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Observation 1: abort factorization

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QAlgorithms.Papers.Independent.Intakethird100.P22.abort_factorization

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).

Let Preal=rPabort+(1−r)PpassP^{\rm real}=rP^{\rm abort}+(1-r)P^{\rm pass}Preal=rPabort+(1−r)Ppass and Pideal=rPabort+(1−r)Pideal,passP^{\rm ideal}=rP^{\rm abort}+(1-r)P^{\rm ideal,pass}Pideal=rPabort+(1−r)Pideal,pass. Then

dNS(Preal,Pideal)=(1−r)dNS(Ppass,Pideal,pass)=k.d_{\rm NS}(P^{\rm real},P^{\rm ideal})=(1-r)d_{\rm NS}(P^{\rm pass},P^{\rm ideal,pass})=k.dNS​(Preal,Pideal)=(1−r)dNS​(Ppass,Pideal,pass)=k.

A common abort branch contributes no distinguishing advantage.

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 p.31, Observation 1, (90): equal abort branches factor out exactly. -/
theorem abort_factorization {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) :
    cdDistance (withAbort r.abortProbability r.abort r.pass)
      (withAbort r.abortProbability r.abort (idealCDKey r.pass)) = securityError r := 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, Observation 1 / equation (90)
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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 have abort probability α∈[0,1]\alpha\in[0,1]α∈[0,1] and pass and abort arrays P,QP,QP,Q. Each array is nonnegative, sums to 111 over a,b,t,ea,b,t,ea,b,t,e for every zzz, and has ∑eRz(a,b,t,e)\sum_eR_z(a,b,t,e)∑e​Rz​(a,b,t,e) independent of zzz. Put

Iz(a,b,t,e)=1{a=b}∑u,v∈KPz(u,v,t,e)∣K∣.I_z(a,b,t,e)=\mathbf1_{\{a=b\}}\frac{\sum_{u,v\in K}P_z(u,v,t,e)}{|K|}.Iz​(a,b,t,e)=1{a=b}​∣K∣∑u,v∈K​Pz​(u,v,t,e)​.

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

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 arbitrary nonnegative real probability distributions on ZZZ and {0,…,m−1}\{0,\ldots,m-1\}{0,…,m−1}, respectively. With C=K×K×TC=K\times K\times TC=K×K×T, the assertion is

d(αQ+(1−α)P, αQ+(1−α)I)=(1−α)d(P,I),d\bigl(\alpha Q+(1-\alpha)P,\ \alpha Q+(1-\alpha)I\bigr) =(1-\alpha)d(P,I),d(αQ+(1−α)P, αQ+(1−α)I)=(1−α)d(P,I),

whose right side is the defined security error of rrr. Both endpoints α=0\alpha=0α=0 and α=1\alpha=1α=1 are included.

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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