At an $\varepsilon$-safe cut of a $4$-AP-free enumeration of $\mathbb{Z}$, nothing before the earlier extreme has the extremes' $2$-adic valuation

Let $f:\mathbb{N}\to\mathbb{Z}$ be a bijection with no monotone $4$-term arithmetic progression, and fix a cut $N$ such that the prefix $f(0),\dots,f(N-1)$ is $\varepsilon$-safe for some $\varepsilon:\mathbb{N}\to\{0,1\}$ (i.e. the completion "prefix, then all remaining integers in the $2$-adic order $\prec_\varepsilon$" has no monotone $4$-AP; here $x\prec_\varepsilon y$ iff the $k$-th two's-complement digit of $x$ equals $\varepsilon_k$, $k=\nu_2(y-x)$). Suppose positions $i<j<N$ carry the two extremes of the prefix (the minimum and the maximum, in either order), and let $k_0=\nu_2(f(j)-f(i))$. Then no element placed before the earlier extreme has the same valuation against it: for every $a<i$, $\nu_2(f(i)-f(a))\neq k_0$ (equivalently $f(a)\not\equiv f(j)\pmod{2^{k_0+1}}$).

Proof idea: the pairs $(a,i)$ and $(i,j)$ are both open at $N$ (their continuations $2y-x$, $3y-2x$ lie beyond the extremes, hence outside the prefix); by the digit-consistency criterion for open pairs, equal valuations $k$ would force the $k$-th digits of $f(i)$ and $f(j)$ to agree, but they differ at $k=\nu_2(f(j)-f(i))$. A necessary condition on every cut of a Ho-style chain with stage-dependent $\varepsilon$ for Erdős Problem 195; it does not decide the problem.