Take the group $S_n$ of permutations of $n$ points and as generator set take $S = \{s_1 := (1,2), s_2 := (2,3),\ldots, s_{n-1} := (n-1,n)\}$ the set of Coxeter generators. Let $t = (i,j) \in S_n$ be a transposition and $w \in S^{*}$ a reduced word for $t$. Then $w$ has odd length, so we can write $w = w_0s_kw_1$ for some $k$ such that $w_0$ and $w_1$ have the same length.
Now let $\pi_0$ and $\pi_1$ be the permutations given by $w_0$ and $w_1$.
Is it true that $\pi_0 = \pi_1^{-1}$?
With some computations I showed that it is true for all $n \leq 7$, but I can't find a general proof. I tried some inductions on $n$ or the length of $w$, but none of them gave the desired result.
We have $$\pi_0s_k\pi_1(i,j)=1$$ By the strong exchange property, we obtain a word for the identity by deleting one letter from $w_0s_kw_1$. The words to the left and right of this deleted letter, say $w_2$ and $w_3$, so that our new word is $w_2w_3$, will both be reduced, and the corresponding permutations $\pi_2$ and $\pi_3$ will be inverses of each other, hence must have the same length. It follows that the deleted letter must be the central $s_k$, hence $$\pi_0=\pi_1^{-1}$$ as desired.
Notice that we used no specifics of the symmetric group here. This works for a reflection in any Coxeter group.