\ldots, v_n^{*})$ such that $P^{2} \subseteq T_2$, $n \geqslant 3$, and $v_1^{*} = f(u_1)$, $v_n^{*} = f(v_1)$. By , we have $$\label{t3.4:e3}
d_{w_2}(v_1^{*}, v_n^{*}) = \sum_{i=1}^{n-1} w_2(\{v_i^*, v_{i+1}^{*}\}) = \sum_{i=1}^{n-1} d_{w_2} (v_i^{*}, v_{i+1}^{*}).$$ Since $f$ is an isometry of $(V(T_1), d_{w_1})$ and $(V(T_2), d_{w_2})$, from it follows that $$\label{t3.4:e4}
\begin{aligned}
d_{w_1}(u_1, v_1) &= d_{w_1}(u_1, f^{-1}(v_2^*)) + d_{w_1}(f^{-1}(v_{n-1}^*), v_1) \\
& \quad + \sum_{i=2}^{n-2} d_{w_1}(f^{-1}(v_{i}^*), f^{-1}(v_{i+1}^*)).
\end{aligned}$$ (For $i = 2$, $\ldots$, $n-2$ we identify the one-point set $f^{-1}(v_{i}^*)$ with the unique point of this set.)

Since $d_{w_1}$ is an additive metric, there are

1.  a path $P_{1}^{1} \subseteq T_1$ joining $f^{-1}(v_1^*) = u_1$ and $f^{-1}(v_2^*)$,

2.  a path $P_{2}^{1} \subseteq T_1$ joining $f^{-1}(v_2^*)$ and $f^{-1}(v_3^*)$,

3.  …,

4.  and a path $P_{n-1}^{1} \subseteq T_1$ joining $f^{-1}(v_{n-1}^*)$ and $f^{-1}(v_{n}^*) = v_1$.

Using  again we can write as $$\label{t3.4:e5}
w_1(\{u_1, v_1\}) = \sum_{i=1}^{n-1} \sum_{e \in E(P_i^{1})} w(e).$$

Let us consider a graph $G$ such that $$V(G) = \bigcup_{i=1}^{n-1} V(P_i^{1}) \quad \text{and} \quad E(G) = \bigcup_{i=1}^{n-1} E(P_i^{1}).$$ It is clear that $G$ is connected, $G \subseteq T_1$, and $u_1$, $v_1 \in V(G)$. Since $T_1$ is a tree, $G$ is a subtree of $T_1$. Consequently, there is a unique path $P_{u, v}$ joining $u_1$ 