The formal determinant tie-break is nondegenerate and consistent

Regard one requested determinant as a polynomial in independent coordinate increments $z_{i,j}$. It is not the zero polynomial: for orientation choose the perturbed points $0,e_1,\dots ,e_d$, whose homogeneous determinant is nonzero. For a lifted determinant choose the first $d+1$ perturbed points $0,e_1,\dots ,e_d$ and the last point $(2,\dots ,2)$. The base sphere has center $(1\mathrm{/}2,\dots ,1\mathrm{/}2)$ and squared radius $d\mathrm{/}4$, while the last point has squared distance $9d\mathrm{/}4$ from that center; [F3] therefore gives a nonzero lifted determinant. These coordinate assignments are available because the $z_{i,j}$ are independent variables.

In every orientation determinant each coordinate increment has degree at most one. In a lifted determinant it has degree at most two, because it can occur squared in one lifted entry and a determinant term selects only one entry from that row. Thus every monomial exponent vector has digits $a_k\in \{0,1,2\}$. Under the substitution in [F2], its $\epsilon$ exponent is ${\sum }_k{a}_k{3}^k$. Uniqueness of base-three expansion makes this map injective, so distinct monomials cannot merge or cancel. The nonzero polynomial from [step 1.1] remains nonzero after substitution.

Write any resulting polynomial as $P(\epsilon )=c_m{\epsilon }^m+{\sum }_{j>m}{c}_j{\epsilon }^j$, with $c_m\mathrm{\ne }0$. If $B={\sum }_{j>m}\mathrm{\mid }{c}_j\mathrm{\mid }$ is positive, then for $0<\epsilon <\min (1,\mathrm{\mid }{c}_m\mathrm{\mid }\mathrm{/}(2B))$ the tail after dividing by ${\epsilon }^m$ has absolute value at most $\epsilon B<\mathrm{\mid }{c}_m\mathrm{\mid }\mathrm{/}2$ by [F5] and [F6]. If $B=0$, its sign is already constant. In either case the sign for sufficiently small positive $\epsilon$ is the sign of $c_m$. If the original determinant is nonzero, its constant coefficient is nonzero and is the first coefficient, so its sign is preserved.

A finite point set has only finitely many ordered orientation and lifted predicate tuples. For each resulting polynomial with tail bound $B>0$ in [step 3.1], form its positive threshold $t_P=\min (1,\mathrm{\mid }{c}_m\mathrm{\mid }\mathrm{/}(2B))$. If the finite set of these thresholds is nonempty, let $t$ be its minimum and choose $\epsilon =t\mathrm{/}2$; then [step 3.1] applies to every such polynomial. Every remaining polynomial has $B=0$, so it is a single nonzero monomial and has its first-coefficient sign for every $\epsilon >0$. If there are no thresholds, choose $\epsilon =1$; the same reasoning covers the case of an empty predicate family. Thus one positive value realizes all returned signs simultaneously and no declared determinant is zero. The algorithm returns exactly these first-coefficient signs without selecting or evaluating that $\epsilon$, so the tie-break is consistent and uses no arbitrary epsilon threshold.