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One-even bridge criterion with root bit 0: odd batches around a single new even letter

Proved
Erdos195.one_even_bridge_false

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

arithmetic-progressionscombinatoricserdos-problems

A tree witness θ:N×Z→{0,1}\theta:\mathbb N\times\mathbb Z\to\{0,1\}θ:N×Z→{0,1} defines a strict total order ≺θ\prec_\theta≺θ​ on Z\mathbb ZZ (milestone framework); its root bit θ(0,0)\theta(0,0)θ(0,0) decides the order between integers of different parity (θ(0,0)=0\theta(0,0)=0θ(0,0)=0: every even integer precedes every odd integer at the root). A finite word WWW of integers is θ\thetaθ-safe if the sequence "letters of WWW in order, then all remaining integers in ≺θ\prec_\theta≺θ​-order" has no monotone 4-term arithmetic progression. Write WevW_{\mathrm{ev}}Wev​, WoddW_{\mathrm{odd}}Wodd​ for the halved even and odd subwords (x↦⌊x/2⌋x\mapsto\lfloor x/2\rfloorx↦⌊x/2⌋), and θ(0)(k,r)=θ(k+1,2r)\theta^{(0)}(k,r)=\theta(k+1,2r)θ(0)(k,r)=θ(k+1,2r), θ(1)(k,r)=θ(k+1,2r+1)\theta^{(1)}(k,r)=\theta(k+1,2r+1)θ(1)(k,r)=θ(k+1,2r+1).

Theorem (one-even bridge criterion, root bit 0). Let θ(0,0)=0\theta(0,0)=0θ(0,0)=0. Let SSS be a word that is safe for some tree witness with root bit 000, let PPP and TTT be words of odd integers with P∩S=∅P\cap S=\emptysetP∩S=∅ and T∩(S∪P)=∅T\cap(S\cup P)=\emptysetT∩(S∪P)=∅, and let y∉Sy\notin Sy∈/S be even. Then S⋅P⋅y⋅TS\cdot P\cdot y\cdot TS⋅P⋅y⋅T is θ\thetaθ-safe if and only if

(S⋅y)ev is θ(0)-safe,(S⋅P⋅T)odd is θ(1)-safe,(S\cdot y)_{\mathrm{ev}}\ \text{is}\ \theta^{(0)}\text{-safe},\qquad (S\cdot P\cdot T)_{\mathrm{odd}}\ \text{is}\ \theta^{(1)}\text{-safe},(S⋅y)ev​ is θ(0)-safe,(S⋅P⋅T)odd​ is θ(1)-safe,

and the following root conditions hold:

  1. for every even a∈Sa\in Sa∈S and b∈Pb\in Pb∈P: 2b−a∈S2b-a\in S2b−a∈S, or 3b−2a∈S∪P3b-2a\in S\cup P3b−2a∈S∪P, or (2b−a≠y2b-a\neq y2b−a=y and 3b−2a∈T3b-2a\in T3b−2a∈T);
  2. for every even a∈S∪{y}a\in S\cup\{y\}a∈S∪{y} and b∈Tb\in Tb∈T: 2b−a∈S∪{y}2b-a\in S\cup\{y\}2b−a∈S∪{y} or 3b−2a∈S∪P∪T3b-2a\in S\cup P\cup T3b−2a∈S∪P∪T;
  3. for every odd a∈S∪Pa\in S\cup Pa∈S∪P with 2y−a∈T2y-a\in T2y−a∈T: 3y−2a∈S3y-2a\in S3y−2a∈S;
  4. for odd aaa before even bbb in SSS with 2b−a∈(P∪T)∖S2b-a\in (P\cup T)\setminus S2b−a∈(P∪T)∖S: 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);
  5. for even aaa before odd bbb in SSS with 2b−a=y2b-a=y2b−a=y: 3b−2a∈S∪P3b-2a\in S\cup P3b−2a∈S∪P.

This is the root-bit-0 companion of the one-even bridge criterion (root bit 1). With root bit 0 every NEW odd letter that follows an old even letter (including yyy) must be closed by 2b−a2b-a2b−a or 3b−2a3b-2a3b−2a (conditions 1 and 2), whereas the even target yyy itself needs no reflections, only condition 3 for odd letters reflected into the tail. After the odd shift x↦x+1x\mapsto x+1x↦x+1 (which flips the root bit and swaps parities), the case T=∅T=\emptysetT=∅ is the root-bit-1 exact lifting criterion for an even batch followed by an odd target; the tail TTT is the new ingredient. Together with the root-bit-1 bridge, this gives exact criteria for both root bits.

Formalization note: words are List ℤ, membership/first-occurrence via List.idxOf, safety is Safe (treeLt θ), halves via filter/map (· / 2).

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem one_even_bridge_false (θ : ℕ → ℤ → Bool) (h00 : θ 0 0 = false) (S P T : List ℤ) (y : ℤ)
    (hS : ∃ θS : ℕ → ℤ → Bool, θS 0 0 = false ∧ 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, a % 2 = 0 → ∀ b ∈ P,
        2 * b - a ∈ S ∨ 3 * b - 2 * a ∈ S ++ P ∨ (2 * b - a ≠ y ∧ 3 * b - 2 * a ∈ T)) ∧
      (∀ a ∈ S ++ [y], a % 2 = 0 → ∀ b ∈ T,
        2 * b - a ∈ S ++ [y] ∨ 3 * b - 2 * a ∈ S ++ P ++ T) ∧
      (∀ a ∈ S ++ P, a % 2 = 1 → 2 * y - a ∈ T → 3 * y - 2 * a ∈ S) ∧
      (∀ 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)) ∧
      (∀ a ∈ S, ∀ b ∈ S, a % 2 = 0 → b % 2 = 1 → S.idxOf a < S.idxOf b →
        2 * b - a = y → 3 * b - 2 * a ∈ S ++ P) := by sorry

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

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