= Solution
Interpret the irreversible limit as a large favorable drop for \b[each forward step], so backward transitions during the observation can be neglected. Let the distinct forward rates be $k_1,k_2>0$. The first waiting time has <probability density> $k_1e^{-k_1t_1}$; conditional on its completion, the second has independent <probability density> $k_2e^{-k_2(t-t_1)}$. Integrating over the first completion time gives the <convolution>
$$
p_{1,2}(t)=\int_0^t k_1e^{-k_1t_1}k_2e^{-k_2(t-t_1)}\,dt_1
=\frac{k_1k_2}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right),\quad t\ge0.
$$
Hence
$$
\boxed{\Pr(t<T_1+T_2<t+dt)=p_{1,2}(t)\,dt+o(dt).}
$$
The expression is symmetric under interchanging the rates, positive for either ordering, and integrates to one. Reverse rate constants do not enter this approximation; merely making all four constants different would not justify eliminating them without the downhill condition.
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