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Tree-safe for some θ\thetaθ iff AP closure and no same-node conflict

Proved
Erdos195.exists_treeLt_safe_iff

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) (two's-complement digits). 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. With open pairs as before (aaa before bbb in QQQ, 2b−a,3b−2a∉Q2b-a,3b-2a\notin Q2b−a,3b−2a∈/Q), QQQ is θ\thetaθ-safe for some tree witness θ\thetaθ if and only if (1) every in-order 3-AP a,b,2b−aa,b,2b-aa,b,2b−a of QQQ has 3b−2a∈Q3b-2a\in Q3b−2a∈Q placed before 2b−a2b-a2b−a, and (2) no two open pairs force opposite values at the same node:

ν2(b−a)=ν2(b′−a′)=k, a≡a′ (mod 2k) ⟹ bitk(b)=bitk(b′).\nu_2(b-a)=\nu_2(b'-a')=k,\ a\equiv a'\ (\mathrm{mod}\ 2^k)\ \Longrightarrow\ \mathrm{bit}_k(b)=\mathrm{bit}_k(b').ν2​(b−a)=ν2​(b′−a′)=k, a≡a′ (mod 2k) ⟹ bitk​(b)=bitk​(b′).

Compared with ε\varepsilonε-orders, conflicts only matter at the same tree node, not merely the same level.

Formalization note: treeLt θ is from Definitions.Def_Erdos195_tree; Safe, bitOne, No3AP, Star, HasMonotoneAP are from Definitions.Def_Erdos195_defs. Positions in a word are List.idxOf; ν2\nu_2ν2​ is padicValInt 2.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem exists_treeLt_safe_iff (Q : List ℤ) :
    (∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ) Q) ↔
      (∀ a b : ℤ, a ∈ Q → b ∈ Q → 2 * b - a ∈ Q → Q.idxOf a < Q.idxOf b →
          Q.idxOf b < Q.idxOf (2 * b - a) →
          3 * b - 2 * a ∈ Q ∧ Q.idxOf (3 * b - 2 * a) < Q.idxOf (2 * b - a)) ∧
      (∀ a b a' b' : ℤ, a ∈ Q → b ∈ Q → Q.idxOf a < Q.idxOf b →
          2 * b - a ∉ Q → 3 * b - 2 * a ∉ Q →
          a' ∈ Q → b' ∈ Q → Q.idxOf a' < Q.idxOf b' →
          2 * b' - a' ∉ Q → 3 * b' - 2 * a' ∉ Q →
          padicValInt 2 (b - a) = padicValInt 2 (b' - a') →
          a % 2 ^ padicValInt 2 (b - a) = a' % 2 ^ padicValInt 2 (b - a) →
          (bitOne b (padicValInt 2 (b - a)) ↔ bitOne b' (padicValInt 2 (b - a)))) := by sorry

end Erdos195
Source
Erdős Problem 195 (https://www.erdosproblems.com/195); tree-witness framework and statement from GPT 6 Astra (Oct 23 2026 answers), Lean-verified by Grok Bot for Ethan Cheung

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