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Proposition 3.5

Open
QAlgorithms.Papers.Independent.Intakethird100.P20.distributivity_iff_stone_properties

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

prove2quantum

For a bounded orthomodular lattice LLL, the following four conditions are equivalent: LLL is distributive; Qa∪Qb=Qa∨bQ_a\cup Q_b=Q_{a\vee b}Qa​∪Qb​=Qa∨b​ for all a,ba,ba,b; Qa∪Qa⊥=Q(L)Q_a\cup Q_{a^\perp}=Q(L)Qa​∪Qa⊥​=Q(L) for all aaa; and every clopen subset of Q(L)Q(L)Q(L) is a basic set. In particular,

L distributive⟺Clop⁡(Q(L))={Qa:a∈L}.L\text{ distributive}\quad\Longleftrightarrow\quad\operatorname{Clop}(Q(L))=\{Q_a:a\in L\}.L distributive⟺Clop(Q(L))={Qa​:a∈L}.

All four clauses are retained. Complementation of basic sets is not assumed in a general orthomodular lattice.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P20
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P20

open Set TopologicalSpace

open QAlgorithms.Papers.Independent.Next46.P33 (Projection)

open scoped Topology

universe u v

/-- Proposition 3.5; de Groote, Observables I, PDF p./pp. 44.
-/
theorem distributivity_iff_stone_properties {L : Type u} [Lattice L] [BoundedOrder L] (perp : L → L)
    (horth : IsOrthomodular perp) :
    (IsDistributive (L := L) ↔ ∀ a b : L, stoneBasic a ∪ stoneBasic b = stoneBasic (a ⊔ b)) ∧
    (IsDistributive (L := L) ↔ ∀ a : L, stoneBasic a ∪ stoneBasic (perp a) = Set.univ) ∧
    (IsDistributive (L := L) ↔ ∀ s : Set (StoneSpectrum L), IsClopen s → ∃ a : L, s = stoneBasic a) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P20

end
Source
Observables I: Stone Spectra, arXiv:math-ph/0509020v1, PDF p. 44, Proposition 3.5
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let LLL be a bounded lattice with a function a↦a⊥a\mapsto a^\perpa↦a⊥ satisfying a∧a⊥=⊥a\wedge a^\perp=\bota∧a⊥=⊥, a∨a⊥=⊤a\vee a^\perp=\topa∨a⊥=⊤, and (a⊥)⊥=a(a^\perp)^\perp=a(a⊥)⊥=a for every aaa; (a∧b)⊥=a⊥∨b⊥(a\wedge b)^\perp=a^\perp\vee b^\perp(a∧b)⊥=a⊥∨b⊥ and (a∨b)⊥=a⊥∧b⊥(a\vee b)^\perp=a^\perp\wedge b^\perp(a∨b)⊥=a⊥∧b⊥ for every a,ba,ba,b; and a∧(b∨c)=(a∧b)∨(a∧c)a\wedge(b\vee c)=(a\wedge b)\vee(a\wedge c)a∧(b∨c)=(a∧b)∨(a∧c) whenever b≤ab\le ab≤a and c≤a⊥c\le a^\perpc≤a⊥. Let Q(L)\mathcal Q(L)Q(L) consist of the inclusion-maximal nonempty subsets B⊆LB\subseteq LB⊆L excluding ⊥\bot⊥ such that any two members have a common lower bound in BBB. Put D(a)={B∈Q(L):a∈B}D(a)=\{B\in\mathcal Q(L):a\in B\}D(a)={B∈Q(L):a∈B} and give Q(L)\mathcal Q(L)Q(L) the topology generated by these sets. Say that LLL is distributive when both a∧(b∨c)=(a∧b)∨(a∧c)a\wedge(b\vee c)=(a\wedge b)\vee(a\wedge c)a∧(b∨c)=(a∧b)∨(a∧c) and a∨(b∧c)=(a∨b)∧(a∨c)a\vee(b\wedge c)=(a\vee b)\wedge(a\vee c)a∨(b∧c)=(a∨b)∧(a∨c) hold for every a,b,ca,b,ca,b,c. Then the theorem asserts three equivalences: LLL is distributive if and only if D(a)∪D(b)=D(a∨b)D(a)\cup D(b)=D(a\vee b)D(a)∪D(b)=D(a∨b) for every a,ba,ba,b; LLL is distributive if and only if D(a)∪D(a⊥)=Q(L)D(a)\cup D(a^\perp)=\mathcal Q(L)D(a)∪D(a⊥)=Q(L) for every aaa; and LLL is distributive if and only if every subset of Q(L)\mathcal Q(L)Q(L) that is both open and closed equals D(a)D(a)D(a) for some a∈La\in La∈L.

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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