Open cycle with a node conflict kills every extension (interlocked obstruction for cycles of any length)

A tree witness $\theta:\mathbb N\times\mathbb Z\to\{0,1\}$ defines a strict total order $\prec_\theta$ on $\mathbb Z$ (milestone framework). A finite word $Q$ of integers is $\theta$-safe if the sequence "letters of $Q$ in order, then all remaining integers in $\prec_\theta$-order" has no monotone 4-term arithmetic progression; positions are first occurrences. For an in-order pair $(a,b)$ of a word ($a$ before $b$), call $2b-a$ its third term and $3b-2a$ its fourth term; the pair is closed if one of them occurs. An open cycle in a word $T$ is a family of in-order pairs $(a_i,b_i)$, $0\le i<m$, of letters of $T$ with

and no third term $2b_i-a_i$ in $T$. Write $k_i=\nu_2(b_i-a_i)$ for the 2-adic level of pair $i$; its node is the residue $a_i \bmod 2^{k_i}$. Theorem (open cycle with a node conflict). Let $(a_i,b_i)_{i<m}$ be an open cycle in $T$, and suppose two of its pairs $i,j$ lie at the same node but point in opposite directions:

Then no word $T\,R$ is $\theta$-safe, for any tree witness $\theta$ and any word $R$ (in particular $T$ itself is not safe).

This generalises the interlocked obstruction, which is the case $m=2$ with pairs $(a,a+d)$, $(a+5d,a+4d)$ and both middle terms absent, to open cycles of any length. For example, any word in which $0$ precedes $1$, $-1$ precedes $1$, and $11$ precedes $8$, while $2,3,5$ are absent, has the 3-cycle $(0,1),(-1,1),(11,8)$, with third terms $2,3,5$ and fourth terms $3,5,2$. Pairs $(0,1)$ and $(11,8)$ are both at level $0$, and $8\not\equiv 1 \pmod 2$, so such a word can never be extended safely. It is a forbidden configuration for any construction in the tree-witness approach to Erdős #195, not a construction or an impossibility result.

Formalization note: words are List ℤ, positions via List.idxOf, the pairs are a b : ℕ → ℤ on indices i < m, the successor index is (i+1) % m; the conclusion is ¬ ∃ θ, Safe (treeLt θ) (T ++ R).