Target-last absorption of $2$ after $(0,1,-1)$ forces a released word

A tree witness is a function $\theta(k,r)\in\{0,1\}$ on the nodes $(k,r)$, $0\le r<2^k$, of the binary tree of residues. For distinct integers $x,y$ with $k=\nu_2(y-x)$, put $x\prec_\theta y$ iff $\mathrm{bit}_k(x)=\theta(k,\,x\bmod 2^k)$. A finite word $Q$ of integers is $\theta$-safe if the sequence "the letters of $Q$ in order, then all remaining integers in $\prec_\theta$-order" contains no monotone $4$-term arithmetic progression. For letters $a$ before $b$ in $Q$, the pair $(a,b)$ is OPEN if neither $2b-a$ nor $3b-2a$ occurs in $Q$. Let $\theta$ be any tree witness and let $A=(0,1,-1,r_1,\dots,r_m,2)$ be a word with $2\notin\{r_1,\dots,r_m\}$. If $A$ is $\theta$-safe, then $\theta(0,0)=0$, and $A$ is RELEASED:

So absorbing the target $2$ as the last letter after the prefix $(0,1,-1)$ is only possible by a released word. The proof (following GPT Astra) compares the extremes of $A$ via the extremal-parity lemma, applied to $A$ and to its normalized odd half.

Formalization note: treeLt θ is from Definitions.Def_Erdos195_tree, and Safe is from Definitions.Def_Erdos195_defs. Positions are List.idxOf; the word is written (0 :: 1 :: -1 :: rest) ++ [2], and the hypothesis $2\notin\mathrm{rest}$ makes $2$ the last letter and its only occurrence.