Graph with Even Vertices Partitions into Cycles

Theorem
Let $$G = \left({V, E}\right)$$ be a graph whose vertices are all even.

Then its edge set $$E$$ can be partitioned into cycles, no two of which share an edge.

Proof
Let $$G = \left({V, E}\right)$$ be a graph whose vertices are all even.

If there is more than one vertex in $$G$$, then each vertex must have degree greater than $$0$$.

Begin at any vertex $$u$$.

Since the graph is connected, there must be an edge $$\{u, u_1\}$$ for some vertex $$u_1 \ne u$$.

Since $$u_1$$ has even degree greater than $$0$$, there is an edge $$\{u_1,u_2\}$$.

These two edges make a trail from $$u$$ to $$u_2$$.

Continue this trail, leaving each vertex on an edge that was not previously used, until we reach a vertex $$v$$ that we have met before.

(Note: $$v$$ may or may not be the same vertex as $$u$$. It does not matter either way.)

The edges of the trail between the two occurrences of $$v$$ must form a cycle.

Call the cycle formed by this process $$C_1$$.

If $$C_1$$ covers all the edges of $$G$$, then we are done.

Otherwise, remove the edges forming $$C_1$$ from the graph, leaving the graph $$G_1$$.

All the vertices in $$G_1$$ are still even. So pick some vertex $$u'$$ in $$G_1$$.

Repeat the same process as before, starting with an edge $$\left\{{u', u'_1}\right\}$$.

By the same argument, we can generate a new cycle $$C_2$$, which has no edges in common with $$C_1$$.

If $$C_2$$ covers all the rest of the edges of $$G$$, then we are done.

Otherwise, remove the edges forming $$C_2$$ from the graph, leaving the graph $$G_2$$, which again contains only even vertices.

We continue in this way until we have used up all the edges of $$G$$.

By this time we have a number of cycles, $$C_1, C_2, \ldots, C_k$$ which between them contain all the edges of $$G$$ but no two of them have an edge in common.