Fourth-term completion of an extremal open pair keeps the forced node value

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. Let $(u,w)$ be an open in-order pair of a finite word $P$, let $k=\nu_2(w-u)$, and suppose $w$ is the numerical maximum or the numerical minimum of the letters of $P$ congruent to $w$ modulo $2^k$. Put $c=3w-2u$. Then in the word $P\cdot c$ the pair $(u,c)$ is in order and open, lies at the same node as $(u,w)$, and forces the same value:

So completing an extremal open pair by its fourth term never releases the node it was forcing; the forced value persists (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 in a word are List.idxOf; P ++ [c] appends c.