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Exact lifting criterion with root bit 1: absorbing an odd target after an even batch

Proved
Erdos195.lifting_criterion_true

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. The halves of a word are the subsequences of its even letters and of its odd letters, each mapped by x↦⌊x/2⌋x\mapsto\lfloor x/2\rfloorx↦⌊x/2⌋. The half witnesses are θE(k,r)=θ(k+1,2r)\theta_E(k,r)=\theta(k+1,2r)θE​(k,r)=θ(k+1,2r) and θO(k,r)=θ(k+1,2r+1)\theta_O(k,r)=\theta(k+1,2r+1)θO​(k,r)=θ(k+1,2r+1).

Theorem (exact lifting criterion, root bit 111). Let θ(0,0)=1\theta(0,0)=1θ(0,0)=1. Let SSS be a θ\thetaθ-safe word, RRR a word of even integers, and y∉Sy\notin Sy∈/S an odd integer. Write P=S⋅RP=S\cdot RP=S⋅R. Then S⋅R⋅yS\cdot R\cdot yS⋅R⋅y is θ\thetaθ-safe if and only if

  1. the even half of PPP is θE\theta_EθE​-safe;
  2. the odd half of S⋅yS\cdot yS⋅y is θO\theta_OθO​-safe;
  3. (closing new even letters) for every b∈R∖Sb\in R\setminus Sb∈R∖S and every odd a∈Sa\in Sa∈S: 2b−a∈S2b-a\in S2b−a∈S, or 2b−a=y2b-a=y2b−a=y, or 3b−2a∈P3b-2a\in P3b−2a∈P;
  4. (no new third terms of open pairs) for all a,b∈Sa,b\in Sa,b∈S with b−ab-ab−a odd and aaa before bbb in SSS: if 2b−a∈R∖S2b-a\in R\setminus S2b−a∈R∖S then 3b−2a∈S3b-2a\in S3b−2a∈S;
  5. (in-order progressions ending at yyy) for every even b∈Pb\in Pb∈P with 2b−y∈S2b-y\in S2b−y∈S preceding bbb in PPP: 2y−b∈P2y-b\in P2y−b∈P.

This is the companion of the root-bit-000 lifting criterion (Erdos195.lifting_criterion). When the root bit is 111, open odd-difference pairs must end at an odd letter. So every new even letter must be closed against every odd letter of SSS (condition 3), and a new even letter must not become the third term of a pair of SSS that is open at the root (condition 4). Odd affine maps x↦mx+sx\mapsto mx+sx↦mx+s with sss odd flip the root bit (Erdos195.safe_odd_affine_iff). The two criteria together therefore cover absorbing a target of either parity after a batch of the opposite parity.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem lifting_criterion_true (θ : ℕ → ℤ → Bool) (h00 : θ 0 0 = true) (S R : List ℤ) (y : ℤ)
    (hS : Safe (treeLt θ) S) (hRe : ∀ r ∈ R, r % 2 = 0) (hy : y % 2 = 1) (hyS : y ∉ S) :
    Safe (treeLt θ) (S ++ R ++ [y]) ↔
      Safe (treeLt (fun k r => θ (k + 1) (2 * r)))
        (((S ++ R).filter (fun x => decide (x % 2 = 0))).map (· / 2)) ∧
      Safe (treeLt (fun k r => θ (k + 1) (2 * r + 1)))
        (((S ++ [y]).filter (fun x => decide (x % 2 = 1))).map (· / 2)) ∧
      (∀ b ∈ R, b ∉ S → ∀ a ∈ S, a % 2 = 1 →
        2 * b - a ∈ S ∨ 2 * b - a = y ∨ 3 * b - 2 * a ∈ S ++ R) ∧
      (∀ a ∈ S, ∀ b ∈ S, (b - a) % 2 = 1 → S.idxOf a < S.idxOf b →
        2 * b - a ∈ R → 2 * b - a ∉ S → 3 * b - 2 * a ∈ S) ∧
      (∀ b ∈ S ++ R, b % 2 = 0 → 2 * b - y ∈ S →
        (S ++ R).idxOf (2 * b - y) < (S ++ R).idxOf b → 2 * y - b ∈ S ++ R) := by sorry

end Erdos195
Source
Erdős Problem 195 (https://www.erdosproblems.com/195); tree-witness framework; companion of Erdos195.lifting_criterion; Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)

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