On the proof of Fleury's algorithm. 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.
 A: $(2)\Rightarrow(1):$ 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 $v$). $\square$
