Bit-flip symmetry: $Q$ is $\varepsilon$-safe iff $-1-Q$ is $(1-\varepsilon)$-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 $\sigma(x)=-1-x$ and let $\bar\varepsilon_k=1-\varepsilon_k$ be the complementary digit sequence. Then $Q$ is $\varepsilon$-safe if and only if the word $\sigma(Q)$ (apply $\sigma$ letterwise) is $\bar\varepsilon$-safe.