Let $B$ be a Brownian motion starting at $0$ and $\tau_c=\inf\{t>0:B_t-t=c\}$. I want to show that
$$ P(\tau_{a+b}<\infty\ |\ \tau_a <\infty)=P(\tau_b<\infty) $$
for $a,b>0$ using the strong Markov property, which is (in this case) defined as follows:
For $Y_t(\omega)$ bounded and jointly measurable on $[0,\infty)\times\Omega$ and a stopping time $\tau$ it holds for every $x\in\mathbb{R}$ that
$$ E^x[Y_{\tau}\circ\theta_{\tau}\ |\ F_{\tau}]=E^{B_{\tau}}[Y_{\tau}]\quad\ P^x-a.s.\text{ on } \{\tau<\infty\} $$
where $\theta_s$ denotes a time shift of $s$, i.e. $(\theta_s\omega)(t)=\omega(t+s)$ and $P^x$ is the distribution of $x+B$ (and $E^x$ the expectation under $P^x$).
I know that this is often used as $$ E^x[Y_{\tau}\circ\theta_{\tau}1_{\{\tau<\infty\}}]=E^x[E^{B_{\tau}}[Y_{\tau}1_{\{\tau<\infty\}}]] $$ and that I have to find suitable $Y_t$ but that's exactly the problem. I guess I have to use $\tau_a$ as $\tau$ in the Markov property and $Y_t$ probably has to be some combination of indicator functions in order to write the expectation as probabilities but I don't know how to choose the right $Y_t$ in order to obtain this conditional probability from the Markov property.
Thanks for any help!