# Query regarding Eulerian Graph proof

I'm going through a proof of the following theorem:

A connected graph $G$ is Eulerian iff $d(x)$ is even $\forall x \in G$.

The proof of the harder direction is roughly as follows:

1. Use induction on $e(G)$. Done if $e(G) = 0$.
2. Given connected $G$ with $e(G) > 0$ and the relevant condition on the degrees of the vertices, suppose $G$ not Eulerian and let $C$ be a largest circuit in $G$ with no edges repeated. Note that $e(C) > 0$ as $G$ is not a tree.
3. Let $H$ be a component of $G - E(C)$ with $e(H) > 0$. Then $H$ is connected and $d_H(x)$ is even for all $x \in H$.
4. etc

My problem is understanding why $d_H(x)$ is even for all x.

My thoughts:

We start with a connected graph in which every vertex has even degree. We remove a circuit C (and I see that $d_C(x)$ is even for all $x \in C$). We then take a component $H$ of the remaining graph $G - E(C)$ that has an edge. Suppose $H$ had a vertex $y$ with odd degree (in $H$). Then if we add $C$ back in to the graph, we see that $y$ must be adjacent to an odd number of vertices in $C$. This means that each of these vertices in $C$ must be adjacent to an odd number of vertices in $G - E(C) - y$.

I'm stuck at this point. Am I overcomplicating things? I feel like this is something that should be obvious.

As you noted, all vertices in the circuit $C$ have even degree. Thus, you are taking the degrees of $G$ (all of which are even) and decreasing them all by an even number (possibly 0) by removing the edges of $C$. The result is that every vertex in $G - E(C)$ has even degree ("even minus even is even"). That's all there is to it.