Escape chains terminate at every node, for every tree witness

A tree witness $\theta:\mathbb N\times\mathbb Z\to\{0,1\}$ defines a strict total order $\prec_\theta$ on $\mathbb Z$: for $x\ne y$ with $k=\nu_2(y-x)$, $x\prec_\theta y$ iff the $k$-th binary digit of $x$ is $1$ exactly when $\theta(k,x \bmod 2^k)=1$. A finite word $T$ of integers is $\theta$-safe if the sequence "letters of $T$ in order, then all remaining integers in $\prec_\theta$-order" has no monotone 4-term arithmetic progression (positions are first occurrences); $T$ is tree-safe if it is $\theta$-safe for some $\theta$. For an in-order pair $(a,b)$ of $T$ ($a$ before $b$) call $2b-a$ its third term and $3b-2a$ its fourth term. An open path in $T$ is a family of in-order pairs $(a_i,b_i)$, $0\le i<m$, $m\ge1$, with $3b_i-2a_i=2b_{i+1}-a_{i+1}$ for $i<m-1$ and no third term $2b_i-a_i$ in $T$. Its sink fourth term $w'=3b_{m-1}-2a_{m-1}$ is a true-sink fourth term of $T$ if $w'\notin T$ and every in-order pair $(x,y)$ of $T$ with $2y-x=w'$ has $3y-2x\in T$ (condition C1 for appending $w'$).

Theorem. Let $Q$ be $\theta$-safe for some tree witness $\theta$, and let $o,w$ be letters of $Q$ with $o$ before $w$ and $o\prec_\theta w$ (a "wrong" in-order pair at the node of $(o,w)$). Put $e_n=o+3^n(w-o)$, so $e_0=w$ and $e_{n+1}=3e_n-2o$. Then there is $n\ge0$ such that $e_0,\dots,e_n$ all occur in $Q$, $e_0,\dots,e_{n-1}$ occur after $o$, and either

1. (reflection) $e_n$ occurs after $o$, and $2e_n-o$ occurs in $Q$ before $e_n$; or
2. (reversal) $n>0$ and $e_n$ occurs before $o$.

Since $\nu_2(e_n-o)=\nu_2(w-o)$ and $e_n\equiv w \pmod{2^{\nu_2(w-o)+1}}$, all pairs $(o,e_n)$ lie at the same node and are wrong for $\theta$; the chain cannot go on forever because the $e_n$ are distinct and $Q$ is finite. This generalizes milestone #142 (the root-node case $\theta(0,0)=1$, $o$ odd, $w$ even) to arbitrary levels, nodes and witnesses.