I try to prove the existence of a stationary distribution of finite Markov chains but I am stuck in the following expression: $$\mathbb{P}_z(X_t=x, X_{t+1}=y, \tau_z^{+}\geq t+1)=\mathbb{P}_z(X_t=x, \tau_z^{+}\geq t+1)P(x,y)$$ where $\tau_z^{+}:=\min\{t\geq1: X_t=z\}$ and $P(x,y)=\mathbb{P}(X_{t+1}=y\mid X_t=x)$
In fact, $X_t$ is a finite Markov chain.
I try to use the conditional probability: $$\mathbb{P}_z(X_t=x, X_{t+1}=y, \tau_z^{+}\geq t+1)=\mathbb{P}_z( \tau_z^{+}\geq t+1\mid X_t=x, X_{t+1}=y)\mathbb{P}(X_t=x, X_{t+1}=y)=\mathbb{P}_z( \tau_z^{+}\geq t+1\mid X_{t+1}=y)\mathbb{P}(X_t=x)P(x,y)$$ But how to get $$\mathbb{P}_z( \tau_z^{+}\geq t+1\mid X_{t+1}=y)\mathbb{P}(X_t=x)=\mathbb{P}_z(X_t=x, \tau_z^{+}\geq t+1)?$$
We can write : $\mathbb{P}_z(X_t=x, X_{t+1}=y, \tau_z^{+}\geq t+1)=\mathbb{P}_z( X_{t+1}=y | X_t=x, \tau_z^{+}\geq t+1)\mathbb{P}_z(X_t=x, \tau_z^{+}\geq t+1)$. But since $\lbrace \tau_z^{+}\geq t+1 \rbrace = \lbrace \tau_z^{+}\leq t \rbrace ^c \in \mathcal{F}_t$, we can use Markov property to show that : $\mathbb{P}_z( X_{t+1}=y | X_t=x, \tau_z^{+}\geq t+1)=\mathbb{P}_z(X_{t+1}=y|X_t=x)=P(x,y)$.