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Cofinal SAFE chains need infinitely many genuinely mixed-parity batches (tree orders)

Proved
Erdos195.mixed_batches_recur

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.

Theorem (cofinal SAFE chains need infinitely many genuinely mixed batches). Let Q0⪯Q1⪯Q2⪯⋯Q_0\preceq Q_1\preceq Q_2\preceq\cdotsQ0​⪯Q1​⪯Q2​⪯⋯ be duplicate-free SAFE words, each a prefix of the next (each stage may use its own witness), whose union is all of Z\mathbb ZZ. Then for every JJJ there is j≥Jj\ge Jj≥J such that the batch of letters of Qj+1Q_{j+1}Qj+1​ not in QjQ_jQj​ contains integers of both parities.

In other words, an exhaustive construction of a permutation of Z\mathbb ZZ through SAFE stages can never eventually proceed by single-parity batches; equivalently it cannot eventually factor into finitely many monochromatic SAFE sub-batches per step. This is a necessary condition on any successful stagewise construction for Erdős Problem 195 in the tree-witness framework, not a construction or an impossibility proof. Theorem of GPT 6 Astra.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem mixed_batches_recur (Q : ℕ → List ℤ)
    (hS : ∀ j, ∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ) (Q j))
    (hnd : ∀ j, (Q j).Nodup) (hpre : ∀ j, Q j <+: Q (j + 1))
    (hcov : ∀ x : ℤ, ∃ j, x ∈ Q j) :
    ∀ J, ∃ j ≥ J, ∃ x ∈ Q (j + 1), ∃ x' ∈ Q (j + 1), x ∉ Q j ∧ x' ∉ Q j ∧ x % 2 ≠ x' % 2 := by sorry

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

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