(W0​,W1​), define its synthesized matrices as follows: put N=2nN=2^nN=2n, let dt(j)d_t(j)dt​(j) be the most-significant-first nnn-bit digits of j<Nj<Nj<N, set cjk=1c_{jk}=1cjk​=1 if every t<nt<nt<n with dt(k)=1d_t(k)=1dt​(k)=1 has dn−1−t(j)=1d_{n-1-t}(j)=1dn−1−t​(j)=1, and set cjk=0c_{jk}=0cjk​=0 otherwise; put vk(u)=∑j<Nujcjk(mod2)v_k(u)=\sum_{j<N}u_jc_{jk}\pmod2vk​(u)=∑j<N​uj​cjk​(mod2). On indices (h,a),(h′,b)(h,a),(h',b)(h,a),(h′,b), where h,h′h,h'h,h′ are Boolean words of length iii and a,ba,ba,b are length-NNN words in the matrix index set, set

Wn,i(x)((h,a),(h′,b))={2−(N−1)∑u∈{0,1}Nui=x,u<i=h∏k<NWvk(u)(ak,bk),h=h′,0,h≠h′.W_{n,i}^{(x)}((h,a),(h',b)) = \begin{cases} 2^{-(N-1)} \displaystyle\sum_{\substack{u\in\{0,1\}^N\\u_i=x,\;u_{<i}=h}} \prod_{k<N}W_{v_k(u)}(a_k,b_k),&h=h',\\ 0,&h\ne h'. \end{cases}Wn,i(x)​((h,a),(h′,b))=⎩⎨⎧​2−(N−1)u∈{0,1}Nui​=x,u<i​=h​∑​k<N∏​Wvk​(u)​(ak​,bk​),0,​h=h′,h=h′.​

Write Fr(X,Y)=Re⁡tr⁡X YXF_{\mathrm r}(X,Y)=\operatorname{Re}\operatorname{tr}\sqrt{\sqrt X\,Y\sqrt X}Fr​(X,Y)=RetrX​YX​​, using continuous functional calculus, and S(X)=−Re⁡tr⁡(f(X))S(X)=-\operatorname{Re}\operatorname{tr}(f(X))S(X)=−Retr(f(X)) for f(t)=tlog⁡2tf(t)=t\log_2tf(t)=tlog2​t. Put I(W)=S((W0+W1)/2)−(S(W0)+S(W1))/2I(W)=S((W_0+W_1)/2)-(S(W_0)+S(W_1))/2I(W)=S((W0​+W1​)/2)−(S(W0​)+S(W1​))/2. The assertion is

lim⁡n→∞#{i<N:1−2−Nβ<Fr(Qn,i(0),Qn,i(1)),Fr(Rn,i(0),Rn,i(1))<2−Nβ}N=I(R)−I(Q).\lim_{n\to\infty} \frac{\#\left\{i<N: \begin{array}{l} 1-2^{-N^\beta}<F_{\mathrm r}(Q_{n,i}^{(0)},Q_{n,i}^{(1)}),\\ F_{\mathrm r}(R_{n,i}^{(0)},R_{n,i}^{(1)})<2^{-N^\beta} \end{array}\right\}}{N} =I(R)-I(Q).n→∞lim​N#{i<N:1−2−Nβ<Fr​(Qn,i(0)​,Qn,i(1)​),Fr​(Rn,i(0)​,Rn,i(1)​)<2−Nβ​}​=I(R)−I(Q).

Both displayed membership inequalities are strict. The sequence includes n=0n=0n=0, where N=1N=1N=1, the bit condition is vacuous, and the transformed word is unchanged.

Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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