[nooonuii]

$\displaystyle 0.5(1+erf(\frac{x}{\sqrt{2}}))=\frac{1}{2}+\frac{1}{2}\cdot\frac{2}{\sqrt{\pi}}\int_0^{x/\sqrt{2}}e^{-t^2}dt$

$\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle = \dfrac{1}{2}+\dfrac{1}{\sqrt{\pi}}\int_0^x e^{-u^2/2}\frac{du}{\sqrt{2}};u=\sqrt{2}t$

$\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle = \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{-u^2/2}du+\dfrac{1}{\sqrt{2\pi}}\int_0^x e^{-u^2/2}du$

$\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-u^2/2}du$

$\quad\quad\quad\quad\quad\quad\quad\quad=F(x)$