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Four-class restricted absorption: a SAFE block in one odd class mod 8 absorbs a target with 2n+42n+42n+4 letters

Proved
Erdos195.restricted_absorption_mod8

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 AAA be a SAFE word all of whose entries are congruent modulo 888 to an odd integer rrr. Let aaa be even and b≡r+2(mod4)b\equiv r+2\pmod 4b≡r+2(mod4), and put Q=(a)⋅A⋅(b)Q=(a)\cdot A\cdot(b)Q=(a)⋅A⋅(b). Let y∉Qy\notin Qy∈/Q be such that c=2y−b∉Qc=2y-b\notin Qc=2y−b∈/Q, c≡r(mod8)c\equiv r\pmod 8c≡r(mod8), both A⋅(c)A\cdot(c)A⋅(c) and (c)⋅A(c)\cdot A(c)⋅A are SAFE, and 7b−6y∉A7b-6y\notin A7b−6y∈/A. 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 entrywise images in the original order. This is GPT Astra's mod-8 generalization of the mod-32 absorption theorem: the block AAA may occupy all four residue classes modulo 323232 inside one odd class modulo 888. No size condition is needed and AAA may be empty.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem restricted_absorption_mod8 (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 % 8 = r % 8)
    (ha : a % 2 = 0) (hb : b % 4 = (r + 2) % 4)
    (hy : y ∉ [a] ++ A ++ [b]) (hcQ : 2 * y - b ∉ [a] ++ A ++ [b]) (hc : (2 * y - b) % 8 = r % 8)
    (hAc : ∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ) (A ++ [2 * y - b]))
    (hcA : ∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ) ([2 * y - b] ++ A))
    (h7 : 7 * b - 6 * y ∉ A) :
    ∃ 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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