 Interval threshold strategy for discrete linear orders

With $k$ moves remaining, preserve corresponding gap sizes exactly when smaller than $2^k-1$, and otherwise require both gaps to have at least that many points. One move splits a gap; $2^k-1=2(2^{k-1}-1)+1$ permits matching both smaller gaps. This strategy proves that $(\mathbb{N},<)$ and $(\mathbb{N}+\mathbb{Z},<)$ agree at every finite quantifier rank despite the second order containing points with infinitely many predecessors.