thue_morse_subwords

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Thue-Morse sequence: Every finite binary word eventually appears as a subword. The morphism t → 01, 1 → 10 generates an overlap-free infinite word. Complexity questions remain open.

Why this node was retired

The posted statement is

import Mathlib

def thueMorseSeq : ℕ → Bool
    | 0 => false
    | n + 1 => !(thueMorseSeq n)

theorem thue_morse_subwords :
    ∀ w : List Bool, 0 < w.length →
      ∃ n : ℕ,
        w = (List.map thueMorseSeq (List.range n)).take w.length ∨
        w = (List.map thueMorseSeq (List.range (n + w.length))).drop n := by
  sorry

The recursive definition is the alternating sequence, not Thue–Morse; the prose additionally claims every binary word occurs, which is false for genuine Thue–Morse too.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Define t(2n)=t(n), t(2n+1)=not t(n), t(0)=false, then ask for its actual factor language/subword-complexity function or overlap-freeness. Do not retain universal occurrence of every word.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.