Proof

Let \(v\) be a cutvertex in the connected graph \(G(V,E,\gamma)\).
Since the subgraph \(G[V\setminus v]\) is disconnected, it is not empty and has at least two vertices \(x,y\in V\), which are in different components of this graph left after the removal of the cutvertex \(v\).
Moreover, every path between \(x\) and \(y\) must pass \(v\).
For if a path between \(x\) and \(y\) existed, which does not pass \(v\), the removal of the vertex \(v\) would not leave the graph \(G\) disconnected, in contradiction to \(v\) being a cutvertex.

Let \(x,y,v\) be different vertices of \(G\) such that every path between \(x\) and \(y\) passes \(v\).
Then the removal of \(v\) cuts all these passes and there is no path between \(x\) and \(y\) in the graph \(G[V\setminus v]\).
Thus, \(G[V\setminus v]\) is disconnected and \(v\) is a cutvertex.
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