In 2009, I posted a calculational proof of the handshaking lemma, a well-known elementary result on undirected graphs. I was very pleased about my proof because the amount of guessing involved was very small (especially when compared with conventional proofs). However, one of the steps was too complicated and I did not know how to improve it.

In June, Jeremy Weissmann read my proof and he proposed a different development. His argument was well structured, but it wasn’t as goal-oriented as I’d hoped for. Gladly, after a brief discussion, we realised that we were missing a great opportunity to use the trading rule (details below)!

I was so pleased with the final outcome that I decided to record and share the new proof.

## Problem statement

In graph theory, the degree of a vertex $A$, $\fapp{d}{A}$, is the number of edges incident with the vertex $A$, counting loops twice. So, considering Graph 0 below, we have $\fapp{d}{A}=3$, $\fapp{d}{B}=3$, $\fapp{d}{C}=1$, $\fapp{d}{D}=3$, and $\fapp{d}{E}=2$.

A well-known property is that **every undirected graph contains an even number of vertices with odd degree**. The result first appeared in Euler’s 1736 paper on the Seven Bridges of Königsberg and is also known as the handshaking lemma (that’s because another way of formulating the property is that the number of people that have shaken hands an odd number of times is even).

As we can easily verify, Graph 0 satisfies this property. There are four vertices with odd degree ($A$,$B$, $C$, and $D$), and 4, of course, is an even number.

Although the proof of this property is simple, all the conventional proofs that I know of are not goal-oriented. My goal is to show you a development of a goal-oriented proof. Also, my proof is completely guided by the shape of the formulae involved, which helps reducing the amount of guessing involved.