A simple question about conditional probability

Question

Let $(X_n)_{n \ge 0}$ be a homogeneous Markov chain with a countable state space $(E,\mathcal B).$

Why we have that $$\mathbf{P} \left ( X_r=i,X_{r-1} \ne j,...,X_1\ne j\mid X_0={i}\right )\cdot\mathbf{P} \left ( X_n=j,X_{n-1} \ne \left\{i,j \right\},...,X_{r+1}\ne \left\{i,j \right\}\mid X_r={i}\right )=$$ $$\mathbf{P} \left ( X_n=j,X_{n-1} \ne \left\{i,j \right\},...,X_{r+1}\ne \left\{i,j \right\},X_r=i,X_{r-1} \ne j,...,X_1\ne j\mid X_0={i}\right )$$

Answer 1

All $\{X_{k}\}_{k=r+1}^n$ are independent of all $\{ X_{p} \}_{p=0}^{r-1}$ given $X_{r}$, by the Markov property. Therefore \begin{align*} &\mathbf{P} \left ( X_n=j,X_{n-1} \ne \left\{i,j \right\},...,X_{r+1}\ne \left\{i,j \right\}\mid X_r={i}\right )\\ ={}&\mathbf{P} \left ( X_n=j,X_{n-1} \ne \left\{i,j \right\},...,X_{r+1}\ne \left\{i,j \right\}\mid X_r={i},X_r=i,X_{r-1} \ne j,...,X_1\ne j, X_0={i} \right ) \end{align*}