Minimal-envelope extension (E): a hole with extends to a hole of size (Erdős 624)
ProvedErdos624.minimal_envelope_extensioncombinatoricserdos-problems
Setting: is a finite pool of inputs (to apply this to a sub-pool , take to be the subtype ), a set of colours, an arbitrary list map (e.g. ), , and with . A set is a hole if for all . (minimal witnesses of size ), , , , , . All of these are written out inline in the Lean statement (no custom definitions).
Theorem (E). If is a hole with , then there is a hole with . (Lean uses natural-number subtraction ; this is harmless since always.)
Context: Erdős Problem 624 asks whether , where is the least such that some with satisfies for all . This is an auxiliary finite statement from Sol's Round 12 Step 17 cross-review / resilience analysis of the two-track (PAIR) case. It does not settle Erdős Problem 624, PAIR or Step 17.
Preamble
import Mathlib open Finset
Formal statement
namespace Erdos624
theorem minimal_envelope_extension {U : Type*} [Fintype U] [DecidableEq U] {C : Type*} [DecidableEq C]
(L : Finset U → Finset C) (k : ℕ) (u : C) (hu : u ∉ L ∅)
(B : Finset U) (hB : #B ≤ k) (hBhole : ∀ A ⊆ B, u ∉ L A) :
∃ B' : Finset U, B ⊆ B' ∧ (∀ A ⊆ B', u ∉ L A) ∧
#B' = min k ((Module.finrank (ZMod 2) ((U → ZMod 2) ⧸
Submodule.span (ZMod 2) {v : U → ZMod 2 | ∃ A : Finset U, 0 < #A ∧ #A ≤ k ∧ u ∈ L A ∧
(∀ A' ⊂ A, u ∉ L A') ∧ v = fun x => if x ∈ A then 1 else 0})) +
(#B - Module.finrank (ZMod 2) (Submodule.span (ZMod 2)
((fun x => (Submodule.span (ZMod 2) {v : U → ZMod 2 | ∃ A : Finset U, 0 < #A ∧ #A ≤ k ∧ u ∈ L A ∧
(∀ A' ⊂ A, u ∉ L A') ∧ v = fun x => if x ∈ A then 1 else 0}).mkQ (Pi.single x 1)) '' (B : Set U))))) := by sorry
end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); Round 12: Sol's 'Step 17: First-Round Cross-Review' and 'Step 17: resilient relations below the top layer'; derived from GPT 6.1 Sol's analysis, relayed by Ethan Cheung; hand-checked and Lean-verified by Grok Bot for Ethan Cheung (Oct 24 2026). Auxiliary result only; does NOT settle PAIR / Step 17 / #624.