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 graph.

We use a Proof by Contraposition.

To that end, suppose that there exists a pair of vertices $u$ and $v$ in $T$ such that there is not exactly one path between them.

If there is no path between $u$ and $v$, $T$ is not connected.

In this case, $T$ is certainly not a tree.

Suppose there is more than one path between $u$ and $v$.

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.

From Rule of Transposition it follows that if $T$ is a tree, there is exactly one path between any pair of vertices.