= Solution
Let $j$ be the common-prefix index from the <first-divergence lemma for minimal walks>. If $j=n$, the walk to $\xi$ has followed the entire walk to $\alpha$ and must make at least one further step, contradicting the assumed equality of lengths. Hence $j<n$.
The current nodes agree, so put $\delta=\beta_j^\xi=\beta_j^\alpha$. Then $C_\delta\cap\xi$ is an initial segment of $C_\delta\cap\alpha$. It is proper because $\beta_{j+1}^\xi\in C_\delta\cap[\xi,\alpha)$. Therefore
$$
\boxed{C_{\beta_j^\xi}\cap\xi\text{ is a proper initial segment of }
C_{\beta_j^\alpha}\cap\alpha.}
$$
In particular, \b[each function $\rho_\beta$ is injective]: equality of two trace sequences would give equal lengths and then contradict this proper-initial-segment conclusion.
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