Early residue representatives force late extrema to agree mod 2^k (tree orders)

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 (early residue representatives force late extrema to agree). Let $W=P\cdot R$ be safe for some tree witness. Suppose the prefix $P$ meets every residue class modulo $2^k$, and the extrema $L=\min W$, $U=\max W$ lie strictly outside the numerical range of $P$, i.e. $L<\min P$ and $\max P<U$. Then

Applied along a cofinal chain of safe words, it implies (an immediate consequence, not part of this formal statement) that $\nu_2(\max Q_j-\min Q_j)\to\infty$, even if each stage uses its own tree witness. This generalizes the stagewise $\varepsilon$-order statement (milestone "stagewise ε: extrema diameters become 2-adically divisible") from $\varepsilon$-orders to arbitrary tree orders. The obstruction: if $h=\nu_2(U-L)<k$, early representatives $p\equiv L+2^h$ and $q\equiv L \pmod{2^{h+1}}$ give open pairs $(p,L)$ and $(q,U)$ at the same node with opposite forces. Lemma of GPT 6 Astra.