(I am trying to understand the proof of Theorem $3.1$ in this paper.)
Denote by $B^{(x)}(t)$ a Brownian bridge process which starts at $0$ at time $0$ and ends at $x$ at time $t$, i.e.
\begin{equation} B^{(x)}(t) = B(t) - t(B(1) - x), \end{equation}
where $B(\cdot)$ is a standard Brownian motion.
I am interested in
\begin{equation} \mathbb{P} \bigl( \,\inf_{0 \le t \le 1} \{ c_1 B_1(t) + c_2 B_2(1 - t)\} \le x \mid B_1(1) = b_1, B_2(1) = b_2 \bigr), \tag{1} \end{equation}
where $B_1(\cdot)$ and $B_2(\cdot)$ are two independent standard Brownian motions. By conditioning on the event, $B_1(\cdot)$ and $B_2(\cdot)$ essentially become two Brownian bridge processes. Consider the case that $x \le \min(b_1c_1,b_2c_2)$, then, with some abuse of notation, we have
\begin{equation} (1) = \mathbb{P} \bigl( \, \inf_{0 \le t \le 1} \{ c_1 B_1^{(b_1)}(t) + c_2 B_2^{(b_2)}(1-t)\} \le x \bigr), \tag{2} \end{equation}
where we can simplify
\begin{align} B_2^{(b_2)}(1-t) &= B(1 - t) - (1 - t)(B(1) - b_2) \\ &= B(1 - t) - B(1) + tB(1) + b_2 - tb_2 \\ &= - \bigl( B(t) - t(B(1) - b_2) \bigr) + b_2 \\ &= - B_2^{(b_2)}(t) + b_2 \end{align}
since $B(1) - B(1 - t)$ is again a Brownian motion (at time $t$). Plugging this back in $(2)$ we obtain
\begin{equation} (2) = \mathbb{P} \bigl( \, \inf_{0 \le t \le 1} \{ b_2 c_2 + c_1 B_1^{(b_1)}(t) - c_2 B_2^{(b_2)}(t)\} \le x \bigr), \tag{3} \end{equation}
which, according to the paper, is equal to
\begin{equation} (3) = \mathbb{P} \bigl( \, \inf_{0 \le t \le 1} \{ b_2 c_2 + \sqrt{c_1^2 + c_2^2} B^{(\frac{b_1 c_1 - b_2 c_2}{\sqrt{c_1^2 + c_2^2}})}(t)\} \le x \bigr). \end{equation}
How does the last equality follow and is the rest correct as well?