An edge coloring of a graph $G$ is to color all the edges in the graph such that adjacent edges receive different colors.
It is acyclic if each cycle in the graph receives at least three colors.
Fiamčik (1978) and Alon, Sudakov and Zaks (2001) conjectured that
every simple graph with maximum degree $\Delta$ is acyclically edge $(\Delta + 2)$-colorable —
the well-known acyclic edge coloring conjecture (AECC).
Despite many major breakthroughs and minor improvements, the conjecture remains open even for planar graphs.
In this paper, we prove that planar graphs are acyclically edge $(\Delta + 5)$-colorable.
Our proof has two main steps:
Using discharging methods, we first show that every non-trivial planar graph must have one of the eight groups of well characterized local structures;
and then acyclically edge color the graph using no more than $\Delta + 5$ colors by an induction on the number of edges.

Acyclic edge coloring; planar graph; local structure; discharging; induction

§.§.§ Acknowledgments.

This research is supported by the NSERC Canada and the China Scholarship Council.

§ INTRODUCTION

We consider simple and $2$-connected graphs in this paper,
i.e., graphs without self-loops, without parallel edges but with at least two vertex-disjoint paths between any pair of vertices,
as explained below.

We use $V(G)$ and $E(G)$ to denote the vertex set and the edge set of a graph $G$, respectively,
and simplify them as $V$ and $E$ respectively when the graph is clear from the context.
This way, $G$ is also denoted as $G = (V, E)$.

For an integer $k \ge 2$, we use the integers $1, 2, \ldots, k$ to represent the $k$ distinct colors.
An edge $k$-coloring of the graph $G$ is a mapping $c: E(G) \to \{1, 2, \ldots, k\}$ such that any two adjacent edges receive different colors.
If such a mapping exists, then $G$ is said edge $k$-colorable.
The chromatic index of $G$, denoted as $\chi'(G)$, is the smallest integer $k$ such that $G$ is edge $k$-colorable.
An edge $k$-coloring of $G$ is called acyclic if each cycle receives at least three different colors,
or equivalently there is no bichromatic cycle in $G$, or equivalently the subgraph of $G$ induced by any two colors is a forest.
Correspondingly, $a'(G)$ denotes the smallest integer $k$ such that $G$ is acyclically edge $k$-colorable, called the acyclic chromatic index of $G$.

One sees from the definition of acyclic edge coloring the reason why we consider simple and connected graphs only.
Furthermore, a connected graph can be represented as a block-cut tree,
in which each block is a $2$-connected component of the graph and two blocks are adjacent if and only if they overlap at a cut-vertex.
Given an acyclic edge $k$-coloring of the graph $G$, after swapping any two colors inside a block it remains as an acyclic edge $k$-coloring.
Therefore, we may constrain ourselves on $2$-connected graphs.
In the sequel, the input graph is simple and $2$-connected.

We use $\Delta(G)$ to denote the maximum degree of the graph $G$, and simply as $\Delta$ when the graph is clear from the context.
It holds trivially that $\Delta \le \chi'(G) \le a'(G)$.
By Vizing's theorem [17], $\chi'(G) \le \Delta + 1$;
it follows from $\chi'(K_3) = 3 = \Delta + 1$ that $\Delta + 1$ is tight for the chromatic index $\chi'(G)$.
For the acyclic chromatic index, one sees that $a'(K_4) = 5 = \Delta + 2$.
Fiamčik [7] and Alon, Sudakov and Zaks [2] independently conjectured that it is upper bounded by $\Delta + 2$,
called the acyclic edge coloring conjecture (AECC):

For any graph $G$, $a'(G) \le \Delta + 2$.

The AECC has been confirmed true for graphs with $\Delta \in \{3, 4\}$ [16, 3, 4, 15, 19].
However, for an arbitrary graph $G$, the conjecture is far away from reached.
Yet, several major milestones bounding $a'(G)$ have been achieved:
Alon, McDiarmid and Reed [1] proved that $a'(G) \le 64 \Delta$ by a probabilistic argument;
and then the upper bound was improved to $16 \Delta$ [9], to $\lceil 9.62 (\Delta - 1) \rceil$ [10], to $4 \Delta - 4$ [6],
and most recently in 2017 to $\lceil 3.74 (\Delta - 1) \rceil + 1$ by Giotis et al. [8] using Lovász local lemma.

A planar graph is one that can be drawn in the two-dimensional plane so that its edges intersect only at their ending vertices.
The AECC on planar graphs also attracts many studies with much tighter upper bounds.
For an arbitrary planar graph $G$, Basavaraju et al. [5] first showed that $a'(G) \le \Delta + 12$;
and then the upper bound was improved to $\Delta + 7$ by Wang, Shu and Wang [21] and the latest to $\Delta + 6$ by Wang and Zhang [18].
For some sub-classes of planar graphs, the AECC has been confirmed true,
for example, planar graphs without $i$-cycles for each $i \in \{3, 4, 5, 6\}$ [14, 20, 13, 22],
and planar graphs without intersecting triangles (i.e., $3$-cycles) [11, 12].

In this paper, we focus on arbitrary planar graphs, aiming to improve the upper bound from $\Delta + 6$ to $\Delta + 5$, stated in the following theorem.

If $G$ is a planar graph, then $a'(G)\leq \Delta + 5$.

Recall that when $\Delta \le 4$, the AECC has been confirmed true.
We thus prove Theorem <ref> assuming $\Delta \ge 5$.
The rest of the paper is organized as follows.
In Section 2, we characterize eight groups of local structures (also called configurations),
and by discharging methods we prove that $G$ must contain at least one of these local structures.
We prove the theorem in Section 3 by an induction on the number of edges in $G$,
where we employ an important known property of an acyclic edge coloring (Lemma <ref>) from [15].
We conclude the paper in Section 4.

§ NOTATIONS AND THE EIGHT GROUPS OF LOCAL STRUCTURES

We fix a simple and $2$-connected planar graph $G$ with $\Delta\ge 5$ for discussion.

Let $d(v)$ denote the degree of the vertex $v$ in $G$.
A vertex of degree $k$ (at least $k$, at most $k$, respectively) is called a $k$-vertex ($k^+$-vertex, $k^-$-vertex, respectively).
For convenience, let $n_k(v)$ ($n_{k^{+}}(v)$, $n_{k^{-}}(v)$, respectively) denote the number of $k$-vertices ($k^{+}$-vertices, $k^{-}$-vertices, respectively) adjacent to $v$ in $G$.

We use $[a, b]$ to denote the set of integers in between $a$ and $b$, inclusive.
Let $S_i$ be a set of integers, $i = 1, 2$;
then an $(S_1, S_2)$ vertex is a $k$-vertex $x$ of $G$ such that $k \in S_1$ and $n_{6^{+}}(x) \in S_2$ (and in this case, we say that $x$ is $(S_1, S_2)$).
When $S_1 = \{k\}$ ($S_2 = \{j\}$, $S_1 = \{k_1, k_2\}$, $S_2 = \{j_1, j_2\}$, respectively), the above notation is simplified as $(k, S_2)$ ($(S_1, j)$, $(k_1/k_2, S_2)$, $(S_1, j_1/j_2)$, respectively).
Similarly, when $S_1 = [k, \infty]$ ($S_1 = [1, k]$, $S_2 = [j, \infty]$, $S_2 = [1, j]$, respectively),
we can define a $(k^+, S_2)$ ($(k^-, S_2)$, $(S_1, j^+)$, $(S_1, j^-)$, respectively) vertex.
For instance, $(9, 5)$ is a degree-$9$ vertex $x$ having exactly five $6^+$-neighbors,
and $(7^+, 4^-)$ is a $7^+$-vertex $x$ with no more than four $6^+$-neighbors.

Some local structures of the groups ($A_2$)–($A_8$) as defined in Theorem <ref>.
In each local structure, filled vertices are distinct from each other and each has its degree determined,
while a non-filled vertex has an undetermined degree (i.e., could be incident with more edges not shown).
In ($A_{6.1}$) and ($A_{6.3}$), a dashed edge indicates the existence of a possible edge incident at the vertex.

The graph $G$ contains at least one of the following local structures ($A_1$)–($A_8$), most of which are depicted in Figure <ref>.

A path $u v w$ with $d(v)= 2$ such that at least one of the following holds:

$u w\not\in E(G)$;

$u w \in E(G)$ and $d(u)\le 7$;

$u w \in E(G)$ and $n_{5^-}(u)\ge d(u) - 6$.

An edge $u v$ with $N(v) = \{u, v_1, v_2\}$ such that at least one of the following holds:

$d(u)\le 7$;

$u$ is $([8, 11], 5^-)$, and $u v_1, u v_2\in E(G)$, $v_1 v_2\not\in E(G)$;

$u$ is $(8, 4^-)$, $v_2$ is $([8, 11], 5^-)$, and $u v_1, v_1 v_2\in E(G)$, $u v_2\not\in E(G)$.

$d(u) = 8$, and $u v_1, u v_2, v_1 v_2\in E(G)$.

A $(9, 5^-)$ vertex $u$ adjacent to a $3$-vertex $v$ with $N(v) = \{u, v_1, v_2\}$ and $u v_1, u v_2, v_1 v_2\in E(G)$ such that at least one of the following holds:

$d(v_i)- n_2(v_i)\ne \Delta$ for each $i\in \{1, 2\}$;

$v_2$ is $([9, 13], 6^-)$, and $d(v_1)- n_2(v_1)\ne \Delta$ or $n_{6^+}(v_1)\le \Delta - d(v_2) + 6$;

$u$ is $(9, 4^-)$ and $v_2$ is $([9, 13], 5^-)$;

$u$ is $(9, 4^-)$, $v_1$ is $(\Delta, 7)$, and $v_2$ is $(\Delta, 6)$, where $\Delta \in [11, 13]$;

$u$ is $(9, 4^-)$, $v_1$ is $(14, 7)$, and $v_2$ is $(13, 7)$ or $(14, 7^-)$;

$u$ is $(9, 5)$ and $v_2$ is $([9, 13], 4^-)$;

$u$ is $(9, 5)$, $v_1$ is $(\Delta, 7)$, and $v_2$ is $(\Delta, 5)$, where $\Delta \in [10, 11]$;

$u$ is $(9, 5)$, $v_1$ is $(\Delta, 7)$, and $v_2$ is $(\Delta, 5/6)$, where $\Delta \in [12, 13]$.

A $(10, 5^-)$ vertex $u$ adjacent to a $3$-vertex $v$ with $N(v)= \{u, v_1, v_2\}$ and $uv_1, uv_2, v_1v_2\in E(G)$ such that at least one of the following holds:

$v_2$ is $(d(v_2), 4^-)$;

$v_2$ is $(9, 6)$ or $([10, 13], 5/6)$, and $n_{6^+}(v_1)\le \Delta - d(v_2) + 6$;

$v_2$ is $(10/11, 5)$ and $d(v_1)- n_2(v_1)\ne \Delta$;

$v_1$ is $(\Delta, 7)$, and $v_2$ is $(\Delta, 5)$, where $\Delta \in [10, 11]$.

A $(11, 5^-)$ vertex $u$ adjacent to a $3$-vertex $v$ with $N(v) = \{u, v_1, v_2\}$ and $u v_1, u v_2, v_1 v_2\in E(G)$ such that at least one of the following holds:

$v_1$ is $([9, 14], 6)$ or $([11, 14], 5^-)$, and $v_2$ is $([9, 12], 6)$ or $([11, 12], 5^-)$;

$v_1$ is $(13^+, 7)$ and $v_2$ is $(11, 5^-)$.

A $4$-vertex $u$ with $N(u) = \{v, u_1, u_2, u_3\}$ such that at least one of the following cases holds:

$d(v)\in \{4, 5\}$ and $\sum_{x\in N(u)}d(x)\le 2\Delta + 13$;

$d(v) = 6$, $uv$ is contained in a $3$-cycle, and $\sum_{x\in N(u)}d(x)\le 2\Delta + 13$;

$d(v)\in \{5, 6\}$, $uv$ is contained in two $3$-cycles, and $\sum_{x\in N(u)}d(x)\le 2\Delta + 14$;

$v$ is $(7, 4^-)$, $u_1 u_2$, $u_1 u_3$, $v u_2$, $v u_3\in E(G)$, and $\min \{d(u_1), d(u_2), d(u_3)\}\le 8$.

A $5$-vertex $u$ adjacent to a $5$-vertex $v$ with $N(u) = \{v, u_1, u_2, u_3, u_4\}$ and $uv$ contained in two $3$-cycles $u v u_3$,
$uvu_4$ such that at least one of the following holds:

$d(u_3)\in \{5, 6\}$ and $\sum_{x\in N(u)\setminus \{v\}}d(x)\le 2\Delta + 13$;

$d(u_3) = 7$, $\min \{d(u_1), d(u_2)\}\le 6$, and $\sum_{x\in N(u)\setminus \{v\}}d(x)\le \Delta + 20$;

$d(u_3) = 8$ and $d(u_1) = d(u_2) = 6$.

A $5$-vertex $u$ adjacent to a $6$-vertex $v$ with $N(u) = \{v, u_1, u_2, u_3, u_4\}$ and $u v$ contained in two $3$-cycles $u v u_3$,
$u v u_4$ such that at least one of the following holds:

$d(u_3)\le 6$, $d(u_4)\le 6$, and $\min\{d(u_1), d(u_2)\}\le 7$;

$d(u_3)\le 6$, $u_4$ is $(7, 4^-)$, and $\min\{d(u_1), d(u_2)\}\le 6$;

$d(u_3)\le 6$, $u_4$ is $(7, 4^-)$, and $d(u_i)\le 7$ for each $i\in \{1, 2\}$;

$u_i$ is $(7, 4^-)$ for each $i\in \{3, 4\}$ and $d(u_1)\le 6$, $d(u_2)\le 7$.

The rest of the section is devoted to the proof of Theorem <ref>, by contradiction built on discharging methods.
To this purpose, we assume to the contrary that $G$ contains none of the local structures ($A_1$)–($A_8$).

§.§ Definitions and notations

Since $G$ is $2$-connected, $d(v)\ge 2$ for every vertex $v \in V(G)$.
Let $G'$ be the graph obtained by deleting all the $2$-vertices of $G$, and let $H$ be a connected component of $G'$.
One sees that every vertex $v \in V(H)$ has its (called original) degree $d(v) \ge 3$ in $G$, and $H$ is clearly also planar.
In what follows, we assume that $H$ is embedded in the two dimensional plane such that its edges intersect only at their ending vertices,
and we refer $H$ to as a plane graph.

For objects in the plane graph $H$, we use the following notations:

* $N_H(v) = \{u \mid u v\in E(H)\}$ and $d_H(v) = |N_H(v)|$ — the degree of the vertex $v\in V(H)$;

* similarly define what a $k$-vertex, a $k^+$-vertex, and a $k^-$-vertex are;

* $n'_k(v)$ ($n'_{k^+}(v)$, $n'_{k^-}(v)$, respectively) — the number of $k$- ($k^+$-, $k^{-}$-, respectively) neighbors of $v$;

* an $(S_1, S_2)_H$ vertex — a vertex $x$ with $d_H(x) \in S_1$ and $n'_{6^{+}}(x)\in S_2$;

* an $(S_1, S_2)_{HG}$ vertex — a vertex $x$ with $d_H(x) = d(x) \in S_1$ and $n'_{6^{+}}(x) = n_{6^{+}}(x) \in S_2$.

Note that we do not discuss the embedding of the graph $G$.
Every face in the proof of Theorem <ref> is in the embedding of the graph $H$, or is simplified as in the embedding of the graph $H$.
We reserve the character $f$ to denote a face in the embedding.

* $F(H)$ — the face set of $H$;

* $V_k(H)$ — the set of $k$-vertices in $V(H)$;

* $V(f)$ — the set of vertices on (the boundary of) the face $f$;

* a vertex $v$ is incident with a face $f$ if $v\in V(f)$;

* an edge $u v$ is incident with a face $f$ if $u, v\in V(f)$;

* $n_k(f)$ ($n_{k^+}(f)$, $n_{k^-}(f)$, respectively) — the number of $k$- ($k^+$-, $k^{-}$-, respectively) vertices in $V(f)$;

* $\delta(f)$ — the minimum degree of the vertices in $V(f)$;

* $d(f)$ — the degree of the face $f$, which is the number of edges on the face $f$ with cut-edges counted twice,
and similarly define what a $k$-face, a $k^+$-face and a $k^-$-face are;

* $F(v) = \{f \in F(H) \mid v\in V(f)\}$ — the set of faces incident with $v$;

* $F_3(v) = \{f \in F(H) \mid v\in V(f), d(f)= 3\}$ — the set of $3$-faces incident with $v$;

* $F(u v) = \{f \in F(H) \mid uv\in E(G), u, v\in V(f)\}$ — the set of faces incident with the edge $uv$;
note that $|F(u v)| = 2$ for any edge $uv$;

* $f_{u v} = F(u v)\setminus \{f\}$ — the face incident with the edge $uv$ and adjacent to the face $f\in F(u v)$;

* $m_k(v)$ ($m_{k^+}(v)$, $m_{k^-}(v)$, respectively) — the number of $k$- ($k^+$-, $k^-$-, respectively) faces in $F(v)$;

* $m_3(u v)$— the number of $3$-faces in $F(u v)$.

§.§ Structural properties

Note from the construction of $H$ that $d_H(u) = d(u)- n_2(u)$ and thus $d_H(u) = d(u)$ if $n_2(u) = 0$.
In this subsection, we characterize structural properties for $5^-$-vertices in $H$, which are used in the discharging.

For any $u \in V(H)$,

if $3 \le d(u) \le 7$, then $d_H(u) = d(u)$;

if $d(u) \ge 8$, then $d_H(u) \ge 7$;

$\delta(H)\ge 3$, $n'_2(u) = 0$, $n'_{k}(u) = n_{k}(u)$, $k\in [3, 6]$, and $n'_{5^-}(u) = n_{5^-}(u) - n_2(u)$, $n'_{6^+}(u) = n_{6^+}(u)$;

if $d(u) \ge 8$, then $n'_{5^-}(u)\le d_H(u) - 7$ and $n'_{6^+}(u)\ge 7$.

The non-existence of ($A_{1.1}$) in $G$ states that $u w\in E(G)$ for each $2$-vertex $v$ with $N(v) = \{u, w\}$.

The non-existence of ($A_{1.2}$) in $G$ states that no $7^-$-vertex is adjacent to any $2$-vertex in $G$;
thus $3 \le d(u) \le 7$ implies $d_H(u) = d(u)$.

The non-existences of $(A_{1.3})$ in $G$ implies that every $8^+$-vertex $u$ is adjacent to at most ($d(u) - 7$) $2$-vertices in $G$;
thus if $d(u) \ge 8$, then $d_H(u) = d(u)- n_2(u) \ge d(u) - (d(u) - 7) = 7$.

Note that all $2$-vertices are deleted from $G$, a $k$-neighbor of $u$ in $G$ remains as a $k$-neighbor of $u$ in $H$, $k\in [3, 6]$,
and a $7^+$-neighbor of $u$ in $G$ remains as a $7^+$-neighbor of $u$ in $H$,
Therefore, $\delta(H)\ge 3$, $n'_2(u) = 0$, $n'_{k}(u) = n_{k}(u)$, for any $k\in [3, 6]$, and $n'_{5^-}(u) = n_{5^-}(u) - n_2(u)$, $n'_{6^+}(u) = n_{6^+}(u)$.

The non-existences of $(A_{1.3})$ in $G$ implies that every $8^+$-vertex $u$ is adjacent to at most ($d(u) - 6$) $5^-$-vertices in $G$;
thus if $d(u) \ge 8$, then $n'_{5^-}(u) = n_{5^-}(u) - n_2(u)\le d(u) - 7 - (d(u) - d_H(u)) = d_H(u) - 7$, and $n'_{6^+}(u) \ge 7$.

By Lemma <ref>, for any $k$-vertex, where $k\in [3, 6]$, we do not distinguish which graph, $G$ or $H$, it is in;
furthermore, we use $d(u)$ instead of $d_H(u)$, use $n_k(u)$ instead of $n'_k(u)$, and use $n_{6^+}(u)$ instead of $n'_{6^+}(u)$.

For any vertex $u \in V(H)$, if $d(u)> d_H(u)\ge 7$, then $n'_{5^-}(u)\le d_H(u) - 7$, $n_{6^+}(u) = n'_{6^+}(u) \ge 7$;
and if $n'_{5^-}(u) = n_{5^-}(u) > 0$, then $d_H(u)\ge 7$.

For any $uv \in E(H)$,

if $d(v) = 3$ with $N(v) = \{u, v_1, v_2\}$, then $d(u)\ge 8$ and $d_H(u)\ge 8$;

if $d(u) = 4$ with $N(u) = \{v, u_1, u_2, u_3\}$, then for each $i = 1, 2, 3$, we have $d_H(u_i)\ge 8$ and

$d(u_i)\ge 10$ if $d(v) = 4$;

$d(u_i)\ge 9$ if $d(v) = 5$;

$d(u_i)\ge 8$ if $d(v) = 6$ and $u v$ is contained in a $3$-cycle;

$d(u_i)\ge 10$ if $d(v) = 5$ and $u v$ is contained in two $3$-cycles;

$d(u_i)\ge 9$ if $d(v) = 6$ and $u v$ is contained in two $3$-cycles;

$d(u_i)\ge 9$ if $v$ is $(7, 4^-)$ and $u v$ is contained in two $3$-cycles $u u_2 v$, $u u_3 v$ with $u_1 u_2, u_1 u_3\in E(G)$.

The non-existence of ($A_{2.1}$) in $G$ states that no $7^-$-vertex is adjacent to a $3$-vertex in $G$;
thus $d(u)\ge 8$ and by Corollary <ref>, $d_H(u)\ge 8$.
The non-existence of ($A_{1.1}$), ($A_{1.2}$), ($A_{2.1}$) and ($A_{6}$) in $G$ and Corollary <ref> imply that (2.1)-(2.6) hold.

Let $u$, $v$, $u_1$, $v_1$ be four $4^-$-vertices in $H$.

If $d(u) = 3$ and $d(v)\le 4$, or $d(u) = d(v) = 4$ with $u v\not\in E(H)$,
then $F_3(u)\cap F_3(v) = \emptyset$.

If $d(u) = d(v) = 4$ with $u v\in E(H)$ and $d(u_1)\le 4$,
then $(F_3(u)\cup F_3(v))\cap F_3(u_1) = \emptyset$.

If $d(u) = d(v) = 4$ with $u v\in E(H)$ and $d(u_1) = d(v_1) = 4$ with $u_1 v_1\in E(H)$,
then $(F_3(u)\cup F_3(v))\cap (F_3(u_1)\cup F_3(v_1)) = \emptyset$. $\Box$

§.§ Discharging to show contradictions

To derive a contradiction, we make use of the discharging methods, which are very similar to an amortized analysis through token re-distribution.
First, by Euler's formula on the embedding $|V(H)| - |E(H)| + |F(H)|= 2$ and the relation $\sum_{v\in V(H)}d_H(v) = \sum_{f\in F(H)}d(f) = 2|E(H)|$,
we have the following equality:

\begin{equation}\label{eq01}
\sum_{u\in V(H)}(4 - d_H(u)) + \sum_{f\in F(H)}(4 - d(f)) = 8.
\end{equation}

Next, we define an initial weight (i.e., token, or charge) function $\omega(\cdot)$ on $V(H)\cup F(H)$ by
setting $\omega(u) = 4 - d_H(u)$ for each vertex $u\in V(H)$ and
setting $\omega(f) = 4 - d(f)$ for each face $f\in F(H)$.
It follows from Eq. (<ref>) that the total weight is equal to $8$.
We define next a set of discharging rules (R1)–(R5) to move portions of weights from $x$ to $z$ (possibly through $y$), that is,
the function $\tau(x \rightarrow z)$ (or $\tau(x \rightarrow z)_y$), where $x, y, z\in V(H)\cup F(H)$.
The effect of this function decreases the charge at $x$ by $\tau(x \rightarrow z)_y$, leaves the charge of $y$ unchanged, and increases the charge of $z$ by $\tau(x \rightarrow z)_y$.
For counting purposes, we say that $x$ sends the charge to $y$, and $z$ receives the charge from $y$.
At the end of the discharging, a new weight function $\omega'(\cdot)$ is achieved and we will show that $\omega'(x) \le 0$ for every element $x \in V(H)\cup F(H)$.
This contradicts the positive total weight.

Before presenting the discharging rules, we introduce an important parameter $\alpha_u$ for a vertex $u\in V(H)$ satisfying $d_H(u) = k\ge 6$ and $n'_{5^-}(u)> 0$:

\begin{equation}
\label{eq02}
\alpha_u = \left\{
\begin{array}{ll}
\frac{k - 4 - \frac 13(k - 2n'_{5^-}(u))}{2n'_{5^-}(u)} 	= \frac{k - 6}{3n'_{5^-}(u)} + \frac 13,	&\mbox{ if }	n'_{5^-}(u)\le \lfloor\frac k2\rfloor;\\
\frac{k - 4}k = 1 - \frac 4k, 						&\mbox{ otherwise.}%        \displaystyle{n'_{5^-}(u)\ge \lceil\frac k2\rceil}.
\end{array}\right.
\end{equation}

The following properties hold for $\alpha_u$, of which some values are given in Figure <ref>.

Let $d_H(u)= k\ge 6$ with $n'_{5^-}(u)> 0$.

$\alpha_u \ge 1 - \frac 4k$.

If $d_H(u) = 6$, then $\alpha_u = \frac 13$;
if $d_H(u)\ge 12$, then $\alpha_u \ge \frac 23$.

If $u$ is $(k, 7^+)_H$, then (i) if $n'_{5^-}(u)\le \lfloor\frac k2\rfloor$, then $\alpha_u \ge \frac 23 + \frac 1{3n'_{5^-}(u)}> \frac 23$;
(ii) $\alpha_u \ge \frac 57> \frac 23$.

If $d(u) > d_H(u)\ge 8$, then $\alpha_u \ge \frac 57> \frac 23$.

If $u$ is $(k, 6^+)_H$, then $\alpha_u \ge \frac 23$.

If $\alpha_u < \frac 23$, then $d(u) = d_H(u)= k\in [7, 11]$, $u$ is $(k, 5^-)$, and if $n'_{3}(u)> 0$, then $k\in [8, 11]$.

By the definition of $\alpha_u$ in Eq. (<ref>), to prove (1) it suffices to assume that $n'_{5^-}(u)\le \lfloor\frac k2\rfloor$.
In this case, we have $\alpha_u - (1 - \frac 4k) = \frac{k - 6}{3n'_{5^-}(u)} - \frac 23 + \frac 4k \ge \frac{k - 6}{3\times \frac{k}{2}} - \frac 23 + \frac 4k = 0$,
i.e., $\alpha_u \ge 1 - \frac 4k$.

For (2), if $d_H(u) = 6$, then by the definition $\alpha_u = \frac 13$;
if $d_H(u) \ge 12 $, then by (1) $\alpha_u \ge \frac 23$.

To prove (3), if $d_H(u)\ge 14$, then by (1), $\alpha_u \ge 1 - \frac 4k \ge \frac 57$.
If $d_H(u)\le 14$, then $n'_{5^-}(u) \le k - 7 \le \lfloor\frac k2\rfloor\le 7$;
thus, $\alpha_u = \frac{k - 6}{3n'_{5^-}(u)} + \frac 13 \ge \frac 1{3n'_{5^-}(u)} + \frac 23 \ge \frac 57$.

By Corollary <ref>, if $d(u) > d_H(u)$, then we have $n'_{5^-}(u)\le d_H(u) - 7$ and thus $u$ is $(k, 7^+)_H$.
It follows from (3) that $\alpha_u \ge \frac 57> \frac 23$.
This proves (4).

To prove (5), if $d_H(u)\ge 12$, then by (2) $\alpha_u \ge \frac 23$.
If $d_H(u)\le 11$, then $n'_{5^-}(u)\le k - 6 \le \lfloor\frac k2\rfloor$;
thus, $\alpha_u  =\frac{k - 6}{3n'_{5^-}(u)} + \frac 13
\ge \frac{k - 6}{3(k - 6)} + \frac 13
= \frac{2}{3}. $

Lastly, if $\alpha_u < \frac 23$, then by (2) and (5) $d(u) = d_H(u) = k\in [7, 11]$ and $u$ is $(k, 5^-)$.
If $n'_{3}(u)> 0$, then by the non-existence of ($A_{2.1}$) in $G$, we have $k\in [8, 11]$.
This proves (6).

The $\alpha_u$ value for a vertex $u$ with $d_H(u) = k\in [7, 15]$ and $n'_{5^-}(u)\le 9$.

We now present the discharging rules (R1)–(R5).
A graphical view of how weights are moved from an element to another is depicted in Figure <ref>.

For each $4^-$-vertex $u$ adjacent to a $6^+$-vertex $v$ with $F(u v) = \{f_1, f_2\}$,
if $d(f_1)\ge 4$ and $d(f_2)\ge 4$, then $\tau(u \rightarrow v)_{f_1} = \tau(u \rightarrow v)_{f_2} = \frac{\alpha_v}2$.

In the other four rules (R2)–(R5), $f= [uvw]$ is a $3$-face.

If $\delta(f)\ge 6$, then for any $x\in V(f)$, $\tau(f \rightarrow x) = \frac 13$.

In the rest three rules (R3)–(R5), $\delta(f) \le 5$ and $d(u)\le 5$.

For any $x\in \{v, w\}$ with $d_H(x)\ge 6$,

$\tau(f \rightarrow x)_f = \alpha_x$;

$\tau(f \rightarrow x)_{f_{ux}} = \begin{cases}
\frac {\alpha_x}2,	&\mbox{ if } d(f_{ux})\ge 4;\\
0,							&\mbox{ if } d(f_{ux}) = 3.
\end{cases}$

If $d(u) = 4$, $d(v) = 5$ and $m_3(u) = 4$ with $f_{u v} = [u v w^*]$, then

$\tau(F(u v) \rightarrow v) = \frac 1{10}$ if $\min\{d(f_{v w}), d((f_{u v})_{v w^*})\} = 3$ and $\max\{d(f_{v w}), d((f_{u v})_{v w^*})\}\ge 4$;

$\tau(F(u v) \rightarrow v) = \frac 25$ if $d(f_{v w}) = d((f_{u v})_{v w^*}) = 3$.

If $\delta(f) = d(u) = 5$ with $n_5(f) = k$, then

if after applying (R3), the face $f$ still has charge $\beta \ge 0$, then $\tau(f \rightarrow u) = \frac{\beta}{k}$;

if $d_H(v)\ge 6$, $d_H(w)\ge 6$, and $\alpha_v + \alpha_w\ge 1$, then $\tau(u \rightarrow f) = \alpha_v + \alpha_w - 1$.

The discharging rules (R1)–(R5): A graphical view of how weights are moved from an element to another by red arrows,
each is labeled with the weight.

In the rest of this section we prove that at the end of discharging the new weight function $\omega'(\cdot) \le 0$, through ten Discharging-Lemmas.
The first of which is obvious.

For every face $f$, if $d(f)\ge 4$, then $f$ does not receive or give any weight and thus $\omega'(f) = \omega(f) = 4 - d(f)\le 0$.

This holds trivially since the discharging rules (R1)–(R5) do not apply to $f$ with $d(f)\ge 4$.

For every vertex $v\in V(H)$ with $d_H(v) = k \ge 6$, $\omega'(v)\le 0$.

Let $f = [\ldots u v w\ldots]$ be a face incident to $v$.
If $d_H(u)\ge 6$ and $d_H(w)\ge 6$, then by (R2), $\tau(f \rightarrow v)\le \frac 13$.
If $d_H(u)\le 5$ and $d(f) = 3$, then by (R3.1), $\tau(f \rightarrow v)_f = \alpha_v$.
If $d_H(u)\le 5$ and $d(f)\geq 4$, then by (R1), (R3.1) and (R3.2), $\tau(u \rightarrow v)_f + \tau(f_{v u} \rightarrow v)_f = \frac {\alpha_v}2$,
and $\tau(w \rightarrow v)_f + \tau(f_{vw} \rightarrow v)_f \le \frac {\alpha_v}2$.

Let $k_1$ be the number of faces $f = [\ldots u v w\ldots]$ incident to $v$ such that $d_H(u)\ge 6$ and $d_H(w)\ge 6$.
The total weight transferred to $v$ is at most $k_1 \times \frac{1}{3} + (d_H(v) - k_1)\times \alpha_v$.
It follows that
$\omega'(v) \le \omega(v) + k_1 \times \frac 13 + (d_H(v) - k_1)\times \alpha_v
= 4 - d_H(v) + k_1 \times \frac 13 + (d_H(v) - k_1)\times \alpha_v
%= 4 - k + k_1 \times \frac 13 + (k - k_1)\times \alpha_v
= 4 - k + k\times \alpha_v - (\alpha_v - \frac 13)\times k_1$.

If $n'_{5^-}(u)\ge \lceil\frac k2\rceil$,
then $\omega'(v)\le 4 - k + k\times \alpha_v %- (\alpha_v - \frac 13)\times 0
= 4 - k + k\times \frac {k - 4}k = 0$.
If $n'_{5^-}(u)\le \lfloor\frac k2\rfloor$, then it follows from $k_1\ge k - 2n'_{5^-}(u)$ that
$\omega'(v)\le 4 - k + k\times \alpha_v - (\alpha_v - \frac 13)\times (k - 2n'_{5^-}(u))
%= 4 - k + k\times \alpha_v - (k - 2n'_{5^-}(u))\times \alpha_v + \frac 13\times (k - 2n'_{5^-}(u))
%= 4 - k + 2n'_{5^-}(u)\times \alpha_v + \frac 13\times k - \frac 13\times 2n'_{5^-}(u)
%= 4 - \frac 23\times k - \frac 23 n'_{5^-}(u) + 2n'_{5^-}(u)\times (\frac{k - 6}{3n'_{5^-}(u)} + \frac 13)
= 4 - \frac 23\times k + \frac 23\times (k - 6)
= 0$.

For every vertex $v\in V(H)$ with $d(v) = 3$, $\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)\le 0$.

Let $u_i v\in E(H)$, and $f_i = [\ldots u_i v u_{(i\bmod3) + 1} \ldots]$, $i\in \{1, 2, 3\}$, be the three faces incident to $v$.
One sees from Lemma <ref>(1) that $d(u_i)\ge 8$,
$d_H(u_i)\ge 8$ and $\alpha_{u_i}\ge \frac 12$ for each $i = 1, 2, 3$.
We distinguish the following four cases for $m_3(v)$.

Case 1. $m_3(v) = 0$.

In this case, by (R1), $\tau(v \rightarrow u_1)_{f_3} + \tau(v \rightarrow u_1)_{f_1} = \alpha_{u_1}$,
$\tau(v \rightarrow u_2)_{f_1} + \tau(v \rightarrow u_2)_{f_2} = \alpha_{u_2}$, and
$\tau(v \rightarrow u_3)_{f_2} + \tau(v \rightarrow u_3)_{f_3} = \alpha_{u_3}$.

\[
\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)
= \omega'(v) = \omega(v) - \sum_{i\in [1, 3]}\tau(v \rightarrow u_i)
%= 4 - 3 - \sum_{i\in [1, 3]}\tau(v \rightarrow u_i)\\
= 1- \sum_{i\in [1, 3]}\alpha_{u_i} \le 1- 3\times \frac 12 < 0.
\]

Case 2. $m_3(v) = 1$ with $d(f_2) = 3$.

In this case, by (R1), (R3.1) and (R3.2), $\tau(v \rightarrow u_1)_{f_3} + \tau(v \rightarrow u_1)_{f_1} = \alpha_{u_1}$,
$\tau(f_2 \rightarrow u_2)_{f_1} = \frac {\alpha_{u_2}}2$,
$\tau(f_2 \rightarrow u_3)_{f_3} = \frac {\alpha_{u_3}}2$, and
for each $i\in [2, 3]$, $\tau(f_2 \rightarrow u_i)_{f_2} = \alpha_{u_i}$.

\begin{equation*}
\begin{aligned}
\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)
& = \sum_{x\in \{v, f_2\}}\omega(x)- \left(\tau(v \rightarrow u_1)
+ \sum_{i\in [2, 3]}\tau(f_2 \rightarrow u_i) + \tau(f_2 \rightarrow u_2)_{f_1} +  \tau(f_2 \rightarrow u_3)_{f_3}\right)\\
& = \sum_{x\in \{v, f_2\}}(4 - 3) - (\alpha_{u_1} +  \alpha_{u_2} + \alpha_{u_3} + \frac 12 \alpha_{u_2} + \frac 12 \alpha_{u_3})
%& = 2\times 1 - (\alpha_{u_1} + \frac 23 \alpha_{u_2} + \frac 23 \alpha_{u_3})
\le 2\times 1 - 4\times \frac 12 = 0.
\end{aligned}
\end{equation*}

Case 3. $m_3(v) = 2$ with $d(f_2) = d(f_3) = 3$.

In this case, by (R3.1) and (R3.2), $\tau(f_3 \rightarrow u_1)_{f_3} = \alpha_{u_1}$,
$\tau(f_3 \rightarrow u_1)_{f_1} = \frac {\alpha_{u_1}}2$,
$\tau(f_2 \rightarrow u_2)_{f_2} = \alpha_{u_2}$,
$\tau(f_2 \rightarrow u_2)_{f_1} = \frac {\alpha_{u_2}}2$,
and $\tau(f_2 \rightarrow u_3)_{f_2} = \tau(f_3 \rightarrow u_3)_{f_3} = \alpha_{u_3}$.

\begin{equation}
\label{eq03}
\begin{aligned}
\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)
& = \sum_{x\in \{v, f_2, f_3\}}\omega(x) - \left(\sum_{i\in [2, 3]}\tau(f_2 \rightarrow u_i)
+ \sum_{i\in \{1, 3\}}\tau(f_3 \rightarrow u_i) + \tau(f_3 \rightarrow u_1)_{f_1} + \tau(f_2 \rightarrow u_2)_{f_1}\right)\\
& = \sum_{x\in \{v, f_2, f_3\}}(4 - 3)- (\alpha_{u_2} + \alpha_{u_3} + \alpha_{u_1} + \alpha_{u_3} + \frac 12 \alpha_{u_1} + \frac 12 \alpha_{u_2})\\
& = 3\times 1 - (2\alpha_{u_3} + \frac 32 \alpha_{u_1} + \frac 32 \alpha_{u_2}).
%= 3 - (2\alpha_{u_3} + \frac 32 \alpha_{u_1} + \frac 32 \alpha_{u_2}).
\end{aligned}
\end{equation}

Below we distinguish two cases where $u_1 u_2\in E(G)$ and $u_1 u_2\not\in E(G)$, respectively.

Case 3.1. $u_1u_2\in E(G)$.

In this case, for each $i\in [1, 3]$, the non-existence of ($A_{2.1}$) and ($A_{2.4}$) in $G$ states that $d(u_i)\ge 9$;
by Corollary <ref>, no $7$-vertex is adjacent to any $5^-$-vertex in $H$ and $d_H(u_i)\ge 8$.
Thus, we have $\alpha_{u_i}\ge \frac 59$ for each $i\in [1, 3]$.
Note that if $d_H(u_i)= 8$ for some $i\in [1, 3]$, then $d(u_i)\ge 9 > d_H(u_i)$, and it follows from Lemma <ref> that $\alpha_{u_i}\ge \frac 57 > \frac23$.

When $\alpha_{u_3}\ge \frac 23$, Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) % = 3 - (2\alpha_{u_3} + \frac 23 \alpha_{u_1} + \frac 32 \alpha_{u_2})
\le 3 - (2\times \frac 23 + \frac 32 \times \frac 59 + \frac 32\times \frac 59) = 0$.

When $\alpha_{u_1}\ge \frac{7}{11}$ and $\alpha_{u_2}\ge \frac 7{11}$, Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)
%& = 3 - (2\alpha_{u_3} + \frac 32\alpha_{u_1} + \frac 32\alpha_{u_2})
\le 3 - (2\times \frac 59 + \frac 32 \times \frac 7{11} + \frac 32 \times \frac 7{11})
%& = 3 - (\frac {10}9 + 3\times \frac 7{11}) =  3 - (\frac {10}9 + \frac {21}{11})
= - \frac 19 + \frac 1{11} < 0$.

When $\alpha_{u_3}< \frac 23$ and $\alpha_{u_i}< \frac 7{11}$ for some $i\in [1, 2]$,
it follows from Lemma <ref> that $d_H(u_3) = d(u_3)\in [9, 11]$, $u_3$ is $(d(u_3), 5^-)$,
and there exists an $i\in [1, 2]$ such that $d_H(u_i) = d(u_i)\in [9, 10]$, $u_i$ is $(d(u_i), 5^-)$.
It follows that there exists a $w\in N(v)$ such that $w$ is $(10, 5^-)$.

Assume that $w_1$ is $(9, 5^-)$.
The non-existence of ($A_{3.3}$) in $G$ implies that $w_1$ is $(9, 5)$;
the non-existence of ($A_{3.6}$) in $G$ implies that there exists a $w_2\in N(v)\setminus \{w_1\}$ such that $w_2$ is $([9, 11], 5)$.
Let $w_3\in N(v)\setminus \{w_1, w_2\}$.
Then by the non-existence of ($A_{3.2}$) in $G$, we have $d_H(w_3)= d(w_3) = \Delta$ and $w_3$ is $(\Delta, 7^+)_{HG}$.
It is clear that $w_3\in \{u_1, u_2\}$, i.e., $w_3 \ne u_3$.
Therefore, $\alpha_{w_1}\ge \frac 7{12}$, $\alpha_{w_2}\ge \frac 7{12}$, and by Lemma <ref> $\alpha_{w_3}\ge \frac 57$.
Hence, Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)
%& = 3 - (2\alpha_{u_3} + \frac 32 \alpha_{u_1} + \frac 32 \alpha_{u_2})
\le 3 - (2\times \frac 7{12} + \frac 32\times \frac 7{12} + \frac 32\times \frac 57)
%\\& = 3 - (\frac 72\times \frac 7{12} + \frac {15}{14}) =  3 - (\frac {49}{24} + \frac {15}{14})
= - \frac 1{24} - \frac 1{14} < 0$.

Assume that there is no $(9, 5^-)$ in $\{u_1, u_2, u_3\}$ and $w_1$ is $(10, 5^-)$.
The non-existence of ($A_{4.1}$) in $G$ implies that there exists a $w_2\in N(v)\setminus \{w_1\}$ such that $w_2$ is $(10/11, 5)$.
Let $w_3\in N(v)\setminus \{w_1, w_2\}$.
Thus, by the non-existence of ($A_{4.3}$) in $G$, we have $d_H(w_3) = d(w_3) = \Delta$ and by the non-existence of ($A_{4.2}$) in $G$, $w_3$ is $(\Delta, 7^+)_{H G}$.
It is clear that $w_3\in \{u_1, u_2\}$, i.e., $w_3 \ne u_3$.
Therefore, $\alpha_{w_1}\ge \frac 35$, $\alpha_{w_2}\ge \frac 35$, and by Lemma <ref> $\alpha_{w_3}\ge \frac 57$.
Hence, Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %& = 3 - (2\alpha_{u_3} + \frac 32\alpha_{u_1} + \frac{3}{2}\alpha_{u_2})
\le 3 - (2\times \frac 35 + \frac 32\times \frac 35 + \frac 32\times \frac 57)
%\\& = 3 - (\frac 72\times \frac 35 + \frac {15}{14}) =  3 - (\frac {21}{10} + \frac {15}{14})
= - \frac 1{10} - \frac 1{14} < 0$.

Case 3.2. $u_1 u_2\not\in E(G)$.

In this case, $d_H(u_3)\ge 12$, or $d(u_3) > d_H(u_3)$,
or $d_H(u_3) = d(u_3)\in [8, 11]$ and by the non-existence of ($A_{2.2}$) in $G$ $u_3$ is $(d(u_3), 6^+)_{HG}$.
Then by Lemma <ref>(2), (5) and (6), $\alpha_{u_3}\ge \frac 23$.

When $\alpha_{u_1}\ge \frac 59$ and $\alpha_{u_2}\ge \frac 59$, Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %= 3 - (2\alpha_{u_3} + \frac 32 \alpha_{u_1} + \frac 32 \alpha_{u_2})
\le 3 - (2\times \frac 23 + \frac 32 \times \frac 59 + \frac 32\times \frac 59) = 0$.

When $\alpha_{u_i}< \frac 59$ for some $i\in [1, 2]$,
it follows from Lemma <ref>(2) and (4) that there exists a $u_i\in \{u_1, u_2\}$ such that $d(u_i) = d_H(u_i) = 8$ and $u_i$ is $(8, 4^-)_{H G}$.
Assuming w.l.o.g. $u_1$ is such a vertex.
From the non-existence of ($A_{2.2}$) in $G$, $d_H(u_2)\ge 12$, or $d(u_2) > d_H(u_2)$, or $d_H(u_2) = d(u_2)\in [8, 11]$ and $u_2$ is $(d(u_2), 6^+)_{HG}$.
We have $\alpha_{u_2}\ge \frac 23$ by Lemma <ref>(2), (5) and (6).
Thus, Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %= 3 - (2\alpha_{u_3} + \frac 32\alpha_{u_1} + \frac 32\alpha_{u_2})
\le 3 - (2\times \frac 23 + \frac 32\times \frac 12 + \frac 32 \times \frac 23) = -\frac 1{12} < 0$.

Case 4. $m_3(v) = 3$.

In this case, by a similar discussion as in Case 3.1 where $u_1 u_2\in E(G)$,
it suffices to assume that for each $i\in [1, 3]$, $\alpha_{u_i}\ge \frac 59$, and either $d(u_i)\ge 9$ or $u_i$ is $(8, 7^+)$.
It follows from (R3.1) that $\tau(f_3 \rightarrow u_1)_{f_3} = \tau(f_1 \rightarrow u_1)_{f_1} = \alpha_{u_1}$,
$\tau(f_1 \rightarrow u_2)_{f_1} = \tau(f_2 \rightarrow u_2)_{f_2} = \alpha_{u_2}$,
$\tau(f_2 \rightarrow u_3)_{f_2} = \tau(f_3 \rightarrow u_3)_{f_3} = \alpha_{u_3}$,
and thus,

\begin{equation}
\label{eq04}
\begin{aligned}
\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)
& = \sum_{x\in \{v, f_1, f_2, f_3\}}\omega(x)- \left(\sum_{i\in \{1, 3\}}\tau(f_i \rightarrow u_1) + \sum_{i\in [1, 2]}\tau(f_i \rightarrow u_2)
+ \sum_{i\in [2, 3]}\tau(f_i \rightarrow u_3)\right)\\
%& = \sum_{x\in \{v, f_1, f_2, f_3\}}(4 - 3)- (2\alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3})\\
& = 4 - 2 (\alpha_{u_1} + \alpha_{u_2} + \alpha_{u_3}).
\end{aligned}
\end{equation}

If $\alpha_{u_i}\ge \frac 23$ for each $i\in [1, 3]$, then Eq. (<ref>) clearly is $\le 0$;
otherwise, there exists an $i_0\in [1, 3]$ such that $\alpha_{u_{i_0}} < \frac 23$, and then by Lemma <ref>(6), $u_{i_0}$ is $([9, 11], 5^-)_{HG}$.
We discuss the following three subcases on the existence of a $(9/10/11, 5^-)$ vertex, which w.l.o.g. is assumed $u_3$.

Case 4.1. $u_3$ is $(9, 5^-)$.
In this case, the non-existence of ($A_{3.1}$) in $G$ states that there exits an $i^*\in [1, 2]$ such that $d(u_{i^*}) = d_H(u_{i^*}) = \Delta$.
If $\alpha_{u_i}\ge \frac {13}{18}$ for each $i\in [1, 2]$, then Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %= 4 - (2\alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3})
\le 4 - 2 (\frac {13}{18} + \frac {13}{18} + \frac 59) = 0$.
Otherwise, there exits an $i_1\in [1, 2]$ such that $\alpha_{u_{i_1}} < \frac {13}{18} < \frac {11}{15}$.
One sees that if $d_H(u_{i_1})\ge 15$, or $u_{i_1}$ is $(14, 8^+)_H$, or $u_{i_1}$ is $([8, 13], 7^+)_H$,
then $\alpha_{u_{i_1}}\ge \frac {13}{18}$.
Therefore, $u_{i_1}$ is $([9, 13], 6^-)_{H G}$, or $u_{i_1}$ is $(14, 7^-)_H$.
We consider them separately below, assuming $i_1 = 2$:

(Case 4.1.1.) $u_2$ is $([9, 13], 6^-)_{H G}$.
In this case, the non-existence of ($A_{3.2}$) in $G$ states that $u_1$ is $(\Delta, (\Delta - d(u_2) + 7)^+)_{H G}$, i.e., $n'_{5^-}(u)\le d(u_2)- 7\le 6$.
Let $\{a, b\} = \{\alpha_{u_1}, \alpha_{u_2}\}$.
When $a\ge \frac 23$ and $b\ge \frac 79$, using $\alpha_{u_3} \ge \frac 59$ Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %= 4 - (2\alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3})
\le 4 - 2 (\frac 79 + \frac 23 + \frac 59) = 0$.
We thus assume below that $a< \frac 23$ or $b < \frac 79$.
Similarly, if $\alpha_{u_3}\ge \frac 7{12}$, then we assume that $a< \frac 23$ or $b< \frac 34$.

$u_3$ is $(9, 4^-)$.
The non-existence of ($A_{3.3}$) in $G$ states that $u_2$ is $([9, 13], 6)_{HG}$.
It follows from $\alpha_{u_2}\ge \frac 23$ that $\alpha_{u_1}< \frac 79$.
If $d(u_1) = \Delta\ge 14$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 6 \le \lfloor\frac k2\rfloor$,
and thus $\alpha_{u_1} = \frac{k - 6}{3n'_{5^-}(u)} + \frac 13 \ge
% \frac{14 - 6}{3\times 6} + \frac 13  = \frac 8{18} + \frac 13 =
\frac 79$, a contradiction.
If $u_1$ is $(9/10, 7^+)_{H G}$, or $u_1$ is $([11, 13], 8^+)_{H G}$, then $\alpha_{u_1}\ge \frac 79$, also a contradiction.
Therefore, $d_H(u_1) = d(u_1)= \Delta\in [11, 13]$, and furthermore $u_1$ is $([11, 13], 7)_{H G}$.
Since $d(u_2) < d(u_1)$ implies $n_{6^+}(u_1)\ge 8$, we have $u_2$ is $(\Delta, 6)_{H G}$,
by which the local structure ($A_{3.4}$) is obtained, a contradiction again.

$u_3$ is $(9, 5)$ and $\alpha_{u_3}\ge \frac 7{12}$.
The non-existence of ($A_{3.6}$) in $G$ states that $u_2$ is $([9, 13], 5/6)_{H G}$ and thus $\alpha_{u_2}\ge \frac 7{12}$.
When $\alpha_{u_1}\ge \frac{5}{6}$, Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %= 4 - (2\alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3})
\le 4 - 2 (\frac 56 + \frac 7{12} + \frac 7{12}) = 0$.
When $\alpha_{u_1} < \frac 56$, one sees that $d_H(u_1) = d(u_1) = k$,
and $u_1$ cannot be $(9^+, (k - 2)^+)_{H G}$, or $(11^+, (k - 3)^+)_{H G}$, or $(12^+, (k - 4)^+)_{H G}$, or $(14^+, (k - 5)^+)_{H G}$.
Since $u_1$ is $(\Delta, (\Delta - d(u_2) + 7)^+)_{H G}$, we have $n'_{5^-}(u)\le d(u_2)- 7\le 6$, and $k\ge 12$ or $u_1$ is $(10/11, 7)_{H G}$.
We discuss the following three subcases for three possible values of $d(u_2)$.

* $u_2$ is $(9, 5/6)_{H G}$.
It follows that $d(u_1) = \Delta\ge 10$, and then $n'_{5^-}(u_1)\le d(u_2)- 7\le 2$.
Thus, $u_1$ is $(10^+, (k-2)^+)_{H G}$.

* $u_2$ is $(10, 5/6)_{H G}$.
If $d(u_1) = \Delta\ge 11$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 3$, and thus $u_1$ is $(11^+, (k - 3)^+)$.
Otherwise, $u_1$ is $(10, 7)_{H G}$.
It follows from $\alpha_{u_1} = \frac 79$ that $\alpha_{u_2}< \frac 23$.
Then $u_2$ is $(10, 5)_{H G}$, and thus ($A_{3.7}$) is obtained, a contradiction.

* $u_2$ is $(11, 5/6)_{H G}$.
If $d(u_1) = \Delta\ge 12$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 4$, and $u_1$ is $(12^+, (k - 4)^+)$.
Otherwise, $u_1$ is $(11, 7)_{H G}$.
It follows from $\alpha_{u_1} = \frac 34$ that $\alpha_{u_2}< \frac 23$.
Then $u_2$ is $(11, 5)_{H G}$, and thus ($A_{3.7}$) is obtained, a contradiction.

* $u_2$ is $(12/13, 5/6)_{H G}$.
It follows from $\alpha_{u_2}\ge \frac 23$ that $\alpha_{u_1}< \frac 34$.
One sees that $d_H(u_1) = d(u_1) = k\ge 12$,
and if $u_1$ is $(12^+, (k - 4)^+)_{H G}$, or $(13^+, (k - 5)^+)_{H G}$, or $(14^+, (k - 6)^+)_{H G}$, then $\alpha_{u_1}\ge \frac 34$.
Hence, $k\ge 14$, or $u_1$ is $(12/13, 7)_{H G}$.

* $u_2$ is $(12, 5/6)_{H G}$.
If $d(u_1) = \Delta\ge 13$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 5$, and thus $u_1$ is $(13^+, (k - 5)^+)$.
Otherwise, $u_1$ is $(12, 7)_{H G}$, and thus ($A_{3.8}$) is obtained, a contradiction.

* $u_2$ is $(13, 5/6)_{H G}$.
If $d(u_1) = \Delta\ge 14$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 6$, and thus $u_1$ is $(14^+, (k - 6)^+)$.
Otherwise, $u_1$ is $(13, 7)_{H G}$, and thus ($A_{3.8}$) is obtained, a contradiction.

(Case 4.1.2.) None of $\{u_1, u_2\}$ is $([9, 13], 6^-)_{H G}$ and $u_2$ is $(14, 7^-)_H$.
In this case, $d_H(u_1)\ge 14$ or $u_1$ is $([9, 13], 7^+)_H$.
It follows from Lemma <ref>(1) and (3) that $\alpha_{u_1}\ge \frac 57$ and $\alpha_{u_2}\ge \frac 57$.
If $u_3$ is $(9, 5)$, then Eq. (<ref>) becomes
$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) %= 4 - (2\alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3})
\le 4 - 2 (\frac 57 + \frac 57 + \frac 7{12})= - \frac 1{42} < 0$.
Otherwise, $u_3$ is $(9, 4^-)$.
Note that $2 - (\alpha_{u_3} + \alpha_{u_2}) = 2 - (\frac 59 +  \frac 57) < \frac {11}{15}$.
One sees that if $d_H(u_1)\ge 15$, or $u_1$ is $(13/14, 8^+)_H$, or $u_1$ is $([9, 12], 7^+)_H$,
then $\alpha_{u_1}\ge \frac {11}{15}$ and thus Eq. (<ref>) becomes $\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) < 0$.
In the other case, $u_1$ is $(13, 7)_H$ or $(14, 7^-)_H$, either of which leads to a structure ($A_{3.5}$), a contradiction.

Case 4.2. None of $\{u_1, u_2, u_3\}$ is $(9, 5^-)$ and $u_3$ is $(10, 5^-)$.
In this case, for each $i\in [1, 3]$, $\alpha_{u_i}\ge \frac 35$.
When $\alpha_{u_i}\ge \frac 7{10}$ for each $i\in [1, 2]$, Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) \le 4 - 2 (\frac 7{10} + \frac 7{10} + \frac 35) = 0$.
When there exits an $i_1\in [1, 2]$ such that $\alpha_{u_{i_1}}< \frac 7{10}$,
by Lemma <ref>(1) and (3), $u_{i_1}$ is $(9, 6)_{HG}$ or $([10, 13], 6^-)_{H G}$.
Assuming w.l.o.g. $i_1 = 2$.
The non-existence of ($A_{4.1}$) in $G$ implies that $u_2$ is $(9, 6)_{H G}$ or $([10, 13], 5/6)_{H G}$;
the non-existence of ($A_{4.2}$) in $G$ implies that $n'_{6^+}(u_1) = n_{6^+}(u_1)\ge \Delta - d(u_2) + 7\ge 7$ and $n'_{5^-}(u_1)\le d(u_2) - 7$.
It follows from Lemma <ref> that $\alpha_{u_1}\ge \frac 57$.
Recall that $\alpha_{u_3}\ge \frac 35$, and $2 - (\alpha_{u_3} + \alpha_{u_1}) = 2 - (\frac 35 +  \frac 57) < \frac 9{13}$.
If $d_H(u_1)\ge 13$, then $\alpha_{u_1}\ge \frac 9{13}$ and thus Eq. (<ref>) becomes $\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) < 0$.
In the other case, $u_1$ is $(9, 6)_{H G}$ or $([10, 12], 5/6)_{H G}$, and we discuss them separately below:

(Case 4.2.1) $u_2$ is $([9, 11], 6)$ or $(12, 5/6)$.
In this case, $\alpha_{u_2}= \frac 23$.
When $d_H(u_1)\ge 15$, or $u_1$ is $(12^-, 7^+)_H$ or $(13/14, 8^+)_H$, $\alpha_{u_1}\ge \frac {11}{15}$, and thus Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) \le 4 - 2(\frac {11}{15} + \frac 23 + \frac 35) = 0$.
In the other case, $u_1$ is $(13/14, 7)_H$ and thus $\Delta\ge d_H(u_1)\ge 13$.
It follows that $7 \ge n'_{6^+}(u_1)\ge \Delta - d(u_2) + 7\ge 13 - 12 + 7 = 8$, a contradiction.

(Case 4.2.2) $u_2$ is $(10/11, 5)$.
The non-existence of ($A_{4.3}$) in $G$ implies that $d_H(u_1) = d(u_1) = \Delta$.
Hence, $u_1$ is $(\Delta, (\Delta - d(u_2) + 7)^+){H G}$ where $\Delta\ge d(u_1)\ge 10$.
When $u_1$ is $(10, 8^+)_{H G}$, $(11^+, (k - 3)^+)_{H G}$, or $(12^+, (k - 4)^+)_{H G}$, we have $\alpha_{u_1}\ge \frac{4}{5}$,
and thus Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) \le 4 - 2(\frac 45 + \frac 35 + \frac 35) = 0$.
In the other case, $d_H(u_1) = d(u_1) = k \ge 12$ or $u_1$ is $(10/11, 7)_{H G}$.
We discuss them separately below to derive contradictions:

$u_2$ is $(10, 5)$.
If $d(u_1) = \Delta\ge 11$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 3$;
thus $u_1$ is $(11^+, (k - 3)^+)_{H G}$, a contradiction.
Otherwise, $u_1$ is $(10, 7)_{H G}$, and thus ($A_{4.4}$) is obtained, a contradiction too.

$u_2$ is $(11, 5)$.
If $d(u_1) = \Delta\ge 12$, then $n'_{5^-}(u_1)\le d(u_2)- 7\le 4$;
thus $u_1$ is $(12^+, (k - 4)^+)_{H G}$, a contradiction.
Otherwise, $u_1$ is $(11, 7)_{H G}$, and thus ($A_{4.4}$) is obtained, a contradiction too.

Case 4.3. None of $\{u_1, u_2, u_3\}$ is $(9/10, 5^-)$ and $u_3$ is $(11, 5^-)$.
In this case, for each $i\in [1, 3]$, $\alpha_{u_i}\ge \frac 7{11}$.
When $\alpha_{u_i}\ge \frac {15}{22}$ for each $i\in [1, 2]$, Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) \le 4 - 2(\frac {15}{22} + \frac {15}{22} + \frac 7{11}) = 0$.

In the other case, there exits an $i_1\in [1, 2]$ such that $\alpha_{u_{i_1}}< \frac{15}{22} < \frac{9}{13}$.
By Lemma <ref>(1) and (3), $u_{i_2}$ is $([9, 12], 6)_{H G}$, or $(12, 5^-)_{H G}$, or $(11, 5^-)_{H G}$.
Assume w.l.o.g. $i_2 = 2$.
We have the following two scenarios:

(Case 4.3.1) $u_2$ is $([9, 12], 6)_{H G}$, or $(12, 5^-)_{H G}$.
In this case, $\alpha_{u_2} = \frac 23$.
When $\alpha_{u_1}\ge \frac {23}{33}$, Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) \le 4 - 2(\frac {23}{33} + \frac 23 + \frac 7{11}) = 0$.
In the other case, $\alpha_{u_1} < \frac{23}{33} < \frac 57$.
One sees that if $d_H(u_1)\ge 14$ or $n'_{6^+}(u_1)\ge 7$ by Lemma <ref>(1) and (3),
then $\alpha_{u_1}\ge \frac 57$ and thus Eq. (<ref>) becomes $\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) < 0$.
It follows that $n_{6^+}(u_1) = n'_{6^+}(u_1)\le 6$, and $u_1$ is $([9, 13], 6)_{H G}$, or $([11, 14], 5^-)$;
but then ($A_{5.1}$) is obtained, a contradiction.

(Case 4.4.2) None of $\{u_1, u_2\}$ is $([9, 12], 6)_{H G}$ or $(12, 5^-)_{H G}$, and $u_2$ is $(11, 5^-)$.
In this case, $\alpha_{u_2} = \frac 23$.
When $d_H(u_1)\ge 15$, or $u_1$ is $(12^-, 7^+)_H$, or $(13/14, 8^+)_H$, we have $\alpha_{u_1}\ge \frac 8{11}$,
and thus Eq. (<ref>) becomes

$\sum_{x\in \{v\}\cup F_3(v)}\omega'(x) \le 4 - 2(\frac 8{11} + \frac 7{11} + \frac 7{11}) = 0$.
In the other case, $d_H(u_1)\le 14$, and $n_{6^+}(u_1) = n'_{6^+}(u_1)\le 6$ or $u_1$ is $(13/14, 7)_H$.
Recall that $d(u_1) = d_H(u_1)$ if $n'_{6^+}(u_1)\le 6$.
The non-existence of ($A_{5.1}$) in $G$ implies that $u_1$ is $(13/14, 7)_H$, that is, $u_1$ is $(13^+, 7)$ in $G$,
which leads to a structure ($A_{5.2}$), a contradiction.
This finishes all cases, and thus proves the lemma.

For every vertex $u\in V(H)$ with $d_H(u) = d(u) = 4$, we have

for each $6^+$-vertex $x$ in $N_H(u)$, the total weight transferred from $u$ and the $3$-faces in $F(u x)$ to $x$ is $(1 + \frac 12 m_3(u x))\alpha_{x}$;

if $u$ is adjacent to some $4$-vertex $v$ in $H$, then $\sum_{x\in \{u, v\}\cup F_3(u)\cup F_3(v)}\omega'(x)\le 0$;

if $d_H(v)\ge 5$ for each $v\in N_H(u)$, then $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)\le 0$.

For each $x\in N_H(u)$ with $d_H(x)\ge 6$, let $\Gamma(u \rightarrow x)$ be the total weight transferred from $u$ and the $3$-faces in $F(u x)$ to $x$.
To prove (1), we discuss the following three subcases for the three possible values of $m_3(u x)$.

* $m_3(u x) = 0$.
By (R1), $\tau(u \rightarrow x)_f = \tau(u \rightarrow x)_{f_{u x}} = \frac 12 \times \alpha_{x}$;
and thus $\Gamma(u \rightarrow x) = \tau(u \rightarrow x)_f + \tau(u \rightarrow x)_{f_{u x}} = \alpha_{x}$.

* $m_3(u x) = 1$ with $d(f_3) = 3$.
By (R3.1) and (R3.2), $\tau(f \rightarrow x)_f = \alpha_{x}$, $\tau(f \rightarrow x)_{f_{u x}} = \frac 12 \times \alpha_{x}$;
and thus $\Gamma(u \rightarrow x) = \tau(f \rightarrow x)_f + \tau(f \rightarrow x)_{f_{u x}} = \alpha_{x} + \frac 12 \times \alpha_{x} = \frac 32\alpha_{x}$.

* $m_3(u x) = 2$.
By (R3.1), $\tau(f \rightarrow x)_f = \tau(f_{ux} \rightarrow x)_{f_{u x}} = \alpha_{x}$;
and thus $\Gamma(u \rightarrow x) = \tau(f \rightarrow x)_f + \tau(f_{ux} \rightarrow x)_{f_{u x}} = \alpha_{x} + \alpha_{x} = 2\alpha_{x}$.

Hence, the total weight transferred from $u$ and the $3$-faces in $F(u x)$ to $x$ is $(1 + \frac 12 m_3(u x))\alpha_{x}$.

Now let $u_1, u_2, v, u_3 $ be all the neighbors of $u$ in clockwise order,
and denote the face containing $u_1 u u_2$ ($u_2 u v$, $v u u_3$, $u_3 u u_1$, respectively) as $f_1$ ($f_2, f_3, f_4$, respectively).
Note that

\begin{equation}
\label{eq05}
\sum_{x\in \{u, v\}\cup F_3(u)\cup F_3(v)}\omega(x)
%& = \sum_{x\in \{u, v\}}\omega(x) + \sum_{x\in F_3(u)\cup F_3(v)}\omega(x)\\
%& = \sum_{x\in \{u, v\}}\times (4 - 4) + \sum_{x\in F_3(u)\cup F_3(v)}\times (4 - 3)\\
= \sum_{x\in F_3(u)\cup F_3(v)}\omega(x) = m_3(u) + m_3(v) - m_3(uv).
\end{equation}

To prove (2), we have $d(v) = d_H(v) = 4$ and let $v_1, v_2, v_3 $ be the other neighbors of $v$ in clockwise order,
and denote the face containing $v_1 v v_2$ ($v_2 v v_3$, respectively) as $h_1$ ($h_2$, respectively).

For each $x\in N_H(u)\cup N_H(v)$, it follows from Lemma <ref> that $d(x)\ge 10$ and $d_H(x) \ge 8$;
and then by Lemma <ref>(1) and (4), we have $\alpha_x\ge \frac 35$.
Using (R1), (R3.1) and (R3.1), we discuss the following three subcases for the three possible values of $m_3(uv)$.

Case 1. $m_3(uv) = 0$.

In this case, $m_3(u)\le 2$ and $m_3(v)\le 2$.
It follows from Eq. (<ref>) that

\begin{equation}
\label{eq06}
\begin{aligned}
\sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)
& = \sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega(x) - \sum_{x\in \{u_1, u_2, u_3, v_1, v_2, v_3\}}\Gamma(u \rightarrow x)\\
%& = m_3(u) + m_3(v) - m_3(uv)- \sum_{x\in \{u_1, u_2, u_3, v_1, v_2, v_3\}}\Gamma(u \rightarrow x)\\
& = m_3(u)+ m_3(v) - \sum_{x\in \{u_1, u_2, u_3, v_1, v_2, v_3\}}\Gamma(u \rightarrow x).
\end{aligned}
\end{equation}

By (1), $\sum_{x\in N_H(u)\setminus \{v\}}\Gamma(u \rightarrow x)\ge \alpha_{u_1} + \alpha_{u_2} + \alpha_{u_3}$,
$\sum_{x\in N_H(v)\setminus \{u\}}\Gamma(u \rightarrow x)\ge \alpha_{v_1} + \alpha_{v_2} + \alpha_{v_3}$.

When $m_3(u)+ m_3(v)\le 3$, Eq. (<ref>) becomes
$\sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)
\le 3 - (\alpha_{u_1} + \alpha_{u_2} + \alpha_{u_3} + \alpha_{v_1} + \alpha_{v_2} + \alpha_{v_3}) \le 3 - 6\times \frac 35 = - \frac 35 < 0$.

In the other case, $m_3(u)+ m_3(v) = 4$, i.e., $m_3(u) = m_3(v)= 2$.
Then by (1), $\sum_{x\in N_H(u)\setminus \{v\}}\Gamma(u \rightarrow x)\ge 2\alpha_{u_1} + \alpha_{u_2} + \alpha_{u_3}$,
$\sum_{x\in N_H(v)\setminus \{u\}}\Gamma(u \rightarrow x)\ge \alpha_{v_1} + 2\alpha_{v_2} + \alpha_{v_3}$, and Eq. (<ref>) becomes
$\sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)
\le 4 - (2\alpha_{u_1} + \alpha_{u_2} + \alpha_{u_3} + \alpha_{v_1} + 2\alpha_{v_2} + \alpha_{v_3}) \le 4 - 8\times \frac 35 = - \frac 45< 0$.

Case 2. $m_3(u v) = 1$ with $d(f_2) = 3$.
In this case, $u_2 = v_1$.

Let $\Gamma(u/v \rightarrow u_2)$ be the total weight transferred from $u$, $v$ and  the $3$-faces in $F(u u_2)\cup F(v u_2)$ to $u_2$.
If $d(f_1) = d(h_1) = 3$,
then $\Gamma(u/v \rightarrow u_2) = \tau(f_1 \rightarrow u_2)_{f_1} + \tau(f_2 \rightarrow u_2)_{f_2} + \tau(h_1 \rightarrow u_2)_{h_1} = 3\alpha_{u_2}$;
if $d(f_1), d(h_1)\ge 4$,
then $\Gamma(u/v \rightarrow u_2) = \tau(f_2 \rightarrow u_2)_{f_1} + \tau(f_2 \rightarrow u_2)_{f_2} + \tau(f_2 \rightarrow u_2)_{h_1} = 2\alpha_{u_2}$;
otherwise, only one of $\{f_1, h_1\}$ is a $3$-face, assuming w.l.o.g. $d(f_1) = 3$,
thus $\Gamma(u/v \rightarrow u_2) = \tau(f_1 \rightarrow u_2)_{f_1} + \tau(f_2 \rightarrow u_2)_{f_2} + \tau(f_2 \rightarrow u_2)_{h_1}= \frac 52\alpha_{u_2}$.
That is,

\begin{equation}
\label{eq07}
\begin{aligned}
\Gamma(u/v \rightarrow u_2) = \begin{cases}
3\alpha_{u_2},				&\mbox{ if } d(f_1) = d(h_1)= 3;\\
2\alpha_{u_2},				&\mbox{ if } d(f_1), d(h_1)\ge 4;\\
\frac 52\alpha_{u_2},	&\mbox{ otherwise}.\end{cases}
\end{aligned}
\end{equation}

By (1) and Eq. (<ref>), $\sum_{x\in \{u_1, u_3\}}\Gamma(u \rightarrow x)\ge \alpha_{u_1} + \alpha_{u_3}$,
$\sum_{x\in \{v_2, v_3\}}\Gamma(u \rightarrow x)\ge \alpha_{v_2} + \alpha_{v_3}$, and

\begin{equation}
\label{eq08}
\sum_{x\in \{u, v\}\cup F_3(u)\cup F_3(v)}\omega'(x) = m_3(u)+ m_3(v)- 1 - \sum_{x\in \{u_1, u_3, v_2, v_3\}}\Gamma(u \rightarrow x) - \Gamma(u/v \rightarrow u_2).
\end{equation}

When $m_3(u)+ m_3(v)\le 4$, Eq. (<ref>) becomes
$\sum_{x\in \{u, v\}\cup F_3(u)\cup F_3(v)}\omega'(x)
\le 3 - (\alpha_{u_1} + \alpha_{u_3} + \alpha_{v_2} + \alpha_{v_3} + 2\alpha_{u_2}) \le 3 - 6\times \frac 35 = - \frac 35< 0$.

In the other case, $5\le m_3(u)+ m_3(v)\le 6$.
It follows that $\max\{m_3(u), m_3(v)\}\ge 3$.
Assume w.l.o.g. $m_3(u) = 3$, i.e., $d(f_1) = d(f_4) = 3$, and $m_3(v) \ge 2$.
Then by (1) and Eq. (<ref>), $\sum_{x\in \{u_1, u_3\}}\Gamma(u \rightarrow x)\ge 2\alpha_{u_1} + \frac 32 \alpha_{u_3}$,
$\Gamma(u/v \rightarrow u_2) \ge 52 \alpha_{u_2}$, $\sum_{x\in \{v_2, v_3\}}\Gamma(u \rightarrow x)\ge \frac 32 \alpha_{v_2} + \alpha_{v_3}$,
and hence Eq. (<ref>) becomes
$\sum_{x\in \{u, v\}\cup F_3(u)\cup F_3(v)}\omega'(x)
\le 6 - 1 - (2\alpha_{u_1} + \frac 32 \alpha_{u_3} + \frac 32 \alpha_{v_2} + \alpha_{v_3} + \frac 52 \alpha_{u_2}) \le 5 - \frac {17}2\times \frac 35 = - \frac 1{10}< 0$.

Case 3. $m_3(u v) = 2$.
In this case, $u_2 = v_1$ and $u_3 = v_3$.
Let $\Gamma(u/v \rightarrow u_3)$ be the total weight transferred from $u$, $v$ and the $3$-faces in $F(u u_2)\cup F(v u_3)$ to $u_3$.
We have the following similar to Eq. (<ref>):

\begin{equation}
\label{eq09}
\begin{aligned}
\Gamma(u/v \rightarrow u_3) = \begin{cases}
3\alpha_{u_3},				&\mbox{ if } d(f_4) = d(h_2) = 3;\\
2\alpha_{u_3},				&\mbox{ if } d(f_4), d(h_2)\ge 4;\\
\frac 52 \alpha_{u_3},	&\mbox{ otherwise}.\end{cases}
\end{aligned}
\end{equation}

By (1), $\Gamma(u \rightarrow u_1)= (1 + \frac{1}{2}m_3(uu_1))\alpha_{u_1}$, $\Gamma(u \rightarrow v_2) = (1 + \frac 12 m_3(v v_2))\alpha_{v_2}$.
Since $m_3(u) + m_3(v) - m_3(u v) = m_3(u u_1) + 2 + m_3(v v_2) + 2 - 2 =  m_3(u u_1) + 2 + m_3(v v_2)$, we have

\begin{equation}
\label{eq10}
\begin{aligned}
&\sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)
= m_3(u) + m_3(v) -2 - \sum_{x\in \{u_1, v_2\}}\Gamma(u \rightarrow x) - \sum_{y\in \{u_2, u_3\}}\Gamma(u/v \rightarrow y)\\
%& = m_3(u u_1) + 2 + m_3(v v_2) - \Gamma(u/v \rightarrow u_2) - \Gamma(u/v \rightarrow u_3)- (1 + \frac 12 m_3(u u_1))\alpha_{u_1} - (1 + \frac 12 m_3(v v_2))\alpha_{v_2}- \\
= \	&(1-\frac 12 \alpha_{u_1})\times m_3(u u_1) + (1-\frac 12 \alpha_{v_2})\times m_3(v v_2) + \alpha_{u_1} + \alpha_{v_2}
+ 2 - \sum_{y\in \{u_2, u_3\}}\Gamma(u/v \rightarrow y)
\end{aligned}
\end{equation}

When $m_3(u u_1) + m_3(v v_2)\le 2$, Eq. (<ref>) becomes
$\sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)
\le (1-\frac 12 \alpha_{u_1})\times m_3(u u_1) + (1-\frac 12 \alpha_{v_2})\times m_3(v v_2) + 2 - \alpha_{u_1} - \alpha_{v_2}- 2\alpha_{u_2}  - 2\alpha_{u_3}
\le (1 - \frac 12 \times \frac 35)\times 2 + 2 - 6\times \frac 35 = - \frac 15< 0$.

In the other case, $3\le m_3(u u_1) + m_3(v v_2) \le 4$.
It follows that $\max\{m_3(u u_1), m_3(v v_2)\} \ge 2$.
Assume w.l.o.g. $m_3(u u_1) = 2$, $m_3(v v_2) \ge 1$ with $d(h_1) = 3$.
Then $\Gamma(u/v \rightarrow u_2) = 3\alpha_{u_2}$.
One sees that if $d(h_2)\ge 4$, $\Gamma(u/v \rightarrow u_3) = \frac 52 \alpha_{u_3}$;
if $d(h_2)= 3$, $\Gamma(u/v \rightarrow u_3) = 3\alpha_{u_3}$, i.e., $\Gamma(u/v \rightarrow u_3) = (\frac 52 + \frac 12 \times (m_3(v v_2) - 1))\alpha_{u_3}$.
Hence, Eq. (<ref>) becomes

\begin{equation*}
\begin{aligned}
&\ \sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)\\
\le 	&\ (1-\frac 12 \alpha_{u_1})\times 2 + (1-\frac 12 \alpha_{v_2})\times m_3(v v_2) + 2 - \alpha_{u_1} - \alpha_{v_2}
- 3\alpha_{u_2} - (\frac{5}{2} + \frac{1}{2}\times (m_3(vv_2) - 1))\times \alpha_{u_3}\\
= 		&\ 4 - 2\alpha_{u_1}- \alpha_{v_2} - 3\alpha_{u_2} - 2\alpha_{u_3} + (1 - \frac 12 \alpha_{v_2} - \frac 12 \alpha_{u_3}) m_3(v v_2)\\
\le 	&\ 4 - 8\times \frac 35 + (1 - \frac 35)m_3(v v_2)= - \frac 45 + (1 - \frac 35)m_3(v v_2)\\
=		&\ - \frac 45 + \frac 25 \times 2 = 0.
\end{aligned}
\end{equation*}

To prove (3), we have

\[
\sum_{x\in \{u\}\cup F_3(u)}\omega(x) = \omega(u) + \sum_{x\in F_3(u)}\omega(x) = \sum_{x\in F_3(u)} 1 = m_3(u) = m_3(uu_1) + m_3(uv)= m_3(uu_2) + m_3(uu_3).
\]

Using (R1), (R3.1), (R3.2), (R4.1) and (R4.2), we discuss the following two cases for the minimum degree of vertices in $N_H(u)$.

Case 1. For each $x\in N_H(u)$, $d_H(x)\ge 6$.

In this case, for each $x\in \{u_1, u_2, u_3\}$, $\alpha_x\ge \frac 13$,
and by (1), $\Gamma(u \rightarrow x) = (1 + \frac 12 m_3(ux))\alpha_{x}$,
$\Gamma(u \rightarrow v) = (1 + \frac 12 m_3(uv))\alpha_{v}$,
and thus

\begin{equation}
\label{eq11}
\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) = m_3(u)- \sum_{x\in N_H(x)}\Gamma(u \rightarrow x) = m_3(u)- \sum_{x\in N_H(x)}(1 + \frac 12 m_3(ux))\alpha_{x}.
\end{equation}

One sees from Lemma <ref>(1) and (4) that if $\alpha_x < \frac 12$, then $d(x) = d_H(x)\in [6, 7]$;
furthermore, by the definition of $\alpha_x$ in Eq. (<ref>), either $d(x) = d_H(x) = 6$, or $x$ is $(7, 4^-)_{H G}$.
We separate the discussion into the following three scenarios:

Case 1.1. For each $x\in N_H(x)$, $\alpha_x \ge \frac 12$.
When $m_3(u) = m_3(u u_1) + m_3(u v) \le 2$, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
\le m_3(u u_1) + m_3(u v) - (1 + \frac 12 m_3(u u_1))\alpha_{u_1} - \alpha_{u_2}- \alpha_{u_3} - (1 + \frac 12 m_3(u v))\alpha_{v} \le -\frac 12< 0$.

In the other case,  $3\le m_3(u) = m_3(u u_1) + m_3(u v)\le 4$, and we assume w.l.o.g. $m_3(u u_1) = 2$;
it follows that $m_3(u v)\ge 1$ and $d(f_2) = 3$.
Furthermore, we have $m_3(u u_2)= 2$, $m_3(u u_3)\ge 1$, and thus $m_3(u) = 3 + m_3(u u_3) - 1 = m_3(u u_3) + 2$ and
$(1 + \frac 12 m_3(u v))\alpha_{v} = (1 + \frac 12 m_3(u u_3))\alpha_{v}$.
This way, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= m_3(u u_3) + 2 - 2\alpha_{u_1} - 2\alpha_{u_2}- (1 + \frac 12 m_3(u u_3))\alpha_{u_3} - (1 + \frac 12 m_3(u u_3))\alpha_{v} \le 0$.

Case 1.2. $d(v) = d_H(v) = 6$ and $d_H(x)\ge 6$ for each $x\in \{u_1, u_2, u_3\}$.

Recall that for each $x\in N_H(x)$, $\Gamma(u \rightarrow x)= (1 + \frac 12 m_3(u x))\alpha_{x}$ and $\alpha_x\ge \frac 13$.
When $m_3(u)\le 1$, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 1- \sum_{x\in N_H(x)}\alpha_x = 1 - 4\times \frac 13 = - \frac 13< 0$.

In the other case, $m_3(u) \ge 2$.
If $m_3(u v) = 0$, i.e., $d(f_1) = d(f_4) = 3$, then $m_3(u) = 2$, and thus Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 2 - (2\alpha_{u_1} +  \frac 32 \alpha_{u_2} + \frac 32 \alpha_{u_3} + \alpha_v) = 2 - 6\times \frac 13 = 0$.
If $m_3(uv) \ge 1$ (and thus $d(f_2)= 3$), it follows from Lemma <ref>(2.3) that for each $x\in \{u_1, u_2, u_3\}$, $d_H(x) \ge 8$ and $d(x)\ge 8$;
and thus, by Lemma <ref>(1) and (4), $\alpha_x\ge \frac 12$.
Therefore, when $m_3(u) = 2$, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
\le 2 - (\alpha_{u_1} + \frac 32 \alpha_{u_2} + \alpha_{u_3} + \frac 32 \alpha_v) = 2 - \frac 72\times \frac 12 - \frac 32 \times \frac 13 = -\frac 14< 0$.
In the other case, $m_3(u) \ge 3$.
If $m_3(u v) = 1$, i.e., $d(f_1) = d(f_4) = 3$, then $m_3(u) = 3$, and Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
\le 3 - (2\alpha_{u_1} +  2 \alpha_{u_2} + \frac 32 \alpha_{u_3} + \frac 32 \alpha_v) = 3 - \frac {11}2 \times \frac 12 - \frac 32 \times \frac 13  = - \frac 12< 0$.
If $m_3(u v) = 2$, then it follows from Lemma <ref>(2.5) that for each $x\in \{u_1, u_2, u_3\}$, $d_H(x)\ge 8$ and $d(x)\ge 9$;
and thus, by Lemma <ref>(1) and (4), $\alpha_x\ge \frac 59$.
Since $m_3(u) \ge 3$, assuming w.l.o.g. $d(f_1) = 3$, we have $m_3(u) = m_3(u u_3) + 2 = m_3(u u_1) + 2$, and Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) = m_3(u u_1) + 2 - ((1 + \frac 12 m_3(u u_1))\alpha_{u_1} + 2\alpha_{u_2} + (1 + \frac 12 m_3(u u_1))\alpha_{u_3} + 2\alpha_v)
\le 2 +  \frac 49 \times m_3(u u_1) - \frac{26}{9}
\le 2 +  \frac 49 \times 2 - \frac{26}{9} = 0$.

Case 1.3. $v$ is $(7, 4^-)$ and $d_H(x)\ge 7$ for each $x\in \{u_1, u_2, u_3\}$.

Recall that for each $x\in N_H(x)$, $\Gamma(u \rightarrow x)= (1 + \frac 12 m_3(u x))\alpha_{x}$ and $\alpha_x\ge \frac 37$.
When $m_3(u)\le 1$, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 1- \sum_{x\in N_H(x)}\alpha_x = 1 - 4\times \frac 37 = - \frac 57< 0$.

In the other case, $m_3(u)\ge 2$.
If $m_3(uv) = 0$, i.e., $d(f_1) = d(f_4) = 3$, then $m_3(u) = 2$, and Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 2 - (2\alpha_{u_1} +  \frac 32 \alpha_{u_2} + \frac 32 \alpha_{u_3} + \alpha_v) = 2 - 6\times \frac 37 = -\frac 47< 0$.
If $m_3(uv) \ge 1$ (and thus $d(f_2) = 3$), then when $m_3(u) \le 2$, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 2 - (\alpha_{u_1} + \frac 32 \alpha_{u_2} + \alpha_{u_3} + \frac 32 \alpha_v) = 2 - 5\times \frac 37 = -\frac 17< 0$;
when $m_3(u) = 3$, no matter which two of $\{f_1, f_4, f_3\}$ are $3$-faces, Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 3 - 7\times \frac 37 = 0$;
when $m_3(u) = 4$, it follows from Lemma <ref>(2.6) that for each $x\in \{u_1, u_2, u_3\}$, $d_H(x)\ge 8$ and $d(x)\ge 9$,
and thus, by Lemma <ref>(1) and (4), $\alpha_x\ge \frac 59$, and Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) = 4 - (2\alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3} + 2\alpha_v) \le 4 - 6\times \frac 59 - 2\times \frac 37 = \frac 4 {21}< 0$.

Case 2. For some $x\in N_H(u)$, $d(x) = d_H(x)= 5$, assuming w.l.o.g. $d(v) = d_H(v) = 5$.

In this case, for each $x\in \{u_1, u_2, u_3\}$, it follows from Lemma <ref> (2.2) that $d_H(x) \ge 8$ and $d_(x) \ge 9$;
and then by Lemma <ref>(1) and (4), $\alpha_x\ge \frac 59$.
Recall that for each $x\in \{u_1, u_2, u_3\}$, $\Gamma(u \rightarrow x) = (1 + \frac 12 m_3(u x))\alpha_{x}$,
and if $m_3(u) \ne 4$ then $\Gamma(u \rightarrow v)= 0$.
When $m_3(u) \le 1$, Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) \le 1- \sum_{x\in N_H(x)}\alpha_x = 1 - 3\times \frac 59 = - \frac 23< 0$.

In the other case, $m_3(u) \ge 2$.
If $m_3(uv) = 0$, assuming w.l.o.g. $d(f_1) = 3$,
then $m_3(u) = m_3(u u_1) = m_3(u u_3) + 1$, $\Gamma(u \rightarrow u_2) = \frac 32 \alpha_{u_2}$,
$\Gamma(u \rightarrow u_1)= (1 + \frac 12 m_3(uu_1))\alpha_{u_1}$,
$\Gamma(u \rightarrow u_3)= (1 + \frac 12 (m_3(uu_1) - 1))\alpha_{u_3}$,
and thus Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) = m_3(uu_1) - (\frac 32 \alpha_{u_2} +  (1 + \frac 12 m_3(uu_1))\alpha_{u_1} + (1 + \frac 12 (m_3(uu_1) - 1))\alpha_{u_3})
= (1 - \frac 12\alpha_{u_1} - \frac 12 \alpha_{u_3})m_3(uu_1) - (\frac 32 \alpha_{u_2} + \alpha_{u_1} + \frac 12 \alpha_{u_3})
\le (1 - \frac 59)\times m_3(uu_1) - 3\times \frac 59 = -\frac 79 < 0$.

If $m_3(uv) \ge 1$ (and thus $d(f_2)= 3$), then when $m_3(u v)\le 1$, at least one of $\{f_1, f_4\}$ is a $3$-face,
and we discuss the following three subcases for three possible values of $m_3(u)$:

* $d(f_1) = d(f_4) = 3$.
Then Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 3 - (2\alpha_{u_1} + 2\alpha_{u_2} + \frac 32\alpha_{u_3}) \le 3 - \frac {11} 2 \times \frac 59 = - \frac 1{18}< 0$.

* $d(f_1) = 3$ and $d(f_4)\ge 4$.
Then Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 2 - (\frac 32 \alpha_{u_1} + 2\alpha_{u_2} + \alpha_{u_3}) \le 2 - \frac {9} 2\times \frac 59 = - \frac 12< 0$.

* $d(f_1) \ge 4$ and $d(f_4) = 3$.
Then Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 2 - (\frac 32 \alpha_{u_1} + \frac 32 \alpha_{u_2} + \frac 32 \alpha_{u_3}) \le 2 - \frac {9} 2\times \frac 59 < 0$.

When $m_3(u v) = 2$, i.e., $d(f_3) = d(f_2) = 3$, it follows from Lemma <ref>(2.4) that for each $x\in \{u_1, u_2, u_3\}$, $d_H(x)\ge 8$ and $d(x)\ge 10$;
and thus, by Lemma <ref>(1) and (4), $\alpha_x\ge \frac 35$.
We discuss the following three subcases for three possible values of $m_3(u)$.

* $m_3(u) = 2$, i.e., $d(f_1)\ge 4$ and $d(f_4)\ge 4$.
Then Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 2 - (\alpha_{u_1} + \frac 32\alpha_{u_2} + \frac 32\alpha_{u_3}) \le 2 - 4\times \frac 35 = - \frac 25< 0$.

* $m_3(u) = 3$, and assuming w.l.o.g. $d(f_1) = 3$ and $d(f_4)\ge 4$.
Then Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 3 - (\frac 32 \alpha_{u_1} + 2\alpha_{u_2} + \frac 32 \alpha_{u_3}) \le 3 - 5\times \frac 35 = 0$.

* $m_3(u) = 4$, i.e., $d(f_1) = d(f_4) = 3$.
We can have the following three scenarios for three possible values of $d((f_2)_{u_2v})$ and $d((f_3)_{u_3v})$:

* $d((f_2)_{u_2 v}) \ge 4$ and $d((f_3)_{u_3 v}) \ge 4$.
Then Eq. (<ref>) becomes $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 4 - (2 \alpha_{u_1} + \frac 52 \alpha_{u_2} + \frac 52 \alpha_{u_3}) \le 4 - 7 \times \frac 35 = - \frac 15< 0$.

* $\min\{d((f_2)_{u_2 v}), d((f_3)_{u_3 v})\} = 3$ and $\max\{d(f_{v w}), d((f_{u v})_{v w^*})\}\ge 4$,
and assuming w.l.o.g. that $d((f_2)_{u_2 v}) = 3$ and $d((f_3)_{u_3 v}) \ge 4$.
Then by (R4.1), $\tau(F(uv) \rightarrow v) = \frac 1{10}$, and thus Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)
= 4 - (2 \alpha_{u_1} + 2\alpha_{u_2} + \frac 52 \alpha_{u_3} + \frac 1{10}) \le 4 - \frac {13}2 \times \frac 35 - \frac 1{10}= 0$.

* $d((f_2)_{u_2 v}) = 3$ and $d((f_3)_{u_3 v}) = 3$.
Then by (R4.2), $\tau(F(u v) \rightarrow v) = \frac 25$, and thus Eq. (<ref>) becomes
$\sum_{x\in \{u\}\cup F_3(u)}\omega'(x) = 4 - (2 \alpha_{u_1} + 2\alpha_{u_2} + 2\alpha_{u_3} + \frac 2{5}) \le 4 - 6 \times \frac 35 - \frac 25 = 0$.

This finishes the proof of the lemma.

Let $V_{34} = V_3(H)\cup V_4(H)$ and $F_{0}= \bigcup_{v\in V_{34}(H)}F_3(v)$. Then
$\sum_{x\in V_{34}\cup F_{0}}\omega'(x)\le 0. $

It holds naturally by Lemma <ref>, Discharging-Lemmas <ref> and <ref>.

For every $3$-face $f = [u v w]\in F(H)$ with $\delta(f)\ge 5$, $\omega'(f) \le 0$.

When $\delta(f)\ge 6$, by (R2), we have for each $x\in V(f)$, $\tau(f \rightarrow v)_f = \frac 13$, and thus
$\omega'(f) = \omega(f) - \sum_{x\in \{u, v, w\}} \tau(f \rightarrow x)_f = 4 - 3 - 3\times \frac 13 = 0$.

When $\delta(f) = 5$, i.e, there exists $x\in \{u, v, w\}$ such that $d(x) = d_H(x) = 5$, and assuming w.l.o.g. $d(u) = d_H(u) = 5$.
If $d_H(v)\ge 6$, $d_H(w)\ge 6$, and $\alpha_v + \alpha_w\ge 1$,
then by (R3.1), (R3.2), and (R5.2), $\tau(f \rightarrow v)_f = \alpha_v$, $\tau(f \rightarrow w)_f = \alpha_w$,
$\tau(u \rightarrow f) = \alpha_v + \alpha_w - 1$,
and thus
$\omega'(f) = \omega(f) + \tau(u \rightarrow f) - \sum_{x\in \{v, w\}} \tau(f \rightarrow x)_f \le 4 - 3 + (\alpha_v + \alpha_w - 1) - (\alpha_v + \alpha_w) = 0$.
Otherwise, by (R3.1), (R3.2) and (R5.1), the total weight transferred from $f$ to $u$, $v$, $w$ is at least $1$,
and thus $\omega'(f) \le \omega(f) - 1 = 4 - 3 - 1 = 0$.

One sees that for each $5$-vertex $x$ in $H$, $\omega(x) = 4 - 5 = -1$, and $d(x) = d_H(x) = 5$.
It remains to validate that $\omega'(u)\le 0$ for every $u\in V(H)$ with $d_H(u) = 5$.

In the sequel we assume that $d(u) = d_H(u) = 5$.
Before validating $\omega'(u)$, using (R3.1), (R3.2), and (R5.1),
we characterize of the weight $\tau(f \rightarrow u)$ transferred from an incident $3$-face $f = [u v w]$ to $u$ in Discharging-Lemma <ref>,
also depicted in Figure <ref>.

The weight $\tau(f \rightarrow u)$ transferred from an incident $3$-face $f = [u v w]$ to $u$.

For every $5$-vertex $u$ in $V(H)$,
let $f = [u v w]$ be a $3$-face,
if $d_H(v)\ge 6$, $d_H(w)\ge 6$, and $\alpha_v + \alpha_w\ge 1$,
then $\tau(u \rightarrow f) = \alpha_v + \alpha_w - 1$;
otherwise, $\tau(f \rightarrow u)\le \frac 13$, with the following details:

if $d_H(v) = d_H(w) = 5$, then $\tau(f \rightarrow u) = \frac 13$;

if $d_H(v)= 5$ and $d_H(w)\ge 6$, then
$$\tau(f \rightarrow u) =\begin{cases}
\frac {1 - \alpha_w}2, 				&\mbox{ if } d(f_{uw}) = d(f_{vw}) = 3, \mbox{ and } \alpha_w\le 1;\\
\frac {1 - \frac 32 \alpha_w}2,	&\mbox{ if } 3\in \{d(f_{uw}), d(f_{vw})\}, \max\{d(f_{uw}), d(f_{vw})\}\ge 4,  \mbox{ and } \alpha_w\le \frac 23;\\
\frac {1 - 2\alpha_w}2,				&\mbox{ if } d(f_{uw})\ge 4, d(f_{vw})\ge 4, \mbox{ and } \alpha_w\le \frac 12;\end{cases}$$

if $d_H(v)\ge 6$, and $d_H(w)\ge 6$, then
$$\tau(f \rightarrow u) =\begin{cases}
1 - \alpha_v - \alpha_w,				&\mbox{ if } d(f_{uv}) = d(f_{uw}) = 3, \mbox{ and } \alpha_v + \alpha_w\le 1;\\
1 - \frac 32\alpha_v - \alpha_w, &\mbox{ if } d(f_{uv})\ge 4, d(f_{uw}) = 3, \mbox{ and } \frac 32\alpha_v + \alpha_w\le 1;\\
1 - \alpha_v - \frac 32\alpha_w, &\mbox{ if } d(f_{uv})= 3, d(f_{uw})\ge 4, \mbox{ and } \alpha_v + \frac 32 \alpha_w\le 1;\\
1 - \frac 32\alpha_v - \frac 32\alpha_w = 0, &\mbox{ if } d(f_{uv})\ge 4, d(f_{uw})\ge 4, \mbox{ and } \alpha_v = \alpha_w = \frac 13.\end{cases}$$

When $d_H(v)\ge 6$, $d_H(w)\ge 6$, and $\alpha_v + \alpha_w\ge 1$,
then it follows from (R5.2) clearly that $\tau(u \rightarrow f) = \alpha_v + \alpha_w - 1$.
In the other case, either $\min \{d_H(v), d_H(w)\} = 5$,
or $d_H(v)\ge 6$, $d_H(w)\ge 6$ and $\alpha_v + \alpha_w < 1$.
We assume that after applying (R3), the face $f$ still has charge $\beta \ge 0$.
Using (R3.1), (R3.2), and (R5.1), we discuss the following three subcases for three possible values of $n_5(f)$.

Case 1. $n_5(f) = 3$, i.e, $d_H(v) = d_H(w) = 5$.

In this case, no weight transferred from $f$ to $u$, $v$, or $w$ by (R3.1) and (R3.2).
It follows that $\beta = 1$, and $\tau(f \rightarrow u) = \frac \beta 3 = \frac 13$.

Case 2. $n_5(f) = 2$, and assuming w.l.o.g. $d_H(v)= 5$ and $d_H(w)\ge 6$.

In this case, $\tau(f \rightarrow w)_f = \alpha_w$ and $\tau(f \rightarrow w)_{f_{uw}} \le \frac 12\alpha_w$,
$\tau(f \rightarrow w)_{f_{vw}} \le \frac 12\alpha_w$.
Recall that $\alpha_w\ge \frac 13$.

* $d(f_{uw}) = d(f_{vw}) = 3$.
It follows that $\beta = 1 - \alpha_w$.
If $\alpha_w\le 1$, then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \frac \beta 2 = \frac {1 - \alpha_w}2 \le \frac {1 - \frac 13}2 = \frac 13$.

* $3\in \{d(f_{uw}), d(f_{vw})\}$ and $\max\{d(f_{uw}), d(f_{vw})\}\ge 4$.
It follows that $\beta = 1 - \alpha_w - \frac 12 \alpha_w = 1 - \frac 32 \alpha_w$.
If $\frac 32 \alpha_w \le 1$, i.e., $\alpha_w\le \frac 23$,
then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \frac \beta 2 = \frac {1 - \frac 32 \alpha_w}2 \le \frac {1 - \frac 32 \times \frac 13}2 = \frac 14$.

* $d(f_{uw})\ge 4$, and $d(f_{vw})\ge 4$.
It follows that $\beta = 1 - \alpha_w - \frac 12 \alpha_w - \frac 12 \alpha_w = 1 - 2 \alpha_w$.
If $2 \alpha_w \le 1$, i.e., $\alpha_w\le \frac 12$,
then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \frac \beta 2 = \frac {1 - 2 \alpha_w}2 \le \frac {1 - 2\times \frac 13}2 = \frac 16$.

Case 3. $n_5(f) = 1$, i.e., $d_H(v)\ge 6$ and $d_H(w)\ge 6$.

In this case, $\tau(f \rightarrow v)_f = \alpha_v$, $\tau(f \rightarrow w)_f = \alpha_w$,
and $\tau(f \rightarrow w)_{f_{uv}} \le \frac 12\alpha_v$,
$\tau(f \rightarrow w)_{f_{uw}} \le \frac 12\alpha_w$.
Recall that $\alpha_v\ge \frac 13$ and $\alpha_w\ge \frac 13$.

* $d(f_{uw}) = d(f_{vw}) = 3$.
It follows that $\beta = 1 - \alpha_v - \alpha_w$.
If $\alpha_v + \alpha_w\le 1$, then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \beta  = 1 - \alpha_v - \alpha_w \le 1 - \frac 13 - \frac 13 = \frac 13$.

* $d(f_{uv})\ge 4$, $d(f_{uw}) = 3$.
It follows that $\beta = 1 - \frac 32 \alpha_v - \alpha_w$.
If $\frac 32\alpha_v + \alpha_w \le 1$,
then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \beta  = 1 - \frac 32 \alpha_v - \alpha_w \le 1 - \frac 32 \times \frac 13 - \frac 13 = \frac 16$.

* $d(f_{uv}) = 3$, $d(f_{uw}) \ge 4$.
It follows that $\beta = 1 - \alpha_v - \frac 32 \alpha_w$.
If $\alpha_v + \frac 32 \alpha_w \le 1$,
then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \beta  = 1 - \alpha_v - \frac 32 \alpha_w \le 1 - \frac 13 - \frac 32 \times \frac 13 = \frac 16$.

* $d(f_{uv}) \ge $, $d(f_{uw}) \ge 4$.
It follows that $\beta = 1 - \frac 32 \alpha_v - \frac 32 \alpha_w$.
If $\frac 32 \alpha_v - \frac 32 \alpha_w\le 1$, i.e., $\alpha_v = \alpha_w = \frac 13$,
then $\beta\ge 0$ and thus $\tau(f \rightarrow u) = \beta  = 1 - \frac 32 \alpha_v - \frac 32 \alpha_w = 1 -\frac 32 \times \frac 13 - \frac 32 \times \frac 13 = 0$.

This finishes the proof of the lemma that $\tau(f \rightarrow u) \le \frac 13$.

Discharging-Lemma <ref> tells that the value of $\tau(f \rightarrow u)$ depends on $n_5(f)$ and some other factors.
One sees that for any $x$ with $d_H(x) = k\ge 6$, $\alpha_x \ge 1 - \frac 4k$.
Hence, we have the following lemma to upper bound $\tau(f \rightarrow u)$.

For every $5$-vertex $u$ in $V(H)$, let $f = [u v w]$ be a $3$-face, and then we have the following:

If $d_H(v)= 5$ and $d_H(w) = k \ge 6$, then
$$\tau(f \rightarrow u) =\begin{cases}
\frac {1 - \alpha_w}2\le \frac 2k,									&\mbox{ if } d(f_{uw}) = d(f_{vw}) = 3, \mbox{ and } \alpha_w\le 1;\\
\frac {1 - \frac 32 \alpha_w}2\le \frac 3k - \frac 14,	&\mbox{ if } 3\in \{d(f_{uw}), d(f_{vw})\}, \max\{d(f_{uw}), d(f_{vw})\}\ge 4,  \mbox{ and } k\le 12;\\
\frac {1 - 2\alpha_w}2\le \frac 4k - \frac 12,				&\mbox{ if } d(f_{uw})\ge 4, d(f_{vw})\ge 4, \mbox{ and } k\le 8.\end{cases}$$

If $d_H(v) = k_1\ge 6$ and $d_H(w) = k _2\ge 6$, then
$$\tau(f \rightarrow u) =\begin{cases}
1 - \alpha_v - \alpha_w\le \frac 4{k_1} + \frac 4{k_2} - 1,
&\mbox{ if } d(f_{uv}) = d(f_{uw}) = 3, \mbox{ and } \frac 4{k_1} + \frac 4{k_2} - 1\ge 0;\\
1 - \frac 32\alpha_v - \alpha_w\le \frac 6{k_1} + \frac 4{k_2} - \frac 32,
&\mbox{ if } d(f_{uv})\ge 4, d(f_{uw}) = 3, \mbox{ and } \frac 6{k_1} + \frac 4{k_2} - \frac 32\ge 0;\\
1 - \alpha_v - \frac 32\alpha_w\le \frac 4{k_1} + \frac 6{k_2} - \frac 32,
&\mbox{ if } d(f_{uv})= 3, d(f_{uw})\ge 4, \mbox{ and } \frac 4{k_1} + \frac 6{k_2} - \frac 32\ge 0;\\
1 - \frac 32\alpha_v - \frac 32\alpha_w\le \frac 6{k_1} + \frac 6{k_2} - 2 = 0,
&\mbox{ if } d(f_{uv})\ge 4, d(f_{uw})\ge 4, \mbox{ and } k_1 = k_2 = 6.\end{cases}$$

Some special values for the upper bounds $\frac 2k$, $\frac 3k - \frac 14$, and $\frac 4k - \frac 12$, where $k \ge 6$.

Some special values for the upper bounds $\frac 4{k_1} + \frac 4{k_2} - 1$ and $\frac 6{k_1} + \frac 4{k_2} - 1$, where $k_1, k_2 \ge 6$.

For convenience, for some small values of $k, k_1, k_2 \ge 6$,
we list the upper bounds stated in Discharging-Lemma <ref> in the two tables depicted in Figure <ref> and Figure <ref>.
In the sequel, we will directly refer to these values.
Let $u_1, u_2, u_3, u_4, u_5$ denote all the neighbors of $u$ in clockwise order,
and denote the face containing $u_1 u u_2$ ($u_2 u u_3$, $u_3 u u_4$, $u_4 u u_5$, $u_5 u u_1$, respectively) as $f_1$ ($f_2, f_3, f_4$, $f_5$ respectively).
Recall that for each $x\in N_H(u)$, $d_H(x)\ge 4$.

For a $3$-face $f = [v u w]$ incident to $u$ with $d_H(v)\ge 6$, $d_H(w)\ge 6$, and $\alpha_v + \alpha_w\ge 1$,
by (R5.2), $\tau(u \rightarrow f) = \alpha_v + \alpha_w - 1\ge 0$.
Then $\tau(f \rightarrow u) = - \tau(u \rightarrow f)\le 0$.

\begin{equation}
\label{eq12}
\begin{aligned}
\omega'(u) = \omega(u) + \sum_{f\in F_3(v)} \tau(f \rightarrow u)
= 4 - 5 + \sum_{f\in F_3(v)} \tau(f \rightarrow u)
= -1 + \sum_{f\in F_3(v)} \tau(f \rightarrow u).
\end{aligned}
\end{equation}

We validate that $\omega'(u) \le 0$ for each $5$-vertex $u$, using Eq. (<ref>), in the following two lemmas.

For every $5$-vertex $u$ adjacent with a $4$-vertex $u_1$ in $V(H)$, $\omega'(u) \le 0$.

One sees that for any $4$-vertex $x\in N_H(u)$, $d(x) = d_H(x) = 4$ and
the average weight transferred from each $3$-face in $F(ux)$ to $u$ is at most $\max\{\frac 12\times \frac 1{10}, \frac 12\times \frac 25\} = \frac 15$.
Thus, together with Discharging-Lemma <ref>,
the average weight transferred from each $3$-face in $F_3(u)$ to $u$ is at most $\frac 13$.

When $m_3(u_1) \ne 4$, or $m_3(u_1) = 4$ and $\min\{d(f_2), d(f_4)\}\ge 4$, no weight is transferred from any face in $F(uu_1)$ to $u$;
since the average weight transferred from each $3$-face in $\{f_2, f_3, f_4\}$ is at most $\frac 13$, Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in \{f_2, f_3, f_4\}} \tau(f \rightarrow u) \le -1 + 3\times \frac 13 = 0$.

When $m_3(u_1) = 4$ and $\min\{d(f_2), d(f_4)\} = 3$, it follows from Lemma <ref> (2.4) that for each $x\in \{u_2, u_5\}$, $d_H(x)\ge 8$ and $d(x)\ge 10$;
thus, by Lemma <ref>(1) and (4), $\alpha_x\ge \frac 35$.
Using (R3), (R4), (R5) and Discharging-Lemma <ref>, we discuss the following three subcases on $d(f_i)$, $i\in \{2, 3, 4\}$.

$\max\{d(f_2), d(f_4)\}\ge 4$, and assuming w.l.o.g. $d(f_2) = 3$ and $d(f_4)\ge 4$.
In this case, $\tau(F(uu_1) \rightarrow u) = \frac 1{10}$.
Since the average weight transferred from each $3$-face in $\{f_2, f_3\}$ is at most $\frac 13$, Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in \{f_1, f_5\}} \tau(f \rightarrow u) \le -1 + \frac 1{10} + 2\times \frac 13 = - \frac 7{30}< 0$.

$d(f_2) = d(f_4) = 3$ and $d(f_3)\ge 4$.
In this case, $\tau(F(uu_1) \rightarrow u) = \frac 25$.
One sees that if $d_H(u_3) = 4$, no weight will be transferred from $f_2$ to $u$;
if $d_H(u_3)= 5$, then $\tau(f_2 \rightarrow u)\le \frac {1 - \alpha_{u_2}} 2 \le \frac {1 - \frac 35} 2 = \frac 15$;
if $d_H(u_3)\ge 6$, since $\frac 32 \times \alpha_{u_3} + \alpha_{u_2}\ge \frac 32 \times \frac 13 + \alpha_{u_2}\ge  \frac 12 + \frac 35 = \frac {11}{10}$,
then we have $\tau(f_2 \rightarrow u)\le 0$.
It follows that $\tau(f_2 \rightarrow u)\le \frac 15$; and likewise, $\tau(f_4 \rightarrow u)\le \frac 15$.
Hence, Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in \{f_1, f_5\}} \tau(f \rightarrow u) + \sum_{f\in \{f_2, f_4\}} \tau(f \rightarrow u) \le -1 + \frac 25 + 2\times \frac 15= - \frac 15< 0$.

$d(f_3) = d(f_2) = d(f_4) = 3$.
In this case, $\tau(F(uu_1) \rightarrow u) = \frac 25$.

* $\min\{d_H(u_3), d_H(u_4)\} = 4$, and assuming w.l.o.g. $d(u_3) = d_H(u_3) = 4$.
It follows that $d_H(u_4)\ge 8$ and $d(u_4)\ge 10$,
and then $\tau(F(uu_3) \rightarrow u) \le \frac 25$ and $\tau(u \rightarrow f_4) = \alpha_{u_5} + \alpha_{u_4} - 1\ge 2\times \frac 35 - 1 = \frac 15$,
i.e., $\tau(f_4 \rightarrow u) \le -\frac 15$.
Hence, Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in \{f_1, f_5\}} \tau(f \rightarrow u) + \sum_{f\in \{f_2, f_3\}} \tau(f \rightarrow u) + \tau(f_4 \rightarrow u)
\le - 1 + 2\times \frac 25  - \frac 15 = -\frac 25 < 0$.

* $\min\{d_H(u_3), d_H(u_4)\} = 5$, and assuming w.l.o.g. $d(u_3) = d_H(u_3) = 5$.
The non-existence of ($A_{7.1}$) in $G$ states that $d_H(u_4)\ge 7$.
Then we have $\tau(f_2 \rightarrow u)\le \frac {1 - \alpha_{u_2}}2 \le  \frac {1 - \frac 35}2 = \frac 15$,
$\tau(f_3 \rightarrow u)\le \frac {1 - \alpha_{u_4}}2 \le  \frac {1 - \frac 37}2 = \frac 27$,
and since $\alpha_{u_5} + \alpha_{u_4}\ge \frac 35 + \frac 37 = \frac {36}{35}$,
$\tau(u \rightarrow f_4) = \alpha_{u_5} + \alpha_{u_4} - 1 \ge \frac {36}{35} - 1 = \frac 1{35}$, i.e., $\tau(f_4 \rightarrow u)\le - \frac 1{35}$.
Hence, Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in \{f_1, f_5\}} \tau(f \rightarrow u) + \tau(f_2 \rightarrow u) + \tau(f_3 \rightarrow u) + \tau(f_4 \rightarrow u)
\le - 1 + \frac 25  + \frac 15 + \frac 27 - \frac 1{35}= -\frac 17 < 0$.

* $d_H(u_3) \ge 6$ and $d_H(u_4) \ge 6$.
Then we have $\tau(f_2 \rightarrow u)= 1 - \alpha_{u_2} - \alpha_{u_3} \le 1 - \frac 35 - \frac 13 = \frac 1{15}$,
$\tau(f_3 \rightarrow u)\le \frac 13$, and $\tau(f_4 \rightarrow u)= 1 - \alpha_{u_5} - \alpha_{u_4} \le 1 - \frac 35 - \frac 13 = \frac 1{15}$.
Hence, Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in \{f_1, f_5\}} \tau(f \rightarrow u) + \tau(f_2 \rightarrow u) + \tau(f_3 \rightarrow u) + \tau(f_4 \rightarrow u)
\le - 1 + \frac 25  + \frac 1{15} + \frac 13 + \frac 1{15} = -\frac 2{15} < 0$.

This proves the lemma.

For every $5$-vertex $u$ in $V(H)$, if $d_H(x)\ge 5$ for each $x\in N_H(u)$, then $\omega'(u) \le 0$.

It follows from Discharging-Lemma <ref> that for each face $f\in F_3(u)$, $\tau(f \rightarrow u)\le \frac 13$.
When $m_3(f)\le 3$, Eq. (<ref>) becomes $\omega'(u) \le -1 + 3\times \frac 13 = 0$.

When there exist two faces $f = [\ldots v_1 u w_1 \ldots\}$ and $h = [\ldots v_2 u w_2]$ such that for any $x\in \{v_1, w_1\}\cup \{v_2, w_2\}$, $\alpha_x\ge \frac 12$,
one sees that if $d(f)$ = 3, then $\tau(u \rightarrow f) = \alpha_{v_1} + \alpha_{w_1} - 1 \ge 2\times \frac 12 - 1\ge 0$, i.e., $\tau(f \rightarrow u)\le 0$.
It follows that $\tau(f \rightarrow u)\le 0$, and likewise $\tau(g \rightarrow u)\le 0$, and Eq. (<ref>) becomes
$\omega'(u) = -1 + \sum_{f\in F_3(v)\setminus \{f, h\}} \tau(f \rightarrow u)\le - 1 + 3\times \frac 13 = 0$.

If there exist $v, w\in N_H(u)$ such that $d_H(v)= 5$ and $d_H(w)\le 6$, i.e., $d(v) = 5$ and $d(w)\le 6$,
$f = [vuw]$ is a $3$-face and $m_3(uv) = 2$,
then the non-existence of ($2.1$), ($6.3$) and ($7.1$) in $G$ imply that for each $x\in N_H(u)\setminus \{v, w\}$, $d(x)\ge 8$,
and since $d(u)= 5$, we have $d_H(x)\ge 8$ and thus $\alpha_x\ge \frac 12$.

Recall that if $d_H(x)\ge 8$ or $x$ is $(7, 5^+)_H$, then $\alpha_x\ge \frac 12$;
if $d_H(x)\le 6$, then $d(x) = d_H(x)\le 6$;
and if $x$ is $(7, 4^-)_H$, then $x$ is $(7, 4^-)_{HG}$.
In the other case, $4\le m_3(f)\le 5$.
In the sequel, we continue the proof with the following propositions which summarize the above argument:

For each $v, w\in N_H(x)$ with $vw\in E(H)$ and $m_3(uv) = 2$, if $d_H(v)= 5$, i.e., $d(v) = 5$, then $d_H(w)\ge 7$ and $\alpha_{w}\ge \frac 37$; if $d_H(w)\le 6$, i.e., $d(w) \le 6$, then $d_H(v)\ge 6$ and $\alpha_{v}\ge \frac 13$;

If $\alpha_x < \frac 12$ for some $x$, then $d(x) = d_H(x)\le 6$ or $x$ is $(7, 4^-)_{HG}$.

Assume w.l.o.g. $d(f_i) = 3$, for any $i = 1, 2, 3, 4$, and $d(f_5)\ge 3$.
Using (R3.1), (R3.2), (R5.1), and (R5.2), we discuss the following two cases for two possible values of $m_3(u)$.

Case 1.  $m_3(u) = 4$, i.e., $d(f_5)\ge 4$.

In this case, Eq. (<ref>) becomes

\begin{equation}
\label{eq13}
\omega'(u) = -1 + \sum_{f\in \{f_1, f_2, f_3, f_4\}} \tau(f \rightarrow u).
\end{equation}

We can have the following three scenarios for three possible values of $d_H(u_1)$ and $d_H(u_5)$.

Case 1.1. $d_H(u_1)\ge 6$ and $d_H(u_5)\ge 6$.

(Case 1.1.1.) $d_H(u_2)\ge 6$ and $d_H(u_4)\ge 6$.
In this case, by Discharging-Lemmas <ref> and <ref>
[We do not calculate $\tau(f \rightarrow u)$ in details but refer to its upper bounds in the two tables in Figures <ref> and <ref>],
$\tau(f_1 \rightarrow u) = 1 - \frac 32\alpha_{u_1} - \alpha_{u_2} = 1 - \frac 32 \times \frac 13 - \frac 13 = 1 - \frac 56 = \frac 16 $,
and likewise $\tau(f_4\rightarrow u)\le \frac 16$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 16 + 2\times \frac 13 = 0$.

(Case 1.1.2.) $d_H(u_2) = d_H(u_4) = 5$, i.e.,  $d(u_2) = d(u_4) = 5$.
It follows from (I) that for each $i\in 1, 3, 5$, $\alpha_{u_i}\ge \frac 37$.
Then $\tau(f_1 \rightarrow u) = \frac {1 - \frac 32 \alpha_{u_1}} 2\le \frac {1 - \frac 32 \times \frac 37} 2\le \frac 5{28}$,
and $\tau(f_2\rightarrow u)\le \frac {1 - \alpha_{u_3}}2\le \frac {1 - \frac 37}2 = \frac 27$.
Likewise, $\tau(f_4 \rightarrow u) \le \frac 5{28}$, and $\tau(f_3\rightarrow u)\le \frac 27$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 5{28} + 2\times \frac 27 = -1 + \frac {13}{14} = -\frac 1{14}$.

(Case 1.1.3.) Assuming w.l.o.g. $d_H(u_2) = 5$ and $d_H(u_4)\ge 6$, i.e.,  $d(u_2)= 5$.
It follows from (I) that for each $i\in \{1, 3\}$, $\alpha_{u_i}\ge \frac 37$.
Then $\tau(f_1 \rightarrow u)\le \frac 5{28}$,
$\tau(f_2\rightarrow u)\le \frac {1 - \alpha_{u_3}}2\le \frac {1 - \frac 37}2 = \frac 27$,
$\tau(f_3\rightarrow u) = 1 - \alpha_{u_3} - \alpha_{u_4}\le 1 - \frac 13 - \frac 37 = \frac 5{21}$,
and $\tau(f_4\rightarrow u)\le \frac 16$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 5{28} + \frac 27 + \frac 5{21} + \frac 16 = - 1 + \frac {73}{84} = - \frac {11}{84}< 0$.

Case 1.2. $d_H(u_1) = d_H(u_5) = 5$, i.e., $d(u_1) = d(u_5) = 5$.

In this case, by (I), for each $i\in \{2, 4\}$, $d(u_i)\ge 6$ and $d_H(u_i)\ge 6$, $\alpha_{u_i}\ge \frac 13$.
When $\alpha_{u_3}\ge \frac 12$, $\tau(f_2\rightarrow u) = 1 - \alpha_{u_2} - \alpha_{u_3} \le \frac 16$,
and likewise $\tau(f_3\rightarrow u)\le \frac 12$, and thus Eq. (<ref>) becomes
$\omega'(u) \le -1 + 2\times \frac 16 + 2\times \frac 13 = - 1 + \frac 56 = -\frac 16< 0$.

In the other case, $\alpha_{u_3}< \frac 12$.
It follows from (II) that $d(u_3) = d_H(u_3)\le 6$ or $u_3$ is $(7,4^-)_{HG}$.

(Case 1.2.1.) $d(u_3)= 5$.
For each $i\in \{2, 4\}$, it follows from (I) that $d_H(u_i)\ge 7$;
the non-existence of ($A_{7.2}$) in $G$ implies that $d(u_i)\ge 8$;
and since $n_{5}(u_i)\ge 3$, we have $d_H(u_i)\ge 8$.
It follows that $\alpha_{u_2}\ge \frac 12$ and $\alpha_{u_4}\ge \frac 12$.
Then for each $i\in \{1, 2, 3, 4\}$, $\tau(f_i \rightarrow u)\le \frac 14$, and Eq. (<ref>) becomes $\omega'(u) \le - 1 + 4\times \frac 14 = 0$.

(Case 1.2.2.) $d(u_3)= 6$.
When for each $i\in \{2, 4\}$, $\alpha_{u_i}\ge \frac 12$,
we have $\tau(f_1 \rightarrow u)\le \frac 14$, $\tau(f_2\rightarrow u)\le \frac 16$,
$\tau(f_3\rightarrow u)\le \frac 16$, and $\tau(f_4\rightarrow u)\le \frac 14$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 14 + 2\times \frac 16 = - \frac 16< 0$.

In the other case, there exists an $i\in \{2, 4\}$ such that $\alpha_{u_i}< \frac 12$.
It follows that $d(u_i) = d_H(u_i) = 6$, or $u_i$ is $(7, 4^-)_{HG}$.
Assuming w.l.o.g. $i = 2$, one sees that if $d(u_2) = 6$ or $d(u_4) = 6$, then ($A_{8.1}$) will be obtained, a contradiction.
Hence, $u_1$ is $(7, 4^-)_{HG}$ and $d_H(u_4)\ge 7$.
If $\alpha_{u_4}\ge \frac 12$, then $\tau(f_1 \rightarrow u)\le \frac 27$, $\tau(f_2\rightarrow u)\le \frac 5{21}$,
$\tau(f_3\rightarrow u)\le \frac 16$, and $\tau(f_4\rightarrow u)\le \frac 14$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 27 + \frac 5{21} + \frac 16 + \frac 14 = - 1 + \frac{79}{84}  = - \frac 5{84} < 0$.
Otherwise, $\alpha_{u_4}< \frac 12$.
It follows from (II) that $d(u_4) = d_H(u_4)\le 6$ or $u_4$ is $(7, 4^-)_{HG}$, which leads to ($A_{8.1}$) or ($A_{8.4}$), a contradiction.

(Case 1.2.3.) $u_3$ is $(7, 4^-)_{HG}$.
When there exists an $i\in \{2, 4\}$ such that $d_H(u_i)\ge 7$, and assuming w.l.o.g. $i = 4$,
then $\tau(f_1 \rightarrow u)\le \frac 13$, $\tau(f_2\rightarrow u)\le \frac 5{21}$,
$\tau(f_3\rightarrow u) \frac 17$, and $\tau(f_4\rightarrow u)\le \frac 27$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 13 + \frac 5{21} + \frac 17  + \frac 27 = 0$.
In the other case, for each $i\in \{2, 4\}$, $d(u_i) = d_H(u_i) = 6$, which leads to ($A_{8.2}$), a contradiction.

Case 1.3. Assuming w.l.o.g. $d(u_1) = d_H(u_1) = 5$ and $d_H(u_5)\ge 6$.

In this case, by (I), $d(u_2)\ge 6$ and $\tau(f_1\rightarrow u)\le \frac 13$.

(Case 1.3.1.) $d_H(u_4)= 5$, i.e., $d(u_4) = 5$.
It follows from (I) that for each $i\in \{3, 5\}$, $d_H(u_i)\ge 7$ and thus $\alpha_{u_i}\ge \frac 37$.
When $\alpha_{u_3}\ge \frac 12$,
$\tau(f_2\rightarrow u)\le \frac 16$,
$\tau(f_3\rightarrow u)\le \frac 14$,
$\tau(f_4\rightarrow u)\le \frac 5{28}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 13 + \frac 16 + \frac 14 + \frac 5{28} = -\frac {13}{84}< 0$.

When $\alpha_{u_5}\ge \frac 12$,
$\tau(f_2\rightarrow u)\le \frac 5{21}$,
$\tau(f_3\rightarrow u)\le \frac 27$,
$\tau(f_4\rightarrow u)\le \frac 18$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 13 + \frac 5{21} + \frac 27 + \frac 18 = -\frac 3{168}< 0$.

In the other case, for each $i\in \{3, 5\}$, $\alpha_{u_i} < \frac 12$;
and it follows from (2) that $u_i$ is $(7, 4^-)_{HG}$, which leads to ($A_{8.4}$), a contradiction.

(Case 1.3.2.) $d_H(u_4)\ge 6$ and $d_H(u_2)\ge 6$.
When $d_H(u_3)\ge 7$,
$\tau(f_2\rightarrow u)\le \frac 5{21}$,
$\tau(f_3\rightarrow u)\le \frac 5{21}$,
$\tau(f_4\rightarrow u)\le \frac 16$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 13 + 2\times \frac 5{21} + \frac 16 = - \frac 1{42}< 0$.

In the other case, $d(u_3) = d_H(u_3)\le 6$.
First, assume that $d(u_3) = 5$.
It follows from (I) that for each $i\in \{2, 4\}$, $\alpha_{u_i}\ge \frac 37$.
Then for each $i\in \{1, 2, 3\}$, $\tau(f_i\rightarrow u)\le \frac 27$,
$\tau(f_4\rightarrow u)\le \frac 1{14}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 3\times \frac 27 + \frac 1{14} = - \frac 1{14}< 0$.

Next, assume that $d(u_3) = 6$.
When $\alpha_{u_5}\ge \frac 12$,
$\tau(f_4\rightarrow u_4)_{f_4}\ge \frac 13$,
$\tau(f_4\rightarrow u_5)_{f_4}\ge \frac 12$,
$\tau(f_4\rightarrow u_5)_{f_1}\ge \frac 14$,
it follows that $\tau(f_4\rightarrow u)= 0$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 3\times \frac 13 = 0$.
In the other case, $\alpha_{u_5}<\frac 12$.
It follows from (II) that $d(u_5) = d_H(u_5) = 6$, or $u_5$ is $(7, 4^-)_{HG}$.

* $d(u_5) = 6$.
It follows from (I) that $d_H(u_4)\ge 6$.
If $d_H(u_4) = 6$, then $d(u_4) = 6$ and ($A_{8.1}$) is obtained, a contradiction.
Otherwise, $d_H(u_4) \ge 7$ and $\alpha_{u_4}\ge \frac 37$.
Then, $\tau(f_3\rightarrow u)\le \frac 5{21}$, $\tau(f_4\rightarrow u)\le \frac 1{14}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 13 + \frac 5{21} + \frac 1{14}  = - \frac 1{42}< 0$.

* $u_5$ is $(7, 4^-)$.
When $d_H(u_4)\ge 7$,
$\tau(f_4\rightarrow u_4)_{f_4}\ge \frac 37$,
$\tau(f_4\rightarrow u_5)_{f_4}\ge \frac 37$,
$\tau(f_4\rightarrow u_5)_{f_1}\ge \frac 3{14}$,
it follows that $\tau(f_4\rightarrow u)= 0$, and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 3\times \frac 13 = 0$.
In the other case, $d_H(u_4)= 6$, i.e., $d(u_4) = d_H(u_4) = 6$, which leads to ($A_{8.2}$), a contradiction.

Case 2.  $m_3(u) = 5$.

In this case, Eq. (<ref>) becomes

\begin{equation}
\label{eq14}
\omega'(u) = -1 + \sum_{f\in \{f_1, f_2, f_3, f_4, f_5\}} \tau(f \rightarrow u).
\end{equation}

When $d_H(x)\ge 7$ for each $x\in N_H(u)$, $\alpha_x\ge \frac 37$,
for each $i\in [1, 5]$, $\tau(f_i\rightarrow u)\le \frac 17$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 5\times \frac 17 = -\frac 27$.

In the other case, there exists an $x\in N_H(x)$ such that $d_H(x)\le 6$ and $d(x) = d_H(x)\le 6$.
We have the following two subcases for two possible values of $\min_{x\in N_H(x)} d_H(x)$.

Case 2.1. $d(u_3)= d_H(u_3) = 5$.

In this case, it follows from (I) that $d_H(u_2)\ge 7$, $d_H(u_4)\ge 7$, and $n_5(u)\le 2$.
We can have the following two scenarios:

(Case 2.1.1.) $n_5(u) = 2$, and assuming w.l.o.g. $d(u_1) = d_H(u_1)= 5$.
It follows from (I) that $d_H(u_5)\ge 7$.
When $\alpha_{u_i}\ge \frac 12$ for each $i\in \{2, 4, 5\}$,
$\tau(f_i \rightarrow u)\le \frac 14$, $i\in \{1, 2, 3, 5\}$,
and $\tau(f_4 \rightarrow u)\le 0$ since $\tau(f_4 \rightarrow u_4)_{f_4}\ge \frac 12$ and $\tau(f_4 \rightarrow u_5)_{f_4}\ge \frac 12$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + 4\times \frac 14 = 0$.

In the other case, there exists a $w\in \{u_2, u_4, u_5\}$ such that $\alpha_{w}< \frac 12$.
It follows from (II) that $w$ is $(7, 4^-)_{HG}$.
The non-existence of ($A_{7.2}$) in $G$ implies that for each $x\in \{u_2, u_4, u_5\}\setminus \{w\}$, $d(x)\ge 9$,
and since $n'_{5}(x)\ge 2$, $d_H(x)\ge 9$ and $\alpha_x\ge \frac 59$.
By symmetry, we assume w.l.o.g. $w\in \{u_2, u_4\}$.

* $u_2$ is $(7, 4^-)_{HG}$, $d_H(u_4)\ge 9$ and $d_H(u_5)\ge 9$.
Then $\tau(f_i \rightarrow u)\le \frac 27$, $i = 1, 2$;
$\tau(f_j \rightarrow u)\le \frac 29$, $j = 3, 5$;
since $\alpha_{u_4}\ge \frac 59$ and $\alpha_{u_5}\ge \frac 59$,
we have $\tau(u \rightarrow f_4) = \alpha_{u_4} + \alpha_{u_5} - 1 \ge 2\times 59 -1 = \frac 19$, i.e., $\tau(f_4 \rightarrow u)\le -\frac 19$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 27 + 2\times \frac 29 - \frac 19 = -\frac 2{21}< 0$.

* $u_4$ is $(7, 4^-)_{HG}$, $d_H(u_2)\ge 9$ and $d_H(u_5)\ge 9$.
Then $\tau(f_i \rightarrow u)\le \frac 29$, $i \in \{1, 2, 5\}$;
$\tau(f_3 \rightarrow u)\le \frac 72$, and $\tau(f_4 \rightarrow u)\le \frac 1{63}$.
Thus, Eq. (<ref>) becomes $\omega'(u) \le - 1 + 3\times \frac 29 + \frac 27 + \frac 1{63} = -\frac 2{63}< 0$.

(Case 2.1.2.) $n_5(u) = 1$, i.e., $d_H(u_1)\ge 6$ and $d_H(u_5)\ge 6$.
When $n'_{7^+}(u) = 4$,
$\tau(f_i \rightarrow u)\le \frac 17$, $i\in \{1, 4, 5\}$,
$\tau(f_j \rightarrow u)\le \frac 27$, $j\in \{2, 3\}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 3\times \frac 17 + 2\times \frac 27 = 0$.

In the other case, $n'_{7^+}(u) \le 3$, i.e., there exists a $w\in \{u_1, u_5\}$ such that $d_H(w) = 6$ and then $d(w) = 6$.
Assume w.l.o.g. that $w = u_1$.
If $\alpha_{u_5}\ge \frac 59$,
then for each $i\in \{2, 3\}$, $\tau(f_1 \rightarrow u)\le \frac 5{21}$,
$\tau(f_i \rightarrow u)\le \frac 27$,
$\tau(f_4 \rightarrow u)\le \frac 1{63}$,
$\tau(f_5 \rightarrow u)\le \frac 19$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 5{21} + 2\times \frac 27 + \frac 1{63} + \frac 19 = -\frac 4{63}< 0$.
Otherwise, $\alpha_{u_5}< \frac 59$.
Hence, $d(u_5) = d_H(u_5) = 6$, or $u_5$ is $(7, 5^-)_{HG}$, or $u_5$ is $(8, 4^-)_{HG}$.

* $d(u_5) = d_H(u_5) = 6$.
The non-existence of ($A_{7.2}$) and ($A_{7.3}$) in $G$ imply that for each $i\in \{2, 4\}$, $d(u_i)\ge 9$,
and since $n'_{5}(u_i) = n_{5}(u_i)\ge 2$, $d_H(u_i)\ge 9$.
It follows that $\alpha_{u_2}\ge \frac 59$ and $\alpha_{u_4}\ge \frac 59$.
Then $\tau(f_i \rightarrow u)\le \frac 19$, $i = 1, 4$,
$\tau(f_j \rightarrow u)\le \frac 29$, $j = 2, 3$,
$\tau(f_5 \rightarrow u)\le \frac 13$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 19 + 2\times \frac 29 + \frac 13 = 0$.

* $u_5$ is $(7, 5^-)_{HG}$.
The non-existence of ($A_{7.2}$) in $G$ implies that for each $i\in \{2, 4\}$,
$d(u_i)\ge 8$ and since $n'_{5}(u_i) = n_{5}(u_i)\ge 2$,  $d_H(u_i)\ge 8$.
It follows that $\alpha_{u_2}\ge \frac 12$ and $\alpha_{u_4}\ge \frac 12$.
Then $\tau(f_1 \rightarrow u)\le \frac 16$,
$\tau(f_i \rightarrow u)\le \frac 14$, $i = 2, 3$,
$\tau(f_4 \rightarrow u)\le \frac 1{14}$,
$\tau(f_5 \rightarrow u)\le \frac 5{21}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 16 + 2\times \frac 14 + \frac 1{14}+ \frac 5{21} = \frac 1{42}< 0$.

* $u_5$ is $(8, 4^-)_{HG}$.
Then $\tau(f_5 \rightarrow u)\le \frac 16$.
When $\alpha_{u_2}\ge \frac 12$,
$\tau(f_1 \rightarrow u)\le \frac 16$,
$\tau(f_2 \rightarrow u)\le \frac 14$,
$\tau(f_3 \rightarrow u)\le \frac 27$,
$\tau(f_4 \rightarrow u)\le \frac 1{14}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + 2\times \frac 16 + \frac 14 + \frac 27+ \frac 1{14} = -\frac 5{84}< 0$.

When $\alpha_{u_4}\ge \frac 12$,
$\tau(f_1 \rightarrow u)\le \frac 5{21}$,
$\tau(f_2 \rightarrow u)\le \frac 27$,
$\tau(f_3 \rightarrow u)\le \frac 14$,
$\tau(f_4 \rightarrow u)\le 0$ since $\tau(f_4 \rightarrow u)_{f_4}\ge \frac 12$ and $\tau(f_4 \rightarrow u)_{f_5}= \frac 12$,
and thus Eq. (<ref>) becomes $\omega'(u) \le - 1 + \frac 5{21} +  \frac 27 + \frac 14 + \frac 16 = -\frac 5{84}< 0$.

In the other case, $\alpha_{u_2}< \frac 12$ and $\alpha_{u_4}< \frac 12$.
It follows from (II) that $u_2$ and $u_4$ are $(7, 4^-)_{HG}$, which leads to ($A_{7.2}$), a contradiction.

Case 2.2. $d(u_3)= d_H(u_3) = 6$ and $n'_{6^+}(u)= 5$.

In this case, Eq. (<ref>) becomes

\begin{equation}
\label{eq15}
\omega'(u) \le - 1 + \sum_{i\in [1, 5]}(1 - \alpha_{u_i} - \alpha_{u_{(i \mod 5) + 1}}) = 4 - 2\sum_{i\in [1, 5]} \alpha_{u_i}.
\end{equation}

When $n'_{7^+}(u) = 4$, $\alpha_{u_i}\ge \frac 37$ for each $i\in \{1, 2, 4, 5\}$,
and thus Eq. (<ref>) becomes $\omega'(u) \le 4 - 2\times (4\times \frac 37 + \frac 13) =  - \frac 2{21}< 0$.

In the other case, $n'_{7^+}(u)\le 3$, i.e., $n_{6}(u) = n'_{6}(u)\ge 2$.
The non-existence of ($A_{8.1}$) in $G$ implies that $n_{6}(u)\le 3$.
By symmetry, we can have the following two scenarios:

(Case 2.2.1.) $n_{6}(u) = 3$.
In this case, $d(u_1) = d(u_4) = 6$ or $d(u_2) = d(u_4) = 6$.
The non-existence of ($A_{8.1}$) and ($A_{8.2}$) in $G$ imply that for each $7^+$-vertex $x\in N(u)$, $d(x)\ge 8$ or $x$ is $(7, 5^+)$;
it follows that $x$ is $(7, 5^+)_H$ or $d_H(u_i)\ge 8$, which implies that $\alpha_{x}\ge \frac 12$.
Thus Eq. (<ref>) becomes $\omega'(u) \le 4 - 2\times (2\times \frac 12 + 3\times \frac 13) = 0$.

(Case 2.2.2.) $n_{6}(u) = 3$.
When there exists a $7^+$-vertex $w\in N_H(u)$ such that $\alpha_w\ge \frac 12$, Eq. (<ref>) becomes
$\omega'(u) \le 4 - 2\times (\frac 12 + 2\times \frac 37 + 2\times \frac 13) = -\frac 1{21} < 0$.
In the other case, for each $7^+$-vertex $x\in N_H(u)$, $\alpha_x < \frac 12$.
It follows from (II) that $x$ is $(7, 4^-)_{HG}$.
Since $d(u_2) = 6$ or $d(u_1) = 6$, ($A_{8.3}$) or ($A_{8.4}$) is obtained in $G$, a contradiction.

The above Discharging-Lemmas <ref>, <ref>, <ref> and <ref> together
contradict the positive total weight of $8$ stated in Eq. (<ref>),
and thus prove Theorem <ref>.

§ ACYCLIC EDGE $(\DELTA+5)$-COLORING

In this section, we show how to derive an acyclic edge $(\Delta+5)$-coloring for a simple $2$-connected planar graph $G$,
by an induction on the number of edges $|E(G)|$.
The following lemma gives the starting point.

([16, 3, 4, 15, 19])

If $\Delta \in \{3, 4\}$, then $a'(G) \le \Delta + 2$, and an acyclic edge $(\Delta + 2)$-coloring can be obtained in polynomial time.
[[16] gives an $O(n)$-time algorithm for $\Delta = 3$;
[4] gives a polynomial-time algorithm for $\Delta = 4$ except $4$-regular;
[15, 19] gives a polynomial-time algorithm for $4$-regular graphs.]

Given a partial acyclic edge $k$-coloring $c(\cdot)$ of the graph $G$ using the color set $C = \{1, 2, \ldots, k\}$,
for a vertex $v\in V(G)$, let $C(v)$ denote the set of colors assigned to the edges incident at $v$ under $c(\cdot)$.
If the edges of a path $P= ux\ldots v$ are alternatively colored $i$ and $j$, we call it an $(i, j)_{(u, v)}$-path.
Furthermore, if $uv\in E(G)$ is also colored $i$ or $j$, we call $ux\ldots vu$ an $(i, j)_{(u, v)}$-cycle
(which implies $c(\cdot)$ is invalid and needs to be revised).

For simplicity, we use $\{e_1, e_2, \ldots, e_m\} \to a$ to state that all the edges $e_1$, $e_2$, $\ldots$, $e_m$ are colored $a$,
use simply $e_1 \to a$ to state that $e_1$ is colored $a$,
use $e_1 \to S$ ($S \ne\emptyset$) to state that $e_1$ is colored with a color in $S$,
and use $(e_1, e_2, \ldots, e_m) \to (a_1, a_2, \ldots, a_m)$ to state that $e_j$ is colored $a_j$, for $j = 1, 2, \ldots, m$.
We also use $(e_1, e_2, \ldots, e_m)_c = (a_1, a_2, \ldots, a_m)$ to denote that $c(e_j) = a_j$, for $j = 1, 2, \ldots, m$;
that is, to emphasize the coloring is by $c(\cdot)$.
Moreover, we use $i_1/i_2/\ldots/i_k\in S$ to state that at least one of $i_1, i_2, \ldots, i_k$ is in $S$, i.e., $S\cap \{i_1, i_2, \ldots, i_k\}\ne \emptyset$.
By a graph $H$ containing an $(S_1, S_2)_{(u, v)}$-path
(an $(a, S)_{(u, v)}$-path, an $(a, i_1/i_2/\ldots/i_k)_{(u, v)}$-path, respectively),
it means that $H$ contains an $(i, j)_{(u, v)}$-path for each pair $i\in S_1$, $j\in S_2$
(an $(a, j)_{(u, v)}$-path for each $j\in S$, an $(a, j)_{(u, v)}$-path for some $j\in \{i_1, i_2, \ldots, i_k\}$, respectively).

The following lemma is obvious (via a simple contradiction):

[15]

Suppose $G$ has an acyclic edge coloring $c(\cdot)$,
and $P = uv_1v_2$-$\ldots$-$v_kv_{k+1}$ is a maximal $(a, b)_{(u, v_{k+1})}$-path with $c(uv_1) = a$ and $b \not\in C(u)$.
Then there is no $(a, b)_{(u, w)}$-path for any vertex $w \not\in V(P)$.

The rest of the section is devoted to the proof of Theorem <ref>, by an induction on the number of edges $|E(G)|$.
The flow of the proof is depicted in Figure <ref>.

The flow of the proof of Theorem <ref> by an induction on $|E(G)|$.

For the base cases in the induction,
one sees that when $|E(G)| \le \Delta + 5$, coloring each edge by a distinct color gives a valid acyclic edge $(\Delta + 5)$-coloring;
when $\Delta \le 4$, an acyclic edge $(\Delta + 5)$-coloring is guaranteed by Lemma <ref>.

§.§ 3.0  Configuration ($A_1$), the simplest cases of the inductive step

In the sequel we consider a simple $2$-connected planar graph $G$ such that $\Delta \ge 5$ and $|E(G)| \ge \Delta + 6$.
We first use some simplest cases to convince the readers that the inductive step works,
then starting Subsection 3.1 the inductive step becomes more complex yet using the similar ideas.
Theorem <ref> states that $G$ contains at least one of the configurations ($A_1$)–($A_8$),
which are the cases to be discussed inside the inductive step.

As the most simplest case of the inductive step, consider the configuration ($A_{1.1}$) in $G$.
Let $G' = G - v + u w$, that is, by merging the two edges $u v$ and $v w$ into a new edge $u w$ while eliminating the vertex $v$.
The graph $G'$ is planar, $\Delta(G') = \Delta$, and contains one less edge than $G$.
Using the inductive hypothesis, $G'$ has an acyclic edge $(\Delta + 5)$-coloring $c(\cdot)$ using the color set $C = \{1, 2, \ldots, \Delta + 5\}$,
in which we assume that $c(u w) = 1$.
To construct an edge coloring for $G$,
we keep the color for every edge of $E(G') \setminus \{u w\}$, $v w \to 1$, and $u v \to C \setminus C(u)$.
One may trivially verify that no bichromatic cycle would be introduced, that is, the achieved edge $(\Delta + 5)$-coloring for $G$ is acyclic.

One sees that for the edge $uv$ in each of the other configurations,
we have the degree $d(v)\in \{2, 3\}$ (in ($A_{1.2}$), ($A_{1.3}$), ($A_{2}$)–($A_5$) respectively),
or the degree $d(u)\in \{4, 5\}$ (in ($A_6$)–($A_8$) respectively).
We pick the edge $uv$ and let $H = G - uv$ (that is, $H$ is obtained by removing the edge $uv$ from $G$, or precisely $H = (V(G), E(G) \setminus\{uv\})$).
The graph $H$ is planar, and $4 \le \Delta - 1 \le \Delta(H) \le \Delta$.
[Note that $H$ might not be $2$-connected, but it does not matter.]
Using the inductive hypothesis, $H$ has an acyclic edge $(\Delta(H) + 5)$-coloring $c(\cdot)$,
and thus a $(\Delta + 5)$-coloring, using the color set $C = \{1, 2, \ldots, \Delta + 5\}$.

We have $d_H(u) + d_H(v) \le \Delta - 1 + 5 - 1 = \Delta + 3 < |C|$, or equivalently $C \setminus (C(u)\cup C(v)) \ne \emptyset$.
One sees that,
if $C(u)\cap C(v) = \emptyset$, then $uv\overset{\divideontimes}\to C\setminus (C(u)\cup C(v))$.
We define the notation $\overset{\divideontimes}\to$ to mean that $uv\to C\setminus (C(u)\cup C(v))$ gives an acyclic edge $(\Delta + 5)$-coloring for $G$ and thus we are done.
Or, if there exists a $j\in C\setminus (C(u)\cup C(v))$ such that, for all $i\in C(u)\cap C(v)$, $H$ contains no $(i, j)_{(u, v)}$-path,
then $uv\overset{\divideontimes}\to j$.
In the remaining case, for each $i\in C(u)\cap C(v)$, we assume that $c(u u_s) = c(v v_t) = i$,
and we define the color set

\[
B_i = \{j \mid \mbox{there is an } (i, j)_{(u_s, v_t)}\mbox{-path in } H\}.
\]

Therefore, for each $j \in B_i$, there is a neighbor $x$ of $u_s$ and a neighbor $y$ of $v_t$, respectively,
such that $c(u_s x) = c(v_t y) = j$, and each of $x$ and $y$ is incident with an edge colored $i$.

One sees that $B_i \subseteq C(u_s)\cap C(v_t)$.
In the sequel we continue the proof with the following Proposition <ref>
[Throughout the paper, we characterize a lot of intermediate properties, but not final conclusions,
that are useful for down-stream proofs into propositions.
These propositions are subsection numbered in four levels.
When citing them, we only use the 4-level labels, without mentioning “Proposition”.],
or otherwise we are done:

$C(u)\cap C(v) \ne \emptyset$, and $C\setminus (C(u)\cup C(v))\subseteq \bigcup _{i\in C(u)\cap C(v)}B_i$.

For each degree $k\in \{2, 3, 4, 5\}$,
we collect the colors of the edges connecting $u$ and its $k$-neighbors ($k^-$-neighbors, respectively) into
$S_k(u) = \{c(u w) \mid w \in N(u) \setminus \{v\}, d(w) = k\}$ ($S_{k^-}(u) = \{c(u w) \mid w \in N(u) \setminus \{v\}, d(w) \le k\}$, respectively).

The second simplest cases would be the configurations ($A_{1.2}$) and ($A_{1.3}$), in which there is a path $u v w$ with $d(v)= 2$.
Let $u_1, u_2, \ldots, u_{d(u) - 2}$ be the other neighbors of $u$.
Using the acyclic edge coloring $c(\cdot)$ for $H = G - u v$,
by <ref>, $c(vw) \in C(u)$, $H$ contains a $(c(v w), i)_{(u, w)}$-path for every $i \in C \setminus C(u)$,
and thus $(C\setminus C(u))\subseteq C(w) \setminus \{c(v w), c(u w)\}$.

Let $c(u u_i) = i$ for each $1 \le i \le d(u) - 2$ and $(v w, u w)_c= (1, d(u) - 1)$.
It follows that ($A_{1.3}$) holds and $[d(u), \Delta + 5]\subseteq C(w)$.
One sees that $|[d(u), \Delta + 5]|\ge \Delta + 5 - (d(u) - 1) = \Delta - d(u) + 6\ge 6$.
Since $n_{5^-}(u)\ge d(u) - 6$, $S_{5^-}(u)\ge d(u) - 7$ and $|S_{5^-}(u)\cup (C\setminus C(u))\cup \{1, d(u) - 1\}|\ge d(u) - 7 + (\Delta - d(u) + 6) + 2 = \Delta + 1$.
It follows that $6\in S_{5^-}(u)\setminus C(w)$.
Then $6\to v w$ and $[d(u), \Delta + 5]\subseteq C(u_6)\to u v$ give rise to an acyclic edge $(\Delta + 5)$-coloring for $G$, and we are done.

We have thus far demonstrated that the inductive step works for the case where $G$ contains the configuration ($A_1$),
i.e., one of the configurations ($A_{1.1}$)–($A_{1.3}$),
and summarize in the following lemma, where the running time refers to the time spent during the inductive step.

If ($A_1$) holds, then in $O(1)$ time an acyclic edge $(\Delta + 5)$-coloring for $G$ can be obtained.

The next two subsections deal with the configurations ($A_{2}$)–($A_{5}$) and ($A_6$)–($A_8$), respectively, and arranged as follows.

In Subsection 3.1, we first deal with ($A_{2.2}$) in Lemma <ref>,
and then $|C(u)\cap C(v)| = 2$ in Lemma <ref>,
which says that in $O(1)$ time either an acyclic edge $(\Delta + 5)$-coloring for $G$ can be obtained,
or an acyclic edge $(\Delta + 5)$-coloring for $H$ can be obtained such that $|C(u)\cap C(v)| = 1$;
next, we deal with ($A_{2.1}$) in Subsection 3.1.1, ($A_{2.3}$) in Subsection 3.1.2, and ($A_{2.4}$), ($A_{3}$)– ($A_{5}$) in Subsection 3.1.3, respectively.

In Subsection 3.2, we first consider the condition where $|C(u)\cap C(v)| = 4$ in Lemma <ref>,
$|C(u)\cap C(v)| = 3$ in Lemma <ref>,
and $|C(u)\cap C(v)| = 1$ in Lemma <ref>,
which says that in $O(1)$ time either an acyclic edge $(\Delta + 5)$-coloring for $G$ can be obtained,
or an acyclic edge $(\Delta + 5)$-coloring for $H$ can be obtained such that $|C(u)\cap C(v)| = 1$;
next, we deal with $d(u) = 4$ in Subsection 3.2.1, $d(u) = 5$ in Subsection 3.2.2, respectively.
In particular, in Subsection 3.2.2., we distinguish two cases for the two possible values $d(v)$.
It takes a long time in Subsection 3.2.2 to verify that if $G$ contains ($A_{8.1}$),
then an acyclic edge $(\Delta + 5)$-coloring for $H$ can be obtained.
We remark that this process is the most interesting.

§.§ Configurations ($A_{2}$)–($A_{5}$)

In this subsection we prove the inductive step for the case where $G$ contains one of the configurations ($A_{2}$)–($A_{5}$).
We have the following revised ($A_{2}$)–($A_{5}$):
An edge $u v$ with $d(v) = 3$ and $d(u)= k\le 11$ such that at least one of the configurations ($A_{2}$)–($A_{5}$) occurs.

If ($A_{2.2}$) holds, then in $O(1)$ time an acyclic edge $(\Delta + 5)$-coloring for $G$ can be obtained.

Let $G' = G - v + v_1 v_2$, that is, merging the two edges $v_1 v$ and $v v_2$ into a new edge $v_1 v_2$ while eliminating the vertex $v$.
The graph $G'$ is a planar graph, $\Delta(G') = \Delta$, and contains two less edges than $G$.
By the inductive hypothesis, $G'$ has an acyclic edge $(\Delta + 5)$-coloring $c(\cdot)$ using the color set $C = \{1, 2, \ldots, \Delta + 5\}$,
in which $C(x)$ is the color set of the edges incident at $x\in \{u, v_1, v_2\}$,
and we assume that $(u v_1, u v_2)_c = (2, 3)$ and $c(u u_i) = i$, $i\in \{1, 4, 5, \ldots, k - 1\}$,
where $u_1, u_4, \ldots, u_{k - 1}$ are the neighbors of $u$ other that $v, v_1, v_2$.

To transform back an edge coloring for $G$,
we keep the color for every edge of $E(G') \setminus \{v_1 v_2\}$, and color $\{u v, v v_1, v v_2\}$ as follows by recoloring some edges incident to $u$ if needed.

Case 1. $c(v_1 v_2)\not\in C(u)$, assuming w.l.o.g. $c(v_1 v_2) = k$.

When $[k + 1, \Delta + 5]\setminus C(v_2) \ne \emptyset$, $v v_1\to k$,
$v v_2\to \alpha\in [k + 1, \Delta + 5]\setminus C(v_2)$,
and $u v\overset{\divideontimes}\to [k + 1, \Delta + 5]\setminus \{\alpha\}$.

In the other case, $[k + 1, \Delta + 5]\subseteq C(v_2)$ and likewise, $[k + 1, \Delta + 5]\subseteq C(v_1)$.
Since $u$ is $([8, 11], 5^-)$, $|S_{5^-}(u)|\ge k - 6$.
One sees that $|S_{5^-}(u)\cup [k + 1, \Delta + 5]|\ge k - 6 + \Delta + 5 - k = \Delta - k -1$.
It follows that there exists an $i_0\in S_{5^-}(u)\setminus C(v_1)$.
Then $(v v_1, v v_2)\to (i_0, k)$ and $u v\overset{\divideontimes}\to [k + 1, \Delta + 5]\setminus C(u_{i_0})$.

Case 2. $c(v_1 v_2)\in C(u)$, assuming w.l.o.g. $c(v_1 v_2) = 1$.

When $[k, \Delta + 5]\setminus C(v_i)\ne \emptyset$ for some $i\in \{1, 2\}$, assuming w.l.o.g. $k\not\in C(v_2)$,
if there is no $(1, j_0)_{(u_1, v_1)}$-path for some $j_0\in [k + 1, \Delta + 5]$,
then $(v v_1, v v_2, u v)\overset{\divideontimes}\to (1, k, j_0)$;

otherwise, $G$ contains a $(1, j)_{(u_1, v_1)}$-path for each $j\in [k + 1, \Delta + 5]\subseteq C(v_1)$.
Since $|[k + 1, \Delta + 5]| = \Delta - k + 5 \ge 5$ and $S_{5^-}(u)\ge k - 6$,
there exists an $j_1\in S_{5^-}(u)\setminus C(v_1)$.
Then $(v v_1, v v_2)\to (j_1, 1)$ and $u v\overset{\divideontimes}\to [k + 1, \Delta + 5]\setminus C(u_{j_1})$.

In the other case, $[k, \Delta + 5]\subseteq C(v_1)\cap C(v_2)$.
Note that $|[k, \Delta + 5]| = \Delta - k + 6\ge 6$.
When there exists $j_2\in S_{5^-}(u)\setminus (C(v_1)\cup C(v_2))$,
let $\beta\in [k, \Delta + 5]\setminus C(u_{j_2})$,
by Lemma <ref>, $(v v_1, v v_2, u v)\to (1, j_2, \beta)$ or $(v v_1, v v_2, u v)\to (j_2, 1, \beta)$ gives rise to an acyclic edge $(\Delta + 5)$-coloring for $G$.

Now we assume that $S_{5^-}(u)\subseteq (C(v_1)\cup C(v_2))$.
Since $2 + |[k, \Delta + 5]| + 2 + |[k, \Delta + 5]| + |S_{5^-}(u)| = 2\Delta + 10 - k \le d(v_1) + d(v_2)\le 2\Delta$, we have $k\ge 10$.
Since $2 + |[k, \Delta + 5]| + |S_{5^-}(u)| = 2 + \Delta + 6 - k + k - 6 = \Delta + 2$,
we have $|S_{5^-}(u)\setminus C(v_1)|\ge 2$ and $|S_{5^-}(u)\setminus C(v_2)|\ge 2$.

For each $i_1\in S_{5^-}(u)\setminus C(v_1)$ and $i\in C\setminus C(v_2)$,
if $G$ contains no $(1, j)_{(v_2, u_1)}$-path for some $j\in [k, \Delta + 5]\setminus C(u_{i_1})$,
then $(v v_1, v v_2, u v)\overset{\divideontimes}\to (i_1, 1, j)$;
otherwise, $G$ contains a $(1, j)_{(v_2, u_1)}$-path for each $j\in [k, \Delta + 5]\setminus C(u_{i_1})$;
if $G$ contains no $(i, j)_{(v_2, u)}$-path for some $j\in [k, \Delta + 5]\setminus C(u_{i_1})$,
then $(v v_1, v v_2, u v)\overset{\divideontimes}\to (1, i, j)$.
In the other case, we proceed with the following proposition:

for each $i_1\in S_{5^-}(u)\setminus C(v_1)$ and $i\in C\setminus C(v_2)$,
$G$ contains a $(1, j)_{(v_2, u_1)}$-path and a $(i, j)_{(v_2, u_{i_2})}$-path for each $j\in [k, \Delta + 5]\setminus C(u_{i_1})$;

for each $i_2\in S_{5^-}(u)\setminus C(v_2)$ and $i\in C\setminus C(v_1)$,
$G$ contains a $(1, j)_{(v_1, u_1)}$-path and a $(i, j)_{(v_1, u_{i_1})}$-path for each $j\in [k, \Delta + 5]\setminus C(u_{i_2})$;

for each $i_1\in S_{5^-}(u)\setminus C(v_1)$ and for each $i_2\in S_{5^-}(u)\setminus C(v_2)$,
$[k, \Delta + 5]\setminus C(u_{i_1})\subseteq C(u_{i_2})$,
$[k, \Delta + 5]\setminus C(u_{i_2})\subseteq C(u_{i_1})$,
and $[k, \Delta + 5]\subseteq C(u_{i_1})\cup C(u_{i_2})$.

For each $i_2\in S_{5^-}(u)\setminus C(v_2)$,
when for some $j\in [k, \Delta + 5]\setminus C(u_{i_2})$, $G$ contains no $(i, j)_{(u_{i_2}, u)}$-path for any $i\in C(u)$,
it follows from <ref> that $G$ contains a $(1, j)_{(v_1, u_1)}$-path.
If $G$ contains no $(1, l)_{(u_1, v_1)}$-path for some $[k, \Delta + 5]\setminus \{j\}$,
then $(v v_1, v v_2, u u_{i_2}, u v)\overset{\divideontimes}\to (1, i_2, j, l)$.

Otherwise, $G$ contains a $(1, l)_{(u_1, v_1)}$-path for any $l\in [k, \Delta + 5]$.
Then $(v v_1, v v_2, u v)\overset{\divideontimes}\to (i_1, 1, l)$.

In the other case, we proceed with the following proposition:

For each $j\in [k, \Delta + 5]\setminus C(u_{i_2})$, where $i_2\in S_{5^-}(u)\setminus C(v_2)$, $G$ contains a $(i, j)_{(u_{i_2}, u)}$-path for some $i\in C(u)$.

For each $j\in [k, \Delta + 5]\setminus C(u_{i_1})$, where $i_1\in S_{5^-}(u)\setminus C(v_1)$, $G$ contains a $(i, j)_{(u_{i_1}, u)}$-path for some $i\in C(u)$.

It follows that $|C(u_{i_1})\cap [k, \Delta + 5]|\le 3$ and $|C(u_{i_2})\cap [k, \Delta + 5]|\le 3$.
Since $|[k, \Delta + 5]|\ge 6$, we have $|C(u_{i_1})\cap [k, \Delta + 5]|= 3$, $|C(u_{i_2})\cap [k, \Delta + 5]|= 3$, and $\Delta = k$.
Further, by Propositions <ref> and <ref>,
we assume that for each $i_2\in S_{5^-}(u)\setminus C(v_2)$, $C(u_{i_2}) = \{a_{i_2}, k, k + 1, k + 2\}$;
and for each $i_1\in S_{5^-}(u)\setminus C(v_1)$, $C(u_{i_1}) = \{a_{i_1}, k + 3, k + 4, k + 5\}$.

We consider below the case where $k = 10$ and $k = 11$, respectively.

Case 2.1. $k = 10$, and $C(v_1) = \{2, 6, 7\}\cup [10, 16]$,
$C(v_2) = \{3, 8, 9\}\cup [10, 16]$.

Then $C(u_6) = \{a_6, 10, 11, 12\}$, $C(u_7) = \{a_7, 10, 11, 12\}$,
and by Propositions <ref>, $G$ contains a $(i, 15)_{(u, v_1)}$-path for any $i\in \{1, 3, 4, 5, 8, 9\}$.
However, by Propositions <ref>, $G$ contains a $(2, 15)_{(v_1, u_6)}$-path and a $(2, 15)_{(v_1, u_7)}$-path, a contradiction.

Case 2.2. $k = 11$.

Assume w.l.o.g. $C(v_1) = \{2, 6, 7, 10\}\cup [11, \Delta + 5]$,
$\{3, 8, 9\}\cup [11, \Delta + 5]\subseteq C(v_2)$.
Then $C(u_6) = \{a_6, 11, 12, 13\}$, $C(u_7) = \{a_7, 11, 12, 13\}$,
and $C(u_8) = \{a_8, 14, 15, 16\}$, $C(u_9) = \{a_9, 14, 15, 16\}$,
where $a_i\in C(u)$ for $i \in [6, 9]$.
It follows from Propositions <ref> that $G$ contains a $(i, 15)_{(u, v_1)}$-path for any $i\in \{1, 3, 4, 5, 8, 9\}$ and a $(1, 11)_{(u_1, v_2)}$-path.

Hence, by Propositions <ref>, $G$ contains a $(i, 15)_{(u, u_6)}$-path and a $(j, 15)_{(u, u_7)}$-path for some $i, j\in \{2, 10\}$.
By Lemma <ref>, we assume w.l.o.g. $G$ contains a $(2, 15)_{(v_1, u_6)}$-path and a $(10, 15)_{(u_{10}, u_7)}$-path,
and $C(u_6) = \{2, 6, 11, 12, 13\}$, $C(u_7) = \{7, 10, 11, 12, 13\}$.
It follows from Propositions <ref> that $G$ contains a $(10, j)_{(u_{10}, u_7)}$-path for each $j\in \{14, 15, 16\}$, and assume w.l.o.g. $11\not\in C(u_{10})$.

If $10\not\in C(v_2)$, then $(v v_1, v v_2, u v)\overset{\divideontimes}\to (1, 10, 11)$.

$C(v_2) = \{3\}\cup [8, 16]$.
It follows from Propositions <ref> that $G$ contains a $(i, 11)_{(v_2, u)}$-path for each $i\in \{1, 2, 4, 5, 6, 7\}$.
However, by Propositions <ref>, $G$ contains a $(3, 11)_{(v_2, u_8)}$-path and a $(3, 11)_{(v_2, u_8)}$-path, a contradiction.

By Lemma <ref>, we consider below the case where $G$ contains the configuration ($A_{2}$)–($A_{5}$) other that ($A_{2.2}$).

Let $u_1$, $u_2$, $\ldots$, $u_{k - 1}$ be the other neighbors of $u$ other than $v$,
and assume that $c(u u_i)= i$ for $i = 1, 2, \ldots, k - 1$, i.e., $C(u) = [1, k - 1]$;
let $v_1$, $v_2$ be the neighbors of $v$ other than $u$.

Before we discuss the detailed configurations,
we do not distinguish $v_1$ and $v_2$ but use $x_1$ to refer to any one of $u_1$ and $u_2$ while $x_2$ to refer to the other ($\{x_1, x_2\} = \{u_1, u_2\}$).
In particular, $u_1 = x_1$ in ($A_{2.3}$), ($A_{2.4}$), and ($A_{3}$)–($A_{5}$), and $u_2 = x_2$ in ($A_{2.4}$), and ($A_{3}$)–($A_{5}$).
It follows from <ref> that $1\le|C(u)\cap C(v)|\le 2$,
and $C\setminus (C(u)\cup C(v))\subseteq \bigcup _{i\in C(u)\cap C(v)}B_i$.

If $|C(u)\cap C(v)| = 2$, then in $O(1)$ time either an acyclic edge $(\Delta + 5)$-coloring for $G$ can be obtained,
or an acyclic edge $(\Delta + 5)$-coloring for $H$ can be obtained such that $|C(u)\cap C(v)| = 1$.

Assume w.l.o.g. that $(v x_1, v x_2)_c= (a, b)$, where $a, b\in [1, k - 1]$.
When there exists a color $j\in [k, \Delta + 5]\setminus C(x_i)$ for some $i \in \{1, 2\}$,
$v x_i\to j$ gives rise to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
In the other case, we proceed with the following proposition:

(1) $[k, \Delta + 5]\subseteq C(u_a)\cup C(u_b)$; (2) $[k, \Delta + 5]\subseteq C(x_1)\cap C(x_2)$.

Case 1. ($A_{2.1}$) holds, i.e., $d(u) = 7$, and $C(u)= [1, 6]$.

Assuming w.l.o.g. $\{a, b\} = \{1, 2\}$.
It follows from <ref> that $C(x_1) = \{1\}\cup [7, \Delta + 5]$ and $C(x_2) = \{2\}\cup [7, \Delta + 5]$.
Since $7\in B_1\cup B_2$, assume w.l.o.g. $7\in B_1$.

If $G$ contains no $(i, 7)_{(x_1, u_i)}$-path for some $i\in \{2, 3\}$,
then $(v x_1, v x_2, u v)\overset{\divideontimes}\to (i, 1, 7)$.

Otherwise, for each $i\in \{2, 3\}$, $G$ contains a $(i, 7)_{(x_1, u_i)}$-path and $7\in C(u_i)$.
Likewise, we have $[7, \Delta + 5]\subseteq C(u_1)\cap C(u_2)\cap C(u_3)$ and thus $C(u_i) = \{i\}\cup [7, \Delta + 5]$ for every $i\in \{1, 2, 3\}$.
By Lemma <ref>, there is neither a $(2, 7)_{(u_2, x_2)}$-path nor a $(3, 7)_{(u_1, x_1)}$-path,
and $(u u_1, u u_3, v x_1, u v)\overset{\divideontimes}\to (3, 1, 3, 7)$.

Case 2. ($A_{2.3}$) holds, i.e., $u$ is $(8, 4^-)$, $v_2$ is $([8, 11], 5^-)$, and $u v_1, v_1 v_2\in E(G)$, $u v_2\not\in E(G)$.

Assuming w.l.o.g. $c(vx_1) = 2$ and $c(v x_2)\in C(u)\setminus \{2\}$.
It follows from <ref> that $C(x_1) = \{1, 2\}\cup [8, \Delta + 5]$ with $c(x_1 x_2) = 8$,
and $\{c(v x_2)\}\cup [8, \Delta + 5]\subseteq C(x_2)$.

When $c(v x_2)\in C(u)\setminus \{1, 2\}$, assuming w.l.o.g. $c(v x_2) = 3$,

since $8\in B_2\cup B_3$ and $3\not\in C(x_1)$, we have $2\in C(x_2)$ and $C(x_2) = \{2, 3\}\cup [8, \Delta + 5]$;
then $(v x_1, u v)\overset{\divideontimes} \to (4, 8)$.

In the other case, $c(v x_2) = 1$.
One sees from $|[3, 7]\cap C(x_2)|\le 1$ that there exists an $i\in [3, 7]\setminus C(x_2)$.
Then $v x_2\to i$ reduces to the case where $c(v x_2)\in C(u)\setminus \{1, 2\}$.

Case 3. ($A_{2.4}$) holds, i.e., $d(u) = 8$, and $u v_1, u v_2, v_1 v_2\in E(G)$, $C(u) = [1, 7]$.

It follows from <ref> that $C(x_1) = \{1, a\}\cup [8, \Delta + 5]$ and $C(x_2) = \{2, b\}\cup [8, \Delta + 5]$ with $c(x_1 x_2) = 8$.
Since $8\in B_a\cup B_b$, we assume w.l.o.g. $a = 2$ and $b = 3$.
Then $(v x_1, u v)\overset{\divideontimes} \to (4, 8)$.

Case 4. ($A_{3}$)–($A_{5}$) holds, i.e., $u$ is $([9, 11], 5^-)$, and $u v_1, u v_2, v_1 v_2\in E(G)$.

Assuming w.l.o.g. $u_i$ be a $6^+$-vertex, $i\in [3, 5]$, and $u_i$ be a $5^-$-vertex, $i\in [6, k - 1]$.
It follows from <ref> that
$[k, \Delta + 5] \subseteq C(x_1)$ and $[k, \Delta + 5] \subseteq C(x_2)$.

For any $i\in \{a, b\}\cap S_{5^-}(u)$, say $6\in \{a, b\}$,
when $C(u_6)\setminus\{6\}\subseteq [k, \Delta + 5]$, $u u_6\to [k, \Delta + 5]\setminus C(u_6)$ and an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$ is obtained.
In the other case, we proceed with the following proposition:

For any $i\in \{a, b\}\cap S_{5^-}(u)$, $|C(u_i)\setminus [k, \Delta + 5]|\ge 2$ and $|C(u_i)\cap [k, \Delta + 5]|\le 3$.

Since $|\{a, b\}\cap \{1, 2\}|\le 1$, $|[3, k - 1]\cap \{a, b\}|\ge 1$ and we discuss the following three
subcases for three possible values of $\{a, b\}$, respectively.

Case 4.1. $a, b\in S_{5^-}(u)$, assuming w.l.o.g. $a = 6$, $b = 7$.

Since $|[k, \Delta + 5]|\ge 6$,
it follows from <ref> that $\Delta = k$ and assume $C(u_6) = \{a_6, k, k + 1, k + 2\}$,
$C(u_7) = \{a_7, k + 3, k + 4, k + 5\}$.

When $7\not\in C(x_1)$, first $v x_1\to 7$,
if $6\not\in C(x_2)$, then $(v x_2, uv)\overset{\divideontimes}\to (6, k)$;
otherwise, $6\in C(x_2)$.
Since $[k, k + 5]\cup \{2\}\subseteq C(x_2)$, there exists $\beta\in [8, k - 1]\setminus C(x_2)$.
Then $v x_2\to \beta$ and $[k, k + 5]\setminus C(u_\beta)\overset{\divideontimes}\to uv$.

In the other case, $7\in C(x_1)$ and likewise, $6\in C(x_2)$.
For some $i\in (\{3, 4, 5\}\cup [8, k - 1])\setminus (C(x_1)\cup C(x_2))$,
if $G$ contains no $(i, j)_{(u_i, x_1)}$-path for some $j\in [k, k + 2]$,
then $(v x_1, u v)\overset{\divideontimes}\to (i, j)$;
if $G$ contains no $(i, j)_{(u_i, x_2)}$-path for some $j\in [k + 3, k + 5]$,
then $(v x_2, u v)\overset{\divideontimes}\to (i, j)$.
Otherwise, $G$ contains a $(i, j)_{(u_i, x_1)}$-path for each $j\in [k, k + 2]$ and a $(i, j)_{(u_i, x_2)}$-path for each $j\in [k + 3, k + 5]$.
Thus, $[k, k + 5]\subseteq C(u_i)$ and $d(u_i)\ge 7$.
It follows that $[8, k - 1]\subseteq C(x_1)\cup C(x_2)$.

(4.1.1.) $c(x_1 x_2)\in \{3, 4, 5\}$, say $c(x_1x_2) = 3$.

Since $2\times |[k, k + 5]| + |\{1, 3, 6, 7\}| + |\{2, 3, 6, 7\}| + |[8, k - 1]| \le 2\Delta = 2k$,
we have $k\ge 12$, a contradiction.

(4.1.2.) $c(x_1 x_2)\in [8, k - 1]$, say $c(x_1 x_2) = 8$.

Since $2\times |[k, k + 5]| + |\{1, 6, 7, 8\}| + |\{2, 6, 7, 8\}| + |[9, k - 1]| \le 2\Delta = 2k$,
we have $k\ge 11$.
It follows that $k = 11$, and $C(x_1) = \{1, 6, 7, 8, 9\}\cup [k, k + 5]$,
$C(x_2) = \{2, 6, 7, 8, 10\}\cup [k, k + 5]$.
Since $G$ contains a $(3, j)_{(u_3, x_1)}$-path for each $j\in [k, k + 2]$ and a $(3, j)_{(u_3, x_2)}$-path for each $j\in [k + 3, k + 5]$,
thus $x_1 x_2\to 3$ reduces to the case where $8\not\in C(x_1)\cup C(x_2)$.

(4.1.3.) $c(x_1 x_2)\in [k, k + 5]$, say $c(x_1x_2) = k$.

Since $k\le 11$ and $2\times |[k, k + 5]| + |\{1, 6, 7\}| + |\{2, 6, 7\}| + |[8, k - 1]| \le 2\Delta = 2k$, i.e., $10 + k\le 2k$,
we have $|\{3, 4, 5\}\cap (C(x_1)\cup C(x_2))|\le 1$.
Assuming w.l.o.g. $3, 4\not\in C(x_1)\cup C(x_2)$.
Then $(v x_1, v x_2, u v)\overset{\divideontimes}\to (3, 4, k)$.

Case 4.2. $\{a, b\}\cap S_{5^-}(u)\ne \emptyset$, assuming w.l.o.g. $a = 6$, $b \in \{1, 3\}$.

When $6\not\in C(x_2)$,
since $[k, \Delta + 5]\cup \{1, 6\}\subseteq C(x_1)$ and $|[k, \Delta + 5]\cup \{1, 6\}| + |[7, k - 1]| = \Delta + 1$, there exists $\gamma\in [7, k - 1]\setminus C(x_1)$.
Then $(v x_2, v x_1)\to (6, \gamma)$ reduces the proof to Case 4.1.

In the other case, $6\in C(x_2)$.
Assume $[7, k - 1]\setminus (C(x_1)\cup C(x_2))\ne \emptyset$, say $7\not\in C(x_1)\cup C(x_2)$,
then $v x_2\to 7$ reduces the proof to Case 4.1.
Now assume that $[7, k - 1]\subseteq C(x_1)\cup C(x_2)$.

One sees that $k\le 11$ and $2\times |[k, \Delta + 5]| + |\{1, 6\}| + |\{2, b, 6\}| + |[7, k - 1]| = 2\Delta + 10 - k \le 2\Delta$.

When $b = 1$, it follows that $\{3, 4, 5\}\setminus (C(x_1)\cup C(x_2))\ne \emptyset$,
say $3\not\in C(x_1)\cup C(x_2)$,
thus $v x_2\to 3$ reduces the proof to the case where $b\in \{3, 4, 5\}$.
Hence, it suffices to assume that $b = 3$.

(4.2.1.) $c(x_1 x_2)\in \{4, 5\}$, say $c(x_1 x_2) = 4$.

Since $2\times |[k, \Delta + 5]| + |\{1, 4, 6\}| + |\{2, 3, 4, 6\}| + |[7, k - 1]| = 2\Delta + 12 - k \le 2\Delta$,
we have $k\ge 12$, a contradiction.

(4.2.2.) $c(x_1 x_2)\in [7, k - 1]$, say $c(x_1 x_2) = 7$.

Since $2\times |[k, \Delta + 5]| + |\{1, 6, 7\}| + |\{2, 3, 6, 7\}| + |[8, k - 1]| = 2\Delta + 11 - k \le 2\Delta$,
we have $k\ge 11$.
It follows that $k = 11$, and $C(x_1) = \{1, 6, 7, 8, 9\}\cup [11, \Delta  + 5]$,
$C(x_2) = \{2, 3, 6, 7\}\cup [10, \Delta  + 5]$.
It follows from <ref> that $|C(u_6)\cap [11, \Delta + 5]|\le 3$.
Let $[14, \Delta + 5]\cap C(u_6) = \emptyset$.
Then $[14, \Delta + 5]\subseteq B_3$.
Let $(v x_1, v x_2)\to (3, 8)$.

If there exists an $i\in [14, \Delta + 5]\setminus C(u_8)$,
then $u v\overset{\divideontimes}\to i$.
Otherwise, $[14, \Delta + 5]\subseteq C(u_8)$ and it follows from <ref> that $\Delta = 11$ and
$|C(u_8)\cap [11, \Delta + 5]| = [14, 16]$.

Hence, $G$ contains a $(3, i)_{(x_1, u_3)}$-path for any $i\in [11, 13]$,
i.e., $G$ contains no $(3, i)_{(x_1, x_2)}$-path for any $i\in [11, 16]$.
Then $(v x_1, x_1 x_2, v x_2)\to (6, 3, 7)$ reduces the proof to Case 4.1.

(4.2.3.) $c(x_1 x_2)\in [k, \Delta + 5]$, say $c(x_1 x_2) = k$.

When $4, 5\not\in C(x_1)\cup C(x_2)$, $(v x_1, v x_2, u v)\overset{\divideontimes}\to (4, 5, k)$.
In the other case, assume that $4\in C(x_1)\cup C(x_2)$.

Since $k\le 11$ and $2\times |[k, \Delta + 5]| + |\{1, 6\}| + |\{2, 3, 6\}| + |[7, k - 1]| + |\{4\}| =
2\Delta  + 11 - k \le 2\Delta$,
we have $k = 11$ and $3\not\in C(x_1)$, $5\not\in C(x_1)\cup C(x_2)$.
Then $(v x_1, v x_2, u v)\overset{\divideontimes}\to (3, 5, k)$.

Case 4.3. $\{a, b\}\cap S_{5^-}(u)= \emptyset$, assuming w.l.o.g. $a = 3$, $b \in \{1, 4\}$.

When $6\not\in C(x_1)\cup C(x_2)$, $v x_2\to 6$ reduces the proof to to Case 4.2.
In the other case, $[6, k - 1]\subseteq C(x_1)\cup C(x_2)$.
One sees that $k\le 11$ and $2\times |[k, \Delta + 5]| + |\{1, 3\}| + |\{2, b\}| + |[6, k - 1]| = 2\Delta + 10 - k \le 2\Delta$.
When $b = 1$, it follows that $\{4, 5\}\setminus (C(x_1)\cup C(x_2))\ne \emptyset$,
say $4\not\in C(x_1)\cup C(x_2)$,
thus $v x_2\to 4$ reduces the proof to the case where $b\in \{4, 5\}$.
Hence, it suffices to assume that $b = 4$.

When $3\not\in C(x_2)$, since $\{1, 3\}\cup [k, \Delta + 5]\subseteq C(x_1)$,
we have $[6, k - 1]\setminus C(x_1)\ne \emptyset$,
and thus $v x_2 \to 3$, $v x_1\to [6, k - 1]\setminus C(x_1)$ reduces the proof to Case 4.2.
In the other case, $3\in C(x_2)$.
Since $k\le 11$ and $2\times |[k, \Delta + 5]| + |\{1, 3\}| + |\{2, 4, 3\}| + |[6, k - 1]| = 2\Delta + 11 - k \le 2\Delta$,
we have $k = 11$, $4\not\in C(x_1)$, $5\not\in C(x_1)\cup C(x_2)$, and $c(x_1 x_2)\in [k, \Delta + 5]$.
Then $(v x_1, u v)\overset{\divideontimes}\to (5, c(x_1 x_2))$.

By Lemma <ref>, hereafter we assume that $|C(u)\cap C(v)| = 1$,
and w.l.o.g. that $(v x_1, v x_2)_c = (a, k)$,
and further from <ref> that $[k + 1, \Delta + 5] = (C\setminus \{1, 2, \ldots, k\})\subseteq B_1 \subseteq C(u_1)\cap C(x_1)$.
We first characterize a useful particularity as follows.

For ($A_{2}$)–($A_{5}$) other than ($A_{2.2}$),
either an acyclic edge $(\Delta + 5)$-coloring for $G$ can be obtained from $H$ in $O(1)$ time,
or $k\in B_a$ and the $(a, k)_{(u_1, x_1)}$-path in $H$ cannot pass through $x_2$.

Assume $a = 1$ and $k\not\in B_1$.
One sees that $|[k + 1, \Delta + 5]| = \Delta + 5 - k$.
When $[k + 1, \Delta + 5]\setminus C(x_2) \ne \emptyset$,
let $S = [k + 1, \Delta + 5]\setminus C(x_2)$;
when $1\not\in C(x_2)$, let $S = C\setminus C(x_2)$;
when there exists $C(u)\setminus (C(x_1)\cup C(x_2))$, let $S = \{\alpha\}$.
Then $v x_2\to S$ and $u v\overset{\divideontimes} \to k$.

In the other case, $[k + 1, \Delta + 5]\cup \{1\} \subseteq C(x_1)\cap C(x_2)$ and $[2, k - 1]\subseteq C(x_1)\cup C(x_2)$.
Note that $2\times |[k + 1, \Delta + 5]\cup \{1\}| + |\{k\}| + |[2, k - 1]| = 2\Delta - k + 11\le d(x_1) + d(x_2)\le 2\Delta.$
It follows that $k = 11$, and assume w.l.o.g. $C(x_1) = [1, 6]\cup [12, \Delta + 5]$,
$C(x_2) = \{1\} \cup [7, \Delta + 5]$.
Since $[k + 1, \Delta + 5] = [12, \Delta + 5]\subseteq B_1 \subseteq C(x_1)$, $d(x_1)\ge 6$.
One sees that the same argument applies if let $v x_1\to [7, 10]$ and thus
$d(x_i)\ge 6$ for each $i\in [7, 10]$.
Since from ($A_{5}$) that $u$ is $(11, 5^-)$,
we have $u$ is $(11, 5)$ and $d(x_2)\le 5$.
Then $(v x_1, v x_2)\to (11, 2)$ and $u u_2\overset{\divideontimes}\to [12, \Delta + 5]\setminus C(u_2)$.

By Lemma <ref>, we assume below $C\setminus C(u) = [k, \Delta + 5]\subseteq C(u_a)\cap C(x_1)$.
Since $|[k, \Delta  + 5]| = \Delta + 6 - k\ge 6$,
we have $a\in C(u)\setminus S_{5^-}(u)$.
That is, if $a\in S_{5^-}(u)$, then we can obtain an acyclic edge $(\Delta + 5)$-coloring for $G$.

§.§.§ Configuration ($A_{2.1}$)

In this case, $C(u) = [1, 7]$, $(v x_1, v x_2)_c = (1, 7)$,
and $C(u_1) = C(x_1) = \{1\}\cup B_1 = \{1\}\cup [7, \Delta + 5]$.

When there exists $b$ such that $C(u_b) = \{b\}\cup [7, \Delta + 5]$, $(u u_1, u u_b, u v)\overset{\divideontimes}\to (b, 1, 8)$.
In the other case, $[7, \Delta + 5]\setminus C(u_i) \ne \emptyset$ for any $i\in [2, 6]$.
When $1\not\in C(x_2)$, $(v x_1, v x_2)\to (2, 1)$ and $u v\overset{\divideontimes} \to [7, \Delta + 5]\setminus C(u_2)$.
One sees that when $G$ contains no $(2, j)_{(x_1, x_2)}$-path for some $j\in \{7\}\cup ([7, \Delta + 5]\setminus C(x_2))$,
it follows from Lemma <ref> that $C(u_2) = \{2\}\cup B_2 = \{2\}\cup [7, \Delta + 5]$ if $(v x_1, v x_2)\to (2, j)$.

In the other case, $G$ contains an $(i, j)_{(x_1, x_2)}$-path for each $i\in [2, 6]$,
$j\in  \{7\}\cup ([7, \Delta + 5]\setminus C(x_2))$, and $[1, 7]\subseteq C(x_2)$.
It follows that $x_1, x_2\not\in N(u)$, i.e., $u v$ is not contained in a $3$-cycle.

When $\Delta = 7$, let $(x_1 w_7, x_1 w_8)_c = (7, 8)$,
it follows that $C(w_7)\setminus \{7\} = C(w_8)\setminus \{8\} = [1, 6]$,
thus by Lemma <ref>, $(x_1 w_7, x_1 w_8, u v)\overset{\divideontimes} \to (8, 7, 8)$.

In the other case, $\Delta \ge 8$ and $u x_1, u x_2\not\in E(G)$.
Let $G' = G - v + \{u x_1, u x_2\}$, that is, merging the two edges $u v$ and $v x_i$ into a new edge $u x_i$, $i = 1, 2$, while eliminating the vertex $v$.
The graph $G'$ is a planar graph, $\Delta(G') = \Delta$, and contains one less edge than $G$.
By the inductive hypothesis, $G'$ has an acyclic edge $(\Delta + 5)$-coloring $c(\cdot)$ using the color set $C = \{1, 2, \ldots, \Delta + 5\}$ and we assume w.l.o.g. that $c(u u_i) = i$, $i\in [1, 6]$,
$(u x_1, u x_2)_c = (7, 8)$.
To transform back an edge coloring for $G$,
we keep the color for every edge of $E(G') \setminus \{u x_1, u x_2\}$, $(v x_1, v x_2)\to (7, 8)$,
and $u v \overset{\divideontimes}\to C\setminus [1, 8]$.

§.§.§ Configuration ($A_{2.3}$)

In this case, $C(u) = [1, 8]$ with $c(u v_1) = 1$.
Let $u_i$ be a $6^+$-vertex, $i\in [2, 4]$ and $u_j$ be a $5^-$-vertex, $j\in [5, 7]$.
When $a = 1$ and assume w.l.o.g. $(v v_1, v v_2)_c = (8, 1)$,
since $[9, \Delta + 5]\subseteq C(v_1)\cap C(v_2)$, there exists an $i\in [2, 7]\setminus (C(v_1)\cup C(v_2))$,
thus $(v v_1, u v)\overset{\divideontimes}\to (i, 8)$[It is clearly that $G$ contains no $(1, 8)_{(v_1, v_2)}$-path that cannot pass through $v_1$,
a contradiction to Lemma <ref>. ].
In the other case, $a\in [2, 4]$, say $a = 2$.
Below we distinguish two cases where $c(v v_2) = 2$ and $c(v v_1) = 2$.

Case 1. $(v v_1, v v_2)_c = (8, 2)$.

In this case, $[8, \Delta + 5]\subseteq C(u_2)\cap C(v_2)$.
If $1\not\in C(v_2)$, then $v v_2\to 1$ reduces the proof to the case where $a = 1$.
Otherwise, $1\in C(v_2)$ and $C(v_2) = \{1, 2\}\cup [8, \Delta + 5]$.
It follows that $c(v_1 v_2)\in [9, \Delta + 5]$, say $c(v_1 v_2) = 9$.
Since $9\in B_2$, we have $2\in C(v_1)$.