Proof. For any $z \in A$, denote $R(z) = \bracs{(\lambda - z)^{-1}|\lambda \in \sigma_A(z)}$. Let $B \subset A$ be the closed subalgebra generated by $1$, $x$, $y$, $R(x)$, $R(y)$, $R(xy)$, $R(x + y)$, then $B$ is a commutative algebra with $\sigma_{B}(x) = \sigma_{A}(x)$, $\sigma_{B}(y) = \sigma_{A}(y)$, $\sigma_{B}(xy) = \sigma_{A}(xy)$, and $\sigma_{B}(x + y) = \sigma_{A}(x + y)$.

By Proposition 29.8.2, for each $z \in B$, $(\Gamma_{B}z)(\Omega(B)) = \sigma_{B}(z)$. Since for any $u, v \in C(\Omega(B); \complex)$,

1. (1)

   $\sigma_{C(\Omega(B); \complex)}(u + v) \subset \sigma_{C(\Omega(B); \complex)}(u) + \sigma_{C(\Omega(B); \complex)}(v)$.

2. (2)

   $\sigma_{C(\Omega(B); \complex)}(uv) \subset \sigma_{C(\Omega(B); \complex)}(u)\sigma_{C(\Omega(B); \complex)}(v)$.

The above holds for $x$ and $y$ with respect to $\sigma_{A}$.$\square$