different hexagons. Therefore, by counting contributions of hexagons and contributions of edges that belong to two hexagons, we get $$\begin{aligned}
M_2(G_1) & = & 24t(G_1) + 24l(G_1) + 25a(G_1) + 27b(G_1) + 9(h(G_1)-1) \\
& = & 24h(G_1) + a(G_1) + 3b(G_1) + 9h(G_1) - 9 \\
& = & 33h(G_1) + s(G_1) + b(G_1) - 10,\end{aligned}$$ where the last equality follows from Lemma \[pomoc\].

In a phenylene $G_2$ there are $4(h(G_1)-1)$ edges of type 3 that appear on a quadrilateral. Therefore, by counting contributions of hexagons and quadrilaterals we get $$\begin{aligned}
M_2(G_2) & = & 24t(G_2) + 24l(G_2) + 25a(G_2) + 27b(G_2) + 36(h(G_2)-1) 
\\ & = & 60h(G_2) + s(G_2) + b(G_2) - 37,\end{aligned}$$ where the computation is done in a similar way as before. This completes the proof.

The following theorem is the main result of this section.

\[Wie\_pol\] If $G_1$ is a catacondensed benzenoid graph and $G_2$ a phenylene, then it holds $$\begin{aligned}
W_p(G_1) & = & 9h(G_1) + s(G_1) + b(G_1) - 7, \\
W_p(G_2) & = & 13h(G_2) + s(G_2) + b(G_2) - 11.\end{aligned}$$

Since $G_1$ does not contain $4$-cycles and $5$-cycles, we have $f(G_1)=|C_4(G_1)|=|C_5(G_1)|=0$ and $|C_6(G_1)| = h(G_1)$. On the other hand, for 