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Show that a simple graph is nonseparable iff for any two given arbitrary edges a circuit can always be found that will include these two edges.

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Suppose I have the following graph

and let those arbitrary edges included in a cycle be ux and vx.

Now, if I remove vertex x, I can still reach from U to V, via the connected component shown.
Similar argument holds if I delete vertex U or V. and I am still able to cover remaining vertices.

Hence, the vertex connectivity of such a graph is not one which makes it nonseparable.

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