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Equation (19): the six directed SSA excesses agree in three pairs

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QAlgorithms.Papers.Independent.Intakethird100.P63.eq19_pairedExcesses

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

prove2quantum

All states below are positive trace-class density operators of trace one on completed complex Hilbert tensor products, represented in countable orthonormal coordinates. Finite dimensional systems are included. Entropies use natural logarithms and the convention 0log⁡0=00\log 0=00log0=0. Every subsystem entropy occurring in an entropy difference is finite; this is the proof-supported finite-entropy reading of the source, and imposes no finite Hilbert dimension.

Let ρ\rhoρ be any tripartite state, and let (a,b,c)(a,b,c)(a,b,c) be any permutation of its three parties. Write I(A:B)=SA+SB−SA∪BI(A:B)=S_A+S_B-S_{A\cup B}I(A:B)=SA​+SB​−SA∪B​ for mutual information of disjoint clusters. Then

I(a:bc)ρ−I(a:b)ρ=I(ab:c)ρ−I(b:c)ρ.I(a:bc)_\rho-I(a:b)_\rho=I(ab:c)_\rho-I(b:c)_\rho.I(a:bc)ρ​−I(a:b)ρ​=I(ab:c)ρ​−I(b:c)ρ​.

This pairs the six directed SSA excesses into three equal pairs. Each displayed side expands to Sab+Sbc−Sb−SabcS_{ab}+S_{bc}-S_b-S_{abc}Sab​+Sbc​−Sb​−Sabc​; the statement does not equate all six excesses.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P63
Formal statement
noncomputable section

open scoped ENNReal BigOperators

namespace QAlgorithms.Papers.Independent.Intakethird100.P63

attribute [local instance] Classical.propDecidable Classical.decEq

-- Reuse the accepted nuclear trace-class, positive, trace-one predicate verbatim.

open QAlgorithms.Papers.Independent.Dependency50.P33 (IsDensity)

open QAlgorithms.Papers.Independent.Dependency50.P13 (finiteRelativeEntropy)

/-- Unnumbered (19), PDF pp.7–8: the six directed SSA excesses pair up under party
permutations. This states every permutation; it does not equate all six excesses. -/
theorem eq19_pairedExcesses (rho : (Matrix (Fin 3 → ℕ) (Fin 3 → ℕ) ℂ)) (hrho : IsDensity rho)
    (hfinite : FiniteSubsystemEntropies rho) (p : Equiv.Perm (Fin 3)) :
    clusterMutualInformation rho {p 0} {p 1, p 2} - clusterMutualInformation rho {p 0} {p 1} =
      clusterMutualInformation rho {p 0, p 1} {p 2} - clusterMutualInformation rho {p 1} {p 2} := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P63

end
Source
On Mutual Information in Multipartite Quantum States and Equality in Strong Subadditivity of Entropy, arXiv:quant-ph/0311193v1, PDF p. 8, Unnumbered Eq. (19): pairwise equality of the six SSA excesses
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let ρ\rhoρ be a complex kernel on N3\mathbb N^3N3 that admits an entrywise expansion ρ(x,y)=∑rur(x)vr(y)‾\rho(x,y)=\sum_r u_r(x)\overline{v_r(y)}ρ(x,y)=∑r​ur​(x)vr​(y)​ with square-summable vectors and ∑r∥ur∥∥vr∥<∞\sum_r\|u_r\|\|v_r\|<\infty∑r​∥ur​∥∥vr​∥<∞, whose every finite quadratic form is real and nonnegative, and whose real diagonal is summable with sum 111. For each C⊆{0,1,2}C\subseteq\{0,1,2\}C⊆{0,1,2}, define ρC(a,b)=∑z:C∁→Nρ(a⊔z,b⊔z)\rho_C(a,b)=\sum_{z:C^\complement\to\mathbb N}\rho(a\sqcup z,b\sqcup z)ρC​(a,b)=∑z:C∁→N​ρ(a⊔z,b⊔z), where a,b:C→Na,b:C\to\mathbb Na,b:C→N and a⊔za\sqcup za⊔z combines their coordinates. For any kernel YYY, define H(Y)∈[0,∞]H(Y)\in[0,\infty]H(Y)∈[0,∞] as the infimum of ∑r[−arlog⁡ar]+\sum_r[-a_r\log a_r]_+∑r​[−ar​logar​]+​ over all entrywise expansions Y=∑rarqrqr∗Y=\sum_r a_rq_rq_r^*Y=∑r​ar​qr​qr∗​ with nonnegative summable weights of sum 111 and square-summable vectors whose positive-weight members are unit and pairwise orthogonal; zero-weight vectors are unrestricted, [t]+=max⁡(t,0)[t]_+=\max(t,0)[t]+​=max(t,0), and an empty infimum is +∞+\infty+∞. Assume H(ρC)≠+∞H(\rho_C)\ne+\inftyH(ρC​)=+∞ for every CCC, including the empty set. Set S(C)=toReal⁡H(ρC)S(C)=\operatorname{toReal}H(\rho_C)S(C)=toRealH(ρC​) and J(A,B)=S(A)+S(B)−S(A∪B)J(A,B)=S(A)+S(B)-S(A\cup B)J(A,B)=S(A)+S(B)−S(A∪B), with the conversion convention toReal⁡(+∞)=0\operatorname{toReal}(+\infty)=0toReal(+∞)=0. For every permutation ppp of {0,1,2}\{0,1,2\}{0,1,2},

J({p(0)},{p(1),p(2)})−J({p(0)},{p(1)})=J({p(0),p(1)},{p(2)})−J({p(1)},{p(2)}).J(\{p(0)\},\{p(1),p(2)\})-J(\{p(0)\},\{p(1)\}) = J(\{p(0),p(1)\},\{p(2)\})-J(\{p(1)\},\{p(2)\}).J({p(0)},{p(1),p(2)})−J({p(0)},{p(1)})=J({p(0),p(1)},{p(2)})−J({p(1)},{p(2)}).
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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