Circuit of Simple Graph has Three Edges or More

Theorem
Let $G$ be a simple graph.

Let $C$ be a circuit in $G$.

Then $C$ has at least $3$ edges.

Proof
By definition, a circuit contains at least one edge.

That is, a single vertex does not comprise a trivial degenerate circuit with no edges.

Suppose $C$ contains only one edge $e$.

As $G$ is a simple graph it contains no loops.

Therefore $e$ starts and ends at two distinct vertices.

Therefore $C$, consisting entirely of $e$ can not be a circuit.

Suppose $C$ contains exactly $2$ edges $e$ and $f$.

Let $e = u v$ where $v$ and $v$ are vertices of $G$.

Then for $C$ to be a circuit it must follow that $f = v u$.

Thus there are two edges between the same vertices $v$ and $u$.

As $G$ is an undirected graph, it must be that $e$ and $f$ are part of a multiple edge between $u$ and $v$.

But $G$ is not a multigraph, and so such a circuit can not exist.

A circuit containing $3$ edges definitely exists, though.

Let $G = \left({V, E}\right)$ where $V = \left\{{u, v, w}\right\}$ and $E = \left\{{\left\{{u, v}\right\}, \left\{{u, w}\right\}, \left\{{v, w}\right\}}\right\}$.

Then $\left({u, w, v, u}\right)$ is a $3$-edge circuit.