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Loss-tolerant steering inequality with a public detection flag

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QAlgorithms.Papers.Independent.Intakethird100.P94.lossTolerantSteering_S1

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

prove2quantum

Let Bob's two binary qubit projective measurements have orthonormal bases {u1a}\{u_{1a}\}{u1a​} and {u2b}\{u_{2b}\}{u2b​}, with a,b∈{0,1}a,b\in\{0,1\}a,b∈{0,1}. Define q=−log⁡2max⁡a,b∣⟨u1a,u2b⟩∣2q=-\log_2\max_{a,b}|\langle u_{1a},u_{2b}\rangle|^2q=−log2​maxa,b​∣⟨u1a​,u2b​⟩∣2. Suppose the observed probabilities P(a,s,b∣i,j)P(a,s,b\mid i,j)P(a,s,b∣i,j) admit a single normalized local-hidden-state ensemble for all setting pairs: Alice's response can depend on iii and the hidden variable, and Bob measures the associated normalized qubit state in basis jjj. Here s=1s=1s=1 means Alice detects, and her outcome at s=0s=0s=0 is predetermined.

Assume Alice's detection probability is the same, ηA\eta_AηA​, at both settings. Let Q2=Pr⁡[A2eB2]Q_2=\Pr[A_2 e B_2]Q2​=Pr[A2​eB2​] use all outputs, including no-click assignments. Write Wi=ηAh(Qips)W_i=\eta_A h(Q_i^{\rm ps})Wi​=ηA​h(Qips​) when ηA>0\eta_A>0ηA​>0, where Qips=Pr⁡[AieBi∣s=1]Q_i^{\rm ps}=\Pr[A_i e B_i\mid s=1]Qips​=Pr[Ai​eBi​∣s=1]; set Wi=0W_i=0Wi​=0 at ηA=0\eta_A=0ηA​=0. The binary entropy hhh is in bits. Then

q≤W1+h(Q2)+(1−ηA).q\le W_1+h(Q_2)+(1-\eta_A).q≤W1​+h(Q2​)+(1−ηA​).

At zero efficiency, q≤1q\le1q≤1. Violation excludes a local-hidden-state description. This is the steering inequality corresponding algebraically to the nonpositive right-hand side of Eq. (8); no operational key-security implication is asserted by this item.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P94
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P94

open MeasureTheory

open scoped ComplexConjugate

/-- arXiv:1109.1435v3, PDF pp.3,6, (S1): loss-tolerant steering inequality.
The printed eta is eta_A from (S6)--(S9). The weighted form includes eta_A=0;
the additional zero branch spells out q<=1 without an undefined postselected error. -/
theorem lossTolerantSteering_S1 {Lambda : Type*} [MeasurableSpace Lambda]
    (U : BobBases) (L : HiddenModel Lambda U)
    (hCommonEfficiency : detectionProbability L false = detectionProbability L true) :
    projectiveOverlapIncompatibility U ≤ weightedDetectedEntropy L false +
      QAlgorithms.binEntropy2 (fullError L true) + (1 - detectionProbability L false) ∧
    (detectionProbability L false = 0 → projectiveOverlapIncompatibility U ≤ 1) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P94

end
Source
One-sided Device-Independent Quantum Key Distribution: Security, feasibility, and the connection with steering, arXiv:1109.1435v3, PDF p. 6, Eq. (S1), derived from (S2)–(S9)
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let B={0,1}\mathbb B=\{0,1\}B={0,1}, representing false and true. For every measurable space Λ\LambdaΛ, two unitary complex matrices U0,U1U_0,U_1U0​,U1​ indexed by B\mathbb BB, and a model LLL, assume LLL consists of a probability measure μ\muμ; positive semidefinite trace-one matrices ρλ\rho_\lambdaρλ​ with measurable entries; nonnegative responses ri(λ,a,s)r_i(\lambda,a,s)ri​(λ,a,s), measurable in λ\lambdaλ, satisfying ∑a,sri(λ,a,s)=1\sum_{a,s}r_i(\lambda,a,s)=1∑a,s​ri​(λ,a,s)=1 for every i,λi,\lambdai,λ; and bits cic_ici​ such that ri(λ,a,0)=0r_i(\lambda,a,0)=0ri​(λ,a,0)=0 whenever a≠cia\ne c_ia=ci​. Set Ej,b(x,y)=Uj(x,b)Uj(y,b)‾E_{j,b}(x,y)=U_j(x,b)\overline{U_j(y,b)}Ej,b​(x,y)=Uj​(x,b)Uj​(y,b)​, qj,b(λ)=Re⁡tr⁡(Ej,bρλ)q_{j,b}(\lambda)=\operatorname{Re}\operatorname{tr}(E_{j,b}\rho_\lambda)qj,b​(λ)=Retr(Ej,b​ρλ​), and pi(a,s,b)=∫Λri(λ,a,s)qi,b(λ) dμ(λ)p_i(a,s,b)=\int_\Lambda r_i(\lambda,a,s)q_{i,b}(\lambda)\,d\mu(\lambda)pi​(a,s,b)=∫Λ​ri​(λ,a,s)qi,b​(λ)dμ(λ), with all products ri(λ,a,s)qj,b(λ)r_i(\lambda,a,s)q_{j,b}(\lambda)ri​(λ,a,s)qj,b​(λ) required to be integrable. Define

ηi=∑a,bpi(a,1,b),ei=∑a≠bpi(a,1,b),Fi=∑a≠b∑spi(a,s,b),Wi={0,ηi=0,ηih2(ei/ηi),ηi≠0,\eta_i=\sum_{a,b}p_i(a,1,b),\qquad e_i=\sum_{a\ne b}p_i(a,1,b),\qquad F_i=\sum_{a\ne b}\sum_s p_i(a,s,b),\qquad W_i=\begin{cases}0,&\eta_i=0,\\ \eta_i h_2(e_i/\eta_i),&\eta_i\ne0,\end{cases}ηi​=a,b∑​pi​(a,1,b),ei​=a=b∑​pi​(a,1,b),Fi​=a=b∑​s∑​pi​(a,s,b),Wi​={0,ηi​h2​(ei​/ηi​),​ηi​=0,ηi​=0,​

where h2(p)=−[pln⁡p+(1−p)ln⁡(1−p)]/ln⁡2h_2(p)=-[p\ln p+(1-p)\ln(1-p)]/\ln2h2​(p)=−[plnp+(1−p)ln(1−p)]/ln2, with 0ln⁡0=00\ln0=00ln0=0, and put C=max⁡a,b∣∑xU0(x,a)‾U1(x,b)∣2C=\max_{a,b}|\sum_x\overline{U_0(x,a)}U_1(x,b)|^2C=maxa,b​∣∑x​U0​(x,a)​U1​(x,b)∣2. Under the additional hypothesis η0=η1\eta_0=\eta_1η0​=η1​, both conclusions hold:

−log⁡2C≤W0+h2(F1)+(1−η0)and(η0=0 ⟹ −log⁡2C≤1).-\log_2 C\le W_0+h_2(F_1)+(1-\eta_0) \qquad\text{and}\qquad \bigl(\eta_0=0\ \Longrightarrow\ -\log_2 C\le1\bigr).−log2​C≤W0​+h2​(F1​)+(1−η0​)and(η0​=0 ⟹ −log2​C≤1).

The equal-efficiency hypothesis permits both efficiencies to be zero.

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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