Seifert matrices of a knot are S-equivalent

At an isotopy step in the finite surface sequence of [L1], [F2] gives equality of the matrices in the transported basis. If the target basis is changed by $P\in {\mathrm{G}\mathrm{L}}_{2g}(\mathbb{Z})$, bilinearity gives $V^{\mathrm{\prime }}=P^{T}VP$, which is one of the defining S-equivalence moves by [F4].

Consider one arc surgery in that sequence. Choose the old basis curves away from the two surgery disks. The added product cylinder contributes a longitudinal class $b$ and a meridian class $c$; together with the old basis these form an integral basis of the new surface because the surgery adds one orientable handle and hence two first-homology generators. In a cross-section $\{t_0\}\times D^2$ of the cylinder, $c=\{t_0\}\times \mathrm{\partial }{D}^2$ bounds the disk $D=\{t_0\}\times D^2$ and the longitudinal class $b$ meets $D$ once. The disk and its parallel copies miss the old curves and their positive push-offs, so both pairings between $c$ and every old basis curve vanish. If the positive normal of the lateral annulus points outward from the cylinder, $b^{+}$ misses $D$ and the slightly larger disk bounded by $c^{+}$ meets $b$ once; hence [F6] gives $V(b,c)=0$ and $V(c,b)=\pm 1$. If that normal points inward, the smaller disk bounded by $c^{+}$ misses $b$, while $b^{+}$ crosses $D$ once, giving $V(c,b)=0$ and $V(b,c)=\pm 1$. Orient $c$ so the single nonzero entry is $+1$. The same disks show $V(c,c)=0$. The tube core has integer self-linking $z=V(b,b)$, and its pairings with the old curves give integer column and row vectors $x$ and $y^T$. By [F7], the difference between the two new cross entries equals the oriented intersection of the same ordered basis curves, so this is exactly one of $E_0(V;x,y,z)$ or $E_1(V;x,y,z)$, up to an integral basis change.

Apply [L1] to obtain a finite sequence of isotopies and arc surgeries or their reverses from $F$ to $F^{\mathrm{\prime }}$. Steps 1.1 and 1.2 turn every isotopy or forward surgery into a defining S-equivalence move, and reverse surgery gives its inverse. Their finite composition, including initial and final basis changes, proves the statement.