With $\varepsilon\equiv1$, every $t\notin\{0,1\}$ extends $(0,1)$ safely: $(0,1,t)$ for odd $t$, $(0,1,2t-1,t)$ for even $t$

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. Here $\varepsilon\equiv1$ (constant digit $1$). Let $t\in\mathbb{Z}\setminus\{0,1\}$. Then

Thus the safe word $(0,1)$ can absorb any prescribed integer with at most one auxiliary entry (the reflection $2t-1$ of $1$ through $t$, placed before $t$). This is the case $j=0$ of the corresponding statement for $(0,2^j)$.

Formalization note: Safe, epsLt are from Definitions.Def_Erdos195_defs; $\varepsilon\equiv1$ is fun _ => true.