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Restricted absorption: a mod-32 sparse SAFE block absorbs any admissible target with 2n+42n+42n+4 letters

Proved
Erdos195.restricted_absorption

by Ethan Cheung Wang · Oct 23, 2026 · Mathlib 0df444a (Lean v4.33.1)

arithmetic-progressionscombinatoricserdos-problems

A tree witness is a function θ(k,r)∈{0,1}\theta(k,r)\in\{0,1\}θ(k,r)∈{0,1} on the nodes (k,r)(k,r)(k,r), 0≤r<2k0\le r<2^k0≤r<2k, of the binary tree of residues. For distinct integers x,yx,yx,y with k=ν2(y−x)k=\nu_2(y-x)k=ν2​(y−x), put x≺θyx\prec_\theta yx≺θ​y iff bitk(x)=θ(k, x mod 2k)\mathrm{bit}_k(x)=\theta(k,\,x\bmod 2^k)bitk​(x)=θ(k,xmod2k). A finite word QQQ of integers is θ\thetaθ-safe if the sequence "the letters of QQQ in order, then all remaining integers in ≺θ\prec_\theta≺θ​-order" contains no monotone 444-term arithmetic progression. QQQ is SAFE if it is θ\thetaθ-safe for some tree witness θ\thetaθ. Let A=(x1,…,xn)A=(x_1,\dots,x_n)A=(x1​,…,xn​) be a SAFE word all of whose entries are congruent modulo 323232 to an odd integer rrr. Let aaa be even, b≡r+2(mod4)b\equiv r+2\pmod 4b≡r+2(mod4), and let y∉Q=(a)⋅A⋅(b)y\notin Q=(a)\cdot A\cdot(b)y∈/Q=(a)⋅A⋅(b) satisfy 2y−b≡r+16(mod32)2y-b\equiv r+16\pmod{32}2y−b≡r+16(mod32). Then

Q⋅(2y−b)⋅(2y−A)⋅(y)⋅(12y−11b)⋅(10y−10b+A)⋅(6y−5b)Q\cdot(2y-b)\cdot(2y-A)\cdot(y)\cdot(12y-11b)\cdot(10y-10b+A)\cdot(6y-5b)Q⋅(2y−b)⋅(2y−A)⋅(y)⋅(12y−11b)⋅(10y−10b+A)⋅(6y−5b)

is SAFE, where 2y−A2y-A2y−A and 10y−10b+A10y-10b+A10y−10b+A are taken entrywise in the original order. No size condition on yyy is needed, AAA may be empty, and SAFE-ness of QQQ itself is not assumed (it follows). This is the mod-32 form of GPT Astra's uniform finite absorption theorem: an infinite class of arbitrarily long SAFE words absorbs every target of a residue class with exactly 2n+42n+42n+4 new entries.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem restricted_absorption (a b y r : ℤ) (A : List ℤ)
    (hA : ∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ) A)
    (hr : r % 2 = 1) (hAr : ∀ x ∈ A, x % 32 = r % 32)
    (ha : a % 2 = 0) (hb : b % 4 = (r + 2) % 4)
    (hy : y ∉ [a] ++ A ++ [b]) (hc : (2 * y - b) % 32 = (r + 16) % 32) :
    ∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ)
      (([a] ++ A ++ [b]) ++ ([2 * y - b] ++ A.map (fun x => 2 * y - x) ++ [y, 12 * y - 11 * b] ++
        A.map (fun x => 10 * y - 10 * b + x) ++ [6 * y - 5 * b])) := by sorry

end Erdos195
Source
Erdős Problem 195 (https://www.erdosproblems.com/195); tree-witness framework; construction and proof sketch by GPT Astra, Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)

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