$\varepsilon$-orders have no endpoints when $\varepsilon$ is not eventually constant

Let $\varepsilon:\mathbb{N}\to\{\mathrm{true},\mathrm{false}\}$ be not eventually constant. Then $\prec_\varepsilon$ has neither a least nor a greatest element:

By contrast, the two's-complement order ($\varepsilon\equiv\mathrm{true}$) has least element $-1$, which blocks Ho's method on $\mathbb{Z}$. Orders without endpoints are the candidate replacement.

Formalization Note epsLt ε x y is x ≠ y ∧ (x / 2^k % 2 = 1 ↔ ε k = true) with k = padicValInt 2 (y - x), i.e. $k=\nu_2(y-x)$; integer division is floor division, so x / 2^k % 2 is the $k$-th two's-complement digit of $x$.