 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 ${\beta }_{j+1}^{\xi }\in C_{\delta }\cap [\xi ,\alpha )$. ThereforeIn particular, each function ${\rho }_{\beta }$ is injective: equality of two trace sequences would give equal lengths and then contradict this proper-initial-segment conclusion.