The Seifert form gives a quadratic enhancement and basis-independent Arf value

If two integral coordinate vectors $x,x^{\mathrm{\prime }}$ represent the same class modulo two, then $x^{\mathrm{\prime }}=x+2y$ for an integral vector $y$, and $(x+2y{)}^{T}V(x+2y)-x^{T}Vx=2x^{T}Vy+2y^{T}Vx+4y^{T}Vy$ is even. Thus $x^{T}Vx(\mathrm{m}\mathrm{o}\mathrm{d}2)$ is well-defined. Expanding the quadratic expression gives $q_F(x+y)-q_F(x)-q_F(y)=x^T(V+V^T)y(\mathrm{m}\mathrm{o}\mathrm{d}2)=x^T(V-V^T)y=x{\cdot }_{F}y,$ where plus and minus agree in ${\mathbb{F}}_2$. By [F2], this alternating form is nondegenerate.

In a symplectic basis, every vector is uniquely $x={\sum }_i({\alpha }_i{a}_i+{\beta }_i{b}_i)$ with ${\alpha }_i,{\beta }_i\in {\mathbb{F}}_2$. The refinement identity gives $q_F(x)={\sum }_i({\alpha }_i{q}_F(a_i)+{\beta }_i{q}_F(b_i)+{\alpha }_i{\beta }_i)\mathrm{.}$ Therefore the finite Gauss sum factors as ${\sum }_{x\in H_1(F;{\mathbb{F}}_2)}(-1{)}^{q_F(x)}={\prod }_i{\sum }_{a,b\in {\mathbb{F}}_2}(-1{)}^{a{q}_F(a_i)+b{q}_F(b_i)+ab}=2^g(-1{)}^{{\sum }_i{q}_F(a_i)q_F(b_i)}\mathrm{.}$ The left side depends only on $q_F$, not on a basis, so the exponent defining the Arf value is basis-independent.

Under either enlargement in [F3], the new basis vectors $u,v$ have $u\cdot v=1$, $q(u)=z(\mathrm{m}\mathrm{o}\mathrm{d}2)$, and $q(v)=0$; moreover $v$ is orthogonal to all old basis vectors because its encircling disk misses them. If $u$ pairs nontrivially with old basis vectors, add a linear combination of the old symplectic basis to $u$ to obtain $u^{\mathrm{\prime }}$ orthogonal to all of them; the nondegenerate old intersection matrix gives the required coefficients, and $u^{\mathrm{\prime }}\cdot v=1$ remains unchanged. The old basis together with $u^{\mathrm{\prime }},v$ is symplectic; its new Arf contribution is $q(u^{\mathrm{\prime }})q(v)=0$, while the old $q$-values are unchanged. More precisely, put $u^{\mathrm{\prime }\mathrm{\prime }}=u^{\mathrm{\prime }}+q(u^{\mathrm{\prime }})v$. The refinement identity and $q(v)=0$, $u^{\mathrm{\prime }}\cdot v=1$ give $q(u^{\mathrm{\prime }\mathrm{\prime }})=q(u^{\mathrm{\prime }})+q(u^{\mathrm{\prime }})=0$, while $u^{\mathrm{\prime }\mathrm{\prime }}\cdot v=1$ and both $u^{\mathrm{\prime }\mathrm{\prime }},v$ remain orthogonal to the old subspace. Thus the enlarged quadratic space is isometric to the old one orthogonally summed with the hyperbolic plane $q_H(s,t)=st$: on $s{u}^{\mathrm{\prime }\mathrm{\prime }}+tv$ the value is $st$. An integral congruence reduces to an invertible mod-two coordinate change and hence an isometry of quadratic spaces. Consequently, the stable isometry class is invariant under congruences, enlargements and their reverses. Here stable isometry means isometry after adjoining finitely many copies of $q_H$ on either side. Thus the Gauss-sum Arf value is unchanged by enlargement or reduction. By the knot-matrix S-equivalence lemma, all spanning surfaces give the same value.