# Graph theory proof

We will show that C does not share two or more common vertices with any other cycle D of G. Suppose otherwise. Then there exists at least two vertices in C that form a cycle with vertices from D. Let C' denote the induced subgraph with the common vertices of C and D, and denote $V(C') = v_{1},...,v_{i}$. Let D' denote the induced subgraph with vertices from D not in C, so $V(D') = V(D) - V(C')$. If |D'| is odd, then there exists an even cycle with $v_{1}$ and $V(D')$ and $v_{i},..., v_{n}$, contradicting the assumption that there exist no cycles. If |D'| is even, then there exists an even cycle with $v_{1},...,v_{i}$ and $V(D')$, again contradicting our assumption.