The setup:
Let two default times be defined as
$$\tau_i=\inf\left\{t: \int_0^t\lambda_i(s) \,ds \geq E_1\right\},\quad i\in\{1,2\}$$
Where $E_1$ is a standard exponential random variable and $\lambda_i$(X) is a stochastic hazard rate related the default of $i$. Let $\tau_1$ and $\tau_2$ be independent. In addition let
$$S_i(t,u)=E\left[1_{\tau_i>u}|F_t\right]=1-Q_i(t,u)$$
With $F_t=\sigma(\left\{\tau\leq u\}:u\leq t\right)$ being the filtration generated by default events.
My question stems from the calculation of the article Everything you Always Wanted to Know About XVA Model Risk but Were Afraid to Ask where the following relation is made: $$\int_t^TE\left[1_{t<\tau_1<\tau_2<T}|F_t\right]du=\int_t^TS_2(t,u)dQ_1(t,u)$$
My first concern is that the LHS is an integrant over a constant (the expectation is not time dependant) and so I think it should actaully be $$\int_t^TE\left[1_{t<\tau_1<\tau_2<u}|F_t\right]du=\int_t^TS_2(t,u)dQ_1(t,u)$$ From here on I still have a hard time figuring out the rewriting, but I was wondering if the would hold $$E\left[1_{t<\tau_1<\tau_2<u}|F_t\right]=E\left[1_{\tau_1<\tau_2<u}|F_t\right]$$ The intuition behind this is that the defaults what be know given at time $t$. From here on I'm confused about how to use the independence of the two default times. Maybe I'm missing a logical argument that when integrating over the survival of the second. But here is my try at the derivation: $$ \begin{align} E\left[1_{t<\tau_1<\tau_2<T}|F_t\right]&=P(t<\tau_1<\tau_2<T) \\ &=\int_t^TE\left[1_{t<\tau_1<\tau_2<u}|F_t\right]du\\ &=\int_t^T\int_t^u dS_2(t,s) dQ_1(t,u)\\ &=\int_t^T S_2(t,u) dQ_1(t,u) \end{align} $$ With the intuition being that the survival of $i=2$ up until some time $u$ depends whether 1 has defaulted or not and then looking at the default probability.
I'm not sure if this totally correct or if I'm missing some argumentation.