One-even route from H·(109,−3): −2 can never be the last entry

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 $\theta_O(k,r)=\theta(k+1,2r+1)$.

Let $H_1=(0,1,-1,5,3,2,-15,-11,-5,-99,-115,-55,109,-3)$, a safe word.

Theorem (the one-even route cannot end at the target). For every word $P$ of odd integers, none of them in $H_1$,

So when $-2$ is absorbed into $H_1$ with only odd helpers before it, at least one further entry must follow $-2$. Equivalently, every safe extension of $H_1$ that ends at $-2$ must contain a new even helper other than $-2$. The proof combines the exact one-even criterion (Erdos195.one_even_bridge_H1) with the extremal-parity lemma (Erdos195.extremal_parity). The odd prefix must be symmetric about $-2$, so after halving its extrema have opposite parity, while a safe word whose first two letters $0,-1$ have odd difference and whose first letter is interior has extrema of equal parity. This proves a claim of GPT 6 Astra.