Prove that these 3 statements are equivalence for a connected graph:
a. degree of all vertices are even
b. graph has a closed Euler path (An Euler circuit that can contain repeated vertices)
c. You can part graph edges into separate circuits in a way that each edge is in exactly 1 circuit.
Any hints how to prove?
I think I would prove $(a)\Rightarrow(c)$ by induction on the number of edges, making one circuit $C$ and then observe that the conditions still hold for $G - E( C)$. Note that this would be a proof that doesn't rely on $G$ being connected.
Then moving on to $(c)\Rightarrow(b)$; given a set of circuits that contain all the edges of a connected graph, show that you can assemble these to make an Eulerian path.
Then finally $(b)\Rightarrow(a)$, given an Eulerian path show that all degrees are even.