Odd-entry demand for root-releasing tree-safe words starting with 0

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. An in-order pair $(a,b)$ of $Q$ ($a$ occurs before $b$) is open if $2b-a\notin Q$ and $3b-2a\notin Q$; it lies at the node $(k,\,a\bmod 2^k)$, $k=\nu_2(b-a)$, and forces $\theta$ there to equal $\mathrm{bit}_k(b)$. A root pair is one with $b-a$ odd (node $(0,0)$); $Q$ is a true root release if every open root pair has even second entry, i.e. no open pair forces the value $1$ at the root. Let $Q$ be a $\theta$-safe word whose first letter is $0$ and which is a true root release. Then for every odd letter $z$ of $Q$:

Moreover, if $Q$ has an odd letter, then some odd letter $z$ is grounded: $3z\notin Q$ and $2z$ occurs before $z$. Thus every odd letter starts a forced chain $z, 3z, 9z,\dots$ that must end at a grounded letter whose double is placed earlier; this is the basic necessary condition for releasing the root after a prefix, e.g. after $(0,1,x)$ the letters $3$ and $3x$ are forced (Erdős Problem 195, tree-witness approach).

Formalization note: treeLt θ is from Definitions.Def_Erdos195_tree; Safe, bitOne are from Definitions.Def_Erdos195_defs. Positions are List.idxOf; "starts with 0" is Q.head? = some 0; the release hypothesis is stated as: every open in-order pair $(a,b)$ with $(b-a) \bmod 2 = 1$ has $b \bmod 2 = 0$.