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rem-3.1

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QAlgorithms.Papers.Independent.Intakethird100.P20.distributive_basic_unions

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

prove2quantum

For a bounded distributive lattice,

Qa∪Qb=Qa∨b.Q_a\cup Q_b=Q_{a\vee b}.Qa​∪Qb​=Qa∨b​.

For any bounded lattice satisfying this union identity, the further identity Q(a∧b)∨(a∧c)=Qa∧(b∨c)Q_{(a\wedge b)\vee(a\wedge c)}=Q_{a\wedge(b\vee c)}Q(a∧b)∨(a∧c)​=Qa∧(b∨c)​ holds for every a,b,ca,b,ca,b,c. This latter consequence does not by itself imply distributivity without order reflection.

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

/-- rem-3.1; de Groote, Observables I, PDF p./pp. 43.
-/
theorem distributive_basic_unions {L : Type u} [Lattice L] [BoundedOrder L] :
    (IsDistributive (L := L) → ∀ a b : L, stoneBasic a ∪ stoneBasic b = stoneBasic (a ⊔ b)) ∧
    ((∀ a b : L, stoneBasic a ∪ stoneBasic b = stoneBasic (a ⊔ b)) →
      ∀ a b c : L, stoneBasic ((a ⊓ b) ⊔ (a ⊓ c)) = stoneBasic (a ⊓ (b ⊔ c))) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P20

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

Let LLL be any bounded lattice. Let Q(L)\mathcal Q(L)Q(L) consist of the inclusion-maximal nonempty subsets B⊆LB\subseteq LB⊆L excluding ⊥\bot⊥ such that for every x,y∈Bx,y\in Bx,y∈B some z∈Bz\in Bz∈B satisfies z≤x,yz\le x,yz≤x,y, and 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}. The theorem asserts both of the following implications. If LLL satisfies both distributive identities 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) for every a,b,c∈La,b,c\in La,b,c∈L, then 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,b∈La,b\in La,b∈L. Conversely, 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,b∈La,b\in La,b∈L, then D((a∧b)∨(a∧c))=D(a∧(b∨c))D((a\wedge b)\vee(a\wedge c))=D(a\wedge(b\vee c))D((a∧b)∨(a∧c))=D(a∧(b∨c)) for every a,b,c∈La,b,c\in La,b,c∈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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