I have not added the full proof of proposition 3.2 because it is not relevant for my problem. In (3.1) they have the cdf $F_X (a-y)$. In (3.2) they have differntiated it with respect to a and they call it $f_X(a-y)$. In proposition 3.2 they use (3.2). It seems they use the normal distribution for $X$ with $\mu=0$ and $var=\sigma ^2$ and with $x=a-y$. Now my problem is if they integrate with respect to y and have integration limits a-y in (3.1) on the normal disitribution. And afterwards differntiate with respect to a in (3.2) and a which clearly is not the same y as a variable. How can they then end up with the normal distribution for $f_X(a-y)$? If you integrate the normal distribution on an online integral calculator you end up with:

$\frac{1}{\sqrt{2 \pi}\sigma} \int_{-\infty}^a e^{-\frac{(x-\mu)^2}{2\sigma^2}} dy=\frac{erf (\frac{x-\mu}{\sqrt{2}\sigma})}{2}$

Where erf is the error function. I dont know how to differentiate the error function because it is defined as an integral.