Four-class restricted absorption, exceptional case (7b−6y ∈ A): reorder the reflections with 4y−3b last

A tree witness $\theta:\mathbb N\times\mathbb Z\to\{0,1\}$ defines a strict total order $\prec_\theta$ on $\mathbb Z$ (milestone framework). A finite word $W$ of integers is SAFE for $\theta$ if the sequence "letters of $W$ in order, then all remaining integers in $\prec_\theta$-order" has no monotone 4-term arithmetic progression. Theorem (exceptional case of the four-class absorption theorem). Let $r$ be odd, let $A$ be a list of integers all $\equiv r \pmod 8$, let $a$ be even and $b\equiv r+2\pmod 4$, and put $Q=(a)\cdot A\cdot(b)$. Let $y\notin Q$ and $c=2y-b\notin Q$ with $c\equiv r\pmod 8$, and assume both $A\cdot(c)$ and $(c)\cdot A$ are SAFE (for some witnesses). Suppose $7b-6y\in A$ but $10y-9b\notin A$. Put $w=6y-5b$ and $t=4y-3b$. Then there is a duplicate-free list $L$ whose entries are exactly the elements of $\{12y-11b\}\cup(10y-10b+A)$ other than $t$, such that

is SAFE for some tree witness.

This is the exceptional case ($7b-6y\in A$) of GPT 6 Astra's four-class restricted absorption theorem. The non-exceptional case is milestone #122. Together they cover Astra's full critical exclusion $\{7b-6y,\,10y-9b\}\not\subseteq A$. When $7b-6y\in A$, the closed-form batch fails because $t=2c-b$ lies in the reflection block. The fix: order the reflection set safely with $t$ last (specified-last-entry ordering, milestone #133) and put $w$ before $t$. The proof uses the two-child decomposition at the root (the even letters are $a,y$; odd-to-$y$ pairs close by reflection about $y$) and at level 1 of the odd class. In the odd class, every cross 3-AP gets its fourth term in time; the only new case is $(b,c,t)$, whose fourth term $w$ now comes before $t$, and $(8y-7b,w,t)$, whose fourth term $c$ is old. The hypothesis $10y-9b\notin A$ rules out the cross triple $(10y-9b,\,8y-7b,\,w)$, whose fourth term would be the delayed $t$.