Two inverted pairs sharing an odd letter: no-escape repairs force translates of the odd part

A tree witness $\theta:\mathbb N\times\mathbb Z\to\{0,1\}$ defines a strict total order $\prec_\theta$ on $\mathbb Z$ (milestone framework); its root bit $\theta(0,0)$ decides the order between integers of different parity. A finite word $Q$ of integers is $\theta$-safe if the sequence "letters of $Q$ in order, then all remaining integers in $\prec_\theta$-order" has no monotone 4-term arithmetic progression; positions are first occurrences. Theorem (two inverted pairs sharing the odd letter). Let $\theta(0,0)=1$. Let $T$ be a word, $b\in T$ odd, and $c\neq c'$ even letters of $T$, both after $b$, such that $t=2c-b$, $w=3c-2b$, $t'=2c'-b$, $w'=3c'-2b$ all lie outside $T$. Let $T\cdot R$ be $\theta$-safe. Suppose that no odd $o$ before $w$ (resp. $w'$) in $T\cdot R$ has $3w-2o\in T\cdot R$ (resp. $3w'-2o\in T\cdot R$). Then $w,w'\in R$. Moreover, if $u$ is the earlier and $u'$ the later of $w,w'$ and $\delta=2(u'-u)$, then

and some odd letter of $R$ lies strictly between $u$ and $u'$.

The first step is w-forcing: with root bit 1, an odd $b$ before an even $c$ with $2c-b,3c-2b\notin T$ forces $3c-2b$ into every safe end-extension. Consequently, a repair of a word with two inversions $(b,c),(b,c')$ that avoids escape letters must contain a translate of the whole odd part of $T$ by $\delta=6(c'-c)$, and these translates come before the later forced letter. So the two-inversion problem does not factor into two independent single-inversion repairs (cf. the single-inversion criterion). It is a structural necessary condition, not a construction or an impossibility result for Erdős #195.

Formalization note: words are List ℤ, positions via List.idxOf, safety is Safe (treeLt θ).