Chains never appending a third term cannot exhaust
ProvedErdos195.no_third_term_chain_not_exhaustivearithmetic-progressionscombinatoricserdos-problems
Let be a finite word of integers and with . Suppose no appended entry is a third term of an in-order pair:
Then is not cofinite in ; in particular such a chain can never exhaust . Both the isolated-minimum completion and the exterior-record insertion for tree-safe words carry exactly this side condition, so in Ho-style constructions of permutations of without monotone -term APs (Erdős Problem 195) chains using only these two moves cannot produce a permutation of .
Formalization note: this is a purely combinatorial statement about integer sequences, phrased with t : ℕ → ℤ (or a chain of lists P n with P (n+1) = P n ++ [w n]); "cofinite" is expressed as existence of a finite set F : Finset ℤ outside of which every integer occurs.
Preamble
import Mathlib import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195
theorem no_third_term_chain_not_exhaustive (P : ℕ → List ℤ) (w : ℕ → ℤ)
(hstep : ∀ n, P (n + 1) = P n ++ [w n]) (hfresh : ∀ n, w n ∉ P n)
(hno : ∀ n u v, u ∈ P n → v ∈ P n → (P n).idxOf u < (P n).idxOf v → w n ≠ 2 * v - u) :
¬ ∃ F : Finset ℤ, ∀ z, z ∉ F → ∃ n, z ∈ P n := by sorry
end Erdos195Source
Erdős Problem 195 (https://www.erdosproblems.com/195); statements from GPT 6 Astra (Oct 23 2026 'tree' answers tree10, generalized), Lean-verified by Grok Bot for Ethan Cheung