Midpoint confinement: an opposite-parity batch against an old root force

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. The halves of a word are the subsequences of its even letters and of its odd letters, each mapped by $x\mapsto\lfloor x/2\rfloor$; the half witnesses are $\theta_E(k,r)=\theta(k+1,2r)$ and $\theta_O(k,r)=\theta(k+1,2r+1)$.

Theorem (midpoint confinement for an opposite-parity batch). Let $Q$ be a word containing an in-order pair $(a,b)$ with $b-a$ odd that is open in $Q$ ($2b-a,\,3b-2a\notin Q$). Suppose the numerical extrema $L=\min Q$ and $U=\max Q$ have the parity of $b$. Let $D$ be a word of entries not in $Q$, all of the opposite parity. If $Q\cdot D$ is $\theta$-safe for a tree witness $\theta$, then every entry $z$ of $D$ satisfies

and if moreover $L\equiv U\pmod 4$, then $D$ is empty.

So an old root force toward the parity of both extrema cannot be overridden by a monochromatic batch of the other parity: such a batch has at most one value, the midpoint. It is the key step in Astra's (not yet formalized) argument that a safe enumeration of $\mathbb Z$ must use genuinely mixed-parity batches infinitely often. The old open pair cannot be retired because its fourth term has the parity of $b$. A new entry beyond the midpoint would form an open root pair with an old extreme that forces the wrong parity. Lemma of GPT 6 Astra.