Connected $k-$regular graph does not have cut-edge for $k \ge 2$

Question

This is a conjecture generalized from the result that "a connected k-regular bipartite graph does not have cut-edge."

Cut–edge is an edge whose deletion will result in a disconnected graph. I think we can prove this by arguing that every two vertex on this graph lies on a cycle. An intuition is that there is no tree like structure in such graph. If we consider a longest path that includes $v_i$ and $v_j$, $i\neq j$, since $\deg(v_i)>1$ for all vertex, the vertex at the end must connect to at least two vertices. This indicates this vertex must connect back to some vertex in this path or it is not the longest path.

How can we formalize this proof? If this conjecture is not true, can anybody give a counterexample?

Answer 1

This is false. Consider a $3-$regular connected graph

which clearly has a cut-edge.

Answer 2

Here's a connected 3-regular graph with a cut-edge. Start with a path on vertices $1,2,3,\dots,14$. Add edges $1-3,1-7,2-5,4-6,8-14,9-11,10-13,12-14$. $7-8$ is a cut-edge.