Prove that a 3-regular graph has a bridge iff it has a cut-vertex.
Here is my work:
Assume a 3-reg graph G has a bridge, say e=uv. Then G-e is disconnected, and so has 2 components, G1 and G2. Since e is a bridge, u and v are in different components. Then either u or v must be a cut-vertex.
Next assume a 3-reg graph G has a cut vertex, say u. Then G-u is disconnected, and so has 2 components, G1 and G2. Since u is a cut-vertex, the edge to which it is adjacent to must be a bridge(?)
I'm not really sure, I have an intuitive picture that is every vertex in a graph has degree of 3, then if it has a cut-vertex, it must also have a bridge, but I am having trouble formally proving this.