The law of total probability: $P(T > t, Z = 1) = \int_t^\infty P(\cap_{j = 2}^k \{ T_j > x \} ) \lambda_1 e^{- \lambda_1 x} \ dx$? I am studying Markov processes with exponential wait times. The following is said:

Assume there are $k$ point events, denoted $w_1, \dots, w_k$, that the waiting time for $w_i$ to occur is $T_i \sim \text{Exp}(\lambda_i) \ (i = 1, \dots, k)$, and that the $T_i$s are independent. Let $T = \min_{1 \le i \le k} T_i$ be the time to the first occurrence, and let $Z = i$ if $T = T_i$, i.e., if $w_i$ is the first event to occur.

The following theorem is then given:

Let $\lambda = \lambda_1 + \dots + \lambda_k$. Then $T \perp Z$, $T \sim \text{Exp}(\lambda)$, and
$$P(Z = i) = \dfrac{\lambda_i}{\lambda}, \ \ \ (i = 1, \dots, k)$$

The proof of this then proceeds as follows:

Consider the case $i = 1$. This simplifies notation, and since the labelling is arbitrary, there is no loss of generality. Compute
$$P(T > t, Z = 1) = P(T_1 > t \ \& \ T_1 < T_j, (j = 2, \dots, k))$$
The right-hand event is that the point event $w_1$ occurs first, and it occurs after time $t$. Recalling that the pdf of $T_1$ is $\lambda_1 e^{- \lambda_1 x}$, the right-hand side is computed by conditioning on $T_1$ and using the law of total probability:
$$P(T > t, Z = 1) = \int_t^\infty P(\cap_{j = 2}^k \{ T_j > x \} ) \lambda_1 e^{- \lambda_1 x} \ dx,$$
where the probability in the integrand is $P(T_1 > t \ \& \ T_1 < T_j, (j = 2, \dots, k) \mid T_1 = x)$.

How exactly was the law of total probability used to get that $P(T > t, Z = 1) = \int_t^\infty P(\cap_{j = 2}^k \{ T_j > x \} ) \lambda_1 e^{- \lambda_1 x} \ dx$? It seems to me that ${\displaystyle P(A)=\int _{-\infty }^{\infty }P(A \mid X = x)f_{X}(x) \ dx}$ might have been used, but I don't understand where the $P(\cap_{j = 2}^k \{ T_j > x \} )$ came from.
 A: The notation here is more complicated than it needs to be.  A simpler way to frame this is to say that:
$$\mathbb{P}(T>t, Z=1) = \mathbb{P}(t<T_1< \min(T_2,...,T_k)).$$
Now, for all values $r \in \mathbb{R}$ we clearly have the event equivalence:
$$\{ r< \min(T_2,...,T_k) \} = \bigcap_{i=2}^k \{ T_i > r \}.$$
Thus, applying the law of total probability to the latter expression (conditioning on $T=r$), we have:
$$\begin{align}
\mathbb{P}(T>t, Z=1) 
&= \mathbb{P}(t<T< \min(T_2,...,T_k)) \\[12pt]
&= \int \limits_t^\infty \mathbb{P}(r < \min(T_2,...,T_k)) f_{T_1}(r) \ dr \\[6pt]
&= \int \limits_t^\infty \mathbb{P} \Bigg( \bigcap_{i=2}^k \{ T_i > r \} \Bigg) \lambda_1 e^{-\lambda_1 r} \ dr. \\[6pt]
\end{align}$$
A: Since no one has attempted an answer, I will say that if $X,Y$ are random variables $P(t < X < Y) = P(X > t, X < Y) = \int_t^\infty P(X < Y \mid X = x)f(x) \ dx$ by an adjusted version of law of total probability (adjusting the limits of $X$). Here we have $P(t < T_1 < T_j ,j = 2 ,\dots, k) = \int_t^\infty P(\cap_{j = 2}^k \{ x < T_j \} ) \lambda_1 e^{- \lambda_1 x} \ dx$ by the adjusted law of total probability.
A: There is a good discussion of this theorem in Wikipedia / Exponential Distribution / Distribution of the minimum of exponential random variables.

How exactly was the law of total probability used ... ?

Total probability in the integral you provide sums over all possible outcomes ("total probability") of the other independent variables $w_i,~ i \neq 1$.
A: The answer is in your question:

where the probability in the integrand is $P(T_1 > t \ \& \ T_1 < T_j, (j = 2, \dots, k) \mid T_1 = x)$.

With your notation, $A = T_1 > t \ \& \ T_1 < T_j, (j = 2, \dots, k)$.
So
$$
P(A\mid T_1 = x) = I(t >x)P(T_j > x, \,\,(j = 2, \dots, k))
$$
Thus,
$$
P(A) = \int_{-\infty}^{\infty}P(A \mid T_1 = x)f_{T_1}(x)\,dx= \int_{t}^{\infty}P(\cap_j T_j > x)f_{T_1}(x)\,dx
$$
Note, $A$ is the event where $T_1$ is smaller than all other $T_j$. With symbols: $A = \cap_{j>1} T_j > T_1$.
