one_third_two_thirds_conjecture

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Proved status carries no information about that problem. Do not import this node or use it as a dependency.

1/3-2/3 conjecture: In any finite partially ordered set that is not a total order, there exist two incomparable elements x, y such that the probability that x precedes y in a uniformly random linear extension is between 1/3 and 2/3.

Why this node was retired

The posted statement is

import Mathlib

-- 1/3-2/3 conjecture for partial orders
-- A linear extension is a permutation σ : Fin n → Fin n respecting the partial order
theorem one_third_two_thirds_conjecture {n : ℕ} [NeZero n]
    (r : Fin n → Fin n → Prop) [DecidableRel r]
    [IsPartialOrder (Fin n) r]
    (hnchain : ∃ x y : Fin n, ¬r x y ∧ ¬r y x) :
    ∃ x y : Fin n, ¬r x y ∧ ¬r y x ∧
      let allExt := Finset.univ.filter (fun σ : Equiv.Perm (Fin n) =>
        ∀ a b : Fin n, r a b → σ a < σ b)
      let extXY := allExt.filter (fun σ => σ x < σ y)
      3 * extXY.card ≤ 2 * allExt.card ∧
      allExt.card ≤ 3 * extXY.card := by
  sorry

The extension filter maps a reflexive relation to strict inequalities even at a=b, making it empty.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Filter permutations by a<b in the strict part of the partial order (r a b ∧ a≠b) implying σa<σb, or use r a b→σa≤σb; preserve the incomparable-pair and counting inequalities.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.