What is $P[B(t)>y, \land_{0 \leq u \leq t} B(u)>0|B(0)=x]$, where $B(t)$ i standard Brownian motion? Here notation $\land_{0 \leq u \leq t}B(u)$ is the minimum of the process between $0$ and $t$.
Is the following attempt correct?
$P[B(t)>y, \land_{0 \leq u \leq t} B(u)>0|B(0)=x] = P[B(t)>y, \lor_{0 \leq u \leq t} B(u)<2x|B(0)=x]$,
using symmetry, where $\lor_{0 \leq u \leq t}B(u)$ is the maximum of the process. Further:
$P[B(t)>y, \lor_{0 \leq u \leq t} B(u)<2x|B(0)=x] = 1-P[B(t)>y, \lor_{0 \leq u \leq t} B(u)\geq2x|B(0)=x)$ $= 1-P[B(t)>y, |B(t)|\geq2x|B(0)=x]$,
where the last equality follows by the reflection principle. Finally, if we subtract the starting position:
$1-P[B(t)>y, |B(t)|\geq2x|B(0)=x] = 1-P[B(t)>y-x, |B(t)|\geq x]$ $= 1-P[-x<B(t)<\max(y-x, x)]$.
First, subtracting the starting position, we are looking for $P(B(t)>y-x, \min_{u\leq t}B(u)>-x)$ or, by symmetry, $P(B(t)<x-y, \max_{u\leq t}B(u)<x).$ Hence,
\begin{align} P\big(B(t)<x-y, \max_{u\leq t}B(u)<x\big)=&P\big(B(t)<x-y\big)-P\big(B(t)<x-y, \max_{u\leq t}B(u)\geq x\big) \\ =& P\big(B(t)<x-y\big)-P\big(B(t)>x+y\big), \end{align} where we used the fact that $P\big(B(t)< a-b, \max_{u\leq t}B(u)\geq a\big)=P\big(B(t)> a+b\big)$ (see, for example, Proposition 2 here).