Interlocked root obstruction: b before c and 5c−4b before 4c−3b kills every SAFE extension
ProvedErdos195.interlocked_root_obstructionA tree witness defines a strict total order on (milestone framework). A finite word of integers is SAFE for if the sequence "letters of in order, then all remaining integers in -order" has no monotone 4-term arithmetic progression. Theorem. Let be a finite word containing an odd before an even , and before , with neither nor in . Then no extension is SAFE for any tree witness.
Here and are interlocked: and . By the previous milestone they stay open forever. Both have odd difference, so both sit at the root node. They demand opposite root bits ( is even, is odd), which is a same-node conflict. This is the excluded case (, ) of GPT 6 Astra's repair lemma for a single interleaving inversion . In Astra's setting follows from the mod-4 class hypotheses. The Lean statement assumes it directly and needs no class or distinctness hypotheses.
import Mathlib import Definitions.Def_Erdos195_tree
namespace Erdos195
theorem interlocked_root_obstruction (T R : List ℤ) (b c : ℤ)
(hb : b ∈ T) (hc : c ∈ T) (hbo : b % 2 = 1) (hce : c % 2 = 0)
(hp : 4 * c - 3 * b ∈ T) (hq : 5 * c - 4 * b ∈ T)
(hbc : T.idxOf b < T.idxOf c) (hqp : T.idxOf (5 * c - 4 * b) < T.idxOf (4 * c - 3 * b))
(htT : 2 * c - b ∉ T) (hwT : 3 * c - 2 * b ∉ T) :
¬ ∃ θ : ℕ → ℤ → Bool, Safe (treeLt θ) (T ++ R) := by sorry
end Erdos195