Three two-helper extensions of $(0,u)$, $u\equiv 3$, by $t\equiv 2 \pmod 4$ are never safe

Fix $\varepsilon:\mathbb{N}\to\{0,1\}$ and the $2$-adic total order on $\mathbb{Z}$ given by $x\prec_\varepsilon y$ iff $x\neq y$ and the $k$-th two's-complement digit of $x$ equals $\varepsilon_k$, where $k=\nu_2(y-x)$. A finite word $Q$ of integers is $\varepsilon$-safe if the sequence "the letters of $Q$ in order, then all remaining integers in $\prec_\varepsilon$-order" contains no monotone $4$-term arithmetic progression. Let $u\equiv 3\pmod 4$, $t\equiv 2\pmod 4$, and put $s=2t-u$, $d=3t-2u$, $h=6t-5u$. Let $Q$ be an $\varepsilon$-safe word (any $\varepsilon$) that begins with $0,u$. Then, regardless of the order of the remaining letters:

These are the three natural two-helper candidates for absorbing a target $t\equiv2\pmod4$ into the safe word $(0,u)$ with $u\equiv 3\pmod 4$ in Ho-style constructions of $4$-AP-free permutations of $\mathbb{Z}$ (Erdős Problem 195); each fails because two open in-order pairs force contradictory values of the same digit $\varepsilon_0$ or $\varepsilon_1$.

Formalization note: "begins with $0,u$" is [0, u] <+: Q; set inclusions are stated letterwise as disjunctions of equalities; Safe, epsLt come from Definitions.Def_Erdos195_defs.