Probability of a point taken from a certain normal distribution will be greater than a point taken from another?

Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then, $$ {\rm P}(X_1 > X_2 ) = {\rm P}(X_1 - X_2 > 0) = 1 - {\rm P}(X_1 - X_2 \le 0). $$ Now, by independence, $X_1 - X_2$ is normally distributed with mean $$ \mu := {\rm E}(X_1 - X_2) = \mu_1 - \mu_2 $$ and variance $$ \sigma^2 := {\rm Var}(X_1 - X_2) = \sigma_1^2 + \sigma_2^2. $$ Hence, $$ \frac{{X_1 - X_2 - \mu}}{{\sigma}} \sim {\rm N}(0,1), $$ and so $$ {\rm P}(X_1 - X_2 \le 0) = {\rm P}\bigg(\frac{{X_1 - X_2 - \mu }}{\sigma } \le \frac{{0 - \mu }}{\sigma }\bigg) = \Phi \Big( \frac{-\mu }{\sigma }\Big), $$ where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Thus, $$ {\rm P}(X_1 > X_2 ) = 1 - {\rm P}(X_1 - X_2 \le 0) = 1 - \Phi \Big( \frac{-\mu }{\sigma }\Big). $$