Path in Tree is Unique

Theorem
Let $$T$$ be a graph.

Then $$T$$ is a tree iff there is exactly one path between any two vertices.

Necessary Condition
Suppose $$T$$ is a tree.

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.

Sufficient Condition
Let $$T$$ be such that between any two vertices there is exactly one path.

Then for a start $$T$$ is by definition connected.

Suppose $$T$$ had a circuit, say $$\left({u, u_1, u_2, \ldots, u_n, v, u}\right)$$.

Then there are two paths from $$u$$ to $$v$$: $$\left({u, u_1, u_2, \ldots, u_n, v}\right)$$ and $$\left({u, v}\right)$$.

Hence, by Modus Tollendo Tollens, $$T$$ can have no circuits.

That is, by definition, $$T$$ is a tree.