 butterfly. 
Since $D$ is a bicycle, $D$ is connected, has no leaves, and  
$\Delta D = 1$. Therefore  $D \in   {\cal CG}_{\bowtie}$ and $\Delta D = 1$, and so $D$ has a cycle $C$ and $\Delta C = 0$.  
Therefore our claim follows from Theorem \ref{EarDcmpConGinfty}, when $G_0$ is a cycle, $G_1 = D$, and $r = 1$.
\hfill $\Box$

\vskip 1.5ex

It turns out that  every graph $G$ in ${\cal G}_{\bowtie}$  has bicycles and, moreover, graph $G$ has some special bicycles.

\begin{claim}
\label{AhasBicy}
 
Suppose that $G$ is a   graph in ${\cal CG}_{\bowtie}$. Then for every two edges $a$, $b$ in $G$ there exists a bicycle in $G$ containing $a$ and $b$. 
 
\end{claim}

\noindent{\bf Proof}  \  
By Claim \ref{BicyStructure}, it is sufficient to prove that $G$ has either a $\Theta$-graph or a dumbbell or a butterfly containing $a$ and $b$. 
\\[1ex]
${\bf (p1)}$
Suppose that $G$ has two cycles $C$, $C'$ such that $a, b \in E(C \cup C')$. We can assume that $\{a, b\} \cap E(C) \ne \emptyset$. 

Suppose first that $|V(C) \cap V(C')| \ge 2$. Then in $C'$ there exists a path $P$ such that $V(C) \cap V(P) = End(P)$ and $D = C \cup P$ contains both $a$ and $b$. Then $D$ is a $\Theta$-graph in $G$ containing $a$ and $b$. 

Now suppose that $|V(C) \cap V(C')| \le 1$. Since $G$ is connected, there exists a path $xPx'$ in $G$ (possibly, $x = x'$) such that $V(C) \cap V(P) = x$ and $V(C') \cap V(P) = x'$. If $|V(P)| = 1$, then $D = C \cup C'$ is a butterfly in $G$ containing $a$ and $b$. If $|V(P)| \ge 2$, then $D= C \cup P \cup C'$ is a dumbbell in $G$ containing $a$ and $b$.
\\[1ex]
${\bf (p2)}$
Next suppose that $G$ has a cycle $C$ containing exactly one of $a, b$, say 
$a$, and $b$ belongs to no cycle in $G$. 
Since $G$ is connected, there exists a path $xPx'$ such that $b \in E(P)$ and $V(C) \cap V(P) = x$. Since 
$G \in {\cal G}_{\bowtie}$, clearly $G$ has no leaves, and therefore $P$ belongs to a lollypop $L$ in $G$ such that $V(C) \cap V(L) = End (L)$. 
Since $b$ belongs to no cycle in $G$, graph $C \cup L$ is a dumbbell containing both $a$ and $b$.   
\\[1ex]
${\bf (p3)}$ 
Finally, suppose that neither $a$ nor $b$ belongs to a cycle in $G$. Since $G$ is connected, there exists a path $P$ in $G$ containing $a$ and $b$. Since 
$G \in {\cal G}_{\bowtie}$, clearly $G$ has no leaves, and therefore $P$ belongs to two lollypops $L$ and $L'$ in $G$. Then $L \cup P \cup L'$ is a dumbbell containing $a$ and $b$.      
\hfill $\Box$


\vskip 1.5ex

Now we will  formulate and prove a Splitter Theorem for graph class 
${\cal G}_{\bowtie}$.


\vskip 1ex

\begin{Theorem} {\sc Splitter theorem for a cacti-graph}
\label{EarDcmpGinfty}

Let  $G, G_0 \in {\cal G}_{\bowtie}$. 
Suppose that $G_0$ is a 
subgraph of $G$.
Then  either $G = G_0$ or there exist  sequences
$(P_1, \ldots, P_r)$ and 
$(G_0, G_1, \ldots, G_r)$  of subgraphs of  $G$ such that $G_r = G$, 
every $G_i \in {\cal G}_{\bowtie}$, every $P_i$ is either a path or a cycle or a lollipop {\em(see Figure \ref{Ears})} or a bicycle, and for every 
$i \in \{1, \ldots, r\}$, we have: $G_i = G_{i-1}\cup P_i$, where 
\\[1ex]
${\bf (p)}$ $V( G_{i-1}) \cap V(P_i) = End(P_i)$ if $P_i$ is a  a path,
\\[1ex]
${\bf (l)}$ $V( G_{i-1}) \cap V(P_i) = End(P_i)$ if $P_i$ is a lollipop,
\\[1ex] 
${\bf (c)}$ $|V( G_{i-1}) \cap V(P_i)| = 1$ if $P_i$ is a cycle, and 
\\[1ex]  
${\bf (b)}$ $V( G_{i-1}) \cap V(P_i) = \emptyset$ if $P_i$ is a bicycle, 
\\[1ex]  
and so  
$G_{i-1} \subset ^s G_i$ and 
$\Delta G_i - \Delta G_{i-1} = 1$.

\end{Theorem}


\noindent{\bf Proof}  \  (uses Theorem \ref{EarDcmpConGinfty})

Suppose, to the contrary, that our claim is not true. Our theorem is vacuously true if $G = G_0$.
Suppose that $G_k$ is a maximal subgraph of $G$ such that our claim is true 
for  $G_k$, and so there exist   sequences $(P_1, \ldots, P_k)$ and 
$(G_0,   \ldots, G_k)$  satisfying all the conditions of our theorem.
Then $G_k \in {\cal G}_{\bowtie}$ and $G_k$ is a proper subgraph of $G$
$($i.e. $G_k \subset ^s G$$)$.
We will get a contradiction by showing that  there exist subgraphs 
 $P_{k+1}$ and $G_{k+1}$ of $G$ such that $G_k \cup P_{k+1} = G_{k+1}$,  
$G_{k+1} \in  {\cal G}_{\bowtie}$, and pair $(G_k, P_{k+1})$  satisfies  one of conditions ${\bf (p)}$,${\bf (l)}$, ${\bf (c)}$, ${\bf (b)}$.
\\[1.5ex]
${\bf (p1)}$
Suppose that $G$ has a component $A$ disjoint from $G_k$.
Since $G \in  {\cal G}_{\bowtie}$, also $A \in  {\cal G}_{\bowtie}$.
By Claim \ref{AhasBicy},  $A$ has a subgraph-bicycle $B$.
Put $P_{k+1} = B$ and $G_{k+1} = G_k \cup  P_{k+1}$. 
Then $G_{k+1} \in {\cal G}_{\bowtie}$, $G_{k+1} \subseteq ^s G$, 
and pair $(G_k, P_{k+1})$  satisfies  condition ${\bf (b)}$.
\\[1.5ex]
${\bf (p2)}$
Now suppose that $G$ has no  component disjoint from $G_k$.
Since $G_k$ is a proper subgraph of $G$,  
there are components $A$ of $G$ and $A_k$ of $G_k$ such that $A_k$ is a proper subgraph of $A$.
Since $G, G_k \in  {\cal G}_{\bowtie}$, also $A,  A_k \in  {\cal G}_{\bowtie}$. 
Therefore $A,  A_k \in  {\cal CG}_{\bowtie}$.