rules.

1.  For every $v\in G-p$, if $v$ is a $2$–vertex, $v$ pulls charge 1 from each incident face.

2.  If $p$ is a $2$–vertex, then $p$ gives charge $\frac{9}{2}$ to each incident face.

Observe that every vertex has nonnegative charge after this discharging process. It remains to show that every face has nonnegative charge.

Let $f$ be a face, and let $r_2$ be the number of 2–vertices on the boundary of $f$, not counting $p$, and consider two cases.

**Case 1**: $d(p)=3$ or $p$ is not adjacent to $f$.

In this case, $p$ does not give charge to $f$, and therefore $f$ has charge $\ell(f) - r_2 - 6$ after discharging. Also, the boundary of $f$ does not contain a path of length 8 containing only vertices of degree 2, thus $r_2 \leq \left\lfloor \frac{7}{8}\ell(f)\right\rfloor$. Since $\ell(f) \geq 41$, we have $$\ell(f) - r_2 - 6 \geq \ell(f) - \left\lfloor \frac{7}{8}\ell(f)\right\rfloor - 6 \geq 0.$$

**Case 2**: $d(p)=2$ and $p$ is adjacent to $f$.

By (R2), $p$ gives charge $\frac{9}{2}$ to $f$, so that $f$ has charge $\ell(f)  - r_2 - \frac{3}{2}$ after discharging. The boundary of $f$ does not contain a path of length 8 containing only vertices 