3.16 "Sum of Gaussians (Optional)" - derivation question

[mryndzionek]

In 3.16 "Sum of Gaussians (Optional)", shouldn't $p(x)=\int_{-\infty}^{\infty}f_{p}(x-z)f_{z}(z)\textbf{dx}$ be
$p(x)=\int_{-\infty}^{\infty}f_{p}(x-z)f_{z}(z)\textbf{dz}$? If we integrate out $x$ then there won't be any $x$
left on the left side, right? Then also the few next formulas need an update.
In the derivation below $z$ turns into $x_{1}$, but then lower we still got $z$. Also, in the first
formula $f_{p}$ has $x - z$ as argument but then below the distribution with $x - z$ uses $\sigma_{z}$
instead of $\sigma_{p}$. I know that convolution is commutative, but I think it would be good to have this
changed for clarity.