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Theorem 1 prime: expanded entropy reformulation

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QAlgorithms.Papers.Independent.Intakethird100.P97.expanded_entropy

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

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Let ρABCD\rho_{ABCD}ρABCD​ be a positive semidefinite complex density matrix of trace one on four finite-dimensional tensor factors A,B,C,DA,B,C,DA,B,C,D, each of positive dimension (dimension one is allowed). Every S(X)S(X)S(X) is the base-two von Neumann entropy of the actual XXX marginal of this same state, with 0log⁡20=00\log_2 0=00log2​0=0. For disjoint groups write S(X∣Y)=S(XY)−S(Y)S(X\mid Y)=S(XY)-S(Y)S(X∣Y)=S(XY)−S(Y), I(X;Y)=S(X)+S(Y)−S(XY)I(X;Y)=S(X)+S(Y)-S(XY)I(X;Y)=S(X)+S(Y)−S(XY), and I(X;Z∣Y)=S(XY)+S(YZ)−S(XYZ)−S(Y)I(X;Z\mid Y)=S(XY)+S(YZ)-S(XYZ)-S(Y)I(X;Z∣Y)=S(XY)+S(YZ)−S(XYZ)−S(Y). Assume

S(AB)+S(BC)−S(B)−S(ABC)=0, S(CA)+S(AB)−S(A)−S(CAB)=0, S(AD)+S(DB)−S(D)−S(ADB)=0.\begin{aligned} S(AB)+S(BC)-S(B)-S(ABC)&=0,\ S(CA)+S(AB)-S(A)-S(CAB)&=0,\ S(AD)+S(DB)-S(D)-S(ADB)&=0. \end{aligned}S(AB)+S(BC)−S(B)−S(ABC)​=0, S(CA)+S(AB)−S(A)−S(CAB)​=0, S(AD)+S(DB)−S(D)−S(ADB)​=0.​

Then

S(ABC)+S(D)≥S(AB)+S(CD),I(ABC;D)≥I(AB;CD).S(ABC)+S(D)\ge S(AB)+S(CD),\qquad I(ABC;D)\ge I(AB;CD).S(ABC)+S(D)≥S(AB)+S(CD),I(ABC;D)≥I(AB;CD).

The two conclusions express the same entropy inequality after cancellation. All reordered subsystem labels denote the same tensor factors.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P97
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P97

attribute [local instance] Classical.propDecidable Classical.decEq

local notation "A" => ({0} : Finset (Fin 4))

local notation "B" => ({1} : Finset (Fin 4))

local notation "C" => ({2} : Finset (Fin 4))

local notation "D" => ({3} : Finset (Fin 4))

/-- PDF p.6, Theorem 1 prime: expanded entropy and mutual-information forms.
Subsystem permutations are unions of the same factors, and no purity is assumed. -/
theorem expanded_entropy (q : FourPartyState)
    (hABC : fourPartyEntropyBits q (A ∪ B) + fourPartyEntropyBits q (B ∪ C) - fourPartyEntropyBits q B - fourPartyEntropyBits q (A ∪ B ∪ C) = 0)
    (hCAB : fourPartyEntropyBits q (C ∪ A) + fourPartyEntropyBits q (A ∪ B) - fourPartyEntropyBits q A - fourPartyEntropyBits q (C ∪ A ∪ B) = 0)
    (hADB : fourPartyEntropyBits q (A ∪ D) + fourPartyEntropyBits q (D ∪ B) - fourPartyEntropyBits q D - fourPartyEntropyBits q (A ∪ D ∪ B) = 0) :
    fourPartyEntropyBits q (A ∪ B) + fourPartyEntropyBits q (C ∪ D) ≤ fourPartyEntropyBits q (A ∪ B ∪ C) + fourPartyEntropyBits q D ∧
    mutualValue (fourPartyEntropyBits q) (A ∪ B) (C ∪ D) ≤ mutualValue (fourPartyEntropyBits q) (A ∪ B ∪ C) D := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P97

end
Source
A new inequality for the von Neumann entropy, arXiv:quant-ph/0406162v1, PDF p. 6, Theorem 1 prime
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let qqq be any four-party state: positive natural-number dimensions did_idi​ for i∈Ω={0,1,2,3}i\in\Omega=\{0,1,2,3\}i∈Ω={0,1,2,3}, allowing dimensions equal to 111, together with a positive semidefinite complex matrix ρ\rhoρ of trace 111 indexed by ∏i{0,…,di−1}\prod_i\{0,\ldots,d_i-1\}∏i​{0,…,di​−1}. For each subset sss, define Hs=−Re⁡tr⁡f(ρs)H_s=-\operatorname{Re}\operatorname{tr}f(\rho_s)Hs​=−Retrf(ρs​), where ρs\rho_sρs​ sums equal row and column coordinates outside sss, and f(u)=ulog⁡2uf(u)=u\log_2uf(u)=ulog2​u is applied by functional calculus with 0log⁡20=00\log_2 0=00log2​0=0. The symbols A,B,C,DA,B,C,DA,B,C,D are the fixed subsets of Ω\OmegaΩ referenced in the declaration, whose elements are not specified in the supplied code; concatenated subscripts denote unions. Assuming all three equalities

HAB+HBC−HB−HABC=0,HAC+HAB−HA−HABC=0,HAD+HBD−HD−HABD=0,H_{AB}+H_{BC}-H_B-H_{ABC}=0,\qquad H_{AC}+H_{AB}-H_A-H_{ABC}=0,\qquad H_{AD}+H_{BD}-H_D-H_{ABD}=0,HAB​+HBC​−HB​−HABC​=0,HAC​+HAB​−HA​−HABC​=0,HAD​+HBD​−HD​−HABD​=0,

the theorem asserts the conjunction

HAB+HCD≤HABC+HDandHAB+HCD−HABCD≤HABC+HD−HABCD.H_{AB}+H_{CD}\le H_{ABC}+H_D \quad\text{and}\quad H_{AB}+H_{CD}-H_{ABCD} \le H_{ABC}+H_D-H_{ABCD}.HAB​+HCD​≤HABC​+HD​andHAB​+HCD​−HABCD​≤HABC​+HD​−HABCD​.
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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