A zero Wada denominator makes that deleted column inadmissible

Under [A1], realize $S^3$ as $\{(z,w)\in {\mathbb{C}}^2:\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2=1\}$ and its standard unknot as $K=\{w=0\}$. Choose $0<a<1\mathrm{/}2$ and the normal-disk tube $N(K)=\{\mathrm{\mid }w\mathrm{\mid }\le a\}$, parametrized by $(v,w)\mapsto (\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2}v,w)$ for $v\in S^1$. By [F3], its exterior is $\{\mathrm{\mid }w\mathrm{\mid }\ge a\}$. Put $b=\sqrt{1-a^2}$. The map $D_b^2\times S^1\to E_K$ given by $(z,u)\mapsto (z,\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}u)$ has inverse $(z,w)\mapsto (z,w\mathrm{/}\mathrm{\mid }w\mathrm{\mid })$, exhibiting a solid torus. Contracting the disk factor retracts it onto its core; a positive normal-disk meridian winds once around the circle factor, oriented accordingly. For a finite simple-spine certificate, triangulate the disk as a cone on a polygon. Collapse its triangles along their outer edges, then its radial edges at their outer vertices, leaving the center. In the product CW structure, perform each disk collapse first over the circle's $1$-cell and then over its vertex. These elementary cellular collapses leave the core circle with one vertex and edge $y$, representing the positive meridian of [F4].

Attaching a $1$-cell $x$ at that vertex and a $2$-cell with attaching word $x$ is an elementary expansion: the new $1$-cell is a free face of the new $2$-cell, so collapsing this pair returns the circle spine. Therefore $⟨x,y\mid x⟩$ is a valid based deficiency-one spine of the same exterior, in the sense of [F1]. The relation kills $x$ in the fundamental group, so [F4] makes the full abelianization send $x$ to $1$ and the positive meridian $y$ to its generator $t$.

The Fox derivatives of $r=x$ are $\mathrm{\partial }r\mathrm{/}\mathrm{\partial }x=1$ and $\mathrm{\partial }r\mathrm{/}\mathrm{\partial }y=0$, so the row Fox matrix is $J=[10]$. By [F2], the $d_1$ blocks are $0$ for $x$ and $t-1$ for $y$, making the localized complex $0\to \mathbb{Q}(t)\stackrel{(1,0)}{\to }\mathbb{Q}(t{)}^2\stackrel{(0,t-1{)}^{\mathsf{T}}}{\to }\mathbb{Q}(t)\to 0$ acyclic. Deleting the $x$ column leaves the zero $1\times 1$ minor and gives denominator $1-1=0$, hence [F1] declares that quotient undefined. Deleting the $y$ column leaves minor $[1]$ and denominator $t-1$, so its quotient is $1\mathrm{/}(t-1)$. This establishes the stated boundary case.