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Proper-parent face inequality (14): Q∘(Y)≤D∘(Y)+(n−∣S∣)+#{u∈S:Y∩(Ru∪Tu)≠∅}Q^\circ(Y)\le D^\circ(Y)+(n-|S|)+\#\{u\in S:Y\cap(R_u\cup T_u)\neq\varnothing\}Q∘(Y)≤D∘(Y)+(n−∣S∣)+#{u∈S:Y∩(Ru​∪Tu​)=∅} (Erdős 624)

Proved
Erdos624.certificate_face_bound

by Ethan Cheung · Oct 24, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricserdos-problems

Let UUU be a set of inputs, CCC a finite set of colours with n=∣C∣n=|C|n=∣C∣, g0,g1:P(U)→Cg_0,g_1:\mathcal P(U)\to Cg0​,g1​:P(U)→C arbitrary and L(A)={g0(A),g1(A)}L(A)=\{g_0(A),g_1(A)\}L(A)={g0​(A),g1​(A)}; let V⊆UV\subseteq UV⊆U be finite and k∈Nk\in\mathbb Nk∈N. Let S⊆C∖L(∅)S\subseteq C\setminus L(\varnothing)S⊆C∖L(∅). For each u∈Su\in Su∈S fix a deletion set Ru⊆UR_u\subseteq URu​⊆U and a parity certificate wu:U→F2w_u:U\to\mathbf F_2wu​:U→F2​ for the proper minimal witnesses of uuu surviving the deletion: for every A⊆V∖RuA\subseteq V\setminus R_uA⊆V∖Ru​ with 0<∣A∣<k0<|A|<k0<∣A∣<k, u∈L(A)u\in L(A)u∈L(A) and u∉L(B)u\notin L(B)u∈/L(B) for all B⊊AB\subsetneq AB⊊A, we have ∑x∈Awu(x)=1\sum_{x\in A}w_u(x)=1∑x∈A​wu​(x)=1. Put Tu={x∈V∖Ru:wu(x)=0}T_u=\{x\in V\setminus R_u: w_u(x)=0\}Tu​={x∈V∖Ru​:wu​(x)=0}. For a kkk-set YYY let J∘(Y)=⋃A⊊YL(A)J^\circ(Y)=\bigcup_{A\subsetneq Y}L(A)J∘(Y)=⋃A⊊Y​L(A), D∘(Y)=n−∣J∘(Y)∣D^\circ(Y)=n-|J^\circ(Y)|D∘(Y)=n−∣J∘(Y)∣, Jodd∘(Y)=⋃A⊊Y, ∣A∣ oddL(A)J^\circ_{\mathrm{odd}}(Y)=\bigcup_{A\subsetneq Y,\,|A|\text{ odd}}L(A)Jodd∘​(Y)=⋃A⊊Y,∣A∣ odd​L(A) and Q∘(Y)=n−∣Jodd∘(Y)∣Q^\circ(Y)=n-|J^\circ_{\mathrm{odd}}(Y)|Q∘(Y)=n−∣Jodd∘​(Y)∣. Then for every kkk-set Y⊆VY\subseteq VY⊆V,

Q∘(Y)  ≤  D∘(Y)+(n−∣S∣)+∣{u∈S: Y∩(Ru∪Tu)≠∅}∣.(14)Q^\circ(Y)\;\le\; D^\circ(Y)+(n-|S|)+\bigl|\{u\in S:\ Y\cap(R_u\cup T_u)\neq\varnothing\}\bigr|.\qquad(14)Q∘(Y)≤D∘(Y)+(n−∣S∣)+​{u∈S: Y∩(Ru​∪Tu​)=∅}​.(14)

Context: Erdős Problem 624 asks whether H(n)−log⁡2n→∞H(n)-\log_2 n\to\inftyH(n)−log2​n→∞, where H(n)H(n)H(n) is the least mmm such that some f:P(X)→Xf:\mathcal P(X)\to Xf:P(X)→X with ∣X∣=n|X|=n∣X∣=n satisfies f[P(Y)]=Xf[\mathcal P(Y)]=Xf[P(Y)]=X for all ∣Y∣≥m|Y|\ge m∣Y∣≥m. By an earlier track-amplification result it suffices to find common holes for fixed numbers of tracks; this is an auxiliary finite inequality from Sol's Round 10 analysis of the two-track case. It does not settle Erdős Problem 624, PAIR or Step 17.

Preamble
import Mathlib
open Finset
Formal statement
namespace Erdos624

theorem certificate_face_bound {U : Type*} [DecidableEq U] {C : Type*} [Fintype C] [DecidableEq C]
    (k : ℕ) (g0 g1 : Finset U → C) (V : Finset U) (S : Finset C)
    (hS : ∀ u ∈ S, u ∉ ({g0 ∅, g1 ∅} : Finset C))
    (R : C → Finset U) (w : C → U → ZMod 2)
    (hw : ∀ u ∈ S, ∀ A ⊆ V \ R u, 0 < #A → #A < k → u ∈ ({g0 A, g1 A} : Finset C) →
      (∀ B ⊂ A, u ∉ ({g0 B, g1 B} : Finset C)) → ∑ x ∈ A, w u x = 1)
    (Y : Finset U) (hYV : Y ⊆ V) (hY : #Y = k) :
    Fintype.card C - #((Y.powerset.filter (fun A => A ≠ Y ∧ #A % 2 = 1)).biUnion
        (fun A => ({g0 A, g1 A} : Finset C))) ≤
      (Fintype.card C - #((Y.powerset.filter (fun A => A ≠ Y)).biUnion
        (fun A => ({g0 A, g1 A} : Finset C)))) + (Fintype.card C - #S) +
      #(S.filter (fun u => ¬ Disjoint Y (R u ∪ (V \ R u).filter (fun x => w u x = 0)))) := by sorry

end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); Round 10 reports (consolidated report §'parity transport', §'many holes one level below'; Step 17 report 'resilient relations below the top layer', eqs. (9)-(26)); derived from GPT 6.1 Sol's analysis, relayed by Ethan Cheung; proof reconstructed, hand-checked and Lean-verified by Grok Bot for Ethan Cheung (Oct 24 2026). Auxiliary result only; does NOT settle PAIR / Step 17 / #624.

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