Let {B(t), t ≥ 0} be a standard Brownian motion. Show that $$ P(B(s)>0, B(t)<0)= \int_{0}^{\infty}\Phi(-\sqrt{s\over{t-s}}y)φ(y)dy$$ for any 0 < s < t, where φ and Φ are the p.d.f. and c.d.f. of standard normal, respectively.
I have tried to approach this problem by suggesting that B(t) follows N(0,t) I have also searched in the textbook and they suggested to change $$Pr(B(s)>0,B(t)<0)=Pr(B(s)>0,B(s)+B(t)-B(s)<0)$$ $$\int_{-\infty}^{\infty}Pr(B(s)>0,B(s)+B(t)-B(s)<0 |B(s)=y) {1\over\sqrt{2π}} \mathrm{e}^{-y^2\over2} dy $$ I would like to know why the second step can be done so?