$\newcommand{\Exp}{\operatorname{Exp}}$Let's suppose
$A\sim \Exp(\lambda_1)$,
$B\sim\Exp(\lambda_2)$,
$C = \min(A,B)$
$D'\sim \Exp(\lambda_3)$
$D = A + D' $ if $C \ne A $ and
$D= C$ if $C=B$
How would I find the distribution of $D$ in either scenario, or even $C$ for that matter?
(We are assuming that $A$, $B$, and $D'$ are independent.) Note that for any $t>0$, $\{C>t\} = \{A>t\}\cap\{B>t\}$. It follows that $$ \mathbb P(C>t) = \mathbb P(A>t)\mathbb P(B>t) = e^{-\lambda_1 t}e^{-\lambda_2t}=e^{-(\lambda_1+\lambda_2)t}, $$ and so $C$ has $\mathsf{Exp}(\lambda_1+\lambda_2)$ distribution. Assuming that $\lambda_3\ne\lambda_1$, we compute the density of $A+D'$ by convolution: \begin{align} f_{A+D'}(t) &= (f_A\star f_{D'})(t)\\ &= \int_0^t f_A(s)f_{D'}(t-s)\ \mathsf ds\\ &= \int_0^t \lambda_1 e^{-\lambda_1s}\lambda_3e^{-\lambda_3(t-s)}\ \mathsf ds\\ &= \lambda_1\lambda_3 e^{-\lambda_3t}\int_0^t e^{-(\lambda_1-\lambda_3)s}\ \mathsf ds\\ &= \frac{\lambda_1\lambda_3}{\lambda_1-\lambda_3}\left( e^{-\lambda_3 t}-e^{-\lambda_1 t}\right). \end{align} (If $\lambda_3=\lambda_1$ then the density is $\lambda_1(\lambda_1 t)e^{-\lambda_1 t} $.) Write $$ D = (A+D')\cdot\mathsf 1_{\{B<A\}} + C\cdot\mathsf 1_{\{B>A\}}, $$ then by symmetry \begin{align} f_D(t) &= \frac12 f_{A+D'}(t) + \frac12 f_C(t)\\ &= \frac{\lambda_1\lambda_3}{2(\lambda_1-\lambda_3)}\left( e^{-\lambda_3 t}-e^{-\lambda_1 t}\right) + \frac12(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)t}. \end{align}