I'm working through the section on Laplace's Equation in Partial Differential Equations by Evans. I'm having trouble following a step in Evans's proof of the symmetry of Green's function (Theorem 2.2.13 on page 35). The statement of the theorem is as follows: For all $x,y \space \epsilon \space U$ with $x \not = y$ we have $$ G(y,x)=G(x,y)$$ The function $G$ is Green's function which is defined as $G(x,y) = \Phi(y-x)-\phi^x(y) $ where $\Phi $ is the fundamental solution of Laplace's equation and $\phi^x(y)$ is constructed such that $\Delta \phi^x = 0 $ in $U$ and $\phi^x = \Phi(y-x)$ on $\partial U $. He begins the proof by defining $v(z) := G(x,z)$ for a fixed $x$ and $ w(z) := G(y,z) $ for a fixed $y$.
In the middle of the proof, he states that $v(z) = \Phi(z-x) - \phi^x(z)$, where $\phi^x$ is smooth in $U$ and that therefore $$(*) \space \space\lim_{\varepsilon \to 0}\int_{\partial B(x,\varepsilon)}\frac{\partial v}{\partial \nu}wdS = \lim_{\varepsilon \to 0}\int_{\partial B(x,\varepsilon)}\frac{\partial \Phi}{\partial \nu}(x-z)w(z)dS = w(x)$$ Here, $\frac{\partial v}{\partial \nu} = \nabla v \cdot \nu $ where $\nu$ is the outward oriented unit normal vector. It is here that I get confused. I understand the second equality in (*) but I don't understand how the first equality works. By my understanding, $$\frac{\partial v}{\partial \nu} = \nabla v \cdot \nu = \nabla \Phi(y-x) \cdot \nu - \nabla \phi^x \cdot \nu$$ and I don't understand what happens to the $\nabla \phi^x$. Why isn't it this instead?: $$\lim_{\varepsilon \to 0}\int_{\partial B(x,\varepsilon)}\frac{\partial v}{\partial \nu}wdS = \lim_{\varepsilon \to 0}\int_{\partial B(x,\varepsilon)}[\frac{\partial \Phi}{\partial \nu}(x-z)-\frac{\partial \phi^x}{\partial \nu}]w(z)dS$$
Any help would be appreciated.