I'm trying to prove that Brownian motion absorbed at the origin is a Markov process with respect to the original filtration $\{\mathcal{F}_{t}\}$. To be more specific, let $(B_{t},\mathcal{F}_{t})_{t \geq 0}$ be a 1-d brownian motion such that $B_{0}=x,\ a.s. \mathbb{P}\ $ for some $x>0.\ $ Let $T_{0}=\inf\{ t \geq 0, \ B_{t}=0\}$ be the hitting time of $0$, and define the Brownian motion absorbed at origin to be the process $\{B_{t \land T_{0}}, \mathcal{F}_{t} \}$. I want to show that such a process is Markov w.r.t. $\mathcal{F_{t}}, \ $ i.e. $\mathbb{P}[B_{t \land T_{0}} \in \Gamma | \mathcal{F}_{s}]=\mathbb{P}[B_{t \land T_{0}} \in \Gamma |B_{s \land T_{0}}]$ for every pair of $0. And further more it has transition density given by: $\mathbb{P}[B_{(t+s) \land T_{0}} \in dy |B_{s \land T_{0}}=z]=\mathbb{P}^{z}[B_{t} \in dy, \ T_{0} >t], \ \forall y,z >0,$ where $\mathbb{P}^{z}$ is the probability measure under which $B$ starts from $z$ almost surely.
I tried to use the strong Markov property for Brownian motion to derive these equalities, but I'm having a hard time dealing with $T_{0}$. For example, if $0 \notin \Gamma$, then$\{B_{t \land T_{0}} \in \Gamma\}=\{B_{t} \in \Gamma, T_{0} >t\}$. But how to apply strong Markov property to compute $\mathbb{P}[B_{t} \in \Gamma, T_{0} >t\ | \mathcal{F}_{s} ]\ $?
Any answer or comment is greatly appreciated, thanks!