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Lemma 3.1

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QAlgorithms.Papers.Independent.Intakethird100.P20.orthomodular_basic_sets_injective

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

prove2quantum

In an orthomodular lattice with its designated orthocomplement, basic Stone sets distinguish lattice elements:

Qa(L)=Qb(L)⟹a=b.Q_a(L)=Q_b(L)\quad\Longrightarrow\quad a=b.Qa​(L)=Qb​(L)⟹a=b.

No distributivity or completeness assumption is added; the one-element lattice is allowed.

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

/-- Lemma 3.1; de Groote, Observables I, PDF p./pp. 43.
-/
theorem orthomodular_basic_sets_injective {L : Type u} [Lattice L] [BoundedOrder L] (perp : L → L)
    (horth : IsOrthomodular perp) {a b : L} :
    stoneBasic a = stoneBasic b → a = b := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P20

end
Source
Observables I: Stone Spectra, arXiv:math-ph/0509020v1, PDF p. 43, Lemma 3.1
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let LLL be a bounded lattice and let a↦a⊥a\mapsto a^\perpa↦a⊥ be a function L→LL\to LL→L satisfying the following identities: for every xxx, x∧x⊥=⊥x\wedge x^\perp=\botx∧x⊥=⊥, x∨x⊥=⊤x\vee x^\perp=\topx∨x⊥=⊤, and (x⊥)⊥=x(x^\perp)^\perp=x(x⊥)⊥=x; for every x,yx,yx,y, (x∧y)⊥=x⊥∨y⊥(x\wedge y)^\perp=x^\perp\vee y^\perp(x∧y)⊥=x⊥∨y⊥ and (x∨y)⊥=x⊥∧y⊥(x\vee y)^\perp=x^\perp\wedge y^\perp(x∨y)⊥=x⊥∧y⊥; and for every x,y,zx,y,zx,y,z with y≤xy\le xy≤x and z≤x⊥z\le x^\perpz≤x⊥, x∧(y∨z)=(x∧y)∨(x∧z)x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)x∧(y∨z)=(x∧y)∨(x∧z). 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, and set D(x)={B∈Q(L):x∈B}D(x)=\{B\in\mathcal Q(L):x\in B\}D(x)={B∈Q(L):x∈B}. For any a,b∈La,b\in La,b∈L, if D(a)=D(b)D(a)=D(b)D(a)=D(b), then a=ba=ba=b.

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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