Failure of the second backward step is located by level

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 $T$ be a finite word, $b\in T$, $a\ne b$, $k=\nu_2(b-a)$, $w=3b-2a\notin T$, $t=2b-a\notin T$, and let $Q=T\,w$ be $\theta$-safe. Assume $t$ passes C1 in $Q$ (every in-order pair $(x,y)$ of $Q$ with $2y-x=t$ has $3y-2x\in Q$). If $Q\,t$ is not tree-safe for any witness, then there exist $x\in T$ with $(x,t)$ fully open in $Q\,t$ ($2t-x,\,3t-2x\notin Q$) and a fully open in-order pair $(a',b')$ of $Q$ (third and fourth terms not in $Q$, neither equal to $t$) at the same level $j=\nu_2(b'-a')=\nu_2(t-x)$ and node $a'\equiv x\pmod{2^j}$ with opposite digits of $b'$ and $t$ at level $j$; moreover exactly one of:

1. (high) $j>k$ and $x\equiv a\pmod{2^{k+1}}$;
2. (level $k$) $j=k$, $x\equiv b\pmod{2^{k+1}}$, and the $k$-th digits of $b'$ and $b$ agree (the blocker has the orientation of $(a,b)$);
3. (low) $j<k$ and $w$ is already released at $x$ in $T$: $2w-x\in T$ or $3w-2x\in T$.

The low case uses that $b,t,w$ agree modulo $2^k$, so below level $k$ the pair $(x,w)$ of $Q$ would sit at the same node with the orientation of $(x,t)$ and conflict with $(a',b')$ unless released. This covers blockers among old pairs of $T$ as well as new pairs $(x',w)$, extending milestone #161 (new pairs only, level $k$) towards a full criterion for the second step of #158–#159.