0 votes 0 votes 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. Graph Theory graph-theory narsinghdeo + – Ayush Upadhyaya asked Jun 8, 2018 Ayush Upadhyaya 479 views answer comment Share Follow See 1 comment See all 1 1 comment reply Ayush Upadhyaya commented Jun 8, 2018 reply Follow Share I made analysis please let me know if I am correct 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. 1 votes 1 votes Please log in or register to add a comment.