Interlocked obstruction at every level: a, a+d before; a+5d before a+4d; middle terms absent

A tree witness $\theta:\mathbb N\times\mathbb Z\to\{0,1\}$ defines a strict total order $\prec_\theta$ on $\mathbb Z$ (milestone framework); it satisfies Ho's identity $(*)$: $x\prec x+r \iff x+2r\prec x+3r$. 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. Theorem (interlocked obstruction, any level). Let $T$ be a finite word, $d\neq 0$, and suppose $a, a+d, a+4d, a+5d\in T$ with $a$ before $a+d$ and $a+5d$ before $a+4d$, while $a+2d\notin T$ and $a+3d\notin T$. Then for every tree witness $\theta$ and every word $R$, the extension $T\,R$ is not $\theta$-safe.

Why: the pairs $(a,a+d)$ and $(a+5d,a+4d)$ have swapped third/fourth terms ($a+2d$, $a+3d$), so by the interlocked-pairs lemma neither term can ever be added. Then both pairs must be $\prec_\theta$-decreasing. But by $(*)$ and totality exactly one of $a\prec a+d$ and $a+5d\prec a+4d$ holds, for any order with $(*)$. This generalises the root-level obstruction (the case $d$ odd) to every 2-adic level and every residue node. It is a structural no-go lemma for Erdős #195, not a construction or an impossibility result.

Formalization note: words are List ℤ, positions via List.idxOf, safety is Safe (treeLt θ).