prove about Graphic Theory

Question

Let $G$ be a graph $7$-connected, $a$, $b$ $\in E (G)$ and $u$, $v \in V (G)$ other vertices. Prove that $G$ contains $2$ cycles $C$ and $C'$ such that you only have $u$ and $v$ as common vertices and none contains $a$ or $b$.

Answer 1

Hint: 6-connected should be sufficient for this problem. Use Menger's Theorem. In a 6-connected graph $G$, there are 6 pairwise internally disjoint paths between any two vertices $u,v \in V(G)$. At least 4 of those do not contain either of the two fixed edges $a,b \in E(G)$. Can you construct $C$ and $C'$ from these paths?