Yes, I'm being stupid (no shock there!). There is a work around since things are Gaussian. If $z\sim \mathcal{N}(\mu, \sigma)$ and $z\sim \mathcal{N}(\mu, \sigma)$, then $x-z \sim \mathcal{N}(0, \sqrt{2} \sigma)$. Then the result follows easily,

$Pr(|x-z|<a) = \text{Erf}(a/(2 \sigma))$.

Yes, I'm being stupid (no shock there!). There is a work around since things are Gaussian. If $z\sim \mathcal{N}(\mu, \sigma)$ and $x\sim \mathcal{N}(\mu, \sigma)$, then $x-z \sim \mathcal{N}(0, \sqrt{2} \sigma)$. Then the result follows easily,

$Pr(|x-z|<a) = \text{Erf}(a/(2 \sigma))$.