I guess it maybe called a "random-truncated normal distribution".

The cumulative distribution function can be calculated by definition as usual:

\( \Pr\{A \leq a|A > B\}

= \frac {\Pr\{A \leq a, A > B\}} {\Pr\{A > B\}} \)

Since \( A \sim \mathcal{N}(\mu_A, \sigma^2_A), B \sim \mathcal{N}(\mu_B, \sigma^2_B) \) and they are independent,

we have \( A - B \sim \mathcal{N}(\mu_A - \mu_B, \sigma^2_A + \sigma^2_B) \)

and thus the denominator

\( \Pr\{A > B\} = \Pr\{A - B > 0\}

= \Phi\left(\frac {\mu_A - \mu_B} {\sqrt{\sigma^2_A + \sigma^2_B}}\right) \)

For the numerator,

\( \Pr\{A \leq a, A > B\} = \int_{-\infty}^a \Pr\{B < x\}f_A(x)dx \)

which at least can be computed numerically.

If you seek the probability density function, by differentiation,

\( f_C(a) =

\Phi\left(\frac {\mu_A - \mu_B} {\sqrt{\sigma^2_A + \sigma^2_B}}\right)^{-1}

\Phi\left(\frac {a - \mu_B} {\sigma_B}\right)\frac {1} {\sigma_A}\phi\left(\frac {a - \mu_A} {\sigma_A}\right) \)