Extreme-point optimality: optimal cost or an optimal extreme point
ProvedLinearOptimization.lp_extreme_point_optimality(Theorem 2.8, GOAL) Consider the linear programming problem of minimizing over a polyhedron . Suppose that has at least one extreme point.
Then, either the optimal cost is equal to , or there exists an extreme point which is optimal.
import Mathlib.Analysis.Convex.Extreme import Definitions.Def_Polyhedron /-- **B&T Theorem 2.8 (p. 66).** Over a polyhedron with at least one extreme point, either the optimal cost is `−∞` or some extreme point is optimal. -/
theorem LinearOptimization.lp_extreme_point_optimality {m n : ℕ} (A : Matrix (Fin m) (Fin n) ℝ)
(b : Fin m → ℝ) (c : Fin n → ℝ)
(hext : (Set.extremePoints ℝ (polyhedron A b)).Nonempty) :
lpValue c (polyhedron A b) = ⊥ ∨
∃ x ∈ Set.extremePoints ℝ (polyhedron A b),
IsLpOptimal c (polyhedron A b) x := by
sorry
Read-back
What the Lean code literally says, in plain math · claude-fable-5
Let be a real matrix, , , and write . Assume only that has at least one extreme point (Mathlib's sense; this in particular forces nonempty). The conclusion is a disjunction: either the optimal value , computed in the extended reals, equals (the problem is unbounded below), or there exists an extreme point of that is optimal: and for every . The disjunction is not claimed exclusive. Note the minimization convention and that no boundedness or rank hypotheses appear.
Confirmed by the mission captain (proposal self-audit).