Every tree-safe word with a fully open pair admits a tree-safe sink step

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 tree-safe and suppose $T$ has a fully open in-order pair $(a_0,b_0)$ (both $2b_0-a_0\notin T$ and $3b_0-2a_0\notin T$). Then there is an open path $(a_i,b_i)_{i<m}$, $m\ge1$, in $T$ whose sink fourth term

is a true-sink fourth term of $T$ and such that $T\,w$ is again tree-safe.

So a deadlock in which every sink step fails cannot occur: one closing step (appending the fourth term of the sink pair, which closes it) is always available. The proof takes $w$ to be the $\prec_\theta$-least true-sink fourth term, which exists because there are finitely many and at least one by milestone #154, and applies the preceding milestone.