Show that for a DTMC $(X_n)_n$ we have $\mathbb{E}[X_n \mid X_A] = \mathbb{E}[X_n \mid X_a], a := \max(A)$

Question

Starting from only the definition, I would like to show that for any $A\subseteq \{0,\dots,n-1\}$ we have that: $$ \mathbb{E}[X_n \mid X_A] = \mathbb{E}[X_n \mid X_a], $$ with $a := \max(A)$. I think this is possible by employing the law of total expectation and sum over all possible values for the $X_b$ with $b \notin A$, but this becomes pretty messy.