Path in Tree is Unique

Theorem
In any tree $$T$$, between any two vertices there is exactly one path.

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

Suppose there is 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.

Hence the result by Proof by Contradiction.