Conditional probability of Brownian bridge

Question

Suppose $B_{0,a}^{T,b}(t)$ is a Brownian bridge such that $B_{0,a}^{T,b}(0)=a$ and $B_{0,a}^{T,b}(T) = b$. The probability density function of $B_{0,a}^{T,b}(t)$ is the conditional probability density function of the Brownian motion $B(t)$ given that $B(0)=a$ and $B(T)=b$. That is $$ f_{B_{0,a}^{T,b}(t)}(x) = f_{B(t)}(x|B(0)=a,B(T)=b).$$ It is equivalent to $$ \mathbb{P}(B_{0,a}^{T,b}(t) \leq x) = \mathbb{P}(B(t) \leq x |B(0)=a,B(T)=b).$$ I want to ask how to express this probability in term of $B(t)$ $$ \mathbb{P}\left(B_{0,a}^{T,b}(t) \leq x \middle| \min_{s \in [0,T]} B_{0,a}^{T,b}(t) \geq y \right) ?$$ Is it equal to $$ \mathbb{P}\left(B_{0,a}^{T,b}(t) \leq x \middle| \min_{s \in [0,T]} B_{0,a}^{T,b}(t) \geq y \right) =\mathbb{P}\left(B(t) \leq x \middle| B(0)=a,B(T)=b, \min_{s \in [0,T]} B(s) \geq y \right) ?$$ May I have some detailed explanation about it?