Path in Tree is Unique/Necessary Condition

Theorem
Let $T$ be a tree.

Then there is exactly one path between any two vertices.

Proof
Suppose $T$ is a tree.

As a tree is by definition connected, there exists at least one path between each pair of vertices.

Aiming for a contradiction, suppose there were more than one path between two vertices $u, v \in T$.

Let two of these paths be:
 * $P_1 = \left({u, u_1, \ldots, u_i, r_1, r_2, \ldots, r_{j-1}, r_j, u_{i+1}, \ldots, v}\right)$;
 * $P_2 = \left({u, u_1, \ldots, u_i, s_1, s_2, \ldots, s_{k-1}, s_k, u_{i+1}, \ldots, v}\right)$.

Now consider the path :
 * $P_3 = \left({u_i, r_1, r_2, \ldots, r_{j-1}, r_j, u_{i+1}, s_k, s_{k-1}\ldots, s_2, s_1, u_i}\right)$

It can be seen that $P_3$ is a circuit.

Thus by definition $T$ can not be a tree.

This contradicts the supposition that there exists more than one path between two vertices of $T$.