Handshaking Lemma I need understand the proof of this Corollary, attributed to Andrew Thomason.
Let $G$ be a multigraph, let $u, v \in V$, and suppose that $d ( w )$ is odd for
each vertex $w \in V-$ { $u,v$ }. Then the number of Hamiltonian paths in $G$ from $u$ to $v$ is even.
I read this 'proof' but I don't understand the end.
 A: The key is to work out why the degrees are as claimed, which is just quite not obvious.
Let $z_0 = y$ and suppose that $P=z_0z_1\ldots z_n$ is a Hamilton path in $G-x$.  Which edges $z_iz_j$ ($j \geq i+2$) can be swapped in to make a new Hamilton path?  We would have to follow $P$ as far as $z_i$, then take the edge $z_iz_j$.  If $j=n$ then we can now come back along $P$ as far as $z_{i+1}$ to obtain a Hamilton path.  But if $j < n$ then we are stuck: we still have to pick up vertices to the left and right of $z_j$ in $P$, but can't do this using only edges of $P$.  So edges $G-x$ can be swapped in to make a new Hamilton path precisely when they are incident on $z_n$.  We can't swap in $z_{n-1}z_n$ since it's already present, so the degree of $P$ in the auxiliary graph where Hamilton paths are joined when they differ by a single edge is $d_{G-x}(z_n)-1$, or
$$
d_G(z_n) - 1 - \begin{cases}1 & z_n \text{ is adjacent to } x \\ 0 & \text{otherwise.}\end{cases}
$$
Since $d_G(z_n)$ is odd, this is odd if and only if $z_n$ is adjacent to $x$, or, equivalently, the Hamilton path in $G-x$ can be extended to a Hamilton path from $y$ to $x$ in $G$.
If you wanted to optimise the proof to minimise space on the page you should define the auxiliary graph to bake in the fact that you only have to think about edges incident on the current endpoint, rather than leaving this as something you have to check.
