Let $(B_t)$ a Brownian motion. For $a>0$ and $b>0$, show that $$p\{B_u\neq 0\text{ for }0\leq u\leq t\mid B_0=a, B_t=b\}=1-e^{-\frac{2ab}{t}}.$$
In the correction they said: Let $X_t=\min_{0\leq u\leq t}B_u$. I denote $p_x(B_t\in A)=p\{B_t\in A\mid B_0=x\}$ Then we have for $c< a\wedge b$ that
\begin{align*} p_a\{B_u\neq c\text{ for }0\leq u\leq t, B_t>b\}&=p_a\{X_t>c, B_t>b\}\\ &= p_a\left\{\max_{0\leq u\leq t}B_u<2a-c, B_t<2a-b\right\}\\ &\underset{(1)}{=} p\left\{\max_{0\leq u\leq t}B_u<a-c, B_t<a-b\right\}\\ &\underset{B\cap A=B\backslash (B\cap A^c)}{=}^\{B_t<a-c\}-p\left\{\max_{0\leq u\leq t}B_u\geq a-c, B_t\leq a-c\right\}\\ &\underset{(2)}{=}p\{B_t\geq b-a\}-p\left\{\max_{0\leq u\leq t}B_u\geq a-c,B_t\geq 2(a-c)-(a-b)\right\}\\ &\underset{(3)}{=}\int_b-a^{+\infty }\rho(u,t)du-\int_{2(a-c)-(a-b)}^{+\infty }\rho(u,t)du, \end{align*}
where $\rho(x,t)$ is the density of an $\mathcal N(0,t)$.
My questions:
$\bullet$ For $(1)$, why does $p_a\{...\}$ becomes $p\{...\}$ ? Shouldn't it be $p_0\{...\}$ ?
$\bullet$ For $(2)$ does the $p\{B_t\geq b-a\}$ comes from the fact that $-B_t$ is also a Brownian motion ? If yes, Shouldn't it be $p\{B_t>b-a\}$ ? (even if it doesn't matter, because $B_t$ is continuous, but it's just to be sure). In $p\left\{\max_{0\leq u\leq t}B_u\geq a-c,B_t\geq 2(a-c)-(a-b)\right\}$, I don't see where $B_t\geq 2(a-c)-(a-b)$ comes form.
$\bullet$ For $(3)$, I don't understand why $$p\left\{\max_{0\leq u\leq t}B_u\geq a-c,B_t\geq 2(a-c)-(a-b)\right\}=\int_{2(a-c)-(a-b)}^{+\infty} \rho(u,t)du,$$
Shouldn't it be $$p\left\{\max_{0\leq u\leq t}B_u\geq a-c,B_t\geq 2(a-c)-(a-b)\right\}$$$$=\int_{a-c}^{+\infty }\int_{2(a-c)-(a-b)}^{+\infty} f_{(\max_{0\leq u\leq t}B_u,B_t)}(s,u\mid B_t=0)duds,$$ where $f_{(\max_{0\leq u\leq t}B_u,B_t)}(s,u\mid B_t=0)$ is the associated density of $(\max_{0\leq u\leq t}B_u,B_t)$ knowing $B_t=0$ ?