Proof of Theorem 2.13 — deletion and contraction of the half-integral node have smaller defect
ProvedLovaszSchrijver.Defect.defect_deletion_contraction_ltLet be a finite graph with no isolated nodes, and let be an inequality with , , valid for , whose defect
is positive. Let be a node with at every vector maximizing (such a node exists by Lemma 2.12). Then both the deletion and the contraction of result in constraints with smaller defect:
- the deletion has a defect, and it is ;
- the contraction has a defect, and it is .
This is the claim on which the induction on in the upper bound of Theorem 2.13 rests.
Formalization Note Deletion and contraction are the zero-extended coefficient vectors on the same graph (see the definition item), and their defects are taken over , as in the paper's proof. The contraction's right-hand side is computed in . The existence of each defect (the maximum being attained) is part of the conclusion.
import Mathlib import Definitions.Def_LovaszSchrijver_Defect_Index
namespace LovaszSchrijver.Defect
theorem defect_deletion_contraction_lt {V : Type} [Fintype V] [DecidableEq V]
(G : SimpleGraph V) [DecidableRel G.Adj] (hG : ∀ v, ∃ w, G.Adj v w)
(a : V → ℕ) (b : ℕ) (hvalid : Valid (STAB G) (fun j => (a j : ℝ)) b)
(r : ℝ) (hr : IsDefect G (fun j => (a j : ℝ)) b r) (hr0 : 0 < r)
(i : V) (hi : ∀ y : V → ℝ, IsFRACMaximizer G (fun j => (a j : ℝ)) y → y i = 1 / 2) :
(∃ r₁ : ℝ, IsDefect G (deletion (fun j => (a j : ℝ)) i) b r₁ ∧ r₁ < r) ∧
(∃ r₂ : ℝ, IsDefect G (contraction G (fun j => (a j : ℝ)) i) ((b : ℝ) - (a i : ℝ)) r₂ ∧
r₂ < r) := by sorry
end LovaszSchrijver.Defect
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What the Lean code literally says, in plain math · claude-opus-5-5
Setting. is a finite type with decidable equality and is a simple graph on with decidable adjacency. The weights and the right-hand side are natural numbers, treated as real numbers wherever they appear below. and .
Hypotheses.
- has no isolated vertices.
- The inequality is valid for , the convex hull of incidence vectors of independent sets.
- is the defect of : , and the maximum is attained. Here .
- The defect is positive: .
- Every that maximizes over has .
Conclusion. Both of the following hold.
- Deletion. Let be with replaced by . There is an that is the defect of :
- Contraction. Let be with and every for adjacent to replaced by . There is an that is the defect of :
The subtraction is ordinary real subtraction, so it can be negative. The graph is still in both parts: nothing is removed from it.
Degenerate cases.
- If is empty, is a single point and the defect is . Hypothesis 4 then fails. In any case, no vertex exists, so the statement is vacuous.
- Hypothesis 1 makes bounded, so maximizers exist and hypothesis 5 is a real constraint.
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.