Two Paths between Vertices in Cycle Graph

Theorem
Let $G$ be a simple graph.

Let $u, v$ be vertices in $G$ such that $u \ne v$.

Then:
 * for any two vertices $u, v$ in $G$ such that $u \ne v$ there exists exactly two paths between $u$ and $v$

iff:
 * $G$ is a cycle graph.

Sufficient Condition
Let $G = \left({V, E}\right) = C_n$ be the cycle graph of order $n$.

Let $V = \left\{{v_1, v_2, \ldots, v_n}\right\}$

Let $u, v \in C_n$.

Then both $u, v$ are on the cycle $C = \left({u, u_1, u_2, \ldots, u_p, v, v_1, v_2, \ldots, v_q, u}\right)$ where $p + q + 2 = n$.

By definition of cycle graph this is the only cycle in $G$.

Let $e$ be the edge $u u_1$ of $G$ (or, if $u$ and $v$ are adjacent, let $e = u v$).

From Size of Cycle Graph equals Order, there are exactly $n$ edges in $G$.

Then the edge deletion $G - e$ has $n - 1$ edges.

By Size of Tree is One Less than Order it follows that $G - e$ is a tree.

Thus by Path in Tree is Unique there is exactly $1$ path from $u$ to $v$ in $G - e$, that is:
 * $P_1 = \left({u, v_q, v_{q-1}, \ldots, v_2, v_1, v}\right)$

By replacing $e$, there exist a second path from $u$ to $v$ in $G$:
 * $P_2 = \left({u, u_1, u_2, \ldots, u_{p-1}, u_p, v}\right)$

or if $u$ and $v$ are adjacent:
 * $P_2 = \left({u, v}\right)$