This question was taken from this article, under the section "Symmetry of independence".
Let $(X_n)_{n=1}^\infty$ be a sequence of random variables, and denote the event $X_n > X_i$ for all $i \in \{1, \ldots, n-1\}$ with $A_n$. I would like to show that $A_n$ is independent of $A_{n+1}$ and then use the symmetry of independence to demonstrate that $A_{n+1}$ is independent of $A_n$ as well.
As we are dealing with sequences, the natural thing to do is to attempt induction.
The base case
As $X_1$ has no preceding elements in the sequence $(X_n)$, $P(A_1) = 1$. Also, as $A_2$ only means that $X_2 > X_1$, we have $P(A_1 \mid A_2) = 1 = P(A_1)$. Therefore the base case is valid.
Induction assumption
$$ P(A_n \mid A_{n+1}) = \frac{ P(A_n \cap A_{n+1}) }{ P(A_{n+1}) } = P(A_n) \quad\text{or}\quad P(A_{n+1})P(A_{n}) = P(A_{n} \cap A_{n+1})\,. $$
Induction step
By the induction assumption we have $P(A_{n+1}) = P(A_{n} \cap A_{n+1}) / P(A_n)$. Then \begin{align} P(A_{n+1} \mid A_{n+2}) &= \frac{ P(A_{n+1} \cap A_{n+2}) }{ P(A_{n+2}) } \\ &= \frac{ P(A_{n+1}) + P( A_{n+2}) - P(A_{n+1} \cup A_{n+2}) }{ P(A_{n+2}) } \\ &= \frac{ \frac{ P(A_{n} \cap A_{n+1}) }{ P(A_n) } + P( A_{n+2}) - P(A_{n+1} \cup A_{n+2}) }{ P(A_{n+2}) } \\ &= \frac{ P(A_{n} \cap A_{n+1}) + P(A_n)P( A_{n+2}) - P(A_n)P(A_{n+1} \cup A_{n+2}) }{ P(A_n) P(A_{n+2}) }\\ &= \cdots \\ &= \frac{ P(A_n \cap A_{n+1} ) }{ P(A_n) } \\ &= P(A_{n+1}) \,. \end{align}
What might the missing steps denoted by $\cdots$ be, or did I take a completely wrong approach in the first place??