TITLE: On the proof of Fleury's algorithm.
QUESTION [5 upvotes]: On pages 42-43 in [1], it says:

We conclude our introduction to Eulerian graphs with an algorithm for
  constructing an Eulerian trail in a give Eulerian graph. The method is
  know as Fleury's algorithm.
THEOREM 2.12 Let $G$ be an Eulerian graph. Then the following construction is always possible, and produces an Eulerian trail of
  $G$.
Start at any vertex $u$ and traverse the edges in an arbitrary manner,
  subject only to the following rules:
(i) erase the edges as they are traversed, and if any isolated
  vertices result, erase them too;
(ii) at each stage, use a bridge only if there is no alternative.
Proof. We show first that the construction can be carried out at each stage.
Suppose that we have just reached a vertex $v$, erasing the edges as
  we go. If $v \neq u$, then the subgraph $H$ that remains is connected
  and has only two vertices of odd degrees, $u$ and $v$. To show that the construction can be carried out, we must show that the removal of the next edge does not disconnect $H$ $-$ or, equivalently, that $v$ is incident with at most one bridge. ...

Please look at the last sentence of the proof. Why the following two statements are equivalent?

The removal of the next edge does not disconnect $H$.
$v$ is incident with at most one bridge.

Thanks in advance.
[1] Robin J. Wilson, Introduction to Graph Theory, 5th ed., Prentice Hall, 2012.

REPLY [4 votes]: $(2)\Rightarrow(1):$ Suppose $u$ is incident with at most one bridge. If $v$ is incident with an edge $e$ which is not a bridge, then the algorithm says that we should remove $e$. Thus, since $e$ is not a bridge, the resulting graph is still connected. On the other hand, if $v$ is only incident with bridge edges, then by assumption $v$ is incident with exactly one edge. Hence, if we remove this edge we isolate $v$. Consequently, $v$ is deleted and the resulting graph is connected.
$(1)\Rightarrow(2):$ This is much more complicated and involves using the fact that $G$ is Eulerian. The algorithm started at vertex $u$ and has now reached vertex $v$ with $u\neq v$. Let $C_{1},...,C_{k}$ be the connected components of $G\setminus\{v\}$, and without loss of generality assume that $u\in C_{1}$. For each $i$, let $d_{i}^{G}(v)$ denote the number of edges of $G$ which are incident with $v$ and a vertex from $C_{i}$. Note that in an Eulerian trail, we must leave $C_{1}$ the same number of times as we enter $C_{1}$. Thus, $d_{1}^{G}(v)$ is even. Similarly, for each $j\neq1$, we must leave $C_{j}$ as often as we enter (so that we start and end at $u$). Thus, $d_{i}^{G}(v)$ is even for all $i$.
We have run this algorithm starting from $u$, deleting edges as we go, and have now arrived at $v$. We started in $C_{1}$ and we are now outside $C_{1}$. So the number of times we left $C_{1}$ must be $1$ greater than the number of times we entered. Hence, the number of edges we have deleted between $v$ and $C_{1}$ is odd. Thus $d_{1}^{G}(v)-d_{1}^{H}(v)$ is odd, so $d_{1}^{H}(v)$ is odd. Let $j\neq1$. We started at $u\in C_{1}$ and ended at $v\notin C_{j}$. Thus, we must have visited $C_{j}$ the same number of times as we left. Hence $d_{j}^{G}(v)-d_{j}^{H}(v)$ is even, so $d_{j}^{H}(v)$ is even.
Suppose $b=(v,x)$ is a bridge of $H$ which is incident with $v$. We have that $x\in C_{i}$ for some $i$. Since $b$ is a bridge, there are no edges between $C_{i}$ and $v$ in $H-b$. Thus, $b$ is the only edge between $C_{i}$ and $v$ in $H$, so $d_{i}^{H}(v)=1$. By the previous paragraph, this means that $i=1$ (since $d_{i}^{H}(v)$ is odd). Therefore, either $v$ isn't incident to any bridges, or $v$ is incident to exactly $1$ bridge (and this occurs only if there is exactly $1$ edge between $C_{1}$ and $v$ in $H$, in which case this edge is the unique bridge incident to $v$). $\square$