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When we refer to a 'cycle in a graph', we are talking about a series of edges and vertices in which you can start at one vertex and follow a connected path that eventually brings you back to the starting vertex without repeating any edge or vertex along the way (except for the starting and ending vertex, which are the same). The presence of a cycle adds redundancy to the graph, providing multiple paths between vertices.

In the context of the exercise, the edge labeled as 'e' is part of a cycle. This means that if you were to trace a path starting from one end of the edge 'e', you could eventually return to the other end of 'e' without using 'e' itself. This concept is crucial in understanding why the graph would remain connected even if the edge 'e' is removed, as there are alternative routes in the cycle that keep the graph intact.

A 'connected graph' is one where there is at least one path between every pair of vertices. No vertex is isolated. This feature is significant in ensuring that information or flow can continue uninterrupted in networks, like social networks or transportation systems, for instance.

The activity presented in the original problem has a graph that is connected, which means that before we remove any edges, there's a way to travel from any vertex to any other vertex. When we are posed with the task of proving that the graph remains connected after removing an edge, we rely on the inherent structure of the connected graph and the redundancies provided by cycles to establish ongoing connectivity. If an edge is removed and the graph still has a path connecting every pair of vertices, it retains its status as a connected graph.

The 'proof of connectivity' after removing an edge, as mentioned in the exercise, revolves around demonstrating that despite the edge's removal, there is still a path for each pair of vertices within the graph. How can this be shown? We need to consider the properties of cycles and connectivity.

In our solved exercise, the crucial point is that the edge 'e' exists in a cycle. We claim that removing 'e' would not disrupt the connectivity because the cycle's presence assures us that an alternative route is available. This alternative is the rest of the cycle that once included 'e'. Therefore, even when 'e' is removed, there is an uninterrupted path between all vertices, proving that the graph remains connected. This logical argument forms the foundation of our proof, illustrating that graph connectivity is not compromised by the removal of an edge when it lies within a cycle.