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Show that an edge in a simple graph is a cut edge if and only if this edge is not a part of any simple circuit in the graph.

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If direction: Suppose the edge e with endpoints (u, v) is not part of any cycle in the graph. Before the removal of e, u and v are in the same connected component. The edge e is the only possible path from u to v. No other path u, x1, x2, . . . , xi , v can exist as then we could create a cycle u, x1, x2, . . . , xi , v, u using the edge e. Thus when we remove e, no path from u to v can exist and so u and v will be in separate connected components. As u and v were initially in the same connected component, removing e has created an additional connected component and so e is a a cut edge. Only If direction: Suppose the edge e is a cut edge. We will use proof by contradition to show that e must not be in a cycle. Assume that e is part of some cycle. Because e is a cut edge, when we remove e, its endpoints must be in seperate connected components. However since e is part of some cycle that must include its endpoints, there must exist another path from u to v that completes this cycle. This means that after removing e a path from u to v must still exist and thus u and v must be in the same connected component. This is a contradiction as by the definition of a cut edge, u and v must be in different components. So e must not be part of any cycle in the graph.
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