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The $\frac 1 3$–$\frac 2 3$ conjecture

Reference: Wikipedia

namespace Conjecture_1_3_to_2_3

Does every finite partially ordered set that is not totally ordered contain two elements $x$ and $y$ such that the probability that $x$ appears before $y$ in a random linear extension is between $\frac 1 3$ and $\frac 2 3$?

The set of all total order extensions is represented as order preserving bijections $P$ of $1, ..., n$.

@[category research open, AMS 6] theorem declaration uses 'sorry'conjecture_1_3_to_2_3 : answer(sorry) (P : Type) [Finite P] [PartialOrder P] (not_total : ¬ Std.Total (α := P) (· ·)) (total_ext : Set <| OrderHom P ) (total_ext_def : σ, σ total_ext Set.range σ = Set.Icc 1 (Nat.card P)), x y : P, ({σ total_ext | σ x < σ y}.ncard / total_ext.ncard : ) Set.Icc (1/3) (2/3) := True (P : Type) [Finite P] [inst : PartialOrder P], (¬Std.Total fun x1 x2 => x1 x2) (total_ext : Set (P →o )), (∀ (σ : P →o ), σ total_ext Set.range σ = Set.Icc 1 (Nat.card P)) x y, {σ | σ total_ext σ x < σ y}.ncard / total_ext.ncard Set.Icc (1 / 3) (2 / 3) All goals completed! 🐙 end Conjecture_1_3_to_2_3