Let $G$ be a directed graph, and let $C$ be a circuit in $G$. Prove there exists a simple circuit $C'$ such that all the edges of $C'$ belong to $C$.

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Let $G$ be a directed graph, and let $C$ be a circuit in $G$. Prove there exists a simple circuit $C'$ such that all of the edges of $C'$ belong to $C$.

I am new to graph theory, my idea was the following: Look at the shortest circuit in $C$. It must be a simple circuit (I don't know how to prove that) and all of its edges belong to $C$. How do I expand on this?

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You have the right idea. If some interior vertex of $C$ is repeated, consider the (necessarily shorter) path between two occurrences of that vertex. What can you say about that path?