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Lemma 1: real versus intermediate distance equals correctness error

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QAlgorithms.Papers.Independent.Intakethird100.P22.intermediate_distance

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

prove2quantum

Let PSASBTE∣ZpassP^{\rm pass}_{S_A S_B T E\mid Z}PSA​SB​TE∣Zpass​ be a normalized classical-device law with a common finite nonempty key alphabet K\mathcal KK and finite nonempty transcript, Eve-output and Eve-input alphabets. The (SA,SB,T)(S_A,S_B,T)(SA​,SB​,T) marginal is independent of ZZZ. The ideal pass law replaces the keys by a uniform perfectly correlated pair, independent of the actual (T,E∣Z)(T,E\mid Z)(T,E∣Z) marginal. The intermediate law copies Alice's key into Bob's register and preserves (SA,T,E∣Z)(S_A,T,E\mid Z)(SA​,T,E∣Z). Let r∈[0,1]r\in[0,1]r∈[0,1] be the abort probability, with an identical normalized abort branch in all three systems. Write dNSd_{\rm NS}dNS​ for the operational NS distance, and s,c,ks,c,ks,c,k for the pass-weighted secrecy, disagreement and security errors, respectively. Thus s=(1−r)dNS(PSATE∣Zpass,PSATE∣Zideal,pass)s=(1-r)d_{\rm NS}(P_{S_A T E\mid Z}^{\rm pass},P_{S_A T E\mid Z}^{\rm ideal,pass})s=(1−r)dNS​(PSA​TE∣Zpass​,PSA​TE∣Zideal,pass​), c=(1−r)Pr⁡[SA≠SB∣pass]c=(1-r)\Pr[S_A\ne S_B\mid\mathrm{pass}]c=(1−r)Pr[SA​=SB​∣pass], and k=(1−r)dNS(Ppass,Pideal,pass)k=(1-r)d_{\rm NS}(P^{\rm pass},P^{\rm ideal,pass})k=(1−r)dNS​(Ppass,Pideal,pass).

For the real and intermediate full output systems,

dNS(Preal,Pint)=c=(1−r)Pr⁡[SA≠SB∣pass].d_{\rm NS}(P^{\rm real},P^{\rm int})=c=(1-r)\Pr[S_A\ne S_B\mid\mathrm{pass}].dNS​(Preal,Pint)=c=(1−r)Pr[SA​=SB​∣pass].

The intermediate system isolates correctness error while keeping Alice and Eve's actual marginal.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P22
Formal statement
noncomputable section

open scoped BigOperators

namespace QAlgorithms.Papers.Independent.Intakethird100.P22

attribute [local instance] Classical.propDecidable Classical.decEq

open QAlgorithms.Papers.Independent.Advance50.P47
  (Box IsCorrelation IsNoSignaling HasLocalModel correlationMixture prBox)

open QAlgorithms.Papers.Independent.Information100.P56 (IsLaw)

open QAlgorithms.Papers.Independent.Information100.P27 (shannon)

open QAlgorithms.Papers.Independent.Gap48.P25 (classicalMI)

/-- PDF pp.31–32, Lemma 1, (94),(104): NS distance to the constructed intermediate
system equals the weighted disagreement probability. -/
theorem intermediate_distance {K T E Z : Type}
    [Fintype K] [Fintype T] [Fintype E] [Fintype Z]
    [Nonempty K] [Nonempty T] [Nonempty E] [Nonempty Z]
    (r : KeyOutput K T E Z) (z : Z) :
    cdDistance (withAbort r.abortProbability r.abort r.pass)
      (withAbort r.abortProbability r.abort (intermediateKey r.pass)) = correctnessError r z := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P22

end
Source
Limitations on device independent key secure against non signaling adversary via the squashed non-locality, arXiv:1903.12154v7, PDF p. 31, Lemma 1 / equations (94), (104)
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What the Lean code literally says, in plain math · gpt-6.1-sol

For all finite nonempty types K,T,E,ZK,T,E,ZK,T,E,Z, let rrr consist of α∈[0,1]\alpha\in[0,1]α∈[0,1] and pass and abort arrays P,QP,QP,Q, both nonnegative and normalized over a,b,t,ea,b,t,ea,b,t,e for each zzz, with their marginals obtained by summing over eee independent of zzz. For every z0∈Zz_0\in Zz0​∈Z, define

Jz(a,b,t,e)={∑v∈KPz(a,v,t,e),a=b,0,a≠b.J_z(a,b,t,e)= \begin{cases} \displaystyle\sum_{v\in K}P_z(a,v,t,e),&a=b,\\ 0,&a\ne b. \end{cases}Jz​(a,b,t,e)=⎩⎨⎧​v∈K∑​Pz​(a,v,t,e),0,​a=b,a=b.​

For arrays R,SR,SR,S on Z×C×EZ\times C\times EZ×C×E, let

d(R,S)=sup⁡m≥1, u,v12∑j=0m−1∣∑c,z,eu(c,z)(Rz(c,e)−Sz(c,e))v(c,z,e,j)∣,d(R,S)= \sup_{\substack{m\ge1,\ u,v}} \frac12\sum_{j=0}^{m-1} \left|\sum_{c,z,e}u(c,z)(R_z(c,e)-S_z(c,e))v(c,z,e,j)\right|,d(R,S)=m≥1, u,v​sup​21​j=0∑m−1​​c,z,e∑​u(c,z)(Rz​(c,e)−Sz​(c,e))v(c,z,e,j)​,

where u(c,⋅)u(c,\cdot)u(c,⋅) and v(c,z,e,⋅)v(c,z,e,\cdot)v(c,z,e,⋅) are nonnegative real probability distributions on ZZZ and {0,…,m−1}\{0,\ldots,m-1\}{0,…,m−1}. With C=K×K×TC=K\times K\times TC=K×K×T, the assertion is

d(αQ+(1−α)P, αQ+(1−α)J)=(1−α)∑a,b∈Ka≠b∑t,ePz0(a,b,t,e).d\bigl(\alpha Q+(1-\alpha)P,\ \alpha Q+(1-\alpha)J\bigr) =(1-\alpha)\sum_{\substack{a,b\in K\\a\ne b}}\sum_{t,e}P_{z_0}(a,b,t,e).d(αQ+(1−α)P, αQ+(1−α)J)=(1−α)a,b∈Ka=b​∑​t,e∑​Pz0​​(a,b,t,e).

The right side is the correctness error at the arbitrarily supplied z0z_0z0​; the left side has no chosen input z0z_0z0​.

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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