§3.3, proof of Theorem 3.1, pp. 17–18 — d = D^µ on acyclic orders with equal symmetric parts
ProvedPrefSemimetric.TopDiff.agree_same_symmPartaxiomatic-characterizationp2o-batch-pfp2ap2o-gran-per-chapterp2o-plan-paperp2o-v1preference-relations
Let be a finite set and let be a semimetric on satisfying Axioms 1 and 3. Let be the measure on with point masses for every . Then for all acyclic orders with the same symmetric parts,
Together with Axiom 4, which lets one replace each order by its strict part plus the diagonal, this yields the "only if" direction of Theorem 3.1.
Formalization Note As in the previous milestone, is given by point weights w with for all . The case is included (both sides are ).
Preamble
import Mathlib import Definitions.Def_PrefSemimetric_TopDiff_Model
Formal statement
namespace PrefSemimetric.TopDiff
theorem agree_same_symmPart {X : Type*} [Fintype X] [DecidableEq X]
(d : (X → X → Prop) → (X → X → Prop) → ℝ) (hd : IsSemimetric d)
(h1 : Axiom1 d) (h3 : Axiom3 d)
(w : X → ℝ) (hw : ∀ a b, a ≠ b → w b = d (prefAB a b) (prefABplus a b))
(R T : X → X → Prop) (hR : IsAcyclicOrder R) (hT : IsAcyclicOrder T)
(hsymm : ∀ x y, symmPart R x y ↔ symmPart T x y) :
d R T = Dmu w R T := by sorry
end PrefSemimetric.TopDiff
Source
Nishimura & Ok, A Class of Dissimilarity Semimetrics for Preference Relations, arXiv:2203.04418v1, pp. 17–18, §3.3, proof of Theorem 3.1 ('Now take any ≿, ⊵ ∈ A(X) and assume that the symmetric parts of these relations are the same.' … 'where the third equality follows from the fact that Dµ satisfies Axiom 1’.')
Human review
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