\[phen\] If $G$ is a phenylene, then $|V(G)|  =  6h(G)$, $|E(G)|  =  8h(G) - 2$, $n_3(G) =  4h(G) - 4$, and $n_2(G)  =  2h(G) + 4$.

Let $G'$ be the hexagonal squeeze of $G$. Since $h(G)=h(G')$ and $G$ has exactly $h(G)-1$ quadrilaterals, $|V(G)| = |V(G')| + 2(h(G)-1)$ and $|E(G)|= |E(G')| + 3(h(G)-1)$. Moreover, $G$ and $G'$ have the same number of vertices of degree two. Therefore, the result follows by Proposition \[gut-ben\].

In the next lemma, we derive relations between the numbers of different types of hexagons and the number of segments.

\[pomoc\] If $G$ is a catacondensed benzenoid graph or a phenylene, then it holds $2b(G) + a(G) = s(G) -1$ and $b(G) + 2 = t(G)$.

Let $T$ be a dualistic tree of $G$. For any $v \in V(T)$ with $\deg(v)=2$ and $N(v)=\lbrace a, b \rbrace$, let $T_v$ be a tree obtained from $T$ by removing $v$, $va$, $vb$ and adding an edge $ab$. In such a case, we say that $T_v$ is obtained from $T$ by *skipping* vertex $v$. Moreover, let $T'$ be a tree obtained from $T$ by successively skipping all the vertices corresponding to the linear hexagons of $G$. Obviously, $|V(T')| = b(G) + a(G) 