Foundation Observable Floor Witness Observable Floor Witness Of Setoid

The setoid bridge

In mathematics, a setoid is a set equipped with an equivalence relation: a rule that declares which elements count as the same for a given purpose. The declaration observableFloorWitness_of_setoid states that if a setoid has at least two elements that its equivalence relation does not identify, then those elements form an observable floor. An observable floor is a pair of states that are not related by the supplied relation, meaning they are distinguishable under that relation. The theorem is a direct restatement of the hypothesis: the existence of a non-identified pair is exactly the witness.

The statement matters because it separates two notions that raw type-theoretic inequality conflates. In a gauge theory, two representatives can be unequal as terms yet physically indistinguishable because they are related by a gauge transformation. The setoid's equivalence relation captures that physical indistinguishability. The theorem says that a physical floor, a pair of states that are not gauge-equivalent, is precisely a pair that the setoid does not identify. This is the quotient-aware repair to an earlier objection: raw inequality alone does not guarantee observable distinguishability, but a non-identified pair under the right relation does.

The framework's machine-checked library of formal theorems proves a companion result: a quotient carrier is non-singleton exactly when the original carrier has an observable distinction relative to the quotienting setoid. This means physical distinguishability lives in the quotient, or equivalently in pairs of representatives not identified by the relation. The setoid bridge is intentionally transparent: its physical content lies in choosing the observational or gauge setoid, not in the theorem itself.

What the declaration does not claim is just as important. It does not assert that any particular relation is the correct physical one; that choice is left to the model builder. It does not say that raw inequality implies observable distinguishability for an arbitrary relation. The library proves the opposite: with the indiscrete relation on the real numbers, where every pair is equivalent, bare inequality holds but no observable floor exists. The theorem is a bridge, not a physical law.

/-- A setoid witness is exactly an observable-floor witness for the setoid's
equivalence relation. This is intentionally a transparent bridge: the physical
content is in choosing the observational/gauge setoid. -/
theorem observableFloorWitness_of_setoid
    {K : Type*} (s : Setoid K) (h : ∃ x y : K, ¬ s.r x y) :
    ObservableFloorWitness K s.r :=
  h

/-- Bare inequality does not imply observable distinguishability for an
arbitrary observational relation. Take the indiscrete relation on `ℝ`, where
every pair is observationally equivalent. -/
theorem bare_distinction_does_not_imply_observable_distinction :
    ∃ (K : Type) (r : K → K → Prop),
      (∃ x y : K, x ≠ y) ∧ ¬ ObservableFloorWitness K r := by
  refine ⟨ℝ, (fun _ _ => True), ⟨?_, ?_⟩⟩
  · exact ⟨0, 1, by norm_num⟩
  · rintro ⟨x, y, hxy⟩
    exact hxy trivial

/-- A quotient carrier is non-singleton exactly when the original carrier has
an observable distinction relative to the quotienting setoid.

This is the quotient-aware repair to the raw-inequality objection: physical
distinguishability lives in `Quotient s`, or equivalently in pairs of
representatives not identified by `s.r`. -/
theorem quotient_nontrivial_iff_observableFloor
    {K : Type*} (s : Setoid K) :
    (∃ a b : Quotient s, a ≠ b) ↔ ObservableFloorWitness K s.r := by
  constructor
  · rintro ⟨a, b, hne⟩
    refine Quotient.inductionOn₂ a b ?_ hne
    intro x y hne'
    refine ⟨x, y, ?_⟩
    intro hxy
    exact hne' (Quotient.sound hxy)
  · rintro ⟨x, y, hxy⟩
    refine ⟨Quotient.mk s x, Quotient.mk s y, ?_⟩
    intro hq
    exact hxy (Quotient.exact hq)

What this page does not claim

The theorem does not specify which equivalence relation is physically correct. The theorem does not say that raw inequality implies observable distinguishability for an arbitrary relation. The theorem does not establish that any particular physical system has an observable floor.

Verify this page

Every tagged claim above names its theorem. To check one yourself rather than trust this page, elaborate the source module with Lean 4 and audit its axiom basis:

$ lake env lean IndisputableMonolith/Foundation/ObservableFloorWitness.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)

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Derived articles

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