# Proof

### "$$\Rightarrow$$"

• Assume, a simple graph $$G(V,E)$$ is semi-Eulerian.
• Then there is an semi-Eulerian tour between a starting and finishing vertices $$v$$ and $$w$$.
• Adding an edge $$e$$ between $$v$$ and $$w$$ creates an Eulerian tour, and thus an Eulerian graph $$G'(V,E')$$ with $$E':=E\cup\{e\}$$.
• By the characterization of Eulerian graphs, $$G'$$ is connected, and each of its vertices has an even degree.
• Whenever this trail passes through a vertex, there is a contribution of $$2$$ to the degree of that vertex.
• By removing the edge $$e$$ again, we see that $$v$$ and $$w$$ are exactly the two vertices with an odd degree.
• Removing the edge $$e$$ does not disconnect the graph $$G'$$, because there is still an open trail between $$v$$ and $$w$$.
• Thus, also $$G$$ is connected.

### "$$\Leftarrow$$"

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### References

#### Bibliography

1. Aldous Joan M., Wilson Robin J.: "Graphs and Applications - An Introductory Approach", Springer, 2000