Lemma 4 — relation (IV) is transitive (except when the middle item is indifferent to both)
ProvedJohnsonFlowShop.ThreeStage.lemma4_transitiveflow-shopjohnson-rulep2o-batch-p200bp2o-gran-per-chapterp2o-plan-paperp2o-v1scheduling
Let be real numbers. If
then
unless both hypotheses hold with equality (item 2 is indifferent to both item 1 and item 3).
This is Lemma 2 (p. 64) for the times and ; it is what allows the definite preferences of (IV) to be arranged into one ordering by successive interchanges.
Formalization Note The page says only "Relation (IV) is transitive. Proof is the same as for Lemma 2", and Lemma 2 carries the exception for an item indifferent to both others. Without the exception the statement is false: take and . No positivity is assumed.
Preamble
import Mathlib
Formal statement
namespace JohnsonFlowShop.ThreeStage
/-- Lemma 4 (p. 67): relation (IV) is transitive, with the same exception as Lemma 2 (item 2
indifferent to both items 1 and 3). For real numbers (no positivity needed): if
`min (A₁ + B₁) (C₂ + B₂) ≤ min (A₂ + B₂) (C₁ + B₁)` and
`min (A₂ + B₂) (C₃ + B₃) ≤ min (A₃ + B₃) (C₂ + B₂)`, then
`min (A₁ + B₁) (C₃ + B₃) ≤ min (A₃ + B₃) (C₁ + B₁)`, or both hypotheses hold with equality. -/
theorem lemma4_transitive (A₁ B₁ C₁ A₂ B₂ C₂ A₃ B₃ C₃ : ℝ)
(h₁₂ : min (A₁ + B₁) (C₂ + B₂) ≤ min (A₂ + B₂) (C₁ + B₁))
(h₂₃ : min (A₂ + B₂) (C₃ + B₃) ≤ min (A₃ + B₃) (C₂ + B₂)) :
min (A₁ + B₁) (C₃ + B₃) ≤ min (A₃ + B₃) (C₁ + B₁) ∨
(min (A₁ + B₁) (C₂ + B₂) = min (A₂ + B₂) (C₁ + B₁) ∧
min (A₂ + B₂) (C₃ + B₃) = min (A₃ + B₃) (C₂ + B₂)) := by sorry
end JohnsonFlowShop.ThreeStage
Source
Johnson, Optimal Two- and Three-Stage Production Schedules with Setup Times Included, NRLQ 1(1) 1954, p. 67, Lemma 4 (with the exception of Lemma 2, p. 64)
Human review
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.