and $v_1$ in $G$. From $G \subseteq T_1$ it follows that $P_{u, v}$ is a path joining $u_1$ and $v_1$ in $T_1$. But the unique path joining $u_1$ and $v_1$ in $T_1$ is $(\{u_1, v_1\}$. Thus, $\{u_1, v_1\} \in E(G)$. Since $w_1(e) > 0$ holds for every $e \in E(T_1)$, equality implies $E(G) = \{\{u_1, v_1\}\}$. The last equality contradicts the definition of the path $P^{2}$.

The validity of follows.

\[c3.1\] Let $(X, d)$ be a finite metric space. Suppose that $T_1 = T_1(w_1)$ and $T_2 = T_2(w_2)$ are weighted trees with strictly positive weights such that $$X = V(T_1) = V(T_2).$$ Then the equalities $d = d_{w_1} = d_{w_2}$ imply $T_1(w_1) = T_2(w_2)$.

It follows from Theorem \[t3.4\] with $f$ satisfying $f(x) = x$ for every $x \in X$.

\[c3.2\] Let $T_1 = T_1(w_1)$ and $T_2 = T_2(w_2)$ be weighted trees with strictly positive weights. Then the equivalence $$\label{c3.2:e1}
(T_1(w_1) \simeq T_2(w_2)) \Leftrightarrow ((V(T_1), d_{w_1}) \simeq (V(T_2), d_{w_2}))$$ is valid.

The concept of isomorphism of weighted trees can be naturally extended to the concept of isomorphism of weighted graphs. Let $G_1 = G_1(w_1)$ and $G_2 = G_2(w_2)$ be connected, nonempty graphs with the weights $w_1 \colon E(G_1) \to \mathbb{R}^{+}$ 