If $X,Y\sim N(0,1)$ independent and I have to compute $\mathbb{P}(X+Y<z)-\mathbb{P}(X+Y<0)$, I write:
$=[[\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-\frac{y^2}{2}}dy]\cdot [\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z-y}e^{-\frac{x^2}{2}}dx]]-[\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-\frac{y^2}{2}}dy]\cdot [\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-y}e^{-\frac{x^2}{2}}dx]=$ ...
Thus, for the first probability:
… = $[…]\cdot [\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z-y}e^{-\frac{x^2}{2}}dx]=[…]\cdot [\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{-\frac{x^2}{2}}dx\cdot \frac{1}{\sqrt{2\pi}}\int_{0}^{z-y}e^{-\frac{x^2}{2}}dx]=[…]\cdot [\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_{0}^{z-y}e^{-\frac{x^2}{2}}dx]$
How can I solve the last integral?
Note that $X+Y\sim N(0,2)$. Then $$ \mathsf{P}(X+Y< z)-\mathsf{P}(X+Y< 0)=\Phi(z/\sqrt{2})-1/2. $$