# Why does the cumulative distribution function of the standard normal distribution have point symmetry in (0,0.5)?

The probability distribution function for the standard normal distribution, $$\phi(z)=\frac{e^{-\frac{1}{2}z^2}}{\sqrt{2\pi}}$$ is even, with line symmetry in the y-axis.

The cumulative distribution function of the standard normal distribution is $$\Phi(z)=\int_{-\infty}^z\phi(t)dt$$.

My textbook said that

Because $$\phi(z)$$ is even, $$\Phi(z)$$ has point symmetry in $$(0,0.5)$$. Which makes sense to me by looking at their graphs (Since the PDF is the derivative of the CDF), but is there a more rigorous mathematical proof for why the CDF has point symmetry in (0,0.5)?

And more importantly, is there some general pattern/mathematical theory that I am missing here which I am supposed to know?

What I mean by this is, obviously, the primitive of $$y=x^2$$ (which is even) is $$y=x^3$$, which has point symmetry in the origin. But I was wondering if there is some greater mathematical theory that says something about 'if a function has line symmetry, its primitive will always have point symmetry' or something like that..

Thanks..

$$\Phi(x) = \int_{-\infty}^x \phi(z) dz \stackrel{z \to -z}{=} \int_{-x}^{+\infty} \phi(-z)dz = \int_{-x}^{+\infty} \phi(z)dz = 1 - \int_{-\infty}^{-x} \phi(z)dz = 1- \Phi(-x)$$