Shield lemma: $(u,w,6w-5u,3w-2u)$ is tree-safe and releases the node $N(u,w)$

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 $\mathrm{bit}_k(x)=\theta(k,\,x\bmod 2^k)$ (two's-complement digits). 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; it is tree-safe if it is $\theta$-safe for some $\theta$. An in-order pair $(a,b)$ of $Q$ ($a$ occurs before $b$) is open if $2b-a\notin Q$ and $3b-2a\notin Q$; an open pair lies at the node $N(a,b)=(k,\,a\bmod 2^k)$, $k=\nu_2(b-a)$, and $\theta$-safety requires $\theta(N(a,b))=\mathrm{bit}_k(b)$ there. For all integers $u\neq w$, the four-letter word

is tree-safe, it closes the pair $(u,w)$ (its fourth term $3w-2u$ is present), and every open in-order pair $(a,b)$ of $S(u,w)$ lying at the node $N(u,w)$ forces the bit opposite to $\mathrm{bit}_k(w)$, $k=\nu_2(w-u)$. (Concretely, the open pairs are $(u,6w-5u)$, $(w,6w-5u)$, $(w,3w-2u)$, at levels $k+1$, $k$, $k+1$; for $(u,w)=(0,1)$ this is the word $(0,1,6,3)$.) Thus a 'shield' releases the value forced at $N(u,w)$ by the pair $(u,w)$: unlike the bare fourth-term completion, it flips the requirement at that node. This is a release lemma for two-letter prefixes in the tree-witness approach to Erdős Problem 195.

Formalization note: treeLt θ is from Definitions.Def_Erdos195_tree; Safe, bitOne are from Definitions.Def_Erdos195_defs. Positions in a word are List.idxOf; P ++ [c] appends c.