Every Planar Graph Without Pairwise Adjacent 3-, 4-, and 5-Cycle is DP-4-Colorable

Abstract

DP-coloring is a generalization of list coloring in simple graphs. Many results in list coloring can be generalized in those of DP-coloring. Kim and Ozeki showed that every planar graph without k-cycles where $k=3,4,5,$ or 6 is DP-4-colorable. Recently, Kim and Yu extended the result on 3- and 4-cycles by showing that every planar graph without triangles adjacent to 4-cycles are DP-4-colorable. Xu and Wu showed that every planar graph without 5-cycles adjacent simultaneously to 3-cycles and 4-cycles is 4-choosable. In this paper, we extend the results on 3-, 4-, and 5-cycles as follows. Let G be a planar graph without pairwise adjacent 3-, 4-, and 5-cycle. We prove that each precoloring of a 3-cycle of G can be extended to a DP-4-coloring of G. As a consequence, each planar graph without pairwise adjacent 3-, 4-, and 5-cycle is DP-4-colorable.

1 Introduction

Every graph in this paper is finite, simple, and undirected. Embedding a graph G in the plane, we let V(G), E(G),  and F(G) denote the vertex set, edge set, and face set of G. For $U \subseteq V(G),$ we let G[U] denote the subgraph of G induced by U. For $X, Y \subseteq V(G)$ where X and Y are disjoint, we let $E_G(X,Y)$ be the set of all edges in G with one endpoint in X and the other in Y.

The concept of choosability was introduced by Vizing [9] and by Erdős et al. [5], independently. A k-assignmentL of a graph G assigns a list L(v) (a set of colors) with $|L(v)|= k$ to each vertex v. A graph G is L-colorable if there is a proper coloring f where $f(v)\in L(v).$ If G is L-colorable for every k-assignment L,  then we say G is k-choosable.

Dvořák and Postle [4] introduced a generalization of list coloring in which they called correspondence coloring. But following Bernshteyn et al. [3], we call it DP-coloring.

Definition 1

Let L be an assignment of a graph G. We call H a cover of G if it satisfies all the followings:

• (i) The vertex set of H is $\bigcup _{u \in V(G)}(\{u\}\times L(u)) =\{(u,c): u \in V(G), c \in L(u) \};$

• (ii) $H[u\times L(u)]$ is a complete graph for every $u \in V(G);$

• (iii) For each $uv \in E(G),$ the set $E_H(\{u\}\times L(u), \{v\}\times L(v))$ is a matching (may be empty).

• (iv) If $uv \notin E(G),$ then no edges of H connect $\{u\}\times L(u)$ and $\{v\}\times L(v).$

Definition 2

An (H, L)-coloring of G is an independent set in a cover H of G with size |V(G)|. We say that a graph is DP-k-colorable if G has an (H, L)-coloring for every k-assignment L and every cover H of G. The DP-chromatic number of G,  denoted by $\chi _{DP}(G),$ is the minimum number k such that G is DP-k-colorable.

If we define edges on H to match exactly the same colors in L(u) and L(v) for each $uv \in E(G),$ then G has an (H, L)-coloring if and only if G is L-colorable. Thus DP-coloring is a generalization of list coloring. This also implies that $\chi _{DP}(G) \ge \chi _l(G).$ In fact, the difference of these two chromatic numbers can be arbitrarily large. For graphs with average degree d,  Bernshteyn [2] showed that $\chi _{DP}(G) = \Omega (d /\log d),$ while Alon [1] showed that $\chi _l(G) = \Omega (\log d).$

Dvořák and Postle [4] showed that $\chi _{DP}(G) \le 5$ for every planar graph G. This extends a major result on list coloring by Thomassen [8]. On the other hand, Voigt [10] gave an example of a planar graph which is not 4-choosable (thus not DP-4-colorable). It is of interest to obtain sufficient conditions for planar graphs to be DP-4-colorable. Kim and Ozeki [6] showed that every planar graph without k-cycles is DP-4-colorable for each $k =3,4,5,6.$ Kim and Yu [7] extended the result on 3- and 4-cycles by showing that every planar graph without triangles adjacent to 4-cycles is DP-4-colorable.

Let ${\mathcal {A}}$ denote the family of planar graphs G without pairwise adjacent 3-, 4-, and 5-cycle. In this paper, we extend the results on 3-, 4-, and 5-cycles as follows.

Theorem 1.1

Let $G \in {\mathcal {A}}.$ Then each precoloring of a 3-cycle can be extended to a DP-4-coloring of G.

The following corollary is immediate.

Corollary 1.2

Every planar graph without pairwise adjacent 3-, 4-, and 5-cycle is DP-4-colorable.

Corollary 1.2 generalizes the aforementioned result by Kim and Yu [7] and the following result by Xu and Wu [11].

Theorem 1.3

[11] Every planar graph without 5-cycles adjacent simultaneously to 3-cycles and 4-cycles is 4-choosable.

2 Preliminaries

First, we introduce some notations and definitions. A k-vertex ($k^+$-vertex, $k^-$-vertex, respectively) is a vertex of degree k (at least k,  at most k,  respectively). The same notations are applied to faces.

A $(d_1,d_2,\dots ,d_k)$-facef is a face of degree k where vertices on f have degree $d_1,d_2,\dots ,d_k$ in a cyclic order. A $(d_1,d_2,\dots ,d_k)$-vertexv is a vertex of degree k where faces incident to v have degree $d_1,d_2,\dots ,d_k$ in a cyclic order. Note that some face may appear more than one time in the order. We say xy is a chord in a cycle C if $x,y\in V(C)$ but $xy\in E(G)- E(C).$ An internal chord of C is a chord inside C while external chord of C is a chord outside C

A graph C(m, n) is obtained from a cycle $x_1x_2\ldots x_{m+n-2}$ with an internal chord $x_1x_m.$ A graph C(l, m, n) is obtained from a cycle $x_1x_2\ldots x_{l+m+n-4}$ with internal chords $x_1x_l$ and $x_1x_{l+m-2}.$ A graph C(m, n, p, q) can be defined similarly. We use int(C) and ext(C) to denote the sets of vertices inside and outside a cycle C, respectively. The cycle C is a separating cycle if int(C) and ext(C) are not empty. We use B(f) to denote a boundary of a face f. It is straightforward to see that if f is a $5^-$-face, then B(f) is a cycle.

3 Structures

Proof of Theorem 1.1

Let G be a minimal counterexample to Theorem 1.1 with |V(G)| minimized and a precolored 3-cycle $C_0.$$\square$

Lemma 3.1

G has no separating 3-cycles.

Proof

Suppose to the contrary that there exists G contains a separating 3-cycle C. Note that C is not necessary $C_0.$ By symmetry, we assume $V(C_0) \subseteq V(C) \cup int(C).$ By the minimality of G,  a precoloring of $C_0$ can be extended to $V(C) \cup int(C).$ After C is colored, then again the coloring on C can be extended to ext(C). Thus we have a DP-4-coloring of G, a contradiction. $\square$

Since $C_0$ is not a separating 3-cycle by Lemma 3.1, we may assume that $C_0$ is the boundary of the outer face D of G in the remaining of the paper.

Definition 3

Let H be a cover of G with a list assignment L. Let $G'= G-F$ where F is an induced subgraph of G. A list assignment $L'$ is a restriction of L on $G'$ if $L'(u) = L(u)$ for each vertex in $G'.$ A graph $H'$ is a restriction of H on $G'$ if $H'= H[\{\{v\} \times L(v): v \in V(G')\}].$ Assume $G'$ has an $(H',L')$-coloring with an independent set $I'$ in $H'$ such that $|I'|= |V(G)|-|V(F)|.$

A residual list assignment$L^*$ of F is defined by

$$\begin{aligned} L^*(x)=L(x)-\bigcup _{ux\in E(G)}\{c'\in L(x) : (u,c)(x,c')\in E(H) \text { and }(u,c)\in I'\} \end{aligned}$$

for each $x \in V(F).$

A residual cover $H^*$ is defined by $H^*=H[\{\{x\} \times L^*(x): x \in V(F)\}].$

From above definitions, we have the following fact.

Lemma 3.2

Assume $I'$ is a $(H',L')$-coloring of $G'.$ A residual cover $H^*$ is a cover of F with an assignment $L^*.$ Furthermore, if F is $(H^*,L^*)$-colorable, then G is (H, L)-colorable.

Proof

One can check from the definitions of a cover and a residual cover that $H^*$ is a cover of F with an assignment $L^*.$

Suppose that F is $(H^*,L^*)$-colorable. Then $H^*$ has an independent set $I^*$ with $|I^*|= |F|.$ It follows from Definition 3 that no edges connect $H^*$ and $I'.$ Additionally, $I'$ and $I^*$ are disjoint. Altogether, we have that $I=I'\cup I^*$ is an independent set in H with $|I|=(|V(G)|-|V(F)|)+|V(F)|=|V(G)|.$ Thus G is (H, L)-colorable. $\square$

Lemma 3.3

Every vertex not on $C_0$ has degree at least 4.

Proof

Suppose to the contrary that G has a vertex x not on $C_0$ with degree at most 3. Let L be a 4-assignment and let H be a cover of G such that G has no (H, L)-coloring. By the minimality of G,  the subgraph $G'=G-x$ an $(H',L')$-coloring where $L'$ (and $H'$) is a restriction of L (and H, respectively) on $G'.$ Thus there is an independent set $I'$ with $|I'|= |G'|$ in $H'.$ Consider a residual list assignment $L^*$ on x. Since $|L(x)| = 4$ and $d(x)\le 3$, we obtain $|L^*(x)|\ge 1.$ Clearly, $\{(x,c)\}$ where $c \in L^*(x)$ is an independent set in $G[\{x\}].$ Thus $G[\{x\}]$ is $(H^*,L^*)$-colorable. It follows from Lemma 3.2 that G is (H, L)-colorable, a contradiction. $\square$

Lemma 3.4

• (a) A 5-cycle has no chords.

• (b) A bounded 3-face f is not adjacent to a 4-face g.

• (c) If bounded 3-faces f and g are adjacent, then $B(f)\cup B(g) = C(3,3).$

• (d) If a bounded 3-face f is adjacent to a 5-face g,  then $B(f)\cup B(g) = C(3,5).$

• (e) If C(3, 5) is obtained from a 6-cycle C with a chord, then C has exactly one chord.

• (f) If bounded 3-faces f and g are adjacent, then f is not adjacent to a bounded $5^-$-face.

Proof

• (a) Let $C= rstuv$ be a 5-cycle. Suppose that rt is a chord. Then we have three pairwise adjacent cycles rst,  rtuv,  and rstuv,  contrary to $G \in {\mathcal {A}}.$

• (b) Let $B(f)=uvw$ and $B(g)=vwxy.$ Suppose that $u = x$ or y. We have that $d(v) =2$ or $d(w)=2,$ contrary to Lemma 3.3. Thus $x\ne u \ne y.$ We obtain a 5-cycle uwxyv with a chord vw,  contrary to (a).

• (c) Let $B(f)=uvw$ and $B(g)=uvx.$ Since both f and g are bounded, we have that $w \ne x.$ Thus $B(f)\cup B(g) = C(3,3).$

• (d) Let $B(f)=uvz$ and $B(g)=uvwxy$. If $z\ \in \{w,x,y\},$ then a 5-cycle B(g) has a chord, contrary to (a).

• (e) Let C be a cycle uzvwxy with a chord uv. Suppose to the contrary that C contains another chord st. By (a) and (b), st is not a chord in a 5-cycle uvwxy. By symmetry, we may assume $s=z$ and $t=x$ or $t=y.$ Then we have three pairwise adjacent cycles uvz,  uvwxy,  and xyuz or uvzy,  contrary to $G \in {\mathcal {A}}.$

• (f) Let $B(f)=uvw$ and $B(g)=vwx$. By (c), $u \ne x.$ Suppose f is adjacent to a k-face h where $k\le 5.$ By (b), $h =3$ or 5.

  • $B(h)=uvz.$

    By (c), $z \ne w.$ Suppose to the contrary that $z =x.$ Then $d(v)=3,$ contrary to Lemma 3.3. Thus $z \ne x.$ Altogether, we have three pairwise adjacent 3-cycles uvw,  uvxw,  zuwxv,  contrary to $G \in {\mathcal {A}}.$

  • $B(h)=rstuv.$

    By (a), $w \notin \{r,s,t\}.$ By (e), $x \notin \{r,s,t\}.$ Altogether, we have three pairwise adjacent cycles uvw,  uvxw,  and rstuv,  contrary to $G \in {\mathcal {A}}.$

Thus h is not a $5^-$-face. $\square$

Lemma 3.4 (f) yields this immediate consequence.

Corollary 3.5

For $k\ge 4,$ a k-vertex v in G is incident to at most $k-2$ 3-faces.

Lemma 3.6

Let $C(l_1,\ldots ,l_k)$ be obtained from a cycle $C=x_1 \ldots x_m$ with k internal chords sharing a common endpoint $x_1$ such that $V(C)\cap V(C_0)= \emptyset .$ Suppose $x_2$ or $x_m$ is not the endpoint of any chord in C. If $d(x_1) \le k+2,$ then there exists $i \in \{2,3,\ldots ,m\}$ such that $d(x_i) \ge 5.$

Proof

By symmetry, let $x_m$ be not an endpoint of any chord in C. Suppose to the contrary that $d(x_i) \le 4$ for each $i =2,3,\ldots ,m.$ Let L be a 4-assignment and let H be a cover of G such that G has no (H, L)-coloring. By the minimality of G,  the subgraph $G'=G-\{x_1,\ldots , x_m\}$ admits an $(H',L')$-coloring where $L'$ (and $H'$, respectively) is a restriction of L (and H, respectively) in $G'.$ Thus there is an independent set $I'$ with $|I'|= |G'|$ in $H'.$

Consider a residual list assignment $L^*$ on F. Since $|L(v)|= 4$ for every $v\in V(G),$ we have $|L^*(x_1)|\ge 3$ and $|L^*(v)| \ge 3$ for each $v\in V(C)$ with an edge $x_1v$ and $|L^*(x_i)| \ge 2$ for each of the remaining vertices $x_i$ in V(C). Let $H^*$ be an residual cover of F. Since $x_m$ is not an endpoint of a chord in C,  we can choose a color c from $L^*(x_1)$ such that $|L^*(x_{m})-\{c':(x_1,c)(x_{m},c')\in E(H^*)\}|\ge 2.$ By choosing colors of $x_2$, $x_3,\dots , x_{m}$ in this order, we obtain an independent set $I^*$ with $|I^*|=m=|F|.$ Thus F is $(H^*,L^*)$-colorable. It follows from Lemma 3.2 that G is (H, L)-colorable, a contradiction. $\square$

Corollary 3.7

For each C(3, 5) such that $V(C(3,5))\cap V(C_0)= \emptyset ,$ there exists a vertex with degree at least 5.

Proof

Let C(3, 5) be obtained from a 6-cycle $C= x_1 \ldots x_6$ with a chord $x_1x_3$ such that $V(C)\cap V(C_0)= \emptyset .$ By Lemma 3.4 (e), C has no other chords. The proof is complete by Lemma 3.6. $\square$

Lemma 3.8

Let $F=C(3,5,3), C(3,5,5)$ or C(5, 3, 5) be obtained from a cycle $C=x_1\ldots x_m$ with two internal chords sharing an endpoint $x_1$ such that $V(C)\cap V(C_0)= \emptyset .$ If $d(x_1)=5,$ then there exists $i\in \{2,\ldots ,m\}$ with $d(x_i) \ge 5.$

Proof

By Lemma 3.6, it suffices to show that $x_2$ or $x_m$ is not an endpoint to a chord in C.

Let $F = C(3,5,3).$ It follows from Lemma 3.4 (e) that $x_2$ is not adjacent to $x_4,x_5,$ or $x_6.$ If $x_2$ is adjacent to $x_7,$ then we have separating 3-cycle $x_1 x_2 x_7,$ contrary to Lemma 3.1. Thus $x_2$ is not an endpoint of any chord of C.

Let $F = C(3,5,5).$ Suppose there exists a chord e of C where $e=x_2x_i,$ for otherwise we have the desired condition. If $x_i= x_9,$ then we have separating 3-cycle $x_1 x_2 x_9,$ contrary to Lemma 3.1. It follows from Lemma 3.4 (e) that $i \notin \{4,5,6\}.$ Then $x_i = x_7$ or $x_8.$ By Lemma 3.4 (a), $x_9$ is not adjacent to $x_6$ or $x_7.$ Thus $x_9$ is not an endpoint of any chord of C.

Let $F_1= C(5,3,5).$ Suppose there exists a chord e of C where $e=x_2x_i,$ for otherwise we have the desired condition. If $x_i= x_9,$ then we have separating 3-cycle $x_1 x_2 x_9,$ contrary to Lemma 3.1. It follows from Lemma 3.4 (e) that $i \notin \{4,5,6\}.$ Then $x_i= x_7$ or $x_8.$ By Lemma 3.4 (a), $x_9$ is not adjacent to $x_6$ or $x_7.$ Thus $x_9$ is not an endpoint of any chord of $C'.$$\square$

Corollary 3.9

Let v be a 5-vertex with incident bounded faces $f_1,\ldots ,f_5$ in a cyclic order. Let $F = B_1 \cup B_2 \cup B_3$ where $B_i$ denote $B(f_i)$ and $V(F)\cap V(C_0)= \emptyset .$ If $(d(f_1),d(f_2),d(f_3))=(3,5,3)$ or (3, 5, 5),  or $(d(f_1),d(f_2),d(f_3),d(f_4))= (5,3,5,3),$ then there exists $w \in V(F)$ such that $d(w) \ge 5$ and $w \ne v.$

Proof

By Lemma 3.8, it suffices to show that $F= C(3,5,3),$C(3, 5, 5),  or C(5, 3, 5).

• $(d(f_1),d(f_2),d(f_3))=(3,5,3).$

  Let $B_1 = rsv,$$B_2= vstuw,$ and $B_3=vwx.$ It follows from Lemma 3.4 (d) that $V(B_1) \cap V(B_2) = \{s, v\}$ and $V(B_2) \cap V(B_3) = \{v,w\}.$ If $r=x,$ then $d(v)=3,$ contrary to Lemma 3.3. Thus $V(B_1) \cap V(B_3) =\{v\}.$ Altogether we have $F=C(3,5,3).$

• $(d(f_1),d(f_2),d(f_3))=(3,5,5).$

  Let $B_1= rsv,$$B_2= vstuw,$ and $B_3=vwxyz.$ It follows from Lemma 3.4 (d) that $V(B_1) \cap V(B_2) = \{s,v\}.$ If $r = z,$ then $d(v) = 3,$ contrary to $d(v)=5.$ It follows from Lemma 3.4 (a) that neither r nor s is in $V(B_3).$ Thus $V(B_1) \cap V(B_3)= \{v\}.$ Now consider $V(B_2) \cap V(B_3).$ It follows from Lemma 3.4 (a) that neither s nor u is in $V(B_3).$ Similarly, neither x nor z is in $V(B_2).$ Then $V(B_2) \cap V(B_3) =\{t=y, v,w\}$ or $\{v,w\}.$ Note that $r\ne z,$ for otherwise rsv is a separating cycle, contrary to Lemma 3.1. Moreover, $r \notin \{x,y\},$ otherwise the cycle vwxyz has a chord, contrary to Lemma 3.4 (a). Thus $V(B_1) \cap V(B_3) = \{v\}.$ If $V(B_2) \cap V(B_3) =\{t=y, v,w\}$, then we have three adjacent pairwise cycles rsv,  stzv,  stuwv,  contrary to $G \in {\mathcal {A}}.$ Thus $V(B_2) \cap V(B_3)= \{v, w\}.$ Altogether we have $F=C(3,5,5).$

• $(d(f_1),d(f_2),d(f_3),d(f_4))= (5,3,5,3).$

  Let $B_1= rstuv,$$B_2= uvw,$$B_3= vwxyz,$ and $B_4=vpz.$ It follows from Lemma 3.4 (e) that $V(B_1) \cap V(B_2) = \{u,v\}$ and $V(B_2) \cap V(B_3) = \{v,w\}.$ Consider $V(B_1) \cap V(B_3).$ It follows from Lemma 3.4 (a) that neither r nor u is in $V(B_3).$ Similarly, neither w nor z is in $V(B_2).$ By Lemma 3.4 (a) that neither r nor u is in $V(B_3).$

  • Suppose $\{s,t\} \subseteq V(B_1) \cap V(B_3).$

    Then $\{x,y\} \subseteq V(B_1) \cap V(B_3)$ and $s=y$ and $t=x.$ Consequently, we have three adjacent pairwise cycles uvw,  uvwt,  rstuv,  contrary to $G \in {\mathcal {A}}.$

  • Suppose $s \in V(B_1) \cap V(B_3)$ but $t \notin V(B_1) \cap V(B_3).$

    Then $\{v,s\} = V(B_1) \cap V(B_3).$ Consequently $s=x$ or $s=y.$ In the former case, we have three adjacent pairwise cycles uvw,  stuw,  vwxyz,  contrary to $G \in {\mathcal {A}}.$ In the later case, we have three adjacent pairwise cycles pvz,  srvz,  vwxyz,  contrary to $G \in {\mathcal {A}}.$

  • Suppose $t \in V(B_1) \cap V(B_3)$ but $s \notin V(B_1) \cap V(B_3).$

    Then $\{v,t\} = V(B_1) \cap V(B_3).$ Consequently $t=x$ or $t=y.$ In the former case, we have three adjacent pairwise cycles uvw,  uvwt,  vwxyz,  contrary to $G \in {\mathcal {A}}.$ In the later case, we have three adjacent pairwise cycles uvw,  uwxy,  vwxyz,  contrary to $G \in {\mathcal {A}}.$

  Thus $V(B_1) \cap V(B_3) = \{v\}.$ Altogether we have $B_1 \cup B_2 \cup B_3 = C(5,3,5).$

$\square$

Corollary 3.10

Let v be a 6-vertex with consecutive incident faces $f_1,\ldots ,f_6.$ Let $F = B_1 \cup B_2 \cup B_3\cup B_4$ where $B_i$ denote $B(f_i)$ and $V(F)\cap V(C_0)= \emptyset .$ If $(d(f_1),d(f_2),d(f_3),d(f_4))= (3,5,3,5),$ then there exists $w \in V(F)$ such that $w \ne v$ and $d(w) \ge 5.$

Proof

By Lemma 3.6, it suffices to show that $F= C(3,5,3,5).$ Similar to the proof of corollary 3.9, one can show that $B_1 \cup B_2 \cup B_3 = C(3,5,3)$ and $B_2 \cup B_3 \cup B_4 = C(5,3,5).$ Let $V(B_1) = \{t,u,v\}$ and $V(B_4) =\{v,w,x,y,z\}$ where $t\in V(B_2)$ and $w \in V(B_3).$ It only remains to show that $u \notin \{x,y,z\}.$ If $u =x$ or y,  then the cycle vwxyz has a chord vx or vy,  contrary to Lemma 3.4 (a). If $u=z,$ then vtz is a separating cycle, contrary to Lemma 3.1. Thus $F= C(3,5,3,5).$$\square$

4 Discharging Process

We are now ready to present a discharging procedure that will complete the proof of Theorem 1.1. Let each vertex $v\in V(G)$ have an initial charge of $\mu (v)=2d(v)-6$, each face $f\ne D$ has an initial charge of $\mu (f)=d(f)-6$ and $\mu (D)=d(D)+6=9$. By Euler’s Formula, $\sum _{x\in V\cup F}\mu (x)=0$. Let $\mu ^*(x)$ be the charge of $x\in V\cup F$ after the discharge procedure. We prove that $\mu ^*(x)\ge 0$ for all $x\in V\cup F$ and $\mu ^*(D)>0$ to get a contradiction.

Let $w(v \rightarrow f)$ be the charge transferred from a vertex v to an incident face f. We say that v is a flaw vertex if v is a $(3,5,3,5^+)$-vertex. The discharging rules are as follows.

(R1) Let f be a 3-face.

(R1.1) For a 4-vertex v not in $C_0,$

$$\begin{aligned} w(v \rightarrow f) = {\left\{ \begin{array}{ll} \frac{3}{5}, &{} \text {if } v \text { is flaw and } f \text { is a } (4,5^+,5^+)\text {-face},\\ \frac{4}{5}, &{} \text {if } v \text { is flaw and } f \text { is a } (4,4,5^+)\text {-face},\\ 1, &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

(R1.2) For a $5^+$-vertex v not in $C_0,$

$$\begin{aligned} w(v \rightarrow f) = {\left\{ \begin{array}{ll} \frac{7}{5}, &{} \text {if } f \text { is a } (4,4,5^+)\text {-face with two incident flaw vertices},\\ \frac{6}{5}, &{} \text {if } f \text { is a } (4,4^+,5^+)\text {-face with exactly one flaw vertex},\\ 1, &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

(R2) Let f be a 4-face.

For a $4^+$-vertex v not in $C_0,$$w(v \rightarrow f) = \frac{1}{2}$.

(R3) Let f be a 5-face.

(R3.1) For a 4-vertex v not in $C_0,$

$$\begin{aligned} w(v \rightarrow f) = {\left\{ \begin{array}{ll} 0, &{} \text {if } v \text { is a flaw vertex with four } 4\text {-neighbors},\\ \frac{1}{10}, &{} \text {if } v \text { is a flaw vertex with exactly one } 5^+\text {-neighbor},\\ \frac{1}{5}, &{} \text {if } v \text { is a flaw vertex with at least two } 5^+\text {-neighbors},\\ \frac{1}{5}, &{} \text {if } v \text { is a } (3,5,4,5)\text {-vertex},\\ \frac{1}{3}, &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

(R3.2) For a 5-vertex v not in $C_0,$

$$\begin{aligned} w(v \rightarrow f) = {\left\{ \begin{array}{ll} \frac{7}{10}, &{} \text {if } f \text { is a } (4,4,4,4,5)\text {-face with five adjacent } 4^-\text {-faces},\\ \frac{3}{5}, &{} \text {if } f \text { is a } (4,4,4,4,5)\text {-face with at least one adjacent } 5^+\text {-face},\\ \frac{2}{5}, &{} \text {if } f \text { is a } (4,4,4,5,5^+)\text {-face}\\ \frac{3}{10}, &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

(R3.3) For a $6^+$-vertex v not in $C_0,$

$$\begin{aligned} w(v \rightarrow f) = {\left\{ \begin{array}{ll} \frac{4}{5}, &{} \text {if } f \text { is a } (4,4,4,4,6^+)\text {-face},\\ \frac{2}{5}, &{}\text {if } f \text { is incident to a } 5^+\text {-vertex other than } v. \end{array}\right. } \end{aligned}$$

(R4) The outerface D gets $\mu (v)$ from each incident vertex v and gives 2 to each 4- or 5-face or 3-face sharing exactly one vertex with D, $\frac{12}{5}$ to each 3-face sharing one edge with D.

It suffices to check that each $x\in V(G)\cup F(G)$ has nonnegative final charge and D has positive final charge. By (R4), we have $\mu ^*(v)=0$ for each $v \in V(C_0).$ Thus we only consider a vertex v not on $C_0.$

Let v be a vertex with neighbors $v_1,v_2,\ldots , v_{d(v)}$ in a cyclic order. Let $f_1,f_2,\ldots ,f_{d(v)}$ be incident faces of v in a cyclic order with $v_i$ and $v_{i+1}$ incident to $f_i$ where $i+1$ is taken in modulo d(v). Thus v is a $(d(f_1),d(f_2),\ldots ,d(f_{d(v)})$-vertex.

Case 1v is a 4-vertex but v is not a flaw vertex.

It follows from Lemma 3.4 (b) that a 3-face is not adjacent to a 4-face, and from Corollary  3.5 that v has at most two incident 3-faces. Thus it suffices to consider that v is a $(4^+ ,4^+ ,4^+ ,4^+ )$-, (3, 5, 4, 5)-, $(3,5^+,4,6^+)$-, $(3,5^+,5^+,5^+)$-, $(3,3, 5^+,5^+)$-, or $(3,5^+,3,5^+)$-vertex.

• v is $(4^+ ,4^+ ,4^+ ,4^+ )$-vertex.

  Then v sends charge at most $\frac{1}{2}$ to each incident face by (R2) and (R3.1). Thus $\mu ^*(v)\ge \mu (v) - 4\times \frac{1}{2} = 0.$

• v is a (3, 5, 4, 5)-vertex.

  Then $w(v \rightarrow f_1) \le 1$ by (R1.1), $w(v \rightarrow f_2) = w(v \rightarrow f_4) = \frac{1}{5}$ by (R3.1), and $w(v \rightarrow f_3) = \frac{1}{2}$ by (R2). Thus $\mu ^*(v)\ge \mu (v) -2\times \frac{1}{5}- 1 -\frac{1}{2}> 0.$

• v is a $(3,5^+,4,6^+)$-vertex.

  Then $w(v \rightarrow f_1) \le 1$ by (R1.1), $w(v \rightarrow f_2) \le \frac{1}{3}$ by (R3.1), and $w(v \rightarrow f_3) = \frac{1}{2}$ by (R2). Thus $\mu ^*(v)\ge \mu (v) - 1 -\frac{1}{2}-\frac{1}{3}> 0.$

• v is a $(3,5^+,5^+,5^+)$-vertex.

  Then $w(v \rightarrow f_1) \le 1$ by (R1.1) and $w(v \rightarrow f_i) \le \frac{1}{3}$ for $2\le i \le 4$ by (R3.1). Thus $\mu ^*(v)\ge \mu (v) - 1 - 3\times \frac{1}{3} = 0.$

• v is a $(3,3,5^+,5^+)$-vertex.

  It follows from Lemma 3.4 (b) that $f_3$ and $f_4$ are $6^+$-faces. Thus $\mu ^*(v) \ge \mu (v) - 2\times 1= 0$ by (R1.1).

• v is a $(3,5^+,3,5^+)$-vertex.

  Since v is not a flaw vertex, $f_2$ and $f_4$ are $6^+$-faces. Thus $\mu ^*(v) \ge \mu (v) - 2\times 1= 0$ by (R1.1).

Case 2v is a flaw vertex, that is v is a $(3,5,3,5^+)$-vertex.

• Each adjacent vertex of v is a 4-vertex.

  Then $\max \{w(v \rightarrow f_1), w(v \rightarrow f_3)\} \le 1$ by (R1.1) and $w(v \rightarrow f_2)=w(v \rightarrow f_4)=0$ by (R3.1). Thus $\mu ^*(v) \ge \mu (v) - 2 \times 1= 0$.

• v is adjacent to exactly one $5^+$-vertex.

  Then $\max \{w(v \rightarrow f_1), w(v \rightarrow f_3)\} \le 1,$$\min \{w(v \rightarrow f_1), w(v \rightarrow f_3)\} \le \frac{4}{5}$ by (R1.1), and $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} = \frac{1}{10}$ by (R3.1). Thus $\mu ^*(v) \ge \mu (v) - 1-\frac{4}{5}-2\times \frac{1}{10} = 0$.

• v is adjacent to at least two $5^+$-vertices and incident to a $(4,5^+,5^+)$-face.

  Then $\max \{w(v \rightarrow f_1), w(v \rightarrow f_3)\} \le 1,$$\min \{w(v \rightarrow f_1), w(v \rightarrow f_3)\} = \frac{3}{5}$ by (R1.1), and $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} = \frac{1}{5}$ by (R3.1). Thus $\mu ^*(v) \ge \mu (v) - 1-\frac{3}{5}-2\times \frac{1}{5}= 0.$

• v is adjacent to at least two $5^+$-vertices but neither $f_1$ nor $f_3$ is a $(4,5^+,5^+)$-face.

  Then $f_1$ and $f_3$ are $(4,4,5^+)$-faces. It follows that $w(v \rightarrow f_1)= w(v \rightarrow f_3)\} = \frac{4}{5}$ by (R1.1) and $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} \le \frac{1}{5}$ by (R3.1). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{4}{5}-2\times \frac{1}{5} = 0.$

Case 3v is a 5-vertex but is not a (3, 5, 3, 5, 5)-vertex. It follows from Corollary 3.5 that v is incident to at most three 3-faces. Since a 3-face is not adjacent to a 4-face [Lemma 3.4 (b)], we may assume v is a $(3^+,4^+,4^+, 4^+, 4^+)$-, $(3,3,4^+, 3^+, 4^+)$-, $(3,6^+,3, 6^+, 5^+)$-, $(3,5,3, 6^+, 6^+)$-, $(3,6^+,3, 5, 5)$-, or $(3,5,3, 5,6^+)$-vertex.

• v is a $(3^+,4^+,4^+, 4^+, 4^+)$-vertex.

  It follows that $w(v \rightarrow f_1) \le \frac{7}{5}$ by (R1.2), (R2), (R3.2).

  • Suppose v is a $(3^+,5^+,5^+, 5^+, 5^+)$-vertex.

    Then each incident 5-face is adjacent to a $5^+$-face. Consequently, $\max \{w(v \rightarrow f_2),\ldots , w(v \rightarrow f_5)\} \le \frac{3}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-4\times \frac{3}{5} > 0.$

  • Suppose there exists $i \in \{2,\ldots ,5\}$ such that $d(f_i)=4.$

    Then $w(v \rightarrow f_i) \le \frac{1}{2}$ by (R2) and $\max \{w(v \rightarrow f_2),\ldots , w(v \rightarrow f_5)\} \le \frac{7}{10}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-\frac{1}{2}-3\times \frac{7}{10} = 0.$

• v is a $(3,3,4^+, 3^+, 4^+)$-vertex.

  It follows from Lemma 3.4 (f) that $f_3$ and $f_5$ are $6^+$-faces. Since both $f_1$ and $f_2$ are 3-faces, each of common incident vertices is not a flaw vertex. Then $\max \{w(v \rightarrow f_1),w(v \rightarrow f_3)\} \le \frac{6}{5}$ and $w(v \rightarrow f_4) \le \frac{7}{5}$ by (R1.2). Thus $\mu ^*(v) \ge \mu (v) -2\times \frac{6}{5}- \frac{7}{5} > 0.$

• v is a $(3,6^+,3, 6^+, 5^+)$- or a $(3,5,3, 6^+, 6^+)$-vertex.

  If v is a $(3,6^+,3, 6^+, 5^+)$-vertex, then $\max \{w(v \rightarrow f_1),w(v \rightarrow f_3)\} \le \frac{7}{5}$ by (R1.2) and $w(v \rightarrow f_5) \le \frac{7}{10}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{7}{5}-\frac{7}{10} > 0.$ The proof is similar for v is a $(3,5,3, 6^+, 6^+)$-vertex.

• v is a $(3,6^+,3, 5, 5)$-vertex.

  Then $\max \{w(v \rightarrow f_1),w(v \rightarrow f_3)\} \le \frac{7}{5}$ by (R1.2) and $\max \{w(v \rightarrow f_4),w(v \rightarrow f_5)\} \le \frac{3}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{7}{5}-2\times \frac{3}{5} = 0.$

• v is a $(3,5,3, 5,6^+)$-vertex.

  If $f_3$ is a $(4^+,5,5^+)$-face, then $f_2$ or $f_4$ is incident to at least two $5^+$-vertex. If $f_3$ is a (4, 4, 5)-vertex, then applying Lemma 3.7 to $f_3$ and $f_4$ yields that $f_4$ is incident to at least two $5^+$-vertex. Consequently, $\max \{w(v \rightarrow f_1),w(v \rightarrow f_3)\} \le \frac{7}{5}$ by (R1.2) and $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} \le \frac{7}{10}$ and $\min \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} \le \frac{2}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{7}{5}-\frac{7}{10}-\frac{2}{5} > 0.$

Case 4v is a (3, 5, 3, 5, 5)-vertex.

• $f_1$ and $f_3$ are (4, 4, 5)-faces.

  Applying Corollary 3.9 to $f_1,f_2,f_3,$ we have that $f_2$ is incident to at least two non-adjacent $5^+$-vertices (including v). Then $\max \{w(v \rightarrow f_1),w(v \rightarrow f_3)\} \le \frac{7}{5}$ by (R1.2) and $w(v \rightarrow f_2) \le \frac{3}{10}$ by (R3.2).

  • Suppose $f_4$ is incident to exactly one $5^+$-vertex.

    Applying Corollary 3.9 to $f_3,f_4,f_5,$ we have that $f_5$ is incident to at least two non-adjacent $5^+$-vertices. Then $w(v \rightarrow f_4) \le \frac{3}{5}$ and $w(v \rightarrow f_5) \le \frac{3}{10}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{7}{5}-2\times \frac{3}{10}-\frac{3}{5} = 0.$

  • Suppose $f_4$ and $f_5$ are incident to at least two $5^+$-vertices. Then $\max \{w(v \rightarrow f_4),w(v \rightarrow f_5)\} \le \frac{2}{5}$ by (R3.2) Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{7}{5}-\frac{3}{10}-2\times \frac{2}{5} > 0.$

• $f_1$ or $f_3$ is a $(5,5^+,5^+)$-face.

  By symmetry, we assume $f_1$ is a $(5,5^+,5^+)$-face. It follows that $f_2$ and $f_5$ are incident to at least two $5^+$-vertices. Then $w(v \rightarrow f_1) \le 1, w(v \rightarrow f_3) \le \frac{7}{5}$ by (R1.2) and $w(v \rightarrow f_4) \le \frac{3}{5}, \max \{w(v \rightarrow f_2),w(v \rightarrow f_5)\} \le \frac{2}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-1-\frac{3}{5}-2\times \frac{2}{5} > 0.$

• $f_1$ and $f_3$ are $(4,5,5^+)$-faces with incident $5^+$-vertex x and y,  respectively, where $x \ne v \ne y.$

  Then $\max \{w(v \rightarrow f_1) , w(v \rightarrow f_3)\} \le \frac{6}{5}$ by (R1.2).

  • Suppose x and y are not incident to $f_2.$

    It follows that $f_4$ and $f_5$ are incident to at least two $5^+$-vertices. Consequently, $w(v \rightarrow f_2) \le \frac{7}{10}, \max \{w(v \rightarrow f_4),w(v \rightarrow f_5)\} \le \frac{2}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{6}{5}-\frac{7}{10}-2\times \frac{2}{5} > 0.$

  • Suppose x and y are incident to $f_2$.

    Then $w(v \rightarrow f_2) = \frac{3}{10}, \max \{w(v \rightarrow f_4),w(v \rightarrow f_5)\} \le \frac{3}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{6}{5}-\frac{3}{10}-2\times \frac{3}{5} > 0.$

  • Suppose x is incident to $f_2$ but y is not.

    Then $f_4$ is incident to at least two $5^+$-vertices. Consequently, $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} \le \frac{2}{5}, w(v \rightarrow f_3) \le \frac{3}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{6}{5}-2\times \frac{2}{5} -\frac{3}{5}> 0.$

• $f_1$ is a (4, 4, 5)-face but $f_3$ is a $(4,5,5^+)$-face with its two incident $5^+$-vertices are also incident to $f_2.$

  It follows that $w(v \rightarrow f_1) \le \frac{7}{5}, w(v \rightarrow f_3) \le \frac{6}{5}$ by (R1.2) and $w(v \rightarrow f_2) \le \frac{2}{5}$ by (R1.2). Applying Corollary 3.9 to $f_1,f_5,f_4,$ we have that $f_4$ or $f_5$ is incident to at least two $5^+$-vertices.

  • Suppose $f_4$ is incident to at least two non-adjacent $5^+$-vertices.

    Then $w(v \rightarrow f_4) \le \frac{3}{10}, w(v \rightarrow f_5) \le \frac{3}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-\frac{6}{5}-\frac{2}{5}-\frac{3}{10}-\frac{3}{5} > 0.$ The proof is similar for $f_5$ is incident to at least two non-adjacent $5^+$-vertices.

  • Suppose a $5^+$-vertex u incident to $f_4$ or $f_5$ is adjacent to v.

    From assumption on $f_1$ and $f_3,$ we have that u is not incident to $f_1$ and is not incident to $f_3.$ It follows that u is incident to $f_4$ and $f_5.$ Then $\max \{w(v \rightarrow f_4,w(v \rightarrow f_5)\} \le \frac{2}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-\frac{6}{5}-3\times \frac{2}{5} > 0.$

• $f_1$ is a (4, 4, 5)-face but $f_3$ is a $(4,5,5^+)$-face with its two $5^+$-vertices are also incident to $f_4.$

  It follows that $w(v \rightarrow f_1) \le \frac{7}{5}, w(v \rightarrow f_3) \le \frac{6}{5}$ by (R1.2). Applying Corollary 3.9 to $f_5,f_1,f_2,$ we have that $f_2$ or $f_5$ is incident to at least two $5^+$-vertices.

  • Suppose $f_2$ is incident to a $5^+$-vertex u where $u \ne v.$

    We have that u is not incident to $f_1$ and $f_3$ by assumption on $f_1$ and $f_3.$ This implies u is not adjacent to v. Then $w(v \rightarrow f_2) \le \frac{3}{10},$$w(v \rightarrow f_4) \le \frac{2}{5},$ and $w(v \rightarrow f_5) \le \frac{3}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-\frac{6}{5}-\frac{3}{10}-\frac{2}{5}-\frac{3}{5} >0.$

  • Suppose $f_5$ is incident to a $5^+$-vertex u where $u \ne v.$

    If u is not incident to $f_4,$ then u is not adjacent to v. It follows that $w(v \rightarrow f_2) \le \frac{7}{10},$$w(v \rightarrow f_4) \le \frac{2}{5},$ and $w(v \rightarrow f_5) \le \frac{3}{10}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-\frac{6}{5}-\frac{7}{10}-\frac{2}{5}-\frac{3}{10} =0.$

    If u is also incident to $f_4,$ then $f_4$ is a $(4^+,4^+,5,5^+,5^+)$-vertex. It follows that $w(v \rightarrow f_2) \le \frac{7}{10},$$w(v \rightarrow f_4) \le \frac{3}{10}$ and $w(v \rightarrow f_5) \le \frac{2}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - \frac{7}{5}-\frac{6}{5}-\frac{7}{10}-\frac{3}{10}-\frac{2}{5} =0.$

Case 5v is a 6-vertex.

From Lemma 3.4 (a) that a 3-face is not adjacent to a 4-face, it suffices to consider v is a $(3,3,k_3,k_4,k_5,k_6)$-, $(3^+,4^+,3^+,4^+,4^+,4^+)$-, $(3^+,4^+,4^+,3^+,4^+,4^+)$-, (3, 5, 3, 5, 3, 5)-, or a $(3, 5^+,3,5^+,3,6^+)$-vertex.

• v is a $(3,3,k_3,k_4,k_5,k_6)$-vertex.

  It follows from Lemma 3.4 (b) that $f_3$ and $f_6$ are $6^+$-faces. Thus $\mu ^*(v) \ge \mu (v) - 4\times \frac{7}{5}>0$ by (R1.2), (R2), and (R3.3).

• v is a $(3^+,4^+,3^+,4^+,4^+,4^+)$-vertex (or $(3^+,4^+,4^+,3^+,4^+,4^+)$-vertex, respectively).

  Then v sends charge at most $\frac{7}{5}$ to $f_1$ and $f_3$ (or $f_4,$ respectively) by (R1.2), (R2), (R3.3), and v sends charge at most $\frac{4}{5}$ to each of the remaining incident faces by (R2), (R3.3). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{7}{5}-4\times \frac{4}{5}= 0.$

• v is a (3, 5, 3, 5, 3, 5)-vertex.

  • Suppose at least two incident 5-faces of v,  say $f_2$ and $f_4,$ are incident to at least two $5^+$-vertices.

    Then $\max \{w(v \rightarrow f_1),w(v \rightarrow f_3),w(v \rightarrow f_5)\} \le \frac{7}{5}$ by (R1.2), $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} = \frac{2}{5}$ and $w(v \rightarrow f_6) \le \frac{4}{5}$ by (R3.3). Thus $\mu ^*(v) \ge \mu (v) - 3\times \frac{7}{5}-2\times \frac{2}{5}-\frac{4}{5}>0.$

  • Suppose two incident 5-faces of v,  say $f_2$ and $f_4,$ are incident to exactly one $5^+$-vertex.

    Applying Lemma 3.10 to $f_1, f_2, f_3, f_4,$ we obtain that $f_1$ is incident to two $5^+$ vertices, say $v_1$ and v,  that are also incident to $f_6.$ Applying Lemma 3.10 to $f_2, f_3, f_4,f_5,$ we obtain that $f_5$ is incident to two $5^+$ vertices, say $v_5$ and v,  that are also incident to $f_6.$ Then $\max \{w(v \rightarrow f_1),w(v \rightarrow f_5)\} \le \frac{6}{5}$ and $w(v \rightarrow f_3) \le \frac{7}{5}$ by (R1.2). Moreover, $\max \{w(v \rightarrow f_2),w(v \rightarrow f_4)\} \le \frac{4}{5}$ and $w(v \rightarrow f_6) \le \frac{2}{5}$ by (R3.2). Thus $\mu ^*(v) \ge \mu (v) - 2\times \frac{6}{5}-\frac{7}{5}-2\times \frac{4}{5}-\frac{2}{5}>0.$

Case 6v is a d-vertex with $d\ge 7.$

• v is a $(3,3,k_3,\ldots , k_d)$-vertex.

  It follows from Lemma 3.4 that $f_3$ and $f_d$ are two $6^+$-faces. Thus $\mu ^*(v) \ge \mu (v) - (d-2)\times \frac{7}{5}=2d-6- (d-2)\times \frac{7}{5}>0$ by (R1.2), (R2), and (R3.3).

• v has no adjacent incident 3-faces.

  It follows that v is incident to at most $\frac{d}{2}$ 3-faces. Since v sends charge at most $\frac{7}{5}$ to each of its incident 3-faces by (R1.2) and v sends charge at most $\frac{4}{5}$ to each of the remaining incident faces by (R2) and (R3.3), we have $\mu ^*(v) \ge \mu (v)-\frac{d}{2}\times \frac{7}{5}-\frac{d}{2}\times \frac{4}{5} =(2d-6)-d\times \frac{11}{10}>0.$

Let f be a face in G. Let $V(f)\cap V(D)\ne \emptyset$. If $d(f)=3$, then f gets $\frac{12}{5}$ from D when f shares an edge with D, 2 from D when f shares exactly one vertex with D. Note that each vertex of f in $int(C_0)$ sends at least $\frac{1}{2}$ to f. It follows that $\mu ^*(f)\ge -3+\min \{\frac{12}{5}+\frac{1}{2}, 2+\frac{1}{2}\times 2\}=0$. If $d(f)\in \{4,5\}$, then it gains 2 from D. Thus $\mu ^*(f)\ge d(f)-6+2\ge 0$. If $d(f)=6$, then $\mu ^*(f)=\mu (f)=0$. If $d(f)\ge 7$, then $\mu ^*(f)\ge (k-6)-k\times \frac{k-6}{k}=0$. Thus we may assume that $V(f)\cap V(D)=\emptyset$ for the remaining of the paper. Let f be a $5^-$-face with vertices $v_1,v_2,\ldots , v_{d(v)}$ in a cyclic order.

Case 7f is a 3-face.

• f is a (4, 4, 4)-face or each vertex of f is not a flaw vertex.

  Then $\mu ^*(f) = \mu (f) +3\times 1=0$ by (R1.1).

• f is a $(4,4,5^+)$-face with exactly one incident flaw vertex, say $v_1.$

  Then $w(v_1 \rightarrow f)=\frac{4}{5},$$w(v_2 \rightarrow f)= 1$ by (R1.1) and $w(v_3 \rightarrow f)= \frac{6}{5}$ (R1.2). Thus $\mu ^*(f) =\mu (f) +\frac{4}{5}+1+\frac{6}{5}=0.$

• f is a $(4,4,5^+)$-face with $v_1$ and $v_2$ are flaw vertices.

  Then $w(v_1 \rightarrow f)=w(v_2 \rightarrow f)= \frac{4}{5}$ by (R1.1) and $w(v_3 \rightarrow f)= \frac{7}{5}$ (R1.2). Thus $\mu ^*(f) = \mu (f) +2\times \frac{4}{5}+\frac{7}{5}=0.$

• f is a $(4,5^+,5^+)$-face and $v_1$ is a flaw vertex.

  Then $w(v_1 \rightarrow f)=\frac{3}{5}$ by (R1.1) and $w(v_2 \rightarrow f)=w(v_3 \rightarrow f)= \frac{6}{5}$ by (R1.2) Thus $\mu ^*(f) = \mu (f) +\frac{3}{5}+2\times \frac{6}{5}=0$.

Case 8f is a 4-face.

We obtain $\mu ^*(f) \ge \mu (f) +4\times \frac{1}{2}= 0$ by (R2).

Case 9f is a 5-face.

• f is incident to at least three $5^+$-vertices.

  It follows that each of its incident 4-vertex is adjacent to at least one $5^+$-vertex. Then each of these 4-vertices sends charge at least $\frac{1}{10}$ to f by (R3.1) and each $5^+$-vertex sends charge at least $\frac{3}{10}$ to f by (R3.2) and (R3.3). Thus $\mu ^*(f)\ge \mu (f)+2\times \frac{1}{10}+ 3\times \frac{3}{10}> 0.$

• f is a $(4,5^+,4,5^+,4)$-face.

  Since $v_1$ and $v_5$ are adjacent to at least one $5^+$-vertex and $v_3$ is adjacent to at least two $5^+$-vertices, we have $\min \{w(v_1 \rightarrow f),w(v_5 \rightarrow f)\} \ge \frac{1}{10}$ and $w(v_3\rightarrow f)\ge \frac{1}{5}$ by (R3.1). We have $\min \{w(v_2 \rightarrow f),w(v_4 \rightarrow f)\} \ge \frac{1}{10}$ by (R3.2) and (R3.3). Thus $\mu ^*(f)\ge \mu (f)+ 2\times \frac{1}{10}+\frac{1}{5}+2\times \frac{3}{10}= 0.$

• f is a $(4,4,4,5^+,5^+)$-face

  Since $v_1$ and $v_3$ are adjacent to at least one $5^+$-vertex, we have $\min \{w(v_1 \rightarrow f),w(v_3 \rightarrow f)\} \ge \frac{1}{10}$ by (R3.1). We have $\min \{w(v_2 \rightarrow f),w(v_4 \rightarrow f)\} \ge \frac{2}{5}$ by (R3.2) and (R3.3). Thus $\mu ^*(f)\ge \mu (f)+2\times \frac{1}{10}+ 2\times \frac{2}{5}= 0.$

• f is a $(4,4,4,4,6^+)$-face.

  Since $v_1$ and $v_4$ are adjacent to at least one $5^+$-vertex, we have $\min \{w(v_1 \rightarrow f),w(v_4 \rightarrow f)\} \ge \frac{1}{10}$ by (R3.1). We have $w(v_5 \rightarrow f) = \frac{4}{5}$ by (R3.3). Thus $\mu ^*(f)\ge \mu (f)+2\times \frac{1}{10}+ \frac{4}{5}= 0.$

• f is a (4, 4, 4, 4, 5)-face with at least one adjacent $5^+$-face $f_i.$

  Let a 4-vertex $v_i$ be in $f_i.$ It follows that a 4-vertex $v_i$ is not a flaw vertex. Then $w(v_i \rightarrow f) = \frac{1}{3}$ by (R3.1), $\min \{w(v_1 \rightarrow f),w(v_4 \rightarrow f)\} \ge \frac{1}{10}$ by (R3.1), and $w(v_5\rightarrow f) = \frac{3}{5}$ by (R3.2), Thus $\mu ^*(f)\ge \mu (f)+\frac{1}{3}+\frac{1}{10}+ \frac{3}{5}> 0.$

• f is a (4, 4, 4, 4, 5)-face with five adjacent $4^-$-faces and a 4-vertex $v_i$ which is not a flaw vertex.

  Then $w(v_i \rightarrow f) = \frac{1}{3}$ by (R3.1) and $w(v_5\rightarrow f) = \frac{7}{10}$ by (R3.2). Thus $\mu ^*(f)\ge \mu (f)+ \frac{1}{3}+\frac{7}{10}> 0.$

• f is a (4, 4, 4, 4, 5)-face with five adjacent $4^-$-faces and four flaw vertices.

  It follows that each adjacent face of f is a 3-face. Let $f_4$ be a 3-face incident to $v_4$ and $v_5$ and let $f_5$ be a 3-face incident to $v_1$ and $v_5.$ Applying Corollary 3.9 to $f, f_4,$ and $f_5,$ we have that $f_4$ or $f_5$ is a $(4,5,5^+)$-face. By symmetry, let $f_5$ be a $(4,5,5^+)$-face. Consequently, $v_1$ is adjacent to at least two $5^+$-vertices and $v_4$ is adjacent to at least one $5^+$-vertex. It follows that $w(v_1\rightarrow f) =\frac{1}{5}$ and $w(v_4\rightarrow f)\ge \frac{1}{10}$ by (R3.1). We have $w(v_5 \rightarrow f) = \frac{7}{10}$ by (R3.2). Thus $\mu ^*(f)\ge \mu (f)+ \frac{1}{5}+\frac{1}{10}+\frac{7}{10}= 0.$

• f is a (4, 4, 4, 4, 4)-face.

  Applying Corollary 3.7 to f and its adjacent 3-face, we have that each adjacent 3-face of f is a $(4,4,5^+)$-face. This implies that each incident flaw vertex of f is adjacent to at least two $5^+$-vertex. If $v_i$ is a flaw vertex, then $w(v_i \rightarrow f) \ge \frac{1}{5},$ otherwise $w(v_i \rightarrow f) \ge \frac{1}{3}$ by (R3.1). Thus $\mu ^*(f)\ge \mu (f)+ 5\times \frac{1}{5}= 0.$

Case 10 Consider the outerface D.

Let $f_3', f'$ be the number of 3-faces sharing exactly one edge with D, 3-faces sharing exactly one vertex with D or 4-or 5-faces sharing vertices with D, respectively. Let $E(C_0, V(G)-C_0)$ be the set of edges between $C_0$ and $V(G)-C_0$ and let $e(C_0, V(G)-C_0)$ be its size. Then by (R4),

$$\begin{aligned} \mu ^*(D)&=3+6+\sum _{v\in C_0} (2d(v)-6)-\frac{12}{5}f_3'-2f' \end{aligned}$$

(1)

$$\begin{aligned}&=9+2\sum _{v\in C_0} (d(v)-2)-2\times 3-\frac{12}{5}f_3'-2f' \end{aligned}$$

(2)

$$\begin{aligned}&=3-\frac{2}{5}f_3'+2(e(C_0,V(G)-C_0)-f_3'-f') \end{aligned}$$

(3)

So we may consider that each edge $e\in E(C_0,V(G)-C_0)$ gives a charge of 2 to D. Since each $5^-$-face is a cycle, it contains two edges in $E(C_0,V(G)-C_0)$. It follows that $e(C_0,V(G)-C_0)-f_3'-f'\ge 0$. Note that $f_3'\le 3$. Thus $\mu ^*(D)>0.$

This completes the proof.