Let $(B_t)_{t \ge 0}$ be a Brownian motion and define its maximal function on a interval $M_a^b = \sup_{a \le s \le b} B_s$. By the reflection principle, I have shown that for $t \in (0,1)$, we have
$$\mathbb{P}(B_1 \le -1, M_{t}^1 \ge 1 \ |\ \mathcal{F}_t) = \mathbb{P}(B_1 \ge 3 \ | \ \mathcal{F}_t) = 1-\phi\left(\frac{3-B_t}{\sqrt{1-t}}\right), $$
which I hope is correct. Now, is it possible to calculate in a similar manner the following probability:
$$\mathbb{P}(B_1 > -1, M_{t}^1 < 1, M_{1}^2 \ge 1 \ |\ \mathcal{F}_t) = \ ?$$
I started by conditioning on $\mathcal{F}_1$ and got to the point $2\mathbb{E}\left((1 - \phi(1-B_1))1_{\lbrace B_1 > -1, M_{t}^1 < 1\rbrace} \ | \ \mathcal{F}_t\right)$, but I don't know how to proceed further.
Edit. I've decided to check the case $t=0$ first. So using the joint distribution of $(B_1, M_0^1)$, omitting the constant $2$ and taking $\phi(1-B_1)$ instead of $1-\phi(1-B_1)$ just for simplicity, I have the following integral to calculate
$$\int_{0}^1 \int_{-1}^b \int_{-\infty}^{1-a} \frac{1}{\pi}(2b-a)e^{-\frac{(2b-a)^2}{2}}e^{-\frac{x^2}{2}} \mbox{d}x\mbox{d}a\mbox{d}b.$$
Is it correct? And is it possible to simplify this expression?