Explanation concerning a proof in probability

Question

In a proof concerning Markov chains:

The sequence $(X_n)_{0 \leq n \leq N}$ has values in the space of states { 0,1}.

\begin{eqnarray} P(X_{n+1} = i_{n+1} | X_n = i_n, \dots , x_0 = i_0) =& \frac{P(X_{n+1} = i_{n+1} \cap X_n = i_n, \dots , x_0 = i_0)}{P (X_n = i_n, \dots , x_0 = i_0)}\\ =& \frac{P(|X_{n+1}-X_n| = |i_{n+1} -i_n| \cap X_n = i_n, \dots , x_0 = i_0)}{P (X_n = i_n, \dots , x_0 = i_0)}\\ =& P(|X_{n+1}-X_n| = |i_{n+1} -i_n|) \end{eqnarray}

Could anyone please explain how we moved from the first inequality to the second one? Based on which probability role we can say $$ P(X_{n+1} = i_{n+1} \cap X_n = i_n, \dots , x_0 = i_0) = P(|X_{n+1}-X_n| = |i_{n+1} -i_n| \cap X_n = i_n, \dots , x_0 = i_0)$$

Answer 1

As your space of states is $\{0,1\}$, you know $X_{n+1}$ by knowing $X_n$ and $|X_{n+1}-X_n|$.

UPDATE:

The values of $X_{n+1}$ and $X_n$ set four different possibilities:

if $(X_{n+1},X_n)=(0,0)$, then $(|X_{n+1}-X_n|,X_n)=(0,0)$;

if $(X_{n+1},X_n)=(0,1)$, then $(|X_{n+1}-X_n|,X_n)=(1,1)$;

if $(X_{n+1},X_n)=(1,0)$, then $(|X_{n+1}-X_n|,X_n)=(1,0)$;

if $(X_{n+1},X_n)=(1,1)$, then $(|X_{n+1}-X_n|,X_n)=(0,1)$.