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One-even bridge criterion: odd batches around a single new even letter (root bit 1)

Proved
Erdos195.one_even_bridge

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 (one-even bridge criterion, root bit 111). Let θ(0,0)=1\theta(0,0)=1θ(0,0)=1. Let SSS be a word that is θS\theta_SθS​-safe for some witness θS\theta_SθS​ with θS(0,0)=1\theta_S(0,0)=1θS​(0,0)=1. Let PPP and TTT be words of odd integers with no letter of PPP in SSS and no letter of TTT in S⋅PS\cdot PS⋅P, and let y∉Sy\notin Sy∈/S be even. Then

S⋅P⋅y⋅T is θ-safeS\cdot P\cdot y\cdot T\ \text{is }\theta\text{-safe}S⋅P⋅y⋅T is θ-safe

if and only if

  1. the even half of S⋅yS\cdot yS⋅y is θE\theta_EθE​-safe and the odd half of S⋅P⋅TS\cdot P\cdot TS⋅P⋅T is θO\theta_OθO​-safe;
  2. (closing pairs into yyy) for every odd a∈S⋅Pa\in S\cdot Pa∈S⋅P: 2y−a∈S⋅P2y-a\in S\cdot P2y−a∈S⋅P or 3y−2a∈S3y-2a\in S3y−2a∈S;
  3. (progressions with third term yyy) for every even a∈Sa\in Sa∈S and odd b∈S⋅Pb\in S\cdot Pb∈S⋅P with 2b−a=y2b-a=y2b−a=y and aaa before bbb: 3b−2a∈S⋅P3b-2a\in S\cdot P3b−2a∈S⋅P;
  4. (new third terms of pairs of SSS) for odd aaa and even bbb in SSS, aaa before bbb, with 2b−a∉S2b-a\notin S2b−a∈/S but 2b−a∈P⋅T2b-a\in P\cdot T2b−a∈P⋅T: either 3b−2a∈S3b-2a\in S3b−2a∈S, or 3b−2a=y3b-2a=y3b−2a=y and 2b−a∉P2b-a\notin P2b−a∈/P.

This is the exact root-level criterion for absorbing a single new even letter into a root-bit-111 word using only odd helpers placed before and after it: nothing is required of the odd batches except odd-half safety, while the letters before yyy must be (almost) closed under reflection a↦2y−aa\mapsto 2y-aa↦2y−a. It complements the lifting criteria Erdos195.lifting_criterion and Erdos195.lifting_criterion_true (odd target after an even batch). Formalization note: SSS itself only needs to be safe for some root-bit-111 witness, not for θ\thetaθ.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem one_even_bridge (θ : ℕ → ℤ → Bool) (h00 : θ 0 0 = true) (S P T : List ℤ) (y : ℤ)
    (hS : ∃ θS : ℕ → ℤ → Bool, θS 0 0 = true ∧ Safe (treeLt θS) S)
    (hPo : ∀ p ∈ P, p % 2 = 1) (hTo : ∀ t ∈ T, t % 2 = 1) (hy : y % 2 = 0) (hyS : y ∉ S)
    (hPS : ∀ p ∈ P, p ∉ S) (hTS : ∀ t ∈ T, t ∉ S ++ P) :
    Safe (treeLt θ) (S ++ P ++ [y] ++ T) ↔
      Safe (treeLt (fun k r => θ (k + 1) (2 * r)))
        (((S ++ [y]).filter (fun x => decide (x % 2 = 0))).map (· / 2)) ∧
      Safe (treeLt (fun k r => θ (k + 1) (2 * r + 1)))
        (((S ++ P ++ T).filter (fun x => decide (x % 2 = 1))).map (· / 2)) ∧
      (∀ a ∈ S ++ P, a % 2 = 1 → 2 * y - a ∈ S ++ P ∨ 3 * y - 2 * a ∈ S) ∧
      (∀ a ∈ S, a % 2 = 0 → ∀ b ∈ S ++ P, b % 2 = 1 → 2 * b - a = y →
        (S ++ P).idxOf a < (S ++ P).idxOf b → 3 * b - 2 * a ∈ S ++ P) ∧
      (∀ a ∈ S, ∀ b ∈ S, a % 2 = 1 → b % 2 = 0 → S.idxOf a < S.idxOf b →
        2 * b - a ∉ S → 2 * b - a ∈ P ++ T →
        3 * b - 2 * a ∈ S ∨ (3 * b - 2 * a = y ∧ 2 * b - a ∉ P)) := by sorry

end Erdos195
Source
Erdős Problem 195 (https://www.erdosproblems.com/195); tree-witness framework; one-even route proposed by GPT 6 Astra; Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)

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