One sees clearly that $w_3 + w_4\ge 8$.
It follows that $c_{3 4}\le 1$ and $t_3 + t_4\ge \Delta - 3$.
we can have the following two subcase.

(2.3.2.1.) $t_3 + t_4 = \Delta - 2$.
Then $C(u_2) = [2, 3]\cup T_3\cup T_4 = [2, 3]\cup T_{u v}$.
It follows from <ref>(1) that $H$ contains a $(5, 3)_{(u_2, u_3)}$-path,
and from <ref>(1) that $H$ contains a $(i, 3)_{(u_3, u_i)}$-path for some $i\in [5, 7]\cap C(u_3)$.
By Lemma <ref>, assume w.l.o.g. $H$ contains a $(3, 6)_{(u_3, v_6)}$-path.
When $1\not\in C(u_4)$,
if there exists an $i\in [5, 6]\setminus C(u_4)$, then $(u u_4, u v)\overset{\divideontimes}\to (i, 4)$.
Otherwise, $W_4 = [2, 6]$ and $W_3 = \{1, 3, 5, 6\}$.
Then $(u u_4, u v)\overset{\divideontimes}\to (7, 4)$.
In the other case, $1\in C(u_4)$.
Then $W_4 = [1, 4]\cup \{b_2\}$, $W_3 = \{1, 3, 5, 6\}$ and $d(u_3) = d(u_4) = \Delta = 7$.
It follows that $d(u_1)\le 6$.
If there exists an $i\in [5, 7]\setminus (C(u_4)\cup C(u_1))$, then $(u u_4, u v)\overset{\divideontimes}\to (i, 4)$.
Otherwise, $\{5, 6, 7\} = \{a_1, a_2, a_3\}\subseteq C(u_1)\cup C(u_4)$.
It follows that $C(u_1) = T_4\cup \{1, a_1, a_2\}$.
Then $(u u_1, u v)\overset{\divideontimes}\to (T_3, T_4)$.

(2.3.2.2.) $t_3 + t_4 = \Delta - 3$.
It follows that $c_{3 4} = 1$, $w_3 + w_4 = 8$ and $d(u_3) = d(u_4) = \Delta = 7$,
$\min\{d(u_1), d(u_2)\}\le 6$.

* $W_3 = \{3, 1, 5, a\}$, $W_4 = \{4, 2, b_2, 3\}$.
When $b_2 = 5$, if there exists an $i\in [6, 7]\setminus (C(u_2)\cup C(u_3))$,
then $(u u_4, u v)\overset{\divideontimes}\to (i, 4)$.
Otherwise, $6, 7\in C(u_2)\cup C(u_3)$ and then assume w.l.o.g. $W_2 = \{2, 3, 6\}$ and $a = 7$.
It follows from <ref>(1) that $H$ contains a $(5, 3)_{(u_2, u_3)}$-path and from <ref> that $H$ contains a $(7, 3)_{(u_3, v_7)}$-path.
Then $(u u_4, u v)\overset{\divideontimes}\to (7, 4)$.

In the other case, assume w.l.o.g. $b_2 = 6$.
If $C(u_2) = \{2, 3, 7\}\cup T_3\cup T_4$,
one sees from <ref>(1) that $H$ contains a $(5, 3)_{(u_2, u_3)}$-path,
then $(u u_4, u v)\overset{\divideontimes}\to (5, 4)$.
Otherwise, $7\not\in C(u_2)$.
If $H$ contains no $(3, 7)_{(u_3, u_4)}$-path, then $(u u_4, u v)\to (7, 4)$;
otherwise, $H$ contains a $(3, 7)_{(u_3, u_4)}$-path.
It follows from <ref>(1) that $H$ contains a $(5, 3)_{(u_3, v_5)}$-path,
and from <ref>(1) that $5/1\in C(u_2)$.
Thus $C(u_2) = \{2, 3, 5/1\}\cup T_3\cup T_4$ and $(u u_2, u u_3, u v)\overset{\divideontimes}\to (12, 2, T_3)$.

* $W_3 = \{3, 1, 5\}$, $W_4 = \{4, 2, b_2, b_3, 3\}$.
It follows from <ref> that $H$ contains a $(5, 3)_{(u_3, v_5)}$-path and from <ref>(1) that $5\in C(u_2)$ or $H$ contains a $(5, i)_{(u_2, u_i)}$-path for some $i\in \{1, 4\}$.
Hence, $C(u_2) = \{2, 3, 1/5\}\cup T_3\cup T_4$ and $T_4\cup T_{3 4}\subseteq C(u_1)$.
One sees that $t_4 = 3$ and $c_{3 4} = 1$ and from <ref> (2) that $b_2/2/3/4\in C(u_1)$.
It follows that $T_3\setminus C(u_1)\ne \emptyset$ and then $C(u_1) = \{1, 4\}\cup T_4\cup C_{3 4}$.
Further, it follows from Corollary <ref>(5.1) that $6, 7\in S_u$.
Hence, $W_4 = \{4, 2, 6, 7, 3\}$ and $W_2 = \{2, 3, 5\}$.
Then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (12, 2, T_3)$.

Case 2.4. $\{i_1, i_2\} = \{1, 3\}$ assume w.l.o.g. $c(v u_3) = 5$.

If $5\not\in C(u_1)$ and $H$ contains no $(5, i)_{(u_1, u_i)}$-path for each $i\in [2, 4]$,
$uu_1\to 5$ reduces the proof to the Case 2.2.
Hence, we proceed with the following proposition, or otherwise we are done:

$5/2/3/4\in C(u_1)$, $5\in S_u\setminus C(u_3)$,
and if $5\not\in C(u_1)$, then $H$ contains a $(5, i)_{(u_1, u_i)}$-path for some $i\in [2, 4]$.

Assume that $H$ contains a $(\rho, 8)_{(u_3, v_\rho)}$-path for some $\rho\in [6, 7]$ and $8\in T_3$.
If $\rho\not\in C(u_1)$ and $H$ contains no $(\rho, i)_{(u_1, u_i)}$-path for each $i\in [2, 4]$,
then $(u u_1, u v)\overset{\divideontimes}\to (\rho, 8)$.
Hence, we proceed with the following proposition, or otherwise we are done:

If $H$ contains a $(\rho, 8)_{(u_3, v_\rho)}$-path for some $\rho\in [6, 7]$ and $8\in T_3$,
then $\rho\in C(u_1)$ or $H$ contains a $(\rho, i)_{(u_1, u_i)}$-path for some $i\in [2, 4]$.

(2.4.1.) $c(v u_4) = 3$. One sees that $T_1\subseteq C(u_3)\cap C(u_4)$, $T_4\subseteq C(u_1)$.

$G$ admits an acyclic edge $(\Delta + 2)$-coloring, or $T_3\ne \emptyset$ and $T_4\ne \emptyset$.

We consider the following two cases on whether $T_3 = \emptyset$ or not.

Case 1. $T_3 = \emptyset$, i.e., $C(u_3) = \{3, 5\}\cup T_{uv}$.
There are two subcase on whether $T_4 = \emptyset$ or not.

Case 1.1. $C(u_4) = \{3, 4\}\cup T_{u v}$.
If $4\not\in C(u_1)$, then $(u u_4, u v)\overset{\divideontimes}\to (6, 4)$.
If $5\not\in C(u_1)$, then $(v u_3, u v)\overset{\divideontimes}\to (2, 5)$.
Otherwise, $4, 5\in C(u_1)$ and then $T_1\ne \emptyset$.
If $H$ contains no $(6, j)_{(u_3, v_6)}$-path for some $j\in T_1$,
then $(u u_3, u v)\overset{\divideontimes}\to (6, j)$.
Otherwise, $H$ contains a $(6, T_1)_{(u_3, v_6)}$-path.
Then $(u u_3, u u_4, u v)\overset{\divideontimes}\to (4, 6, T_1)$.

Case 1.2. $T_4 \ne \emptyset$.
One sees that $H$ contains a $(1, T_4)_{(u_1, v_1)}$-path.
For some $j\in T_4$, if $H$ contains no $(2, j)_{(u_2, u_4)}$-path, then $(u u_3, u u_4)\overset{\divideontimes}\to (4, j)$;
if $H$ contains no $(i, j)_{(u_4, v_i)}$-path for each $i\in [6, 7]$, then $(v u_4, v u_3, u u_3)\to (j, 3, 5)$, these reduce to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{1\}$.
Otherwise, $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path and a $(6/7, T_4)_{(u_4, v)}$-path.
It follows $2\in C(u_2)$, $T_4\subseteq C(u_2)$, and there exists a $a_3\in [6, 7]\cap C(u_4)$.
Assume w.l.o.g. $6\in C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path, where $8\in T_4$.
One sees from <ref> that $5\in S_u\setminus C(u_3)$.

If $3\not\in C(u_1)$ and $H$ contains no $(2, 3)_{(u_1, u_2)}$-path,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (3, 1, 8)$.
If $6\not\in C(u_1)$ and $H$ contains no $(i, 6)_{(u_1, u_i)}$-path-path for each $i\in \{2, 4\}$,
then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
If $4\not\in C(u_2)$ and $H$ contains no $(1, 4)_{(u_1, u_2)}$-path,
then $(u u_2, u u_3, u u_4)\to (4, 2, 8)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{1\}$.
Otherwise, $3\in C(u_1)$ or $H$ contains a $(2, 3)_{(u_1, u_2)}$-path;
$6\in C(u_1)$ or $H$ contains a $(6, 2/4)_{(u_1, u)}$-path;
and $4\in C(u_2)$ or $H$ contains a $(1, 4)_{(u_1, u_2)}$-path.
Since $(v u_3, u u_3)\to (2, 5)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{1, 2\}$, $T_1\subseteq C(u_2)$ and it follows that for each $j\in T_{u v}$,
mult$_{C(u_1)\cup C(u_2)\cup C(u_4)}(j)\ge 2$.

One sees that $2t_{uv} + |\{1\}| + |\{2\}| + |\{4, 3, 2, 6\}| + |\{3, 5\}| = 2\Delta - 4 + 8 = 2\Delta + 4$.
When $T_1 = \emptyset$, i.e., $w_1\le 2$,
it follows that $C(u_1) = [1, 2]\cup T_{u v}$ and $[3, 6]\subseteq C(u_2)$,
thus, $d(u_1) + d(u_2) + d(u_4)\ge 2\Delta + 4 + |\{2\}| + |\{4, 6\}| = 2\Delta + 7$.

In the other case, $T_1 \ne \emptyset$.
It follows that $2/4\in C(u_1)$.
If $6\in C(u_1)\cup C(u_2)$,
then $d(u_1) + d(u_2) + d(u_4)\ge 2\Delta + 4 + |\{2/4\}| + |\{1/4\}| + |\{6\}| = 2\Delta + 7$.
Otherwise, $6\not\in C(u_1)\cup C(u_2)$.
It follows that $4\in C(u_1)$, $H$ contains a $(4, 6)_{(u_1, u_4)}$-path.
If $2\not\in C(u_1)$, then $(u u_2, u u_4, u v)\overset{\divideontimes}\to (6, 8, 2)$.
Otherwise, $2\in C(u_1)$ and thus $d(u_1) + d(u_2) + d(u_4)\ge 2\Delta + 4 + |\{2, 4\}| + |\{1/4\}|= 2\Delta + 7$.

One sees clearly that none of ($A_{8.1}$)–($A_{8.4}$) holds.

Case 2. $C(u_4) = [3, 4]\cup T_{uv}$.
Then $(v u_4, u u_4)\to (4, 5)$ reduces the proof to Case 1.

By Lemma <ref>, hereafter we assume $T_3\ne\emptyset$ and $T_4\ne \emptyset$.

(2.4.1.1.) When there exists a $j\in T_{u v}\setminus (C(u_3)\cup C(u_4))$,
say $8\not\in C(u_3)\cup C(u_4)$,
then $2\in C(u_3)$, $6\in C(u_4)$,
and $H$ contains $(2, 8)_{(u_2, u_3)}$-path, a $(6, 8)_{(u_4, v_6)}$-path.
It follows from Corollary <ref>(5.2) that $H$ contains a $(2, T_3)_{(u_2, u_3)}$-path,
from Corollary <ref>(5.1) that $7\in S_u$,
and from Corollary <ref>(5.2) and (3) that $4\in B_1$, and $4\in C(u_2)$, or $H$ contains a $(2, 4)_{(u_2, u_3)}$-path.

If $3\not\in C(u_2)$ and $H$ contains a $(3, 1)_{(u_1, u_2)}$-path,
then $(u u_2, u u_3)\to (3, 8)$, and if $2\not\in C(u_1)$, $u v\to 2$,
or otherwise, $u v\overset{\divideontimes}\to T_1$.
If $6\not\in C(u_1)$ and $H$ contains a $(6, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$ if $H$ contains no $(4, 6)_{(u_1, u_4)}$-path,
or otherwise, $(u u_1, u u_4, u v)\overset{\divideontimes}\to (6, 8, 4)$.
In the other case, we proceed with the following proposition, or otherwise we are done:

$3\in C(u_2)$ or $H$ contains a $(3, 1)_{(u_1, u_2)}$-path.

$6\in C(u_1)$ or $H$ contains a $(6, i)_{(u_1, u_i)}$-path for some $i\in [2, 3]$.

One sees from Corollary <ref>(7.1) that $2\in C(u_1)$ or $1\in S_u$.
Then $|\{1, 4\}| + |\{2\}| + |\{2, 3, 5\}| + |\{4, 3, 6\}| + |\{4, 3, 6, 7, 5, 1/2\}| + 2t_{u v} + t_0 = 2\Delta - 4 + 15 + t_0 = 2\Delta + 11 + t_0$.

* $H$ contains no $(7, 8)_{(u_3, v_7)}$-path.
It follows from Corollary <ref>(4) that $5\in B_1\subseteq C(u_1)$, and $H$ contains a $(5, i)_{(u_3, u_i)}$-path for some $i\in \{2, 4\}$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + t_0 + |\{5\}| = 2\Delta + 12 + t_0$.
One sees from $1, 4, 5, 6/2/3\in C(u_1)$ that $t_1\ge 2$.
Hence, $T_1\setminus T_0\ne \emptyset$ and then $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path.
It follows from Corollary <ref>(5.2) that $2\in C(u_1)\cup C(u_4)$,
and from Corollary <ref>(5.1) that $1\in C(u_4)$, or $H$ contains  a $(1, i)_{(u_4, u_i)}$-path for some $i\in \{2, 3\}$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge \sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 4, 5\}| + |\{2\}| + |\{3, 5, 2\}| + |\{3, 4, 6\}| + |\{4, 3, 6, 7, 5, 1, 2\}| + 2t_{u v} + t_0= 2\Delta + 13 + t_0$.
It follows that $\sum_{x\in N(u)\setminus \{v\}}d(x) = 2\Delta + 13$, $t_0 = 0$,
mult$_{S_u}(i) = 1$, $i = 1, 7$, mult$_{S_u}(j) = 2$, $i = 2, 3, 6$, and mult$_{S_u}(5) = 3$.

If $5\not\in C(u_2)$, one sees that $H$ contains a $(4, 5)_{(u_3, u_4)}$-path,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (5, 8, T_1)$.
Otherwise, $5\in C(u_2)\setminus C(u_4)$.
When $H$ contains a $(7, 4)_{(u_4, v_7)}$-path, one sees that $7\not\in S_u\setminus C(u_4)$,
if $H$ contains no $(7, j)_{(u_3, v_7)}$-path for some $j\in T_1$,
then $(u u_3, u v)\to (7, j)$; otherwise, $(u u_2, u u_3, u v)\to (7, T_3, T_1)$.
In the other case, $H$ contains no $(7, 4)_{(u_4, v_7)}$-path.
If $H$ contains no $(4, 6)_{(u_4, v_6)}$-path, then $(v u_4, u u_4, u v)\to (4, 8, T_1)$.
Otherwise, $H$ contains a $(4, 6)_{(u_4, v_6)}$-path.
We distinguish the following three scenarios.

* $6\in C(u_3)\setminus (C(u_1)\cup C(u_2))$.
It follows from <ref>(2) that $3\in C(u_1)$,
and then from <ref>(1) that $1\in C(u_2)$.
Hence, $1\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (6, 1, 8)$.

* $6\in C(u_2)\setminus (C(u_1)\cup C(u_3))$.
One sees from <ref>(2) that $2\in C(u_1)$ and $H$ contains $(2, 6)_{(u_1, u_2)}$-path.
It follows that $1\in C(u_4)$ or $H$ contains a $(1, 3)_{(u_3, u_4)}$-path,
i.e., $1\not\in C(u_2)$, and $1\not\in C(u_3)$ or $H$ contains a $(1, 3)_{(u_3, u_4)}$-path.
Then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (1, 6, 2, 8)$.

* $6\in C(u_1)\setminus (C(u_2)\cup C(u_3))$.
If $H$ contains no $(1, 6)_{(u_i, u_1)}$-path,
then if $H$ contains no $(6, j)_{(u_3, v_6)}$-path for some $j\in T_1$,
then $(u u_3, u v)\to (6, j)$; otherwise, $(u u_2, u u_3, u v)\to (6, T_3, T_1)$.
Otherwise, $H$ contains a $(6, 1)_{(u_i, u_1)}$-path for some $i\in [2, 3]$.

If $1\in C(u_3)\setminus (C(u_2)\cup C(u_4))$,
one sees that $H$ contains a $(1, 6)_{(u_1, u_3)}$-path and from <ref>(1) that $3\in C(u_2)\setminus C(u_1)$,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (3, 1, 6, 8)$.
Otherwise, $1\in C(u_2)\setminus (C(u_3)\cup C(u_4))$ and $H$ contains a $(1, 6)_{(u_1, u_2)}$-path.
It follows that $2\in C(u_4)\cap B_3$ and $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
Then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (2, 6, 1, 8)$.

* $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.
One sees clearly that $7\in C(u_3)$.
When $7\not\in S_u\setminus C(u_3)$,
if $1\not\in S_u$, then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (7, 1, 8)$.
if $H$ contains no $(3, 7 )_{(u_1, u_3)}$-path, then $(u u_3, u v)\overset{\divideontimes}\to (7, 8)$.
Otherwise, $1\in S_u$, $3\in C(u_1)$ and $H$ contains a $(3, 7)_{(u_1, u_3)}$-path.
If $H$ contains no $(2, j)_{(u_1, u_2)}$-path for some $j\in T_1$,
then $(u u_4, u u_1, u v)\to (7, j, 8)$.
If $H$ contains no $(4, j)_{(u_1, u_4)}$-path for some $j\in T_1$,
then $(u u_2, u u_1, u v)\to (7, j, 8)$.
Otherwise, $2\in C(u_1)$ and $T_1\subseteq C(u_2)\cap C(u_4)$.
Since, $1, 4, 3, 2\in C(u_1)$, $t_0\ge t_1\ge 2$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge \sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 4, 2, 3\}| + |\{2\}| + |\{3, 5, 2, 7\}| + |\{3, 4, 6\}| + |\{4, 6, 5, 1\}| + 2t_{u v} + 2 = 2\Delta + 14$.

In the other case, $7\in S_u\setminus C(u_3)$.
One sees from <ref> (2) that $6/2/3\in C(u_1)$.
It follows that $t_1\ge 1$.
One sees that $|\{1, 4\}| + |\{2\}| + |\{2, 3, 5, 7\}| + |\{4, 3, 6\}| + |\{4, 3, 6, 7, 5, 1/2\}| + 2t_{u v} + t_0 = 2\Delta - 4 + 16 + t_0 = 2\Delta + 12 + t_0$.
It follows that $t_0\ge t_1\ge 1$, or if $T_1\setminus T_0\ne \emptyset$,
then $1\in S_u$ and $2\in S_u\setminus C(u_3)$.
Hence, mult$_{S_u\setminus C(u_3)}(5) = 1$, mult$_{S_u}(3) = 1$, mult$_{S_u\setminus C(u_4)}(6) = 1$,
and then from <ref> that $3\in C(u_1)$, $5\in C(u_4)$ and $H$ contains a $(3, 5)_{(u_1, u_4)}$-path,
from <ref>(1) that $1\in C(u_2)$ and $H$ contains a $(1, 3)_{(u_1, u_2)}$-path.

* $T_1\subseteq T_0$, it follows that $t_1 = 1$, $2\not\in C(u_1)\cup C(u_4)$,
$C(u_3) = \{1, 3, 4\}\cup T_{uv}$ and from <ref>(2) that $6\in C(u_3)$ and $H$ contains a $(3, 6)_{(u_1, u_3)}$-path.
One sees clearly that $6\not\in C(u_1)\cup C(u_2)$.
Then $(u u_2, u u_4, u v)\overset{\divideontimes}\to (6, 8, 2)$.

* $T_1\ne \emptyset$.
It follows from Corollary <ref>(5.2) and (3) that $2\in B_1\cup B_3$ and $2\in C(u_1)\cup C(u_4)$,
$1\in C(u_4)$ or $H$ contains a $(1, i)_{(u_4, u_i)}$-path for some $i\in [2, 3]$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3, 4\}| + |\{2, 1\}| + |\{3, 5, 2, 7\}| + |\{4, 3, 6, 5\}| + |\{4, 6, 7, 2\}| + 2t_{u v} + t_0 = 2\Delta - 4 + 17 + t_0 = 2\Delta + 13 + t_0$.
It follows that $\sum_{x\in N(u)\setminus \{v\}}d(x) = 2\Delta + 13$, $t_0 = 0$, mult$_{S_u}(1) = 1$.
Further, we have $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
One sees that $T_1\cap C(u_2) = \emptyset$.
If $H$ contains no $(4, 5)_{(u_2, u_4)}$-path,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (5, T_3, T_1)$. Otherwise, $H$ contains a $(4, 5)_{(u_2, u_4)}$-path.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (5, 1, 8)$.

(2.4.1.2.) $T_{u v} \subseteq C(u_3)\cup C(u_4)$.
Let $C(u_3) = W_3\cup T_4\cup C_{34}$, $C(u_4) = W_4\cup T_3\cup C_{34}$.
Then $T_3\cup T_4\subseteq B_1\subseteq C(u_1)$.
One sees that if $2\not\in C(u_4)$ or $H$ contains no $(2, j)_{(u_2, u_4)}$-path,
then $4\in B_1$ and mult$_{S_u}(4)\ge 2$,
if for some $j\in T_3$, $H$ contains no $(i, j)_{(u_3, v_i)}$-path for each $i\in [6, 7]$,
then $5\in B_1$ and mult$_{S_u}(5)\ge 2$.
If $H$ contains no $(4, i)_{(u_4, v_i)}$-path for each $i\in C(v)$ and contains no $(2, j)_{(u_2, 4_4)}$-path for some $j\in T_4$,
then $(v u_4, u u_4)\to (4, j)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v)= \{1\}$.
Otherwise, we proceed with the following proposition:

$H$ contains a $(4, i)_{(u_4, v_i)}$-path for some $i\in C(v)$ or a $(2, T_4)_{(u_2, u_4)}$-path.

When $T_1 = \emptyset$, i.e., $W_1\le 2$. One sees from <ref> that $W_1 = \{1, 5\}$,
or $W_1 = \{1, i\}$ and $H$ contains a $(5, i)_{(u_1, u_i)}$-path, where $i\in \{2, 3, 4\}$.

Assume $C(u_1) = \{1, i\}\cup T_{u v}$.
We assume w.l.o.g. $6\in C(u_3)$, $H$ contains a $(6, 8)_{(u_3, v_6)}$-path, where $8\in T_3$.
If $H$ contains no $(i, 6)_{(u_1, u_i)}$-path,
then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
If $7\not\in S_u$, then $(u u_1, u v)\overset{\divideontimes}\to (7, 8)$,
or $(u u_i, u u_1, u v)\overset{\divideontimes}\to (7, 5, 8)$.
If $1\not\in C(u_i)$ and $H$ contains no $(1, j)_{(u_i, u_j)}$-path for each $j\in C(u)$,
then $(u u_i, u u_1, u v)\overset{\divideontimes}\to (1, 5, 8)$.
Otherwise, we proceed with the following proposition:

If $C(u_1) = \{1, i\}\cup T_{uv}$, then $H$ contains a $(i, 6)_{(u_1, u_i)}$-path, $1, 7\in S_u$,
and if $1\not\in C(u_i)$, then $H$ contains a $(1, j)_{(u_i, u_j)}$-path for some $j\in C(u)$.

* $W_1 = \{1, 5\}$.
Since $4\not\in C(u_1)$, $2\in C(u_4)$,
and $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path, i.e., $T_4\subseteq C(u_2)$.
It follows that $t_0\ge t_4\ge 2$.
Since $u u_1\to i\in [6, 7]$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{3, i\}$, $T_3\cup T_4\subseteq B_6\cap B_7$.
It follows that $b_3 = 5$ and $H$ contains a $(5, T_4)_{(u_3, u_4)}$-path.
If $1\not\in C(u_2)$ and $H$ contains no $(1, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$,
then $(u u_2, u u_1, u v)\overset{\divideontimes}\to (1, 2, T_3)$.
If $1\not\in C(u_4)$ and $H$ contains no $(1, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
then $(u u_4, u u_1, u v)\overset{\divideontimes}\to (1, 4, T_3)$.
If $1\not\in C(u_3)$ and $H$ contains no $(1, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
then $(u u_3, u u_1, u v)\overset{\divideontimes}\to (1, 3, T_4)$.
It follows that mult$_{S_u}(1)\ge 2$.
One sees that the same argument applies if $u u_1\to 6/7$.
Hence, $_{S_u}(i)\ge 2$, $i = 6, 7$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 5\}| + |\{2\}| + |\{a_3, 3, 5\}| + |\{4, 3, 2, 5\}| + 2|\{1, 6, 7\}| + 2t_{u v} + t_0 \ge 16 + 2t_{u v} + 2 = 2\Delta + 14$.

* $W_1 = \{1, 2\}$ and $H$ contains a $(2, 5)_{(u_1, u_2)}$-path, $5\in C(u_2)$.
Since $4\not\in C(u_1)$, $2\in C(u_4)$, and $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path, i.e., $T_4\subseteq C(u_2)$.
It follows that $t_0\ge t_4\ge 2$.
It follows from <ref> that $6\in C(u_2)$, $1, 7\in S_u$, $1/3/4\in C(u_2)$,
and if $1\not\in C(u_2)$, then $H$ contains a $(1, i)_{(u_2, u_i)}$-path for some $i\in [3, 4]$.

One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge\sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 2\}| + |\{2, 5, 6\}| + |\{3, 5, a_3, 6\}| + |\{4, 3, b_3, 2\}| + |\{1\}| + 2t_{u v} + 2 = 2\Delta + 12$.
Since $t_3\ge 2$, $T_3\ne\emptyset$.
It follows that $4\in C(u_3)$ and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.
If $5\not\in C(u_4)$, then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (5, T_3\cap T_2, 4)$.
Otherwise, $5\in C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge \sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 2\}| + |\{2, 5, 6\}| + |\{4, 6, 3, 5\}| + |\{4, 3, 5, 2\}| + |\{1, 7\}| + 2t_{u v} + 2 = 2\Delta + 13$.
It follows that $1\in C(u_2)\setminus (C(u_3)\cup C(u_4))$ and $4\not\in C(u_2)$.
Then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (5, 4, 1, 8)$.

* $W_1 = \{1, 3\}$ and $H$ contains a $(3, 5)_{(u_1, u_4)}$-path, $5\in C(u_4)$.
Since $4\not\in C(u_1)$, $2\in C(u_4)$, and $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path, i.e., $T_4\subseteq C(u_2)$.
It follows that $t_0\ge t_4\ge 2$.
If $4, 5\not\in C(u_2)$, $(u u_2, u v)\to (5, 2)$ or $(u u_2, u u_1, u v)\to (5, 2, T_3)$.
Otherwise, $4/5\in C(u_2)$.
It follows from <ref> that $1, 7\in S_u$ and $H$ contains a $(3, 6)_{(u_1, u_3)}$-path.
If $6\not\in C(u_2)\cup C(u_4)$, then $(u u_4, u v)\overset{\divideontimes}\to (6, 4)$,
or $(u u_4, u u_1, u v)\overset{\divideontimes}\to (6, 4, T_4)$.
Otherwise, $6\in C(u_2)\cup C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge \sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 3\}| + |\{2, 4/5\}| + |\{a_3, 6, 3, 5\}| + |\{4, 3, 5, 2\}| + |\{1, 6, 7\}| + 2t_{u v} + 2 = 2\Delta + 13$.
It follows that $t_0 = t_4 = 2$ and $T_3\setminus C(u_2)\ne \emptyset$.
Hence $a_3 = 4$, and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.
One sees that $2\not\in C(u_3)$. Then $(u u_2, u v)\overset{\divideontimes}\to (T_3, 2)$.

* $W_1 = \{1, 4\}$ and $H$ contains a $(4, 5)_{(u_1, u_4)}$-path, $5\in C(u_4)$.

It follows from <ref> that $1, 7\in S_u$, $6\in C(u_4)$, $H$ contains a $(4, 6)_{(u_1, u_4)}$-path,
and if $1\not\in C(u_4)$, then $H$ contains a $(1, i)_{(u_4, u_i)}$-path for some $i\in [2, 3]$.
It follows that $1\in C(u_3)\cup C(u_4)$ or $2\in C(u_4)$.
One sees clearly that $7\in C(u_2)\setminus (C(u_3)\cup C(u_4))$,
and if $2\not\in C(u_4)$, then $4\in B_1$.
It follows from <ref> that $2\in C(u_4)$, $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path and $T_4\subseteq C(u_2)$.
One sees that $t_0\ge t_4\ge 3$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge \sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 4\}| + |\{2, 1, 7\}| + |\{a_3, 6, 3, 5\}| + |\{4, 3, 5, 6, 2\}| + 2t_{u v} + 3 = 2\Delta + 13$.
It follows that $t_0  = t_4 = 3$ and $a_3 = 4$, $2\not\in C(u_3)$.
Then $(u u_2, u v)\overset{\divideontimes}\to (T_3, 2)$.

In the other case, $T_1\ne\emptyset$.
Since $T_1\subseteq C(u_3)\cap C(u_4)\subseteq C_{34}$, it follows that $c_{34}\ge t_1\ge 1$.
One sees from $\Delta + 7\ge d(u_3) + d(u_4)\ge w_3 + w_4 + t_{u v} + c_{34}$ that $w_3 + w_4\le 9 - c_{34}\le 8$ and $w_3, w_4\le 5$.
One sees from Corollary <ref>(7.2) that if $1\not\in C(u_4)$ and $H$ contains no $(1, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
then $2\in C(u_1)$ and $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path, $T_1\subseteq C(u_2)$.

If $1\not\in C(u_3)$, $H$ contains no $(1, i)_{(u_i, u_i)}$-path for each $i\in \{2, 4\}$,
and $3\not\in S_u\setminus C(u_4)$,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (3, 1, T_4)$.
Otherwise, we proceed with the following proposition:

If $1\not\in C(u_3)$ and $H$ contains no $(1, i)_{(u_i, u_i)}$-path for each $i\in \{2, 4\}$,
then $3\in C(u_1)\cup C(u_2)$.

$1\in S_u$ or $3\in S_u\setminus C(u_4)$.

Assume that $d(u_1) = \Delta$ and $C(u_1)\setminus \{1\} = B_1 = T_3\cup T_4\cup \{4, 5\}$ or $B_1 = T_3\cup T_4\cup \{4, 5, 2\}$.
One sees that $C(v_1) = C(u_1)$, and it follows from Lemma <ref> that $H$ contains a $(5, C_{34})_{(u_3, v_1)}$-path.
If for some $\gamma\in [6, 7]$, $H$ contains no $(j, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
one sees that the same argument applies if $u u_1\to \gamma$, and thus $C(v_1)\setminus \{1\} = C(v_\gamma)\setminus \{\gamma\} = C(u_1)\setminus \{1\}$,
then $(u u_1, v v_\gamma, u v)\overset{\divideontimes}\to (\gamma, C_{34}, T_3)$.
Hence, we proceed with the following proposition, or otherwise we are done:

If $d(u_1) = \Delta$, $C(u_1)\setminus \{1\} = B_1 = T_3\cup T_4\cup \{4, 5\}$ or $B_1 = T_3\cup T_4\cup \{4, 5, 2\}$, then for each $j\in [6, 7]$, $H$ contains a $(i, j)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.

We distinguish the following two scenarios:

* For some $\alpha\in T_3$, $H$ contains no $(i, \alpha)_{(u_3, v_i)}$-path for each $i\in [6, 7]$.
It follows from Corollary <ref>(4) that $5\in B_1\subseteq C(u_1)\cap C(v_1)$,
and $H$ contains a $(5, i)_{(u_3, u_i)}$-path for some $i\in \{2, 4\}$.

When $H$ contains no $(2, \beta)_{(u_2, u_4)}$-path for some $\beta\in T_4$.
It follows from Corollary <ref>(4) that $4\in B_1\subseteq C(u_1)\cap C(v_1)$,
and if $4\not\in C(u_3)$, then $H$ contains a $(2, 4)_{(u_2, u_3)}$-path.

* $1\not\in C(u_3)\cup C(u_4)$ and $2\not\in C(u_4)$.
It follows that $2\in C(u_1)$, $T_1\subseteq C(u_2)$, $c_{34}\ge t_1\ge 2$.
One sees that $|\{1, 2, 4, 5\}| + |\{2\}| + |\{3, 5\}| + |\{3, 4\}| + 2t_{u v} = 2(\Delta - 2) + 9 = 2\Delta + 5$.

* $T_3\subseteq C(u_2)$. One sees that $t_0\ge t_1 + t_3\ge 2 + 1 = 3$.
First assume $2, 4\in C(u_3)$. Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 5 + |\{2, 4, 5, 1/3\}| + t_0\ge 2\Delta + 9 + t_1 + t_3\ge 2\Delta + 9 + 2 + 2 = 2\Delta + 13$,
it follows that $t_1 = t_3 = 2$, $d(u_1) = \Delta = 7$, $C(u_1) = \{1, 5, 4, 2\}\cup T_3\cup T_4$, and $b_3 = 5$, $5\not\in C(u_2)$, $6, 7\not\in S_u$.
If $2\not\in B_1$, then $(u u_2, v u_3, u v)\overset{\divideontimes}\to (5, \alpha, 2)$.
Otherwise, $2\in B_1$.
Hence, by <ref>, we are done.
Next, assume that $|C(u_3)\cap \{2, 4\}| = 1$.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 5 + |\{5, 1/3, 6, 7, a_3\}| + t_0\ge 2\Delta + 10 + t_1 + t_3\ge 2\Delta + 10 + 2 + 1 = 2\Delta + 13$,
it follows that $t_1 = 2$, $t_3 = 1$,
and since $4\in C(u_2)\cup C(u_3)$, we have $a_3 = 4$, $2\not\in C(u_3)\cup C(u_4)$,
and $H$ contains a $(4, 5)_{(u_3, u_4)}$-path, $5\in C(u_4)\setminus C(u_2)$.
Then $(u u_2, v u_3)\to (5, \alpha)$,
and $u v \overset{\divideontimes}\to 2$, or $(uu_3, uv)\overset{\divideontimes}\to (2, C_{34})$.

* $T_3\setminus C(u_2)\ne \emptyset$.
Then $a_3 = 4$ and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.
It follows from Corollary <ref>(5.2) and (3) that $2\in B_1\cap C(u_3)$,
from Corollary <ref>(5.1) that $6, 7\in S_u$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 5 + |\{5, 1/3, 6, 7, 2, 4\}| + t_0\ge 2\Delta + 11 + t_1 \ge 2\Delta + 11 + 2 = 2\Delta + 13$,
it follows that $c_{34} = t_1 = 2$, $d(u_1) = \Delta = 7$, $C(u_1) = \{1, 5, 4, 2\}\cup T_3\cup T_4$, and $W_3 = \{3, 5, 2, 4\}$, $W_4 = \{4, 3, b_3\}$.
Hence, it follows from <ref>(5.1) that $H$ contains a $(2, i)_{(u_2, u_3)}$-path for each $i\in [6, 7]\setminus \{b_3\}$.
Thus, by <ref>, we are done.

* $1\in C(u_3)\cup C(u_4)$ or $2\in C(u_4)$.
One sees that $|\{1, 4, 5\}| + |\{2\}| + |\{3, 5\}| + |\{3, 4\}| + 2t_{u v} = 2(\Delta - 2) + 8 = 2\Delta + 4$.
When $2, 4\in C(u_3)$, $w_3 + w_4 \ge |\{2, 4, 3, 5\}| + |\{4, 3, b_3\}| + |\{1/2\}| = 8$,
it follows that $W_4\subseteq \{4, 3, 1/2, b_3\}$, $c_{34} = t_1 = 1$,
and $d(u_1) = \Delta = 7$, $C(u_1) = \{1, 5, 4\}\cup T_3\cup T_4$.
One sees that there exists an $i\in [6, 7]\setminus C(u_4)$.
Thus, by <ref>, we are done.
Hence, $C(u_3)\cap \{2, 4\} = a_3$.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$.

* $a_3 = 2$. Then $H$ contains a $(2, T_3\cup \{4, 5\})_{(u_2, u_3)}$-path.
If $3\not\in C(u_1)\cup C(u_2)$,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (3, T_3, T_1)$.
Otherwise, $3\in C(u_1)\cup C(u_2)$.
First, assume that $T_1\subseteq C(u_2)$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 4 + |\{4, 5, 2, 6, 7, 3, 1/2\}| + t_0\ge 2\Delta + 11 + t_3 + t_1\ge 2\Delta + 11 + 1 + 1 = 2\Delta + 13$,
it follows that $t_3 = 1$, $c_{34} = t_1 = 1$,
$d(u_1) = \Delta = 7$, $C(u_1) = \{1, 5, 4\}\cup T_3\cup T_4$.
It follows from <ref> that $H$ contains a $(4, i)_{(u_1, u_4)}$-path for each $i\in [6, 7]$ and $W_4 = \{3, 4, 6, 7, 1/2\}$.
Thus, by <ref>, we are done.
Next, assume that $T_1\setminus C(u_2)\ne \emptyset$.
It follows from Corollary <ref>(5.2) that $2\in C(u_1)\cup C(u_4)$ and $1\in S_u$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 4 + |\{4, 5, 2, 6, 7, 3, 1, 2\}| + t_0\ge 2\Delta + 11 + t_3\ge 2\Delta + 12 + 1 = 2\Delta + 13$,
it follows that $t_3 = 1$, $W_3 = \{3, 5, 2\}$, $1/2\in C(u_4)$, $b_3\in [6, 7]$ and mult$_{S_u}(1) = 1$, mult$_{S_u}(i) = 2$, $i\in [2, 3]$.
One sees that if $1\not\in C(u_4)$ and $H$ contains no $(1, 2)_{(u_2, u_4)}$-path,
then it follows from Corollary <ref>(7.2) that $2\in C(u_1)$ and $T_1\subseteq C(u_2)$.
Hence, assume that $1\in C(u_4)$ or $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
One sees clearly that $H$ contains no $(1, 2)_{(u_2, u_3)}$-path.
If $2, 3\not\in C(u_1)$, then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (3, 1, T_4)$.
Otherwise, $2/4\in C(u_1)$.
It follows that $c_{34} = t_1 = 2$, $W_1 = \{1, 5, 4, 2/3\}$, $W_4 = \{3, 4, 6, 2\}$ and $6\not\in C(u_2)$.
It follows from <ref> that $H$ contains a $(4, 6)_{(u_4, v_6)}$-path.
Then $(u u_1, u v)\overset{\divideontimes}\to (6, T_4)$.

* $a_3 = 4$. Then $H$ contains a $(4, T_3\cup \{5\})_{(u_3, u_4)}$-path.
First, assume $2\in C(u_1)\cap C(u_4)$.
It follows that $c_{34} = t_1 = 2$, $W_1 = \{1, 4, 5, 2\}$, $W_4 = \{4, 3, 5, 2\}$,
and $6, 7\in C(u_2)$.
One sees from Corollary <ref>(5.1) that $H$ contains a $(2, 6)_{(u_2, u_4)}$-path.
If $3\not\in C(u_2)$, then $(u u_1, u u_3, u v)\to (3, 1, T_4)$.
Otherwise, $3\in C(u_2)$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 4 + |\{2, 6, 7, 3, 4, 2, 5\}| + t_0\ge 2\Delta + 11 + t_0$ and $t_1 + t_3\ge 2 + 1 = 3$,
it follows that $(T_3\cup T_1)\setminus C(u_2)\ne \emptyset$.
Hence, $2\in B_1$.
Thus, by <ref>, we are done.
Next, assume that $2\not\in C(u_1)\cap C(u_4)$.
It follows that $T_1\cup T_3\subseteq C(u_2)$.
If $5\not\in C(u_2)$, then $(u u_2, v u_3)\to (5, \alpha)$,
and by Lemma <ref>(2), we are done.
Otherwise, $5\in C(u_2)$.
If $H$ contains no $(3, j)_{(u_2, u_3)}$-path for some $j\in T_4\cap T_2$,
then $u u_2\to j$ and by Lemma <ref>(2), we are done.
Otherwise, $T_4\subseteq C(u_2)$ or $3\in T_4$.
One sees that $2\Delta + 4 + |\{5, 6, 7, 4, 5, 1/2\}| + t_0 = 2\Delta + 10 + t_0$ and $t_3 + t_4\ge 3$.
If $T_4\subseteq C(u_2)$, then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0\ge 2\Delta + 10 + t_1 + t_3 + t_4\ge 2\Delta + 10 + 1 + 3 = 2\Delta + 14$.
Otherwise, $T_4\setminus C(u_2)\ne \emptyset$ and $3\in C(u_2)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0 + |\{3\}|\ge 2\Delta + 10 + t_1 + t_3 \ge 2\Delta + 11 + 1 + 1 = 2\Delta + 13$.
It follows that $t_1 = t_3 = 1$, $d(u_1) = \Delta = 7$, $C(u_1) = \{1, 5, 4\}\cup T_3\cup T_4 = B_1\cup \{1\}$, and $1/2\in C(u_4)$.
One sees that $[6, 7]\setminus C(u_4)\ne \emptyset$.
Thus, by <ref>, we are done.

In the other case, $2\in C(u_4)$ and $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path.
It follows that $T_4\subseteq C(u_2)$ and $t_0\ge t_4\ge 2$.
If $2, 4\in C(u_1)\cap C(u_3)$, then $c_{34}\ge t_1\ge 2$ and $d(u_3) + d(u_4)\ge w_3 + w_4 + t_{uv} + c_{u v}\ge |[2, 5]| + |\{2, 3, 4, b_3\}| + \Delta - 2 + c_{34} = \Delta + 6 + c_{34}\ge \Delta + 6 + 2 = \Delta + 8$.
Otherwise, $\{2, 4\}\cap (C(u_1)\cap C(u_3))\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$.
One sees that $|\{1, 5, a_1\}| + |\{2\}| + |\{3, 5, 2/4\}| + |\{2, 3, 4\}| + |\{5, 6, 7, 1/3\}| + 2t_{u v} = 2\Delta + 10$ and $t_1\ge 1$, $t_3\ge 1$.
It follows that $(T_3\cup T_1)\setminus C(u_2)\ne \emptyset$ and from Corollary <ref>(5.2) that $2\in C(u_1)\cup C(u_3)$, say $2\in C(u_{i_1})$.
Then $4\in C(u_{i_1})$ or $T_{i_1}\subseteq C(u_2)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_4 + |4/T_{i_1}|\ge 2\Delta + 13$.
It follows that $t_4 = 2$, $W_4 = \{2, 3, 4, b_3\}$, $\{2, 4\}\cap C(u_{i_2}) = \{4\}$,
mult$_{S_u}(1) +  $mult$_{C(u_2)\cup C(u_1)}(3) = 1$.

* $2\in C(u_3)$ and $\{2, 4\}\cap C(u_1) = 4$.
It follows from Corollary <ref>(5.1) that $1\in S_u$ and then $3\not\in C(u_1)\cup C(u_2)$.
If $H$ contains no $(4, j)_{(u_3, u_4)}$-path,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (3, j, C_{34})$.
otherwise, $4\in C(u_3)$ and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.
It follows that $c_{34} = t_1 = 1$, $W_1 = \{1, 4, 5\}$, $W_3 = [3, 5]$, $W_4 = \{2, 3, 4, b_3\}$.
One sees that $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (3, 1, T_4)$.

* $2\in C(u_1)$ and $\{2, 4\}\cap C(u_3) = 4$.
One sees from Corollary <ref>(7.1) that it $4\not\in C(u_1)$, then $1\in S_u$,
and from Corollary <ref>(5.1) that if $T_1\setminus C(u_2)\ne \emptyset$, then $1\in S_u$.
Hence, $1\in S_u$, and $3\not\in S_u\setminus C(u_4)$.
It follows from <ref> that $1\in C(u_3)$ or $H$ contains  a $(1, 4)_{(u_3, u_4)}$-path.
One sees that $1\not\in C(u_4)$, or $H$ contains a $(1, 4)_{(u_3, u_4)}$-path.
It follows that $c_{34} = t_1 = 1$ and $4\not\in C(u_1)$.
Then $(u u_2, u u_1, u v)\overset{\divideontimes}\to (1, T_1, T_3)$.

* For each $j\in T_3$, $H$ contains a $(6/7, j)_{(u_3, v)}$-path.
Assume w.l.o.g. $H$ contains a $(7, 8)_{(u_3, v_6)}$-path, where $8\in T_3$.
If $7\not\in C(u_1)$ and $H$ contains no $(7, i)_{(u_1, u_i)}$-path for each $i\in [2, 4]$,
then $(u u_1, u v)\to (7, 8)$.
Otherwise, $7\in C(u_1)$ or $H$ contains a $(7, i)_{(u_1, u_i)}$-path for some $i\in [2, 4]$.

One sees that if $\{4\}\cup T_3\cup T_4\subseteq C(v_1)$ and $C_{u v} = \{\mu\}$,
then it follows from Lemma <ref>(1.2) that $C(v_1) = \{1, 4, \mu/5/6/7\}\cup T_3\cup T_4$, i.e., $3\not\in C(v_1)$.
One sees from <ref> that $5\in S_u\setminus C(u_3)$,
and if $C(u_3)\cap \{2, 4\} = \{a_3\}$, then it follows from Corollary <ref>(5.1) that $6\in S_u$.

* $W_3 = 5$, or $c_{34} = 2$.
Since $w_3 + w_4 + c_{34}\le 9$,
it follows that $\Delta = 7$, $d(u_1) + d(u_2)\le 13$, $W_3 = \{3, 4, b_3\}$.
Then $4\in B_1$ and it follows from <ref> that $H$ contains a $(b_3, 4)_{(u_4, v)}$-path.
One sees that if $c_{34} = 2$, $H$ contains no $(1, 3)_{(u_3, u_4)}$-path,
and if $W_3 = 5$, one sees that $t_3 + t_4 = \Delta - 3$, then $3\not\in C(v_1)$.
Hence, it follows from Corollary <ref>(7.2) that $2\in C(u_1)$, $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path and $C_{34}\subseteq C(u_1)\cup C(u_2)$.

* $W_3 = 5$. Then $C(u_1) = \{1, 2, 4\}\cup T_3\cup T_4$ and $T_1\cup \{5, 7\}\subseteq C(u_2)$.
One sees from $2\not\in C(u_4)\cup C(v_1)$ that $2\not\in B_1\cup B_3$.
Hence, it follows from Corollary <ref>(5.2) and (3) that $T_3\subseteq C(u_2)$.
Thus, $d(u_1) = d(u_2) = \Delta$, a contradiction.

* $c_{34} = 2$.
Then $5\in S_u\setminus C(u_3)$ and $6\in S_u$.
One sees that if $a_3 = 2$, then $T_3\subseteq C(u_2)$,
and if $a_3 = 4$, then it follows from Corollary <ref>(5.2) that $T_3\subseteq C(u_2)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 2, 4\}| + |\{2\}| + |\{a_3, 7, 3, 5\}| + |\{4, 3\}| + |\{5, 6, 1/3\}| + 2t_{u v} + t_0 \ge 2\Delta + 9 + t_3 + c_{34}\ge 2\Delta + 9 + 2 + 2 = 2\Delta + 13$.
It follows that $7\not\in S_u\setminus C(u_3)$, and then $3\in C(u_1)$, $1\not\in S_u$.
Then $(uu_1, uu_3, uv)\overset{\divideontimes}\to (1, 7, 8)$.

* In the other case, $W_3 = \{3, 5, a_3, 7\}$, $T_1 = C_{34}$ and $c_{34} = t_1 = 1$.
One sees that $5\in S_u\setminus C(u_3)$ and $6\in S_u$.
If $7\not\in S_u\setminus C(u_3)$ and $H$ contains no $(1, a_3)_{(u_3, u_{a_3})}$-path,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (7, 1, 8)$.
Otherwise, we proceed with the following proposition:

If mult$_{S_u}(7) = 1$, then $H$ contains a $(1, a_3)_{(u_3, u_{a_3})}$-path and $1\in C(u_{a_3})$.

When $H$ contains no $(2, \beta)_{(u_2, u_4)}$-path for some $\beta\in T_4$.
It follows from Corollary <ref>(4) that $4\in B_1\subseteq C(u_1)\cap C(v_1)$,
and if $4\not\in C(u_3)$, then $H$ contains a $(2, 4)_{(u_2, u_3)}$-path.
One sees that if $a_3 = 2$, or if $a_3 = 4$, since $2\not\in B_1\cup B_3$,
then it follows from Corollary <ref>(1.2) or Corollary <ref>(5.2) and (3) that $T_3\subseteq C(u_2)$.
One sees that $7\in S_u\setminus C(u_3)$ or $3\in C(u_1)$,
and it follows from Corollary <ref>(7.1) that $2\in C(u_1)$ or $1\in S_u$.

* $2\in C(u_3)\cap C(u_4)$. Then $4\in C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 4\}| + |\{2, 4\}| + |\{2, 7, 3, 5\}| + |\{4, 3, 2\}| + |\{5, 6, 7/3, 1/2\}| + 2t_{u v} + t_0\ge 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_3 \ge 2\Delta + 13$.
It follows that $T_1\setminus C(u_2)\ne \emptyset$, mult$_{S_u}(3) + $mult$_{S_u}(7) = 3$,
and from Corollary <ref>(5.1) that $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
Further, by <ref> that $3\in C(u_1)\cup C(u_2)$.
Thus, $7\not\in S_u\setminus C(u_3)$, and by <ref>, we are done.

* $2\not\in C(u_3)\cap C(u_4)$.
It follows from Corollary <ref> (5.2) and (4) that $C_{34}\subseteq C(u_2)$,
and if $3\not\in C(u_2)$, then $T_4\subseteq C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 4\}| + |\{2, T_4/3\}| + |\{a_3, 7, 3, 5\}| + |\{4, 3\}| + |\{5, 6, 7/3, 1/2\}| + 2t_{u v} + t_0\ge 2\Delta + 10 + t_3 + c_{34}\ge 2\Delta + 10 + 2 + 1 = 2\Delta + 13$.
It follows that $a_3 = 4$, $2\not\in C(u_4)$, $4\not\in C(u_2)$,
either $2\in C(u_1)$, or $1\in S_u$,
and either $3\in C(u_1)$ or $7\in S_u\setminus C(u_1)$.
If $2\in C(u_1)$, one sees that $1\not\in S_u$,
then $(u u_2, u u_1, u v)\to (1, \{5, 7\}\setminus \{\rho\}, 8)$.
Otherwise, $2\not\in C(u_1)$ and then $1\in S_u$.
It follows from Corollary <ref> (7.1) that $1\in C(u_4)$.
If $H$ contains no $(3, j)_{(u_1, u_3)}$-path for some $j\in \{5, 7\}\setminus C(u_1)$,
then $(u u_2, u u_1, u v)\to (1, j, 8)$.
Otherwise, $3\in C(u_1)$ and $H$ contains a $(3, j)_{(u_1, u_3)}$-path for each $j\in \{5, 7\}\setminus C(u_1)$, i.e., $b_3 = 5$.
Then $(u u_3, u v)\overset{\divideontimes}\to (6, C_{34})$,
or $(u u_4, u u_3, uv)\overset{\divideontimes}\to (6, T_3, C_{34})$[Or since $6\not\in C(u_1)\cup C(u_3)\cup C(u_4)$, by Corollary <ref> (5.1), we are done. ].

In the other case,
$2\in C(u_4)$ and $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path.
It follows that $W_3 = \{3, 5, a_3, 7\}$, $W_4 = \{4, 3, 2, b_3\}$,
$T_4\subseteq C(u_2)$ and $t_0\ge t_4\ge 2$.
One sees that $|\{1, a_1\}| + |\{2\}| + |\{3, 5, 7, a_3\}| + |\{4, 3, 2\}| + |\{5, 6\}| + 2t_{u v} = 2\Delta - 4 + 12 = 2\Delta + 8$.

* $T_3\subseteq C(u_2)$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{1/4\}| + t_0\ge 2\Delta + 9 + t_3 + t_4 = 2\Delta + 13$,
we have $7\not\in S_u\setminus C(u_3)$ and then $3\in C(u_1)$, $1\not\in S_u$.
Thus, by <ref>, we are done.

* $T_3\setminus C(u_2)\ne \emptyset$.
It follows from Corollary <ref> (1.2) that $a_3 = 4$,
and from Corollary <ref>(5.2) that $2\in C(u_1)$.
One sees clearly that $H$ contains no $(1, 4)_{(u_3, u_4)}$-path.
It follows from <ref> that $3\in C(u_1)\cup C(u_2)$.
One sees from Corollary <ref>(7.1) that $4\in C(u_1)$ or $1\in S_u$,
and from Corollary <ref> (1.2) and (5.1) that if $C_{34}\setminus C(u_2)\ne \emptyset$,
then $4\in C(u_1)$ and $1\in S_u$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{3, 1/4\}| + t_0\ge 2\Delta + 10 + t_4 + c_{34} = 2\Delta + 13$ if $C_{34}\subseteq C(u_2)$,
or else, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{3, 1, 4\}| + t_0\ge 2\Delta + 11 + t_4 = 2\Delta + 13$,
we have $7\not\in S_u\setminus C(u_3)$.
Thus, by <ref>, we are done.

(2.4.2.) $c(v u_4) = 1$.
Since $T_{u v}\subseteq C(u_3)\cup C(u_4)$,
let $C(u_3) = W_3\cup T_4\cup C_{3 4}$, $C(u_4) = W_4\cup T_3\cup C_{3 4}$.

If $C(u_3) = \{3, 5\}\cup T_{u v}$, then $(v u_3, u u_3)\to (4, 5)$ reduces the proof to (2.2.1.).
If $C(u_4) = \{1, 4\}\cup T_{u v}$, then $(v u_4, u u_4)\to (4, 5)$ reduces the proof to (2.1.1.).
Otherwise, $T_3\ne \emptyset$, $T_4\ne \emptyset$,
and there exists a $a_3\in \{2, 4\}\cap C(u_3)$, $b_1\in [5, 7]\cap C(u_4)$.
Assume w.l.o.g. $8\in T_3$.

Assume that $1\not\in C(u_3)$ and $H$ contains no $(1, 2)_{(u_2, u_3)}$-path.

If $5\not\in C(u_1)$ and $H$ contains no $(5, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (5, 1, T_3)$.
If for each $j\in [6, 7])\setminus C(u_1)$, $H$ contains no $(j, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $(u u_1, u u_3)\to (j, 1)$ reduces the proof to (2.4.1.).
One sees from Corollary <ref>(2.1) that $3\in C(u_1)$ or $H$ contains a $(3, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.
Hence, we proceed with the following proposition:

If $1\not\in C(u_3)$ and $H$ contains no $(1, 2)_{(u_2, u_3)}$-path,
then for each $j\in \{3, 5, 6, 7\}$,
$j\in C(u_1)$ or $H$ contains a $(j, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.

Assume that $1\not\in C(u_3)$, $H$ contains no $(1, 2)_{(u_2, u_3)}$-path,
and $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$.
One sees from <ref> that for each $i\in \{3, 5, 6, 7\}\setminus C(u_1)$,
$H$ contains a $(2, i)_{(u_1, u_2)}$-path.
If $3\not\in C(u_1)\cup C(u_4)$,
then $(u u_4, u u_3)\to (3, j)$ reduces the proof to (2.2.2.).
If $5\not\in C(u_1)\cup C(u_4)$,
then $(u u_4, u u_3)\to (5, j)$ reduces the proof to (2.2.1.).
If there exists a $i\in [6, 7]\setminus (C(u_1)\cup C(u_3)\cup C(u_4))$,
then $(u u_4, u u_3)\to (i, j)$ reduces the proof to (2.2.2.).
Hence, we proceed with the following proposition:

If $1\not\in C(u_3)$, $H$ contains no $(1, 2)_{(u_2, u_3)}$-path,
and $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
then $3, 5\in C(u_1)\cup C(u_4)$ and $6, 7\in C(u_1)\cup C(u_4)\cup C(u_3)$.

Assume that $H$ contains a $(\rho, T_3)_{(u_3, v_\rho)}$-path for some $\rho\in [6, 7]\cap C(u_3)$.
If $\rho\not\in S_u\setminus C(u_3)$,
one sees from <ref> that $H$ contains a $(3, \rho)_{(u_1, u_3)}$-path:
if $4\not\in B_3$,
then $(u u_4, u v)\overset{\divideontimes}\to (\rho, 4)$;
if $2\not\in B_1\cup B_3$,
then $(u u_2, u v)\overset{\divideontimes}\to (\rho, 2)$;
if $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
then $(u u_4, u u_3)\to (\rho, j)$ reduces the proof to (2.2.2.).
If $1\not\in C(u_3)$, $H$ contains no $(1, 2)_{(u_2, u_3)}$-path,
$H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
and $\rho\not\in C(u_1)\cup C(u_4)$,
one sees from <ref> that $H$ contains a $(2, \rho)_{(u_1, u_2)}$-path,
then $(u u_4, u u_3)\to (\rho, j)$ reduces the proof to (2.2.2.).
Hence, we proceed with the following proposition:

Assume that $H$ contains a $(\rho, T_3)_{(u_3, v_\rho)}$-path, $\rho\in [6, 7]\cap C(u_3)$.

if $\rho\not\in S_u\setminus C(u_3)$,
then $4\in B_3$, $2\in B_1\cup B_3$, and $H$ contains a $(2, T_3)_{(u_2, u_3)}$-path, $2, 4\in C(u_3)$, $T_3\subseteq C(u_2)$.

if $1\not\in C(u_3)$, $H$ contains no $(1, 2)_{(u_2, u_3)}$-path, and $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$, then $\rho\in C(u_1)\cup C(u_4)$.

If $5\not\in C(u_4)$, $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
and $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (5, j, T_4)$.
If $5\not\in C(u_2)$, $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 4\}$,
and $H$ contains no $(4, j)_{(u_4, u_3)}$-path for some $j\in T_3$,
then $(u u_2, u u_3)\to (5, j)$ reduces the proof to (2.3).
One sees from <ref> that if $5\not\in S_u\setminus C(u_3)$, then $H$ contains a $(3, 5)_{(u_1, u_3)}$-path.
In the other case, we proceed with the following proposition:

If $5\not\in C(u_4)$, and $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then $H$ contains a $(2, T_3)_{(u_2, u_3)}$-path and $T_3\subseteq C(u_2)$.

If $5\not\in C(u_2)$, and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 4\}$,
then $H$ contains a $(4, T_3)_{(u_4, u_3)}$-path and $T_3\subseteq C(u_4)$.

If $5\not\in S_u\setminus C(u_3)$,
then $2, 4\in C(u_3)\cap C(u_1)$, $H$ contains a $(2, T_3)_{(u_2, u_3)}$-path, a $(4, T_3)_{(u_4, u_3)}$-path,
and $(4, T_1)_{(u_1, u_4)}$-path, a $(2, T_1)_{(u_2, u_1)}$-path.

One sees from Corollary <ref> (1.2) that if $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path,
then $T_4\subseteq C(u_1)$.

$4\in C(u_3)$.

Assume that $4\not\in C(u_3)$.
It follows that $2\in C(u_3)$, $H$ contains a $(2, T_3)_{(u_2, u_3)}$-path,
$2\in C(u_4)$, $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path,
and then $T_4\subseteq C(u_1)$,
and it follows from Corollary <ref>(5.1) that $3/1/4\in C(u_2)$, $6, 7\in S_u$.
One sees from $T_3\cup T_4\subseteq C(u_1)\cap C(u_2)$ that $t_0\ge t_3 + t_4$.

When for some $j\in T_3$, $H$ contains no $(j, i)_{(u_3, v_i)}$-path for each $i\in [6, 7]$,
it follows from Corollary <ref>(4) that $5\in B_1$ and $H$ contains a $(2, 5)_{(u_2, u_3)}$-path,
i.e., $5\in C(u_i)$, $i\in \{1, 2, 4\}$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge\sum_{i\in [1, 4]}w_i + 2t_{uv} + t_0\ge |\{1, 5\}| + |\{2, 5, 3/1/4\}| + |\{3, 5, 2\}| + |\{4, 1, 2, 5\}| + |\{6, 7\}| + 2t_{u v} + t_3 + t_4 = 2\Delta + 10 + 1 + 2 = 2\Delta + 13$,
we have $\Delta = 7$, $t_0 = t_3 + t_4$, where $t_3 = 1$, $t_4 = 2$, and $2, 4\not\in C(u_1)$.
Thus, $C_{34}\ne \emptyset$ and $T_1\ne \emptyset$, a contradiction to Corollary <ref>(1.2).

In the other case, for each $j\in T_3$, $H$ contains a $(j, 6/7)_{(u_3, v)}$-path,
assume w.l.o.g. $H$ contains a $(7, 8)_{(u_4, v_7)}$-path.
We distinguish whether or not $H$ contains a $(7, T_3)_{(u_3, v_7)}$-path.

* $H$ contains a $(7, T_3)_{(u_3, v_7)}$-path.
It follows from <ref>(3) that $5\in S_u\setminus C(u_3)$,
and from <ref>(1) that $7\in S_u\setminus C(u_3)$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge\sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1\}| + |\{2, 3/1/4\}| + |\{3, 5, 2, 7\}| + |\{4, 1, 2\}| + |\{5, 6, 7\}| + 2t_{u v} + t_3 + t_4 = 2\Delta + 9 + 2 + 2 = 2\Delta + 13$,
we have $\Delta = 7$, $t_0 = t_3 + t_4$, where $t_3 = t_4 = 2$, and $2, 4\not\in C(u_1)$.
Thus, $C_{34}\ne \emptyset$ and $T_1\ne \emptyset$, a contradiction to Corollary <ref>(1.2).

* $6\in C(u_3)$. Then $w_3\ge 5$, $w_4\ge 4$ and $t_3\ge 3$, $t_4\ge 2$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge\sum_{i\in [1, 4]}w_i + 2t_{u v} + t_0\ge |\{1, 5/2/3/4\}| + |\{2, 3/1/4\}| + w_3 + w_4 + 2t_{u v} + t_3 + t_4 = 2\Delta + 9 + 3 + 2 = 2\Delta + 14$.

Assume $5\not\in C(u_2)$ and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 3, 4\}$.
If $2\not\in B_1\cup B_3$, then $(u u_2, u v)\overset{\divideontimes}\to (5, 2)$.
If $2\not\in C(u_1)$ and $H$ contains no $(2, 4)_{(u_1, u_4)}$-path,
then $(u u_2, u u_1, u v)\overset{\divideontimes}\to (5, 2, T_3)$.
If $2\not\in C(u_3)$ and $H$ contains no $(2, 4)_{(u_3, u_4)}$-path,
then $(u u_2, u u_3)\to (5, 2)$ reduces the proof to (2.3.).
Hence, we proceed with the following proposition, or otherwise we are done:

If $5\not\in C(u_2)$ and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 3, 4\}$,
then $2\in B_1\cup B_3$, for each $i\in \{1, 3\}$, $2/4\in C(u_i)$,
and if $2\not\in C(u_i)$, then $H$ contains a $(2, 4)_{(u_i, u_4)}$-path and $4\in C(u_i)$.

By Corollary <ref>(5.2) and (3), we proceed with the following proposition:

If $(T_3\cup T_1)\setminus C(u_2)\ne \emptyset$,
then $2\in B_1\cup B_3$, for each $i\in \{1, 3\}$, $2/4\in C(u_i)$,
and if $2\not\in C(u_i)$, then $H$ contains a $(2, 4)_{(u_i, u_4)}$-path and $4\in C(u_i)$.

There exists a $j\in T_4$ such that $H$ contains no $(2, j)_{(u_2, u_4)}$-path.

$4\in B_3$, $2/4\in C(u_1)$, and if $4\not\in C(u_1)$, then $H$ contains $(2, 4)_{(u_1, u_2)}$-path.

There exists a $\gamma_0\in [5, 7]\cap C(u_4)$ such that $H$ contains a $(\gamma_0, 4)_{(u_4, v_{\gamma_0})}$-path.

For (1), assume that $H$ contains a $(2, T_4)_{(u_2, u_4)}$-path.
Then $2\in C(u_4)$, $T_4\subseteq C(u_2)\cap C(u_1)$.
One sees from <ref> that $5/2/3/4\in C(u_1)$.

When $W_3 = 5$, it follows that $\Delta = 7$, $t_4 = 2$, $t_3 = 3$, and $T_3\cup T_4 = T_{uv}$.
Then $d(u_1) = \Delta$ and $d(u_2)\le 6$.
If $2, 4\not\in C(u_1)$,
one sees from <ref> that $5/1/3/4\in C(u_2)$ and then $T_3\setminus C(u_2)\ne \emptyset$,
then by <ref>, we are done.
Otherwise, $2/4\in C(u_1)$.
Then assume w.l.o.g. $H$ contains a $(7, 8)_{(u_4, v_7)}$-path.
If $4\in C(u_1)$, then $(u u_1, u v)\overset{\divideontimes}\to (\{5, 7\}\setminus C(u_4), 8)$.
Otherwise, $2\in C(u_1)$ and it follows from <ref> that $H$ contains a $(2, 5)_{(u_1, u_2)}$-path,
from <ref> that $H$ contains a $(2, 7)_{(u_1, u_2)}$-path, and $5, 7\in C(u_2)$.
One sees clearly that $T_3\setminus C(u_2)\ne \emptyset$.
It follows from Corollary <ref>(7.2) that $3\in C(u_2)$ and then $C(u_2) = \{2, 3, 5, 7\}\cup T_4$.
Then $(u u_4, u v)\to (\{5, 7\}\setminus C(u_4), 4)$ reduces the proof to (2.1.).

When $1, 2\not\in C(u_3)$,
it follows from <ref> that $3, 5\in C(u_1)\cup C(u_4)$ and $6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$.
Let $|\{6, 7\}\cap C(u_3)| = \tau_{67}$.
One sees that $t_3 = \Delta - 2 - (d(u_3) - w_3) = \Delta - d(u_3) + w_3 - 2\ge w_3 - 2 = 3 + \tau_{67} - 2$.
Then $d(u_1) + d(u_4)\ge |\{1\}| + |\{4, 1, 2\}| + |\{3, 5\}| + |\{6, 7\}\setminus C(u_3)| + t_{u v} + t_0\ge 6 + 2 - \tau_{67} + \Delta - 2 + 3 + \tau_{67} - 2 = \Delta + 7$.
It follows that mult$_{C(u_1)\cup C(u_4)}(5) = 1$ and for each $i\in [6, 7]$, $i\not\in C(u_3)\cap (C(u_1)\cup C(u_4))$.
Thus, $H$ contains a $(7, 8)_{(u_3, v_7)}$-path and by <ref>(2), we are done.

In the other case, $W_3 = \{3, 5, 4, 1\}$ or $W_3 = \{3, 5, 4, 2\}$.
One sees clearly from Corollary <ref>(4)
that $5\in B_1$ and $H$ contains a $(5, i)_{(u_3, u_i)}$-path for some $i\in \{2, 4\}\cap C(u_3)$.
One sees that $|\{1, 5\}| + |\{2\}| + |\{4, 1/2, 3, 5\}| + |\{4, 1, 2, 5\}| + 2t_{u v} = 2\Delta + 7$,
and $t_0\ge t_4\ge 2$.

It follows from Corollary <ref> (5.1) that if $2\not\in C(u_3)$ or $T_3\setminus C(u_2)\ne \emptyset$,
then $3, 6, 7\in S_u$ and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 7 + |\{3, 6, 7\}| + t_0\ge 2\Delta + 10 + t_4\ge 2\Delta + 10 + 2 = 2\Delta + 12$.

* $W_3 = \{3, 5, 4, 1\}$.
Then $3, 6, 7\in S_u$ and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12$.
It follows that $T_3\setminus C(u_2)\ne \emptyset$ and from <ref> that $2\in B_1$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{2\}| = 2\Delta + 13$.
It follows that $t_0 = t_4$, $4\not\in C(u_1)$ and $t_3 + t_4 = \Delta - 3$.
Then $(u u_1, u v)\overset{\divideontimes}\to (C_{34}, T_3)$[Or, since $T_1 = C_{34} \ne \emptyset$ and $4\not\in C(u_1)$, $T_1\setminus C(u_2)\ne \emptyset$, a contradiction to Corollary <ref> (1.2). ].

* $W_3 = \{3, 5, 4, 2\}$.
When $T_3\setminus C(u_2)\ne \emptyset$,
it follows from <ref> that $2/4\in C(u_1)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{2/4\}| = 2\Delta + 13$.
It follows that $t_0 = t_4$, $c_{34} = 1$, $C(u_1) = \{1, 5, 2/4\}$, $C(u_2) = \{2, 3, 6, 7\}\cup T_4$.
Since $C_{34}\setminus (C(u_1)\cup C(u_2))\ne \emptyset$, $4\in C(u_1)$.
Thus, by <ref>, we are done.
In the other case, $T_3\subseteq C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 7 + t_0\ge 2\Delta + 7 + t_3 + t_4\ge 2\Delta + 7 + 2  + 2 = 2\Delta + 11$.
One sees clearly that $\{3, 6, 7\}\setminus S_u\ne \emptyset$.
It follows from <ref> that $1\in C(u_2)$ and $H$ contains a $(1, 2)_{(u_2, u_3))}$-path.
If $5\not\in C(u_2)$ and $2, 4\not\in C(u_1)$, then $u u_2\to 5$, $v u_3\in \alpha\in T_3$,
and $u v\to 2$, or $(u u_1, u v)\to (2, T_3\setminus \{\alpha\})$.
Otherwise, $5\in C(u_2)$ or $2/4\in C(u_1)$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{1\}| + |\{5/2/4\}| = 2\Delta + 13$.
It follows that $\Delta = 7$, $t_0 = t_3 + t_4 = 2 + 2 = 4$ and then $4\in C(u_1)$, $C(u_2) = \{1, 2\}\cup T_3\cup T_4$, $6\not\in S_u$.
Then $(u u_4, u u_1)\to (6, C_{34})$ reduces the proof to (2.1.2.) and (1) holds.

By (1), it follows from Corollary <ref>(5.2), (4) that (2) holds naturally.
For (3), if $H$ contains no $(4, i)_{(u_4, v_i)}$-path for each $i\in [5, 7]\cap C(u_4)$,
then $(v u_4, u u_4, u v)\overset{\divideontimes}\to (4, j, T_3)$.
Otherwise, there exists a $\gamma_0\in [5, 7]\cap C(u_4)$ such that $H$ contains a $(\gamma_0, 4)_{(u_4, v_{\gamma_0})}$-path.

Assume $6\not\in C(u_1)\cup C(u_3)$ and $T_1\cup T_3 = T_{u v}$.
If $H$ contains no $(6, i)_{(u_i, u_j)}$-path for each $i\in \{2, 4\}$, $j\in \{1, 3\}$,
one sees that $u u_1\to 6$ or $u u_3\to 6$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 2$,
we may further assume that $H$ contains a $(6, T_3)_{(u_1, v_6)}$-path,
and a $(6, T_1\cup \{4\})_{u_3, v_6}$-path,
i.e., $C(v_6) = \{4, 6\}\cup T_{uv}$.
Then $(v v_6, u v)\overset{\divideontimes}\to (2, 6)$.
Hence, we proceed with the following proposition,

For each $\eta\in \{6, 7\}$, if $\eta\not\in C(u_1)\cup C(u_3)$ and $T_1\cup T_3 = T_{u v}$,
then $H$ contains a $(\eta, i)_{(i, j)}$-path for some $i\in \{2, 4\}$, $j\in \{1, 3\}$.

$T_4\setminus C(u_1)\ne \emptyset$.

$2\in C(u_1)$, and $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path.

$1\in C(u_2)$ or $H$ contains a $(1, 3)_{(u_2, u_3)}$-path.

Assume that $T_4\subseteq C(u_1)$.
One sees from Lemma <ref>(2) that $2/4\in C(u_1)$.

When $1, 2\not\in C(u_3)$,
it follows from <ref> that $3, 5\in C(u_1)\cup C(u_4)$ and $6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$.
Let $|\{6, 7\}\cap C(u_3)| = \tau_{67}$.
One sees that $t_3 = 3 + \tau_{67} - 2 = \tau_{67} - 1$.
Then $d(u_1) + d(u_4)\ge |\{1, 2/4\}| + |\{4, 1\}| + |\{3, 5\}| + |\{6, 7\}\setminus C(u_3)| + t_{uv} + t_0\ge 6 + 2 - \tau_{67} + \Delta - 2 + 3 + \tau_{67} - 2 = \Delta + 7$.
It follows that mult$_{C(u_1)\cup C(u_4)}(5) = 1$ and for each $i\in [6, 7]$, $i\not\in C(u_3)\cap (C(u_1)\cup C(u_4))$.
Thus, $H$ contains a $(7, 8)_{(u_3, v_7)}$-path and by <ref>(2), we are done.
In the other case, $1/2\in C(u_3)$.
One sees from  $w_3 + w_4\ge 7$ that $c_{34}\le 2$ and $t_3 + t_4\ge \Delta - 4$, $w_1\le 4$.

Case 1. There exists a $j\in T_3$ such that $H$ contains no $(i, j)_{(u_3, v_i)}$-path for each $i\in [6, 7]$.

It follows from Corollary <ref>(3) that $5\in B_1$ and $H$ contains a $(5, i)_{(u_i, u_)}$-path for some $i\in \{2, 4\}\cap C(u_3)$.
Since $1, 5, 2/4\in C(u_1)$, $c_{34}\ge 1$, $\{6, 7\}\setminus C(u_1)\ne \emptyset$,
and $w_3 + w_4\le 8$, $w_3\le 5$.

Assume that $W_4 = \{1, 4, 5\}$.
If $H$ contains no $(2, j)_{(u_1, u_2)}$-path for some $j\in T_1$,
then $(u u_4, u u_1)\to (\{6, 7\}\setminus C(u_1), j)$ reduces the proof to (2.1.2.).
If $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
then $(u u_4, u u_3)\to (\{6, 7\}\setminus C(u_1), j)$ reduces the proof to (2.2.2.).
Otherwise, $2\in C(u_1)\cap C(u_3)$ and $T_3\cup T_1 \subseteq C(u_2)$.

When $w_3 = 5$, it follows that $c_{34} = t_1 = 1$, $W_4 = \{4, 1, 5\}$,
$C(u_1) = \{1, 5, 2\}\cup T_3\cup T_4$, $d(u_2)\le 6$, $4\in C(u_2)$, $T_3\cup T_1\subseteq C(u_2)$,
and it follows from Corollary <ref>(5.1) that $1/3\in C(u_2)$.
Thus, $d(u_2)\ge |\{2, 4, 1/3\}\cup T_3\cup T_1| = 7$, a contradiction.
In the other case, $w_3 = 4$ and $W_3 =\{3, 5, 4, 2/1\}$.

When $w_1 = 4$, it follows that $c_{34} = t_1 = 2$, $W_4 = \{4, 1, 5\}$, $2\in C(u_1)$,
and $T_3\cup T_1\subseteq C(u_2)$, $d(u_2)\le 6$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 2, 5\}| + |\{2\}| + |\{3, 5, 4, 2\}| + |\{1, 4, 5\}| + |\{4\}|+ 2t_{u v} + t_0 = 2\Delta + 8 + t_0\ge 2\Delta + 8 +  t_3 + c_{34}\ge 2\Delta + 8 + 2 + 2 = 2\Delta + 12$.
One sees that if $4\not\in C(u_1)$, then $1, 6, 7\in S_u$,
and if $H$ contains no $(1, 2)_{(u_2, u_3)}$-path, then $3, 6, 7\in S_u$.
It follows that $C(u_1) = \{1, 2, 4, 5\}\cup T_3\cup T_4$ and $C(u_2) = \{1, 2\}\cup T_3\cup C_{34}$.
Then $(u u_2, u u_3, u u_4)\to (5, T_3, 3)$ reduces the proof to (2.3.).

Hence, we assume $W_1 = \{1, 5, 2/4\}$.
We distinguish whether or not $2\in C(u_3)$:

* $W_3 = \{3, 5, 4, 1\}$.
If $H$ contains no $(2, i)_{(u_2, u_4)}$-path for some $i\in \{3, 6, 7\}\setminus C(u_4)$,
then $(u u_4, u u_3)\to (i, T_3)$ reduces the proof to (2.2.2.).
Otherwise, $2\in C(u_4)$.
It follows that $W_4 = \{4, 1, 5, 2\}$, $C_{34} = T_1$, $c_{34} = t_1 = 1$, $d(u_i) = \Delta$, $i = 1, 3, 4$,
$d(u_2)\le 6$, and $H$ contains a $(2, \{3, 6, 7\})_{(u_2, u_4)}$-path, $3, 6, 7\in C(u_2)$.
One sees that $(T_1\cup T_3)\setminus C(u_2)\ne \emptyset$.
It follows from <ref> that $2\in B_1$.
Then $C(u_1) = \{1, 5, 2\}\cup T_3\cup T_4$ and $C(u_2) = \{2, 3, 4, 6, 7\}\cup C_{34}$.
Thus, by <ref>, we are done.

* $W_3 = \{3, 5, 4, 2\}$.
Since $|\{1, 5, 2/4\}| + |\{2\}| + |\{3, 5, 4, 2\}| + |\{4, 1, 5\}| + |\{6, 7, 3/1\}| + 2t_{uv} = 2\Delta + 10$, $(T_1\cup T_3\cup \{5\})\setminus C(u_2)\ne \emptyset$.
It follows from <ref> and <ref> that $2\in C(u_1)$ or $H$ contains a $(2, 4)_{(u_1, u_4)}$-path.

* $4\in C(u_1)$. One sees that $2\in C(u_4)$.
It follows that $C(u_1) = \{1, 4, 5\}\cup T_3\cup T_4$, $C(u_4) = \{4, 1, 2, 5\}\cup T_3\cup C_{34}$, $c_{34} = t_1 = 1$.
If $H$ contains no $(2, 6)_{(u_1, u_4)}$-path,
then $(u u_4, u u_1)\to (6, T_1)$ reduces the proof to (2.1.2.).
Otherwise, $H$ contains a $(2, 6)_{(u_1, u_4)}$-path.
Thus, by <ref>, we are done.

* $2\in C(u_1)$.
It follows from Corollary <ref>(5.1) that $1\in C(u_2)$.
Since $|\{1, 5, 2\}| + |\{2, 1\}| + |\{3, 5, 4, 2\}| + |\{4, 1, 5\}| + |\{6, 7, 4\}| + 2t_{uv} + t_0= 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1\ge 2\Delta + 11 + 1 = 2\Delta + 12$, $T_3\setminus C(u_2)\ne \emptyset$.
If $H$ contains no $(2, i)_{(u_2, u_4)}$-path for some $i\in \{3, 6, 7\}\setminus C(u_4)$,
then $(u u_4, u u_3)\to (i, T_3\cap T_2)$ reduces the proof to (2.2.2.).
Otherwise, $2\in C(u_4)$.
It follows that $W_4 = \{4, 1, 5, 2\}$, $C_{34} = T_1$, $c_{34} = t_1 = 1$, $d(u_i) = \Delta$, $i = 1, 3, 4$, $d(u_2)\le 6$, and $H$ contains a $(2, \{3, 6, 7\})_{(u_2, u_4)}$-path, $3, 6, 7\in C(u_2)$.
Thus, $d(u_2)\ge |[1, 4]\cup [6, 7]\cup T_1|\ge 7$, a contradiction.

Case 2. $H$ contains a $(6/7, T_3)_{(u_3, v)}$-path, and assume w.l.o.g. $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.

When $w_3 = 6$, since $w_3 + w_4\ge 6 +3 = 9$,
$t_3 + t_4 = \Delta - 2$, $W_4 = \{4, 1, b_1\}$, $C(u_1) = \{1, a_1\}\cup T_3\cup T_4$, $d(u_i) = \Delta$, $i = 1, 3, 4$, and $d(u_2)\le 6$.
It follows from <ref> and <ref> that $a_1 = 2$, $4\in C(u_2)$, and $5, 7\in C(u_2)$.
Since $2, 4, 5, 7\in C(u_2)$, $T_3\setminus C(u_2)\ne \emptyset$.
Then $(u u_4, u u_3, T_4)\overset{\divideontimes}\to (3, T_3\cap T_2, T_4)$.
In the other case, $W_3 =\{3, 5, 4, 7, 2/1\}$, and $H$ contains a $(7, T_3)_{(u_3, v_7)}$-path.

When $w_1 = 3$, it follows that $c_{34} = t_1 = 1$, $W_4 = \{4, 1, b_1\}$,
$d(u_i) = \Delta$, $i = 1, 3, 4$, and $d(u_2)\le 6$.
If $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
then $(u u_4, u u_3, u v)\to (\{3, 5, 7\}\setminus (C(u_1)\cup C(u_4)), j, T_4)$.
If $H$ contains no $(2, j)_{(u_2, u_1)}$-path for some $j\in T_1$,
then $(u u_4, u u_1)\to ([5, 7]\setminus (C(u_1)\cup C(u_4)), j)$ reduces the proof to (2.1.).
Otherwise, $2\in C(u_1)\cap C(u_3)$ and $H$ contains a $(2, T_1)_{(u_2, u_1)}$-path, $(2, T_3)_{(u_2, u_3)}$-path.
If $4\in C(u_1)$, one sees that $H$ contains a $(4, \gamma_0)_{(u_4, v)}$-path,
it follows from <ref> and <ref> that $5, 7\in C(u_2)$,
and thus $d(u_2)\ge |\{2, 5, 7\}\cup T_3\cup T_1|\ge 7$.
Otherwise, $4\in C(u_2)\setminus C(u_1)$.
It follows from Corollary <ref>(5.1) that $1\in C(u_2)$,
and thus $d(u_2)\ge |\{2, 4, 1\}\cup T_3\cup T_1|\ge 7$.
Hence, we assume $W_1 = \{1, 2/4\}$.
We distinguish whether or not $2\in C(u_1)$:

* $W_1 = \{1, 4\}$.
One sees that $|[5, 7]\cap C(u_4)\le 2|$ and $H$ contains a $(4, \gamma_0)_{(u_4, v)}$-path.
Thus, by <ref> and <ref>, we are done.

* $W_3 = \{1, 2\}$.
Then $4\in C(u_2)$ and it follows from <ref> and <ref> that $H$ contains a $(2, i)_{(u_1, u_2)}$-path for each $i\in \{5, 7\}\subseteq C(u_2)$.
One sees that $|\{1, 2\}| + |\{2, 4, 5, 7\}| + |\{3, 5, 4, 7, 2\}| + 2t_{u v} = 2\Delta + 7$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 7 + |\{4, 1, b_1\}| + |\{1/3\}| = 2\Delta + 11$,
$T_3\setminus C(u_2)\ne \emptyset$.
It follows Corollary <ref>(5.1) that $3, 6\in S_u$.
If there exists an $i\in \{5, 7\}\setminus C(u_4)$,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (i, T_3\cap T_2, T_4)$.
Otherwise, $W_4 = \{4, 1, 5, 7\}$, $d(u_i) = \Delta$, $i = 1, 3, 4$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge  2\Delta + 7 + |\{4, 1, 5, 7\}| + |\{3, 6\}| = 2\Delta + 13$,
$1\not\in C(u_2)$ and $3\not\in C(u_1)\cup C(u_4)$.
Thus, by <ref>, we are done.

Hence, $T_4\setminus C(u_1)\ne \emptyset$.
It follows from Corollary <ref> (1.2) that $2\in C(u_1)$ and $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path,
and from Corollary <ref> (5.1) that $1\in C(u_2)$ or $H$ contains a $(1, 3)_{(u_2, u_3)}$-path.

If $1\not\in C(u_3)$, $3\not\in C(u_1)\cup C(u_2)$, and $H$ contains no $(4, 3)_{(u_2, u_4)}$-path,
then $(u u_1, u u_2, u u_3)\to (T_4\cap T_1, 3, 1)$ reduces the proof to (2.4.1.).

Assume that $1, 2\not\in C(u_3)$.
It follows from <ref> that $3, 5\in C(u_1)\cup C(u_4)$ and $6, 7\in C(u_1)\cup C(u_4)\cup C(u_3)$.
Let $i\in \{3\}\cup (\{6, 7\}\setminus C(u_3))$ with mult$_{S_u}(i) = 1$.
If $i\in C(u_4)$, one sees from <ref> that $H$ contains a $(4, i)_{(u_1, u_4)}$-path,
then $(u u_1, u u_2, u u_3)\to (T_4\cap T_1, i, 1)$ reduces the proof to (2.4.1.).
Hence, we proceed with the following proposition:

$1\in C(u_3)$, or $3\in C(u_1)\cup C(u_2)$, or $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.

If $1, 2\not\in C(u_3)$, then for each $i\in \{3\}\cup (\{6, 7\}\setminus C(u_3))$ with mult$_{S_u}(i) = 1$,
$i\in C(u_1)$.

$T_3\setminus C(u_2)\ne \emptyset$.

$2\in B_1\cup B_3$, and if $2\not\in C(u_3)$, then $H$ contains a $(2, 4)_{(u_3, u_4)}$-path.

$3\in C(u_4)$, or $H$ contains a $(3, i)_{(u_4, u_i)}$-path for some $i\in \{1, 2\}$.

Assume that $T_3\subseteq C(u_2)$.
By Lemma <ref>, $T_4\setminus C(u_2)\ne\emptyset$, i.e., $T_1\setminus T_0\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1\in C(u_2)\cup C(u_3)$, $6, 7\in S_u$,
and if $1\not\in C(u_2)$, then $H$ contains a $(1, 3)_{(u_2, u_3)}$-path,
from Corollary <ref>(7.1) that if $2\not\in C(u_3)$, then $3\in S_u$.
One sees that $|\{1, 2\}| + |\{2\}| + |\{4, 3, 5\}| + |\{4, 1\}| + |\{4, 1, 6, 7\}| + 2t_{uv} = 2\Delta + 8$.

Case 1. There exists a $j\in T_3$ such that $H$ contains no $(i, j)_{(u_3, v_i)}$-path for each $i\in [6, 7]$.

It follows from Corollary <ref>(3) that $5\in B_1$ and $H$ contains a $(5, i)_{(u_i, u_)}$-path for some $i\in \{2, 4\}\cap C(u_3)$,
from <ref> that $5\in C(u_2)$ or $2\in C(u_3)\cup C(u_4)$.

* $2\not\in C(u_3)$.
It follows from Corollary <ref>(5.1) that $3\in S_u$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{5, 5, 5/2, 3\}| + t_0 \ge 2\Delta + 12 + 1 = 2\Delta + 13$.
It follows that $t_0 = t_3 = 1$, $W_3 = \{3, 5, 4\}$, $1\in C(u_2)$, mult$_{S_u}(i) = 1$, $i = 3, 6, 7$,
and from <ref>(2) that $3, 6, 7\in C(u_1)$.
Thus, $d(u_1) + d(u_2) \ge |\{1, 2, 5, 3, 6, 7\}| + |\{2, 1\}| + |\{4\}| + t_{u v} + t_0\Delta + 8$, a contradiction.

* $2\in C(u_3)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{5, 5, 2\}| + t_0 \ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $t_0 = t_3 = 2$, $1\not\in C(u_3)$ and $3\not\in S_u$, $2\not\in C(u_4)$.
Thus, by <ref>(1), we are done.

Case 2. $H$ contains a $(6/7, T_3)_{(u_3, v)}$-path, and assume w.l.o.g. $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.

* $1, 2\not\in C(u_3)$.
It follows from <ref> that $3, 5\in C(u_1)\cup C(u_4)$, $6\in C(u_1)\cup C(u_4)\cup C(u_3)$,
from <ref>(2) that $7\in C(u_1)\cup C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{3, 5, 7\}| + t_0 \ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $t_0 = t_3 = 2$, $W_3 = \{3, 5, 4, 7\}$, $1\in C(u_2)$,
and mult$_{S_u}(i) = 1$, $i = 3, 6$, mult$_{S_u}(j) = 2$, $i = 5, 7$.
It follows from <ref>(2) that $3, 6\in C(u_1)$.
One sees that $d(u_1) + d(u_2)$ $\ge |\{1, 2, 3, 6\}| + |\{2, 1\}| + |\{4\}| + t_{uv} + t_0\ge 7 + \Delta - 2 + t_0 = \Delta + 7$.
It follows that $5, 7\in C(u_4)$, and from <ref> that $H$ contains a $(4, \{5, 7\})_{(u_1, u_4)}$-path.
Thus, $H$ contains a $(4, i)_{(u_4, v_i)}$-path for some $i\in \{5, 7\}$, a contradiction.

* $1/2\in C(u_3)$.
When $3\not\in C(u_1)$, one sees from <ref> and <ref> that $5, 7\in C(u_1)\cup C(u_2)\cup C(u_4)$,
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 8 + |\{5, 7\}| + t_0 \ge 2\Delta + 10 + 3 = 2\Delta + 13$.
It follows that $t_0 = t_3 = 3$, $3\not\in S_u$,
and then from Corollary <ref>(7.1) that $2\in C(u_3)$.
Then $1\not\in C(u_3)$, and thus, by <ref>(1), we are done.
Now we assume $3\in C(u_1)$.
Since $1/3\in C(u_1)$, $1/2\in C(u_3)$,
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 2, 3\}| + |\{2, 1/3\}| + |\{4\}| + |\{3, 5, 4, 7, 1/2\}| + |\{4, 1\}| +  |\{6\}| + t_{uv} + t_0\ge 2\Delta + 10 + t_0\ge 2\Delta + 13$,
and $d(u_1) + d(u_2)\ge |\{1, 2, 3\}| + |\{2, 1/3\}| + |\{4\}| + t_{uv} + t_0\ge \Delta + 4 + t_0\ge \Delta + 7$.
It follows that $t_0 = t_3 = 3$, $d(u_i) = \Delta = 7$, $i\in [1, 3]$, and $C(u_4) = \{1, 4, 6\}\cup T_3$.
One sees that $H$ contains a $(4, 6)_{(u_4, v_6)}$-path.
Thus, by <ref>, we are done.

Hence, $T_3\setminus C(u_2)\ne \emptyset$.
It follows from Corollary <ref> (5.2) and (4) that $2\in B_1\cup B_3$,
and if $2\not\in C(u_3)$, then $H$ contains a $(2, 4)_{(u_3, u_4)}$-path,
and from Corollary <ref> (5.1) that $3\in C(u_4)$, or $H$ contains a $(3, i)_{(u_4, u_i)}$-path for some $i\in \{1, 2\}$.

Hereafter, we assume w.l.o.g. $8\in T_3\cap T_2$.
If $H$ contains a $(7, 8)_{(u_3, v_\rho)}$-path,
$7\not\in C(u_4)$ and $H$ contains no $(7, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (7, T_3\cap T_2, T_4)$.
If $5\not\in C(u_4)$ and $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (5, T_3\cap T_2, T_4)$.
If $j\in [6, 7]\setminus C(u_3)\cup C(u_4)$ and $H$ contains no $(j, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (j, T_3\cap T_2, T_4)$.
In the other case, together with <ref>, we proceed with the following proposition:

$5\in C(u_4)$, or $H$ contains a $(5, i)_{(u_4, u_i)}$-path for some $i\in [1, 2]$.

For each $j\in [6, 7]$, $i\in C(u_3)\cup C(u_4)$, or $H$ contains a $(j, i)_{(u_4, u_i)}$-path for some $i\in [1, 2]$.

If $H$ contains a $(7, 8)_{(u_3, v_j)}$-path,
then $7\in C(u_4)$ or $H$ contains a $(7, i)_{(u_4, u_i)}$-path for some $i\in [1, 2]$.

Assume $1, 2\not\in C(u_3)$.
If mult$_{S_u\setminus C(u_3)}(5) = 1$, then $5\in C(u_1)\cup C(u_4)$;
and if $H$ contains a $(7, 8)_{(u_3, v_j)}$-path and mult$_{S_u\setminus C(u_3)}(7) = 1$,
then $7\in C(u_1)\cup C(u_4)$.

One sees that $2\in C(u_1)\cap (C(u_3)\cup C(u_4))$, $4\in C(u_3)\cap (C(u_1)\cup C(u_2))$, $1\in C(u_2)\cup C(u_3)$,
and $3, 6, 7\in S_u$.
It follows from <ref>(1) that $5\in S_u\setminus C(u_3)$.
Then $|\{1, 2\}| + |\{2\}| + |\{3, 4, 5\}| + |\{4, 1\}| + |\{2, 4, 1, 3, 5, 6, 7\}| + 2t_{u v} = 2\Delta + 11$.
Since either $5\in B_1$, or $H$ contains a $(7, 8)_{(u_3, v_7)}$-path and it follows from <ref>(3) that $7\in S_u\setminus C(u_3)$,
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{5/7\}| + t_0 \ge 2\Delta + 12 + c_{34}\ge 2\Delta + 12$.
Hence, $c_{34}\le 1$ and $\Delta \le 7$.

Since $4\in C(v_3)$, it follows from Corollary <ref>(6) that $R_3\ne \emptyset$ and then $5/6/7\in C(v_3)$.
Particularly, we proceed with the following proposition:

If $T_{uv} = T_3\cup T_4$, then $T_3\setminus C(v_3) \ne \emptyset$.

If $1/2\in C(v_3)$, then $r_3\ge 2$ and $T_3\setminus C(v_3)\ne \emptyset$.

For some $j\in [6, 7]\setminus (C(u_1)\cup C(u_3))$,
if $uu_1\to 6$ and $(u u_1, u u_3)\to (3, 6)$ reduce to an acyclic edge $(\Delta + 5)$-coloring for $H$, then we are done by Corollary <ref>(2.1).

When $6, 7\in C(u_3)$, since $2\in C(u_3)\cup C(u_4)$,
it follows that $W_3\cup W_4 = [3, 7]\cup \{1, 4, b_1\}\cup \{2\}$, $T_{uv} = T_3\cup T_4$,
and mult$_{C(u_3)\cup C(u_4}(2) = 1$.
We distinguish whether or not $2\in C(u_3)$:

* $2\in C(u_3)$.
It follows from <ref>(3) that $3\in C(u_1)$,
and from <ref>(1) that $b_1 = 5$, i.e., $C(u_1)= [1, 3]\cup T_3$, $C(u_4) = \{1, 4, 5\}\cup T_3$. One sees that $5\not\in B_1$.
Then $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.
Thus, by <ref>(3) we are done.

* $2\in C(u_4)$.
One sees from <ref> that $6, 7\in S_u\setminus C(u_3)$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{6, 7\}| + t_0 = 2\Delta + 13 + t_0$,
it follows that mult$_{S_u\setminus C(u_3)}(i) = 1$, $i\in [5, 7]$, mult$_{S_u}(3) = 1$, $t_0 = 0$,
and $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.
Further, by <ref>(2), $3\in C(u_1)$, and by <ref>(4), $5, 7\in C(u_1)\cup C(u_4)$.
Then $W_1 = [1, 3]\cup \{\alpha\}$ and $W_4 = \{1, 2, 4\}\cup \{\beta\}$, where $\{\alpha, \beta\} = \{5, 7\}$. Thus, by <ref>, we are done.

In the other case, $\{6, 7\}\setminus C(u_3)\ne \emptyset$ and assume w.l.o.g. $6\not\in C(u_3)$.

If $T_3\cup T_4 = T_{uv}$, then

for each $j\in [6, 7]\not\in C(u_3)$,
$H$ contains a $(j, i)_{(u_3, u_i)}$-path for some $i\in \{1, 2, 4\}$, and mult$_{S_u}(j)\ge 2$.

$7\in C(u_3)$, $T_3\cap C(u_2) = \emptyset$, and mult$_{S_u}(i) = 2$, $i = 1, 2, 6$,
mult$_{S_u}(5) + $mult$_{S_u}(7) = 4$.

For (1), assume that $H$ contains no $(6, i)_{(u_3, u_i)}$-path for each $i\in \{1, 2, 4\}$.
One sees from <ref>(1) that $T_3\setminus C(v_3)\ne \emptyset$,
and it follows from Lemma <ref>(1.2) that there exists a $b_3\in [6, 7]\cap C(v_3)$ such that $H$ contains a $(b_3, j)_{(v_3, v_{b_6})}$-path for some $j\in T_3\cap R_3$.
One sees that the same argument applies if $uu_3\to 6$,
and thus $\{4\}\cup T_4\subseteq C(v_6)$, $T_3\setminus C(v_6)\ne \emptyset$,
and it follows from Lemma <ref>(1.2) that there exists a $b_6\in \{3, 7\}\cap C(v_6)$ such that $H$ contains a $(b_6, j)_{(v_6, v_{b_6})}$-path for some $j\in T_3\cap R_6$.
One sees from Lemma <ref>(1.2) and Lemma <ref> that $T_3\subseteq C(v_3)\cup C(v_6)$.

* $2\not\in C(u_4)$.
One sees that $2\in B_3$ and thus $2\in C(v_3)\cap C(v_6)$.
Since $d(v_3) + d(v_6)\ge t_{u v} + t_4 + |\{3, 4, 6/7, 2\}| + |\{6, 4, 3/7, 2\}| = \Delta + 6 + t_4\ge \Delta + 6 + 1 = \Delta + 7$,
it follows that $t_4 = 1$, $W_4 = \{1, 4, b_1\}$, and $1\not\in C(v_3)$.
Thus, by Lemma <ref> (3), we are done.

* $2\in C(u_4)$.
One sees from Corollary <ref> (6) and Lemma <ref> that $2\in C(v_3)\cup C(v_6)$ or $5\in C(v_3)\cup C(v_6)$.
Since $d(v_3) + d(v_6)\ge t_{u v} + t_4 + |\{3, 4, 6/7\}| + |\{6, 4, 3/7\}| + |\{2/5\}| = \Delta + 5 + t_4\ge \Delta + 5 + 2 = \Delta + 7$,
it follows that $t_4 = 2$, $W_4 = \{1, 4, b_1, 2\}$, and $1\not\in C(v_3)\cup C(v_6)$.
Recall that $H$ contains a $(b_1, 4)_{(u_4, v_{b_1})}$-path.
Hence, $b_1\in \{5, 7\}$ and it follows that $3, 6\in C(u_2)$ and $H$ contains a $(2, i)_{(u_2, u_4)}$-path for each $i\in \{3, 6\}$.
One sees that $|\{3, 6\}\cap C(u_1)|\le 1$.
Then $(u u_1, u v)\to (6, T_3\cap R_6)$ if $6\not\in C(u_1)$,
and $(u u_1, u u_3, uv)\to (3, 6, T_3\cap R_3)$ if $3\not\in C(u_1)$.

Now assume that mult$_{S_u}(j) = 1$.
If $H$ contains a $(j, i)_{(u_3, u_i)}$-path for some $i\in [1, 2]$,
then by <ref>, we are done.
Otherwise, $j\in C(u_4)$ and $H$ contains a $(4, 6)_{(u_3, u_4)}$-path.
Then $(u u_1, u v)\overset{\divideontimes}\to (6, j)$ if $H$ contains no $(6, j)_{(u_1, v_6)}$-path for some $j\in T_3$, or else, $(u u_1, u u_2, u v)\overset{\divideontimes}\to (6, T_4\cap T_1, T_3)$.

For (2), since mult$_{S_u}(6)\ge 2$, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{5/7\}| + |\{6\}| + t_0 = 2\Delta + 13 + t_0$.
One sees that if $7\not\in C(u_3)$, then $5\in C(u_1)\cap C(u_4)$, mult$_{S_u}(7)\ge 2$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{5, 7, 6\}| = 2\Delta + 14$.
Hence, $7\in C(u_3)$, $t_0 = 0$, and (2) holds.

(2.4.2.1.) There exists a $j\in T_3$ such that $H$ contains no $(i, j)_{(u_3, v_i)}$-path for each $i\in [6, 7]$.
Then $5\in B_1\subseteq C(u_1)\cap C(u_4)$ and $H$ contains a $(5, i)_{(u_i, u_)}$-path for some $i\in \{2, 4\}\cap C(u_3)$.

When $c_{34} = 1$, it follows that $\Delta = 7$, mult$_{S_u}(i) = 1$, $i = 3, 6, 7$,
and mult$_{S_u}(j) = 2$, $j = 1, 2, 4$.
Since $2\in C(u_3)\cup C(u_4)$ and $w_3 + w_4\le 8$,
$\{6, 7\}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$ and $6\not\in C(u_3)\cup C(u_4)$.
First, assume that $H$ contains a $(2, 6)_{(u_2, u_4)}$-path.
Then $2\in C(u_4)\setminus C(u_3)$.
One sees that $H$ contains no $(i, 6)_{(u_i, u_2)}$-path for each $i\in \{1, 3\}$.
If follows from <ref> that $1\in C(u_3)$,
from Lemma <ref>(3) that $3\in C(u_2)$,
and then from Lemma <ref>(3) that $H$ contains a $(2, 3)_{(u_2, u_4)}$-path.
Thus, by <ref>(3), we are done.
Next, by <ref>(2), $6\in C(u_1)$,
and for each $i\in [6, 7]\setminus (C(u_3)\cup C(u_4))$,
assume that $i\in C(u_1)$ and $H$ contains a $(1, i)_{(u_1, u_4)}$-path.
One sees that $H$ contains no $(6, i)_{(u_i, u_3)}$-path for each $i\in \{1, 2, 4\}$,
and the same argument applies if $uu_3\to 6$.
We further assume that $3\not\in C(u_2)$ and it follows from Lemma <ref>(3) that $1\in C(u_2)$.

* $3\in C(u_4)$.
It follows from <ref>(1) that $4\in C(u_2)$ and $H$ contains a $(4, 3)_{(u_2, u_4)}$-path,
and then from <ref> that $2\in C(u_3)$.
Hence, $C(u_3) = [2, 5]\cup T_4\cup C_{34}$, $W_4 = \{1\}\cup [3, 5]$, and $C(u_1) = [1, 2]\cup [5, 7]\cup T_3$.
One sees clearly that $T_4\cup T_{34}\subseteq B_3\subseteq C(v_3)$ and $T_3\setminus C(v_3) \ne \emptyset$.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (3, 6, T_3\cap R_3)$.

* $3\in C(u_1)$ and $H$ contains a $(1, 3)_{(u_1, u_4)}$-path.
If $2\in C(u_3)$, then $\{2\}\cup T_4\cup C_{34}\subseteq B_3$ with $t_4 + c_{34} = 3$,
thus $d(v_3)\ge |\{3, 1, 4, 2, 5/6/7\}\cup T_4\cup C_{34}|\ge 8$.
Otherwise, $2\in C(u_4)\setminus C(u_3)$.
One sees that $1, 2\not\in C(u_3)$ and from <ref>(2) that $7\in C(u_1)$.
Then $T_4\cup C_{34}\subseteq B_3$ with $t_4 + c_{34} = 4$,
thus $d(v_3)\ge |\{3, 1, 4, 5/6/7\}\cup T_4\cup C_{34}|\ge 8$.

In the other case, $t_3 + t_4 = \Delta - 2$ and by <ref>(2) that $T_3\cap C(u_2) = \emptyset$, mult$_{S_u}(i) = 2$, $i = 1, 2, 6$, mult$_{S_u}(5) = 3$ and mult$_{S_u}(7) = 1$.
If $H$ contains no $(3, 7)_{(u_2, u_3)}$-path, then $(u u_2, u u_1, u v)\to (7, T_4\cap T_1, T_3)$.
Otherwise, $H$ contains a $(3, 7)_{(u_2, u_3)}$-path and $3\in C(u_2)$.
It follows from Lemma <ref>(2) that $H$ contains a $(2, 3)_{(u_2, u_3)}$-path, $2\in C(u_4)$,
and from <ref> that $1\in C(u_3)$, i.e., $W_3 = \{1, 7\}\cup [3, 5]$, $W_4 = \{1, 2, 4, 5\}$.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (6, 3, T_3\cap R_3)$.

(2.4.2.2.) $H$ contains a $(7, T_3)_{(u_3, v)}$-path.
When $c_{34} = 1$, it follows that $W_3\cup W_4 = \{3, 5, 4, 7\}\cup \{1, 4, b_1\}\cup \{2\}$.
If $2\in C(u_3)$, one sees from Lemma <ref>(3) that $3\in C(u_1)$, and $[5, 7]\setminus (C(u_1)\cup C(u_4))\ne \emptyset$,
then by <ref> (1–3), we are done.
Otherwise, $2\in C(u_4)$.
One sees that $1, 2\not\in C(u_3)$.
It follows from <ref>(2) that $3, 5, 6\in C(u_1)$,
and $W_1 = \{1, 2, 3, 5, 6\}$, $b_1 = 7$.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (7, 1, 8)$.

In the other case, $t_3 + t_4 = \Delta - 2$ and by <ref>(2) that $T_3\cap C(u_2) = \emptyset$, mult$_{S_u}(i) = 2$, $i = 1, 2, 6$, mult$_{S_u}(5) = 2$ and mult$_{S_u}(7) = 2$.
One sees clearly that $6\in C(u_1)\cup C(u_4)$ and $H$ contains a $(6, T_3\cap R_3)_{(v_3, v_6)}$-path.
If $6\not\in C(u_1)$ and $H$ contains no $(6, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $(u u_1, u v)\to (6, T_3\cap R_3)$.
Otherwise, $6\in C(u_1)$ or $H$ contains a $(6, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.

* There exists a $j\in \{5, 7\}\cap C(u_2)$.
One sees from <ref> (1) and (3) that $H$ contains a $(2, j)_{(u_2, u_4)}$-path, $2\in C(u_4)$.
It follows from <ref> or <ref> that $3\in C(u_1)$,
and from <ref>(3) that $1\in C(u_2)$.
One sees that $1, 2\not\in C(u_3)$ and $H$ contains no $(j, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$.
Thus, by <ref>, we are done.

* $5, 7\in C(u_1)\cup C(u_4)$.
If $H$ contains a $(2, 3)_{(u_2, u_4)}$-path, i.e., $3\in C(u_2)$ and $2\in C(u_4)$,
it follows from <ref> that $1\in C(u_3)$,
and from <ref> and <ref> that $W_1 = [1, 2]\cup \{5, 7\}$,
$W_4 = \{1, 2, 4, 6\}$,
thus by <ref> (3), $H$ contains a $(4, 6)_{(u_4, v_6)}$-path,
and by Lemma <ref>(1), we are done.
Otherwise, $3\in C(u_1)\cup C(u_4)$ and it follows from <ref> that $1\in C(u_2)$.

* $2\not\in C(u_3)$.
It follows from <ref>(2) that $3\in C(u_1)$.
One sees from <ref> that for each $j\in \{5, 7\}$, $j\not\in C(u_4)$,
or $H$ contains a $(4, j)_{(u_1, u_4)}$-path.
It follows from <ref> (3) that $H$ contains a $(4, 6)_{(u_4, v_6)}$-path.
Thus, by Lemma <ref>(1), we are done.

* $2\in C(u_3)$.
It follows that $C(u_1) = \{1, 2, \alpha_1, \alpha_2\}\cup T_3$, $C(u_4) = \{1, 4, \beta_1, \beta_2\}\cup T_3$, $C(u_2) = \{1, 4, 6\}\cup T_4$,
where $\{\alpha_1, \alpha_2, \beta_1, \beta_2\} = \{3\}\cup [5, 7]$.
If $5, 7\in C(u_1)$, one sees that $H$ contains a $(2, 6)_{(u_1, u_2)}$-path,
from <ref> (3) that $H$ contains a $(4, 6)_{(u_4, v_6)}$-path.
then, by Lemma <ref>(1), we are done.
Otherwise, there exists a $j\in \{5, 7\}\setminus C(u_1)$,
and then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (j, T_3, 1, T_4)$.

(2.4.3.) $c(vu_4) = 6$.
Let $C(u_1) = W_1\cup T_3\cup C_{13}$ and $C(u_3) = W_3\cup T_1\cup C_{13}$.
If $6\not\in C(u_1)$ and $H$ contains no $(6, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $u u_1\to 6$ reduces the proof to (2.4.2.).
If $6\not\in C(u_3)$ and $H$ contains no $(6, i)_{(u_3, u_i)}$-path for each $i\in [1, 2]$,
then $u u_3\to 6$ reduces the proof to (2.4.1.).
Otherwise, we proceed with the following proposition:

$6/2/3\in C(u_1)$,
and if $6\not\in C(u_1)$, then $H$ contains a $(6, i)_{(u_1, u_i)}$-path for some $i\in [2, 3]$.

$6/1/2\in C(u_3)$,
and if $6\not\in C(u_3)$, then $H$ contains a $(6, i)_{(u_3, u_i)}$-path for some $i\in [1, 2]$.

$6\in C(u_1)\cup C(u_3)$, $T_3\ne \emptyset$ and there exists a $a_3\in \{2, 4\}\cap C(u_3)$.

Assume $5\not\in C(u_4)$ and $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$.
If $H$ contains no $(2, j)_{(u_2, u_3)}$-path for some $j\in T_3$,
then $(u u_4, u u_3)\to (5, j)$ reduces the proof to (2.4.1.).
Further, assume $H$ contains no $(5, 3)_{(u_4, u_i)}$-path.
If $4\not\in B_1\cup B_3$, then $(u u_4, u v)\overset{\divideontimes}\to (5, 4)$;
if $4\not\in C(u_3)$ and $H$ contains no $(2, 4)_{(u_2, u_3)}$-path,
then $(u u_4, u u_3)\to (5, 4)$ reduces the proof to (2.4.1.);
if $4\not\in C(u_1)$ and $H$ contains no $(2, 4)_{(u_2, u_1)}$-path,
then $(u u_4, u u_1)\to (5, 4)$ reduces the proof to (2.1.1.).
Otherwise, we proceed with the following proposition:

Assume $5\not\in C(u_4)$ and $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$.

$H$ contains a $(2, T_3)_{(u_2, u_3)}$-path and $2\in C(u_3)$, $T_3\subseteq C(u_2)$.

If $H$ contains no $(5, 3)_{(u_4, u_3)}$-path, then $4\in B_1\cup B_3$,
and for each $i\in \{1, 3\}$, $2/4\in C(u_i)$, and if $4\not\in C(u_i)$,
then $H$ contains a $(2, 4)_{(u_2, u_i)}$-path.

Assume that $5\not\in C(u_2)$ and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 3, 4\}$.
Let $u u_2\to 5$.
If $2\not\in B_1\cup B_3$, then $uv \overset{\divideontimes}\to 2$.
If $2\not\in C(u_1)$ and $H$ contains no $(2, 4)_{(u_1, u_2)}$-path,
then $(u u_1, u v)\overset{\divideontimes}\to (2, T_3)$.
If $H$ contains no $(4, j)_{(u_3, u_4)}$-path for some $j\in T_3\cup (\{2\}\setminus C(u_3))$ and there exists a $l\in T_1\setminus (C(u_2)\cap C(u_3))$,
then $(u u_3, u v)\overset{\divideontimes}\to (j, l)$.
Otherwise, we assume the following proposition:

Assume that $5\not\in C(u_2)$ and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 3, 4\}$.
Then $2\in B_1\cup B_3$, and

$2\in C(u_1)$ or $H$ contains a $(2, 4)_{(u_1, u_2)}$-path;

if $H$ contains no $(j, 4)_{(u_3, u_4)}$-path for some $j\in T_3\cup (\{2\}\setminus C(u_3))$,
then $T_1\subseteq C(u_2)\cap C(u_3)$.

If $5\not\in C(u_1)\cup C(u_2)$, $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$,
and $H$ contains no $(4, j)_{(u_1, u_4)}$-path for some $j\in T_1$,
then $(u u_2, u u_1, u v)\overset{\divideontimes}\to (5, j, T_3)$.
If $5\not\in C(u_4)\cup C(u_1)$, $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
and $H$ contains no $(2, j)_{(u_1, u_2)}$-path for some $j\in T_1$,
then $(u u_4, u u_1, u v)\overset{\divideontimes}\to (5, j, T_3)$.
Otherwise, we proceed with the following proposition:

If $5\not\in C(u_2)\cup C(u_1)$ and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$,
then $4\in C(u_1)$, $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path, and $T_1\subseteq C(u_4)$.

If $5\not\in C(u_4)\cup C(u_1)$ and $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
then $2\in C(u_1)$, $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path, and $T_1\subseteq C(u_2)$.

Assume that $5\not\in S_u\setminus C(u_3)$.
One sees from <ref> that $3\in C(u_1)$ and $H$ contains a $(3, 5)_{(u_1, u_3)}$-path.
It follows from <ref>(2) that $4\in B_1\cup B_3$, $2/4\in C(u_1)$.
Since $1, 3, 2/4 \in C(u_1)$, $T_1\ne \emptyset$.
Together with <ref>, <ref>(1–2), and <ref>(1),

If $5\not\in S_u\setminus C(u_3)$, then $T_1\ne \emptyset$, $2, 4\in (B_1\cup B_3)\cap C(u_1)$,
$H$ contains a $(4, T_1)_{(u_1, u_4)}$-path, a $(2, T_1)_{(u_1, u_2)}$-path, $T_1\subseteq C(u_2)\cap C(u_4)$,
and $2\in C(u_3)$, $H$ contains a $(2, T_3)_{(u_3, u_4)}$-path, $T_3\subseteq C(u_2). $

Assume that $6\not\in C(u_1)$ and $H$ contains no $(6, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$.
Let $u u_1\to 6$.
If $1\not\in C(u_3)$ and $H$ contains no $(1, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
then $u u_3\to 1$;
if $7\not\in C(u_3)\cup C(u_1)$ and $H$ contains no $(7, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
then $u u_3\to 7$,
these reduce the proof to (2.4.2.).
Assume that $6\not\in C(u_3)$ and $H$ contains no $(6, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$.
Let $u u_3\to 6$.
If $3\not\in C(u_1)$ and $H$ contains no $(1, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $u u_1\to 3$;
if $7\not\in C(u_3)\cup C(u_1)$ and $H$ contains no $(7, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $u u_1\to 7$;
these reduces the proof to (2.4.1.).
Otherwise, we proceed with the following proposition:

If $6\not\in C(u_1)$, $H$ contains no $(6, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$, then

$1\in C(u_3)$ or $H$ contains a $(1, i)_{(u_3, u_i)}$-path for some $i\in \{2, 4\}$;

$7\in C(u_3)\cup C(u_1)$ or $H$ contains a $(7, i)_{(u_3, u_i)}$-path for some $i\in \{2, 4\}$.

If $6\not\in C(u_3)$, $H$ contains no $(6, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$, then

$3\in C(u_1)$ or $H$ contains a $(1, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$;

$7\in C(u_3)\cup C(u_1)$ or $H$ contains a $(7, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.

Assume mult$_{S_u}(6) = 1$.
One sees from <ref>(1) that if $6\in C(u_3)$,
then $3\in C(u_1)$ and $H$ contains a $(3, 6)_{(u_1, u_3)}$-path,
or else, $6\in C(u_1)$, and from <ref>(2) that $1\in C(u_3)$ and $H$ contains a $(1, 6)_{(u_1, u_3)}$-path.
If $T_4 = \emptyset$, one sees that $C(u_4) = \{4, 6\}\cup T_{u v}$,
then $(u u_4, v u_4, u v)\overset{\divideontimes}\to (5, 4, 6)$.
Otherwise, $T_4\ne \emptyset$ and it follows from Corollary <ref>(4) that for each $j\in T_4$,
$H$ contains a $(j, 1/3/5/7)_{(u_4, v)}$-path.
One sees from <ref>(1.2) or (2.2) that $7\in S_u$.
Let $u u_2\to 6$.
If $2\not\in B_1\cup B_3$, then $uv\overset{\divideontimes}\to 2$.
If $2\not\in C(u_1)$ and $H$ contains no $(2, 4)_{(u_2, u_4)}$-path,
then $u u_1\to 2$ reduces the proof to (2.4.2.).
If $H$ contains no $(4, j)_{(u_3, u_4)}$-path for some $j\in T_3\cup (\{2\}\setminus C(u_3))$ and there exists a $l\in T_1\setminus (C(u_2)\cap C(u_4))$,
then $(u u_3, u v)\overset{\divideontimes}\to (j, l)$.
Assume that $6\in C(u_3)$. One sees from $3, 2/4, 1\in C(u_1)$ that $T_1\ne \emptyset$.
If $H$ contains no $(4, j)_{(u_1, u_4)}$-path for some $j\in T_1$,
then $u u_1\to j$ reduces the proof to (2.4.2.).
Otherwise, we proceed with the following proposition:

Assume mult$_{S_u}(6) = 1$. Then

$7\in S_u$, $2\in B_1\cup B_3$,
and $2\in C(u_1)$ or $H$ contains a $(2, 4)_{(u_1, u_4)}$-path;

if $H$ contains no $(j, 4)_{(u_3, u_4)}$-path for some $j\in T_3\cup (\{2\}\setminus C(u_3))$,
then $T_1\subseteq C(u_2)\cap C(u_4)$;

if $6\in C(u_3)$, then $\emptyset\ne T_1\subseteq C(u_4)$, $4\in C(u_1)$,
and $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path.

$T_4\ne \emptyset$ and for each $j\in T_4$, $H$ contains a $(j, 1/3/5/7)_{(u_4, v)}$-path.

If $4\not\in C(v_1)$ and $H$ contains no $(4, i)_{(v_1, v_i)}$-path for each $i\in \{3, 5, 6, 7\}$,
then $v v_1\to 4$ reduces the proof to (2.1.2.).
Otherwise, we proceed with the following proposition:

If $2\in C(v_1)$, then $R_1\ne\emptyset$; and if $4\in C(u_1)$, then $T_1\ne \emptyset$.

$4/3/5/6/7\in C(v_1)$, and if $4\not\in C(v_1)$, then $H$ contains a $(4, 3/5/6/7)_{(v_1, v)}$-path.

Assume that $4\not\in C(u_3)$ (or $T_3\cap T_4\ne \emptyset$) and $3\not\in C(u_2)$.
Let $u u_2\to 3$, $u u_3\to T_3$ (or $u u_3\to T_3\cap T_4$).
If $H$ contains no $(3, i)_{(u_2, u_i)}$-path for each $i\in \{1, 4\}$,
then $u v\to 2$ or $uv\to R_1$.
If $H$ contains no $(3, 4)_{(u_2, u_4)}$-path, $2\not\in C(u_1)$,
and $H$ contains no $(2, 4)_{(u_1, u_4)}$-path,
then $u u_1\to 2$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{3\}$.
Assume that $2\not\in C(u_3)$ (or $T_3\cap T_2\ne \emptyset$) and $3\not\in C(u_4)$.
Let $u u_4\to 4$, $u u_3\to T_3$ (or $u u_3\to T_3\cap T_2$).
If $H$ contains no $(3, i)_{(u_4, u_i)}$-path for each $i\in \{1, 2\}$,
then $u v\to 4$ or $u v\to T_1$.
If $H$ contains no $(2, 4)_{(u_2, u_4)}$-path, $4\not\in C(u_1)$, $H$ contains no $(2, 4)_{(u_1, u_2)}$-path,
then $u u_1\to 4$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{3\}$.
Otherwise, we proceed with the following proposition:

If $4\not\in C(u_3)$ (or $T_3\cap T_4\ne \emptyset$) and $3\not\in C(u_2)$, then

$H$ contains a $(3, i)_{(u_2, u_i)}$-path for some $i\in \{1, 4\}$,

if $H$ contains no $(3, 4)_{(u_2, u_4)}$-path,
then $2\in C(u_1)$ or $H$ contains a $(2, 4)_{(u_2, u_4)}$-path.

If $2\not\in C(u_3)$ (or $T_3\cap T_2\ne \emptyset$) and $3\not\in C(u_4)$, then

$H$ contains a $(3, i)_{(u_4, u_i)}$-path for some $i\in \{1, 2\}$;

if $H$ contains no $(3, 2)_{(u_2, u_4)}$-path,
then $4\in C(u_1)$ or $H$ contains a $(2, 4)_{(u_1, u_2)}$-path.

If $4\not\in C(u_3)$, $H$ contains no $(4, i)_{(u_3, v_i)}$-path for each $i\in \{1, 7\}$,
and $H$ contains no $(5, j)_{(u_3, u_j)}$-path for each $j\in [1, 2]$,
then $(v u_3, u u_3)\to (4, 5)$ reduces the proof to (2.4.1.).
Otherwise, we proceed with the following proposition:

If $4\not\in C(u_3)$ and $H$ contains no $(4, i)_{(u_3, v_i)}$-path for each $i\in \{1, 7\}$,
then $H$ contains a $(5, j)_{(u_3, u_j)}$-path for some $j\in [1, 2]$.

Assume that recoloring some edges incident to $u$ reduces to an acyclic edge $(\Delta + 5)$-coloring $c'$ for $H$ such that $C(u)\cap C(v) = \{1, 3\}$, where $(u u_a, u u_b)_{c'} = (1, 3)$, and $a\ne 1$, $b\ne 3$.
One sees from Corollary <ref> (2.1) that $T_{u v}\subseteq C(v_1)\cap C(v_3)$.
If $(T_3\cup T_1)\setminus C(u_2)\ne \emptyset$ and $2\not\in C'(u)$,
one sees from <ref> that $2\not\in C(v_1)$ and from Corollary <ref> (3) that $2\in B_3$,
then on the basic coloring of $c'$, $u v\overset{\divideontimes}\to 2$.
If $(T_3\cup T_1)\setminus C(u_4)\ne \emptyset$ and $4\not\in C'(u)$,
one sees from Corollary <ref> (3) that $4\in B_1\cup B_3$ and if $4\not\in C(v_i)$ for some $i\in \{1, 3\}$,
then on the basic coloring of $c'$, $u v\overset{\divideontimes}\to 4$.
thus $C(v_i) = \{i, 4\}\cup T_{u v}$, $i = 1, 3$ and by Corollary <ref> (2.2), we are done.
Hence, we proceed with the following propositions:

Assume that recoloring some edges incident to $u$ reduces to an acyclic edge $(\Delta + 5)$-coloring $c'$ for $H$ such that $C(u)\cap C(v) = \{1, 3\}$, where $(u u_a, u u_b)_{c'} = (1, 3)$, and $a\ne 1$, $b\ne 3$.
We are done if $(T_3\cup T_1)\setminus C(u_2)\ne \emptyset$ and $2\not\in C'(u)$,
or $(T_3\cup T_1)\setminus C(u_4)\ne \emptyset$ and $4\not\in C'(u)$.

Assume that $(u u_1, u u_3)\to (3, 1)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$.
One sees from Corollary <ref>(2.1) and Corollary <ref>(5.2)(4) that
if $(T_3\cup T_1)\setminus C(u_i)\ne \emptyset$ for some $i\in \{2, 4\}$,
then $(u u_1, u u_3, u u_i)\to (3, 1, (T_3\cup T_1)\setminus C(u_i))$ reduces to an acyclic edge $(\Delta + 5)$-coloring $c'$ for $H$ such that $C(u)\cap C(v) = \{1, 3\}$,
where $(u u_3, u u_1)_{c'} = (1, 3)$ and $i\not\in C'(u)$.
Thus, it follows from <ref> that we may proceed with the following propositions:

If $(u u_1, u u_3)\to (3, 1)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$,
then $T_3\cup T_1\subseteq C(u_2)\cap C(u_4)$,
i.e., if $(T_3\cup T_1)\setminus C(u_i)\ne \emptyset$ for some $i\in \{2, 4\}$,
then $1/3\in S_u$, and $1\in C(u_3)$ or $H$ contains a $(1, 2/4)_{(u_3, u)}$-path,
or $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path.

Assume that $1\not\in C(u_4)$ and $H$ contains no $(1, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$.
Let $(u u_4, u v)\to (1, T_3)$.
If $H$ contains no $(2, j)_{(u_1, u_2)}$-path for some $j\in T_1$,
then $u u_1\overset{\divideontimes} \to j$;
if $4\not\in C(u_1)$ and $H$ contains no $(4, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $u u_1\overset{\divideontimes} \to 4$;
if $5\not\in C(u_1)$ and $H$ contains no $(5, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $u u_1\overset{\divideontimes} \to 5$.

Assume $1\not\in C(u_4)$, $H$ contains no $(1, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$. Then

$H$ contains a $(2, T_1)_{(u_1, u_2)}$-path, $2\in C(u_1)$, $T_1\subseteq C(u_2)$;

For each $j\in \{4, 5\}$, either $j\in C(u_1)$ or $H$ contains a $(j, i)_{(u_1, u_i)}$-path for some $i\in [2, 3]$.

Assume that $1\not\in C(u_2)$ and $H$ contains no $(1, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$.
Let $(u u_2, u v)\to (1, T_3)$.
If $H$ contains no $(4, j)_{(u_1, u_4)}$-path for some $j\in T_1$,
then $u u_1\overset{\divideontimes} \to j$;
if $2\not\in C(u_1)$ and $H$ contains no $(2, i)_{(u_1, u_i)}$-path for each $i\in [3, 4]$,
then $u u_1\overset{\divideontimes} \to 2$;
if $5\not\in C(u_1)$ and $H$ contains no $(5, i)_{(u_1, u_i)}$-path for each $i\in [3, 4]$,
then $u u_1\overset{\divideontimes} \to 5$.
Hence, we proceed with the following proposition:

If $1\not\in C(u_2)$ and $H$ contains no $(1, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$,

$H$ contains a $(4, T_1)_{(u_1, u_4)}$-path, $4\in C(u_1)$, $T_1\subseteq C(u_4)$;

for each $j\in \{2, 5\}$, either $j\in C(u_1)$ or $H$ contains a $(j, 3/4)_{(u_1, u)}$-path.

If $1\not\in C(u_3)$, $H$ contains no $(1, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
and $5\not\in C(u_1)$, $H$ contains no $(5, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (5, 1, T_3)$.
Otherwise, together with <ref>(1.1),
we proceed with the following proposition:

If $1\not\in C(u_3)$ and $H$ contains no $(1, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
then for each $j\in [5, 6]$, $j\in C(u_1)$, or $H$ contains a $(j, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.

Now we introduce several lemmas.

$w_3\ge 4$ and $t_3\ge 2$.

One sees from <ref>(3) that $T_3\ne \emptyset$ and there exists a $a_3\in \{2, 4\}\cap C(u_3)$,
and from <ref>(2) that $6/1/2\in C(u_3)$.
Hence, it suffices to assume that $W_3 = \{3, 5, 2\}$.
Then $H$ contains a $(2, T_3\cup \{6\})_{(u_2, u_3)}$-path, $T_3\cup \{6\}\subseteq C(u_2)$,
from <ref>(3) that $6\in C(u_1)$,
and from Corollary <ref>(5.1) that $3, 7\in S_u$.
One sees from Corollary <ref>(4) that $5\in B_1$ and $H$ contains a $(2, 5)_{(u_2, u_3)}$-path.
Since $5, 6\in C(u_1)$, $T_1\ne\emptyset$ and there exists a $a_1\in \{2/4\}\cap C(u_1)$.
It follows that $t_1\ge 2$.
If $T_1\setminus T_0\ne \emptyset$,
one sees from Corollary <ref>(5.2) and (3) that if there exists a $i\in \{2, 4\}\setminus C(u_1)$,
then mult$_{S_u}(i)\ge 2$,
and from Corollary <ref>(5.1) that $1\in S_u$,
then mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_1)}(2) + $ mult$_{S_u\setminus \{a_1\}}(4)\ge 2$.
One sees that $|\{1, a_1, 5, 6\}| + |\{2, 5, 6\}| + |\{2, 3, 5\}| + |\{4, 6\}| + |\{3, 7\}| + 2t_{uv} = 2\Delta + 10$.

Case 1. $(T_3\cup \{5\})\subseteq C(u_4)$.
Since either $t_0\ge t_1\ge 2$,
or mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_1)}(2) + $ mult$_{S_u\setminus \{a_1\}}(4)\ge 2$,
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |T_3\cup \{5\}| + 2\ge 2\Delta + 14$.

Case 2. $(T_3\cup \{5\})\setminus C(u_4)\ne \emptyset$.
It follows from Corollary <ref>(3) and <ref>(2) that $4\in B_1$ and $H$ contains a $(2, 4)_{(u_2, u_3)}$-path.
One sees from  Corollary <ref>(7.1) that $2\in C(u_1)$ or $1\in S_u$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{4\}| + |\{1/2\}| + t_0 = 2\Delta + 12 + t_0$.
It follows that $T_1\setminus T_0\ne \emptyset$ and then $1\in S_u$.
One sees from Corollary <ref>(1.2) that if $T_1\setminus C(u_4)\ne \emptyset$,
then $2\in C(u_1)$,
and from Corollary <ref>(5.2) that if $T_1\setminus C(u_2)\ne \emptyset$, then $2\in C(u_1)\cup C(u_4)$.
Hence, $2\in C(u_1)\cup C(u_4)$ and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{4\}| + |\{1, 2\}| + t_0 = 2\Delta + 13 + t_0$.
It follows that $\Delta = 7$, $t_0 = 0$, mult$_{S_u}(i) = 1$, $i = 1, 3, 7$, mult$_{S_u}(2) = 2$.

* $T_1\setminus C(u_4)\ne \emptyset$.
Since $2\in C(u_1)$, $T_1\subseteq C(u_2)$ and $t_1 + t_3 + |\{2, 4, 5, 6\}|\le d(u_2)\le 7$,
we have $t_1 = 1$, $t_3 = 2$ and $c_{13} = 2$.
Thus, $d(u_1)\ge |[1, 2]\cup [4, 6]\cup T_3\cup C_{13}|\ge 5 + 1 + 2 = 8$.

* $T_1\setminus C(u_2)\ne \emptyset$.
Since $4\in C(v_1)$, it follows from Corollary <ref>(6) that $R_1\ne \emptyset$.
One sees from Corollary <ref>(5.1) that $1\in C(u_4)$ or $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
It follows from <ref> that $3\in C(u_1)$ or $H$ contains a $(3, i)_{(u_1, u_i)}$-path for some $i\in \{2, 4\}$.
One sees from Corollary <ref>(5.1) that $3\in C(u_2)$ or $H$ contains a $(3, i)_{(u_2, u_i)}$-path for some $i\in \{1, 4\}$.
It follows that $3\in C(u_2)$, $2\in C(u_1)$, $1\in C(u_4)$,
or $3\in C(u_1)$, $1\in C(u_2)$, $2\in C(u_4)$.
Then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to ([1, 3]\setminus C(u_1), [1, 3]\setminus C(u_2), T_3, [1, 2]\setminus C(u_4), R_1)$.

Hence, $w_3\ge 4$ and then $t_3\ge 2$.

$w_1\ge 3$, $T_1\ne \emptyset$, there exists a $a_1\in \{2, 4\}\cap C(u_1)$, and $t_1\ge 1$, $t_1 + t_3\ge 3$.

Otherwise, assume that $w_1\le 2$.

One sees from <ref> and <ref>(1) that $W_1 = \{1, i\}$,
$i\in [2, 3]$ and $H$ contains a $(i, [5, 6])_{(u_1, u_i)}$-path, $5, 6\in C(u_i)$.

Case 1. $W_1 = \{1, 3\}$.
It follows from <ref>(1) that $6\in C(u_2)$,
from <ref>(2) that $5\in C(u_4)$ or $H$ contains a $(2, 5)_{(u_2, u_4)}$-path,
and from <ref> that $5\in C(u_2)$ or $H$ contains a $(4, 5)_{(u_2, u_4)}$-path.

If $T_3\setminus C(u_2)\ne \emptyset$,
then $u u_2\to \alpha\in T_3\cap T_2$, $u v\to 2$ or $(u u_1, u v)\to (2, T_3\setminus \{\alpha\})$.
If $T_3\setminus C(u_4)\ne \emptyset$,
then $u u_4\to \beta\in T_3\cap T_4$, $u v\to 4$ or $(u u_1, u v)\to (4, T_3\setminus \{\beta\})$.
Otherwise, and $T_3\subseteq C(u_2)\cap C(u_4)$.

One sees from <ref> that $1\in C(u_3)$ or $H$ contains a $(1, 2/4)_{(u_3, u)}$-path.
It follows that $1\in S_u$ and likewise $7\in S_u$.
One sees from <ref>(1.2) that if $4\not\in C(u_3)$,
then $3\in C(u_2)$ or $H$ contains a $(3, 4)_{(u_2, u_4)}$-path,
and from  <ref>(2.2) that if $2\not\in C(u_3)$,
then $3\in C(u_4)$ or $H$ contains a $(2, 3)_{(u_2, u_4)}$-path.
It follows that $2, 4\in C(u_3)$ or $3\in C(u_2)\cup C(u_4)$.

One sees from $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3\}| + |\{2, 6, 4/5\}| + |\{4, 6, 2/5\}| + |\{3, 5, 6, 2/4\}| + |\{1, 7, 3/\{2, 4\} \setminus \{a_3\}\}| + 2t_{u v}+ t_0 = 2\Delta + 11 + t_0\ge 2\Delta + 13$ that
$t_3 = 2$, $W_3 = \{3, 5, 6, a_3\}$, $1, 7\in C(u_{a_3})$ and $3\in C(u_{a_3})$,
or $H$ contains a $(3, \{2, 4\}\setminus \{a_3\})_{u_{a_3}, u}$-path.
Hence, we have the following two scenarios:

* $W_3 = \{3, 5, 2, 6\}$ and $W_3 = \{2, 6, 1, 7, 4\}$, $W_4 = \{4, 6, 3, 5\}$.
Then $(u u_1, u u_4, u v)\overset{\divideontimes}\to (4, 1, T_3)$.

* $W_3 = \{3, 5, 4, 6\}$ and $W_3 = \{2, 6, 3, 5\}$, $W_4 = \{4, 6, 1, 7, 2\}$.
Then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 1, T_3)$.

Case 2. $W_1 = \{1, 2\}$.
It follows from <ref>(1) that $6\in C(u_3)$.
One sees from <ref>(1.1) that if $4\not\in C(u_3)$,
then $3\in C(u_2)$ or $H$ contains a $(3, 4)_{(u_2, u_4)}$-path,
and from  <ref>(2.1) that if $2\not\in C(u_3)$,
then $3\in C(u_4)$ or $H$ contains a $(2, 3)_{(u_2, u_4)}$-path.
It follows that $2, 4\in C(u_3)$ or $3\in S_u$.
One sees that $H$ contains a $(2, 5)_{(u_1, u_2)}$-path.
It follows from <ref> that $4\in C(u_3)$ or $H$ contains a $(4, 1/7)_{(u_3, v)}$-path.
One sees that the same argument applies if $7\not\in C(u_2)$ and $u u_1\to 7$.
It follows from <ref> (2) that

we may proceed with the following proposition:

For each $i\in \{1, 7\}$, $i\in C(u_2)$ or $H$ contains a $(i, 3/4)_{(u_2, u)}$-path.

One sees that $|\{1, 2\}| + |\{2, 5, 6\}| + |\{3, 5, 6, a_3\}| + |\{4, 6\}| + |\{1, 7\}| + 2t_{u v} = 2\Delta + 9$.
We distinguish the following on whether $T_3\subseteq C(u_2)\cap C(u_4)$ or not.

* $T_3\subseteq C(u_2)\cap C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + |\{3/\{2, 4\}\setminus\{a_3\}\}| + t_0\ge 2\Delta + 10 + t_3\ge 2\Delta + 10 + 2 = 2\Delta + 12$.
It follows that $\Delta\le 7$, $t_3\le 3$, and if $t_3 = 3$, then mult$_{S_u}(i) = 1$, $i = 1, 7$,
and mult$_{S_u}(3) + $ mult$_{S_u}(4) = 1$.
When $2\in C(u_3)$, one sees from $1/7/4\in C(u_)$ that $t_3 = 3$.
Since $T_3\cup \{2, 5, 6\}\sqsubseteq C(u_2)$, it follows from <ref> that $3/4\in C(u_2)$.
One sees that $\{1, 7\}\setminus C(u_3)\ne \emptyset$ and if $4\in C(u_2)$,
then $\{1, 7\}\setminus C(u_4)\ne \emptyset$.
Thus, by <ref>, we are done.
In the other case, $2\not\in C(u_3)$.
Then $4\in C(u_3)$, $3\in S_u$ and it follows from <ref>(1) that $5\in C(u_4)$.
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0 + |\{5\}|\ge 2\Delta + 11 + t_0\ge 2\Delta + 13$.
It follows that $t_3 = 2$, $2\not\in S_u$, and follows from <ref> (2.1) that $3\in C(u_4)$,
from <ref> that $C(u_2) = [1, 2]\cup [5, 9]$.
Then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (5, 3, 1, 2)$[Or, by <ref>, we are done. ].

* $T_3\setminus C(u_2)\ne \emptyset$.
Then $4\in C(u_3)$ and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.
It follows from Corollary <ref> (5.2) and (3) that $2\in B_1\cup B_3$, and $2\in C(u_3)\cap C(u_4)$,
from Corollary <ref> (2.1) that $3\in C(u_4)$ or $H$ contains a $(3, 2)_{(u_4, u_2)}$-path,
and from <ref>(1) that $5\in C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + |\{3, 5, 2\}| + t_0 = 2\Delta + 12 + t_0\ge 2\Delta + 12$.
It follows that $\Delta\le 7$.

* $2\in C(u_3)\cap C(u_4)$.
Then $W_2 = [1, 3]\cup [5, 7]$, $W_3 = [2, 6]$, $W_3 = \{2, 4, 5, 6\}$.
It follows from <ref>(2.1) that $H$ contains a $(2, 3)_{(u_2, u_4)}$-path,
and the from <ref> that $H$ contains a $(1, 2)_{(u_2, u_3)}$-path.
Thus, by <ref>, we are done.

* $2\in C(u_2)\setminus C(u_4)$.
One sees from <ref>(2.1) that $3\in C(u_4)$, $W_3 = [2, 6]$, and $W_4 = [3, 6]$.
It follows from <ref> that $4\in C(u_2)$, and from <ref> that $C(u_2) = [1, 2]\cup [4, 7]$.
Then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (4, 3, 1, 2, T_3)$.

* $2\in C(u_4)\setminus C(u_2)$.
When $3\in C(u_4)$, one sees that $W_3 = [3, 6]$, $W_4 = [2, 6]$,
and it follows from <ref> that $3\in C(u_2)$,
from <ref> that $C(u_2) = [1, 3]\cup [5, 7]$,
then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (5, 4, 2, 1, T_3)$.
In the other case, $3\not\in C(u_4)$, and it follows from <ref>(2.1) that $H$ contains a $(2, 3)_{(u_2, u_4)}$-path, from <ref> that $1\in C(u_3)$, or $H$ contains a $(1, 4)_{(u_3, u_4)}$-path.
Hence, $7\not\in C(u_3)\cup C(u_4)$.
If $H$ contains no $(2, 7)_{(u_2, u_4)}$-path, then $(u u_4, u u_3, u v)\to (7, T_3\cap T_2, 4)$.
Otherwise, $H$ contains a $(2, 7)_{(u_2, u_4)}$-path.
One sees that $u u_1\to 7$ and $(u u_1, u u_3)\to (3, 7)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $C(u)\cap C(v) = \{3, 7\}$.
Thus, by <ref>, we are done.

* $T_3\setminus C(u_4)\ne \emptyset$.
It follows that $2\in C(u_3)$, $T_3\subseteq C(u_2)$, and
from Corollary <ref> (5.2) and (3) that $4\in B_3\cap C(u_2)$.

Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + |\{2, 4, 3\}| + t_0\ge 2\Delta + 12$ and $d(u_2)\ge |\{2\}\cup [4, 6]| + t_3\ge 7$,
it follows that $d(u_2) = \Delta = 7$, $t_3 = 3$, $W_1 = [2, 6]$, $1, 3\not\in C(u_2)$,
and from <ref> that $H$ contains a $(1, 4)_{(u_2, u_4)}$-path.
Hence, by <ref>(2), we are done.

Hence, $T_1\ne \emptyset$ and it follows from Corollary <ref>(1.2) that there exists a $a_1\in \{2, 4\}\cap C(u_1)$.

If $T_1\setminus C(u_2)\ne \emptyset$, $T_3\setminus C(u_4)\ne \emptyset$,
and mult$_{S_u}(1) = $ mult$_{S_u}(3) = 1$, mult$_{S_u}(2) = $ mult$_{S_u}(4) = 2$,
then we can obtain an acyclic edge $(\Delta + 5)$-coloring for $H$ or reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $a_{uv} \le 1$.

One sees that $2\in C(u_3)$, $4\in C(u_1)$,
and $H$ contains a $(2, T_3)_{(u_2, u_3)}$-path, a $(4, T_1)_{(u_1, u_4)}$-path.
It follows from Corollary <ref> (5.2) and (3) that $2, 4\in B_1\cup B_3$,
and from Corollary <ref> (5.1) that $3\in C(u_2)$, or $H$ contains a $(3, 1/4)_{(u_2, u)}$-path,
and $1\in C(u_4)$, or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path.
We discuss the following three subcases due to $1\in C(u_i)$, $i\in [2, 4]$, respectively.

Case 1. $1\in C(u_2)$.
Then $2\in C(u_4)\setminus C(u_1)$, and $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
One sees from <ref> that $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path.
It follows that $3\in C(u_1)$.
Since $(u u_1, u u_2, u u_3, u u_4)\to (2, 3, T_3\cap T_4, 1)$ is an acyclic edge $(\Delta + 5)$-coloring for $H$ with $4\not\in C(u)$,
we are done by <ref>.

Case 2. $1\in C(u_3)$.
Then $3\in C(u_4)$, $H$ contains a $(1, 3)_{(u_3, u_4)}$-path,
and then $4\in C(u_2)$, $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.
Since $(u u_1, u u_2, u u_3, u u_4)\to (T_1\cap T_2, 3, 4, 1)$ is an acyclic edge $(\Delta + 5)$-coloring for $H$ with $2\not\in C(u)$,
we are done by <ref>.

Case 3. $1\in C(u_4)$. Then $3\in C(u_2)\cup C(u_4)$.
When $3\in C(u_4)$, it follows that $4\in C(u_2)\setminus C(u_3)$ and $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.
one sees that $(u u_1, u u_3)\to (3, 1)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$,
then we are done by <ref>.
In the other case, $3\in C(u_2)$.
It follows from <ref> that $2\in C(u_1)$ or $4\in C(u_3)$.
If $2\in C(u_1)$, then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes} \to (3, 1, T_3, 2, T_1\cap T_2)$.
Otherwise, $4\in C(u_3)$, and  $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 4, 1, 3)$ reduces to (2.1.2.).

Assume that there exists a $j\in T_3$ such that $H$ contains no $(j, i)_{(u_3, v)}$-path for each $i\in \{6, 7\}$.
One sees from Corollary <ref>(4) that $5\in B_1$,
and $H$ contains a $(5, i)_{(u_3, u_i)}$-path for some $i\in \{2, 4\}\cap C(u_3)$.
If $T_3\setminus C(u_4)\ne \emptyset$ and $H$ contains no $(2, 5)_{(u_2, u_3)}$-path,
then $(u u_4, u u_3, v u_3)\to (T_3\cap T_4, 5, j)$ reduces the proof to (2.4.1.) or to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
Assume that $T_3\setminus C(u_2)\ne \emptyset$ and $H$ contains no $(4, 5)_{(u_3, u_4)}$-path.
If $|T_3\cap T_2|\ge 2$, then $(u u_2, u u_3, v u_3)\to ((T_3\cap T_2)\setminus \{j\}, 5, j)$;
if $6, 7\not\in C(u_3)$, then $u u_2\to \alpha\in T_3\cap T_2$, $(u u_3, v u_3)\to (5, T_3\setminus \{\alpha\})$;
these reduce to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
Hence, we proceed with the following proposition, or otherwise we are done:

Assume that there exists a $j\in T_3$ such that $H$ contains no $(j, i)_{(u_3, v)}$-path for each $i\in \{6, 7\}$.

If $T_3\setminus C(u_4)\ne \emptyset$, then $H$ contains a $(2, 5)_{(u_2, u_3)}$-path.

If $T_3\setminus C(u_2)\ne \emptyset$, and $|T_3\cap T_2|\ge 2$, or $6, 7\not\in C(u_3)$,
then $H$ contains a $(4, 5)_{(u_3, u_4)}$-path.

Assume that $H$ contains a $(7, j)_{(v, v_7)}$-path for some $j\in T_3$.
If $7\not\in C(u_1)$ and $H$ contains no $(7, i)_{(u_1, u_i)}$-path for each $i\in [2, 4]$,
then $(u u_1, u v)\overset{\divideontimes}\to (7, j)$.
Let $(u u_2, u v)\to (7, j)$.
Assume that $7\not\in C(u_2)\cup C(u_1)$ and $H$ contains no $(7, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$.
If $H$ contains no $(4, l)_{(u_1, u_4)}$-path for some $l\in T_1$,
then $u u_1\overset{\divideontimes} \to l$;
if $2\not\in C(u_1)$ and $H$ contains no $(2, i)_{(u_1, u_i)}$-path for each $i\in [3, 4]$,
then $u u_1\overset{\divideontimes} \to 2$;
if $5\not\in C(u_1)$ and $H$ contains no $(5, i)_{(u_1, u_i)}$-path for each $i\in [3, 4]$,
then $u u_1\overset{\divideontimes} \to 5$.
Hence, we proceed with the following proposition:

Assume that $H$ contains a $(7, j)_{(v, v_7)}$-path for some $j\in T_3$. Then

$7\in C(u_1)$ or $H$ contains a $(7, i)_{(u_1, u_i)}$-path for some $i\in [2, 4]$.

If $7\not\in C(u_2)\cup C(u_1)$, and $H$ contains no $(7, i)_{(u_2, u_i)}$-path for each $i\in [3, 4]$, then

$H$ contains a $(4, T_1)_{(u_1, u_4)}$-path, $4\in C(u_1)$, $T_1\subseteq C(u_4)$;

for each $j\in \{2, 5\}$, either $j\in C(u_1)$ or $H$ contains a $(j, 3/4)_{(u_1, u)}$-path.

Assume $7\not\in C(u_4)$ and $H$ contains no $(7, i)_{(u_3, u_4)}$-path for each $i\in [2, 3]$.
If $4\not\in C(u_1)$ and $H$ contains no $(4, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $(u u_1, u u_4)\to (4, 7)$ reduces the proof to (2.1.2.).
If $7\not\in C(u_1)$ and $H$ contains no $(2, j)_{(u_1, u_2)}$-path for some $j\in T_1$,
then $(u u_4, u u_1)\to (7, j)$ reduces the proof to (2.1.2.).
Hence, we proceed with the following proposition:

Assume $7\not\in C(u_4)$, $H$ contains no $(7, i)_{(u_i, u_4)}$-path for each $i\in [2, 3]$. Then

$4\in C(u_1)$ or $H$ contains a $(4, i)_{(u_1, u_i)}$-path for some $i\in [2, 3]$;

if $7\not\in C(u_1)$, then $2\in C(u_1)$, $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path and $T_1\subseteq C(u_2)$.

$T_3\subseteq C(u_4)$.

Assume that $T_{uv}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$, i.e., $T_3\cap T_4\ne \emptyset$.
It follows that $2\in C(u_3)$, $H$ contains a $(2, T_3)_{(u_3, u_4)}$-path,
and from Corollary <ref>(3) that $4\in B_1\cup B_3$ and mult$_{S_u}(4)\ge 2$.

Case 1. There exists a $j\in T_3\cap T_4$ such that $H$ contains no $(7, j)_{(u_3, u_4)}$-path.
It follows from Corollary <ref>(4) that $5\in B_1$, and $H$ contains a $(5, 2/4)_{(u_3, u)}$-path.
When $C(v_1) = \{1, 5\}\cup T_{uv}$,
it follows from <ref>(2) that $H$ contains a $(4, 5)_{(u_3, v_1)}$-path.
Then $H$ contains a $(2, 5)_{(u_2, u_3)}$-path.
Let $vv_1\to 2$.
If $H$ contains no $(2, j)_{(u_2, v_1)}$-path, then $uv\overset{\divideontimes}\to j$;
otherwise, $(vu_3, uv)\overset{\divideontimes}\to (j, 5)$.
In the other case, $R_1\ne \emptyset$.
It follows from Corollary <ref>(3) that $1/3\in S_u$.
One sees from <ref> that $2\in C(u_1)$ or $t_1\ge 2$,
and from Corollary <ref>(7.1) that $2\in C(u_1)$ or $1\in S_u$.
Further, One sees that if $2\not\in C(u_1)$, then $1\in S_u$ and $t_1\ge 2$,
and if $2\in C(u_1)$, then $t_1\ge 2$ if $4\in C(u_1)$, or else $1\in S_u$.
It follows that mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_1\ge 3$.

Case 1.1. $H$ contains neither a $(3, j)_{(u_4, v)}$-path nor a $(7, j)_{(u_4, v_7)}$-path.
It follows from Corollary <ref>(4) that $6\in B_1\cup B_3$, $6\in C(u_3)\cup C(u_4)$,
mult$_{S_u}(6)\ge 2$, and $2/6\in C(u_i)$, $i = 3, 4$.
One sees that $|\{1, 5\}| + |\{2\}| + |\{2, 3, 5\}| + |\{4, 6\}| + |\{4, 4, 6, 6, 5\}| + |\{1/3\}| + 2t_{uv} = 2\Delta + 10$.

(1.1.1.) $T_3\subseteq C(u_4)$. Then $t_0\ge t_3\ge 2$.
If $T_1\subseteq T_0$, one sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0\ge 2\Delta + 10 + t_3 + t_1\ge 2\Delta + 13$, then $t_1 = 1$ and $2\not\in C(u_1)$, a contradiction.

Otherwise, $T_1\setminus T_0\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1, 7\in S_u$,
and from Corollary <ref>(9) that $2\in C(u_1)$ or $2\in C(u_4)$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{7, 2\}| + t_0\ge 2\Delta + 12 + 2 = 2\Delta + 14$.

(1.1.2.) $T_3\setminus C(u_4)\ne\emptyset$.
It follows from Corollary <ref>(5.1) that $3, 7\in S_u$,
and from <ref>(1) that $5\in C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{7, 1/2\}| + t_0\ge 2\Delta + 12 + t_0$.
One sees that $2\in C(u_1)$ or $6\in C(u_1)$.
When $T_1\subseteq T_0$, then $t_1 = 1$ and thus $1\in S_u$, $2\not\in C(u_1)$, a contradiction.

In the other case, $T_1\setminus T_0 \ne \emptyset$.

If $T_1\subseteq C(u_2)$, then $2\in C(u_1)$ and thus $d(u_1) + d(u_2) + d(u_3)\ge 2t_{uv} + |\{1, 2, 5\}| + |\{2, 5\}| + |\{3, 2, 5\}| + |\{4, 4, 6, 6\}| = 2\Delta + 8$, a contradiction.
Otherwise, $T_1\setminus C(u_2)\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1\in S_u$,
and from Corollary <ref>(9) that mult$_{S_u}(2)\ge 2$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{1/3, 7, 2\}| + t_0 = 2\Delta + 13 + t_0$,
and mult$_{S_u}(1) = $ mult$_{S_u}(3) = 1$, mult$_{S_u}(2) = $ mult$_{S_u}(4) = 2$.
Thus, by Lemma <ref>, we are done.

Case 1.2. $3\in C(u_3)$ and $H$ contains a $(3, j)_{(u_4, v)}$-path.
One sees that $|\{1, 5\}| + |\{2\}| + |\{2, 3, 5\}| + |\{4, 6, 3\}| + |\{4, 4, 6, 5\}| + |\{1/2\}| + 2t_{uv} = 2\Delta + 10$.

(1.2.1.) $T_1\subseteq T_0$.
If $T_3\subseteq T_0$, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0\ge 2\Delta + t_1 + t_3\ge 2\Delta + 1 + 2 = 2\Delta + 13$,
one sees that $t_1 = 1$ and thus $2\not\in C(u_1)$, a contradiction.
Otherwise, $T_3\setminus T_0\ne \emptyset$, i.e., $T_3\setminus C(u_4)\ne \emptyset$.
It follows from <ref>(1) that $5\in C(u_2)$,
and from Corollary <ref>(5.1) that $7\in S_u$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + |\{7\}| + $mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_0\ge 2\Delta + 10 + $mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_1\ge 2\Delta + 13$,
it follows that mult$_{S_u}(3) = $ mult$_{S_u}(6) = 1$,
and mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_1 = 3$.
Hence, $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.
Then it follows from <ref>(2.1) that $6\in C(u_3)$,
and then $uu_1\overset{\divideontimes}\to 6$[Or, by <ref>(1), we are done. ].

(1.2.2.) $T_1\setminus T_0\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1, 7\in S_u$,
and from Corollary <ref>(9) that $2\in S_u\setminus C(u_3)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{7, 1/2\}| +t_0 = 2\Delta + 12 + t_0$.
One sees clearly that $T_3\setminus T_0 \ne \emptyset$,
and it follows from <ref> that mult$_{S_u}(6)\ge 2$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{6\}| + t_0\ge 2\Delta + 13$.
It follows that mult$_{S_u}(1) = $ mult$_{S_u}(3) = 1$, mult$_{S_u}(2) = $ mult$_{S_u}(4) = 2$.
Thus, by Lemma <ref>, we are done.

Case 1.3. $7\in C(u_3)$ and $H$ contains a $(7, j)_{(u_4, v)}$-path.
One sees that $|\{1, 5\}| + |\{2\}| + |\{2, 3, 5\}| + |\{4, 6, 7\}| + |\{4, 4, 6, 5\}| + |\{1/3\}| + 2t_{uv} = 2\Delta + 10$.

(1.3.1.) $T_3\subseteq T_0$.
If $T_1\subseteq T_0$, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0\ge 2\Delta + t_1 + t_3\ge 2\Delta + 1 + 2 = 2\Delta + 13$,
one sees that $t_1 = 1$ and thus $2\not\in C(u_1)$, a contradiction.
Otherwise, $T_1\setminus T_0\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1\in S_u$,
and Corollary <ref>(9) that $2\in S_u\setminus C(u_3)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{2\}| + t_0\ge 2\Delta + 13 + t_0\ge 2\Delta + 13 + t_3$.
It follows that $t_3 = 2$, $3\not\in S_u$, and mult$_{S_u}(6) = 1$.
Thus, $W_3 = [2, 5]$ and then $uu_3\overset{\divideontimes}\to 6$[Or, by <ref>(2), we are done. ].

(1.3.2.) $T_3\setminus T_0\ne \emptyset$, i.e., $T_3\setminus C(u_4)\ne \emptyset$.
It follows from <ref>(1) that $5\in C(u_2)$,
and from Corollary <ref>(5.1) that $3\in S_u$.
When $7\not\in S_u\setminus C(u_4)$,
it follows from <ref>(1) that $4\in C(u_1)$,
$H$ contains a $(4, 7)_{(u_1, u_4)}$-path, and then $t_1\ge 2$.
If there exists a $\alpha\in T_1\setminus (C(u_3)\cap C(u_4))$,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (7, T_3\cap T_3, \alpha)$.
Otherwise, $T_1\subseteq C(u_2)\cap C(u_4)$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{1/2\}| + t_0\ge 2\Delta + 11 + t_1$.
It follows that $t_1 = 2$, mult$_{S_u}(6) = 1$,
and from <ref>(1) that mult$_{S_u}(2) \ge 2$.
Hence, $1\not\in S_u$ and then $W_1 = \{1, 2, 4, 5\}$.
Thus, $u u_1\overset{\divideontimes}\to 6$[Or, by <ref>(1), we are done. ].
In the other case, $7\in S_u\setminus C(u_4)$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{7\}| + $mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_0 = 2\Delta + 11 + $mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_0$.
Recall that mult$_{S_u}(1) + $ mult$_{S_u\setminus C(u_3)}(2) + t_1\ge 3$.
It follows that $T_1\setminus T_0\ne \emptyset$ and then $1\in S_u$ and $2\in C(u_1)\cup C(u_4)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{7, 1, 2\}| = 2\Delta + 13$.
One sees clearly that mult$_{S_u}(1) = $ mult$_{S_u}(3) = 1$, mult$_{S_u}(2) = $ mult$_{S_u}(4) = 2$.
If $T_1\subseteq C(u_2)$, then $2\in C(u_1)$ and thus $d(u_1) + d(u_2) + d(u_3)\ge 2t_{uv} + |\{1, 2, 5\}| + |\{2, 5\}| + |\{3, 2, 5\}| + |\{4, 4, 6, 7\}| = 2\Delta + 8$, a contradiction.
Otherwise, $T_1\setminus C(u_2)\ne \emptyset$.
Thus, by Lemma <ref>, we are done.

Case 2. $H$ contains a $(7, T_3\cap T_4)_{(u_3, u_4)}$-path.
When for some $l\in \{5, 7\}$, $l\not\in S_u\setminus C(u_3)$,
it follows from <ref> or <ref>(1) that $3\in C(u_1)$,
and from <ref>, or <ref>(2.1) and <ref>(2) that $2, 4\in C(u_1)$ and $T_1\subseteq C(u_2)\cap C(u_4)$.
Recall that $6\in C(u_1)\cup C(u_3)$.
Then $w_1 + w_3 = 9$ and $\Delta = 7$, $t_1 + t_3 = 5$.
One sees that $d(u_1) + d(u_2) + d(u_3)\ge |[1, 4]| + |\{2\}| + |\{2, 7, 3, 5\}| + |\{4, 6\}| + 2t_{uv} = 2\Delta + 7$.
It follows that $4\in C(u_2)\setminus C(u_3)$ and $1\not\in S_u\setminus C(u_4)$.
Thus, $(uu_3, uu_1, uv)\overset{\divideontimes}\to (1, l, j)$[Or, by <ref>, we are done. ].

In the other case, $5, 7\in S_u\setminus C(u_3)$.
One sees from Corollary <ref>(5.2) that $3\in C(u_4)$ or mult$_{S_u}(6)\ge 2$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1\}| + |\{2\}| + |\{2, 7, 3, 5\}| + |\{4, 6\}| + |\{4, 4, 6, 5, 7, 2/1, 3/6\}| + 2t_{u v} + t_0 = 2\Delta + 11 + t_0$.
If $T_3\subseteq T_0$, one sees that $t_3 = 2$ and $T_1\cap T_0 = \emptyset$,
it follows from Corollary <ref>(5.1) that $1\in S_u$,
and from Corollary <ref>(9) that $2\in C(u_1)\cup C(u_4)$, a contradiction.
Otherwise, $T_3\setminus T_0 \ne \emptyset$ and from Corollary <ref>(5.1) that $3\in C(u_2)$,
or $H$ contains a $(3, 1/4)_{(u_2, u)}$-path.

Case 2.1. mult$_{S_u}(6)\ge 2$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{6\}| + t_0 = 2\Delta + 12 + t_0$.

Case 2.1.1. $T_1\subseteq T_0$.
It follows that $t_1 = 1$, mult$_{S_u}(i) = 1$, $i = 3, 5$,
and $1\in S_u$ or $2\in S_u\setminus C(u_3)$.

* $2\in S_u\setminus C(u_3)$.
Then $1\not\in S_u$.
It follows from Corollary <ref>(7.1) that $W_1 = \{1, 2, 4\}$.
One sees that $d(u_1) + d(u_2) + d(u_3)\ge |\{1, 2, 4\}| +  |\{2\}| +  |\{2, 7, 3, 5\}| +  |\{4, 6, 6\}| + 2t_{u v} = 2\Delta + 7$.
Then $1, 3\not\in S_u\setminus C(u_4)$, $4\in C(u_2)\setminus C(u_3)$,
and $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.
Thus, by <ref>, we are done.

* $2\not\in S_u\setminus C(u_3)$.
It follows from <ref>(1) that $W_1 = \{1, 4, 3/6\}$.
When $W_1 = \{1, 4, 3\}$, it follows from <ref>(1) that $H$ contains a $(3, 6)_{(u_1, u_3)}$-path.
One sees $3\not\in C(u_4)$ and $H$ contains no $(j, i)_{(u_4, v)}$-path.
We are done by Corollary <ref>(3), we are done.
In the other case, $W_1 = \{1, 4, 6\}$.
It follows from <ref> that $5\in C(u_4)$ and $H$ contains a $(4, 5)_{(u_1, u_4)}$-path,
and then from <ref>(1) that $3\in C(u_2)$.
One sees that $d(u_1) + d(u_2) + d(u_3)\ge |\{1, 4, 6\}| +  |\{2, 3\}| +  |\{2, 7, 3, 5\}| +  |\{4, 6\}| + 2t_{u v} = 2\Delta + 7$.
It follows from <ref> that $1\in C(u_4)$ and $H$ contains a $(4, 1)_{(u_3, u_4)}$-path.
Thus, $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (3, 1, T_3, 2)$.

Case 2.1.2. $T_1\setminus T_0\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1\in S_u$,
and from Corollary <ref>(9) that $2\in S_u\setminus C(u_3)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{1/2\}| +t_0 = 2\Delta + 13 + t_0$.
It follows that $t_0 = 0$, mult$_{S_u}(1) = $ mult$_{S_u}(3) = 1$, mult$_{S_u}(2) = $ mult$_{S_u}(4) = 2$.
If $T_1\subseteq C(u_2)$, one sees that $d(u_1) + d(u_2) + d(u_3) = |\{1, 2\}| + |\{2\}| + |\{3, 5, 2, 7\}| + |\{4, 4, 6, 6\}| + 2t_{u v} = 2\Delta + 7$,
then $1, 3\not\in S_u\setminus C(u_4)$, $4\in C(u_2)\setminus C(u_3)$, and $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.
Thus, by <ref>, we are done.
Otherwise, $T_1\setminus C(u_2)\ne \emptyset$.
Thus, by Lemma <ref>, we are done.

Case 2.2. mult$_{S_u}(6) = 1$.
One sees from <ref>(1–2) that $1\in C(u_3)$ or $3\in C(u_1)$,
and from Corollary <ref>(5.2) that if there exists a $j\in T_3\cap T_4$ such that $H$ contains no $(3, j)_{(u_4, v_3)}$-path, then mult$_{S_u}(6)\ge 2$.
Hence, $H$ contains a $(3, T_3\cap T_4)_{(u_4, v_3)}$-path and $3\in C(u_4)$.
It follows from <ref>(1) that $2\in C(u_1)$, or $H$ contains a $(4, 2)_{(u_1, u_4)}$-path,
and from <ref>(2) that $T_1\subseteq C(u_2)\cap C(u_4)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1\}| + |\{2\}| + |\{2, 7, 3, 5\}| + |\{4, 6\}| + |\{4, 4, 6, 5, 7, 2, 3, 1/3\}| + 2t_{u v} + t_0 = 2\Delta + 12 + t_0\ge 2\Delta + 12 + t_1\ge 2\Delta + 12 + 1 = 2\Delta + 13$.
It follows that $t_1 = 1$, mult$_{S_u}(i) = 1$, $i = 5, 7$, mult$_{S_u}(1) + $mult$_{S_u\setminus C(u_4)}(3) = 1$,
and $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.

(2.2.1.) $6\in C(u_3)$.
Then $3\in C(u_1)$ and $1\not\in S_u$.
It follows from <ref>(3) that $4\in C(u_1)$ and then $W_1 = \{1, 3, 4\}$.
Then $(u u_1, u u_3)\to (6, 1)$ reduces the proof to (2.4.2.)
[Or, by <ref>, we are done.].

(2.2.2.) $6\in C(u_1)$.
Then $1\in C(u_3)$ and $(u u_3, u u_1)\to (6, 3)$ reduces the proof to (2.4.1.)
[Or, by <ref>(2.1), we are done.].

By Lemma <ref>, hereafter we assume that $T_3\subseteq C(u_4)$.
One sees that $t_3\ge 2$ and $w_1 + w_4 + t_{u v} + t_3 \le d(u_1) + d(u_4)\le \Delta + 7$.
Hence, $w_1 + w_4 + t_3 \le 9$ and $w_1 + w_4\le 7$.

$5\in S_u\setminus C(u_3)$, $7\in S_u$, and mult$_{S_u}(7) +  $mult$_{S_u\setminus C(u_3)}(5)\ge 2$.

If $7\in C(u_3)$ and $H$ contains a $(7, j)_{(u_3, v)}$-path for some $j\in T_3$,
then $7\in S_u\setminus C(u_3)$.

mult$_{S_u}(6)\ge 2$.

If mult$_{S_u}(7) + $mult$_{S_u\setminus C(u_3)}(5) = 2$, then $6\in C(u_3)$ and $H$ contains a $(6, T_3)_{(u_3, u_4)}$-path.

$T_4\ne \emptyset$.

For (1), assume that $5\not\in S_u\setminus C(u_3)$.
It follows from <ref> that $3\in C(u_1)$,
from <ref>(1) that $2\in C(u_3)$, $T_3\subseteq C(u_2)$,
from <ref>(2) that $4\in C(u_2)\cup C(u_3)$,
and from <ref> that $2, 4\in C(u_1)$, $T_1\subseteq C(u_2)\cap C(u_4)$.
If there exists a $j\in \{1, 7\}\setminus S_u$, then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (5, 7, T_3)$.
Otherwise, $1, 7\in S_u$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |[1, 4]| + |\{2\}| + |\{3, 5, 2\}| + |\{4, 6\}| + |\{4, 6, 1, 7\}| + 2t_{u v} + t_0\ge 2\Delta + 10 + t_1 + t_3\ge 2\Delta + 10 + 2 + 2 = 2\Delta + 14$, a contradiction.
Hence, $5\in S_u\setminus C(u_3)$.

Assume that $7\not\in S_u$.
It follows from Corollary <ref>(5.1) that $2, 4\in C(u_1)\cap C(u_3)$ and $T_1\cup T_3\subseteq C(u_2)\cap C(u_4)$,
and from <ref>(1) that mult$_{S_u}(6)\ge 2$.
Since $6\in C(u_1)\cup C(u_3)$, $t_1 + t_3\ge w_1 + w_3 - 4\ge 8 - 4 = 4$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 2, 4\}| + |\{2\}| + |[2, 5]| + |\{4, 6\}| + |\{5, 6, 6\}| + 2t_{u v} + t_0\ge 2\Delta + 9 + t_1 + t_3\ge 2\Delta + 9 + 4 = 2\Delta + 13$.
It follows that $1, 3\not\in S_u$, mult$_{S_u\setminus C(u_3)}(5) = 1$, and $5\in C(u_i)$ for some $i\in \{2, 4\}$.
Then $(u u_1, u u_i, u v)\overset{\divideontimes}\to (5, 1, T_3)$.

Hence, $7\in S_u$ and then mult$_{S_u}(7) +  $mult$_{S_u\setminus C(u_3)}(5)\ge 2$.

For (2), assume that $7\in C(u_3)$, $H$ contains a $(7, 8)_{(u_3, v)}$-path, and $7\not\in S_u\setminus C(u_3)$.
It follows from <ref>(1) that $3\in C(u_1)$ and $H$ contains a $(3, 7)_{(u_1, u_3)}$-path,
and the from <ref>(2.1) that $4\in C(u_1)$, $T_1\subseteq C(u_4)$,
from <ref>(2) that $2\in C(u_1)$, $T_1\subseteq C(u_2)$.
If $2\not\in B_1\cup B_3$, then $(u u_2, u v)\overset{\divideontimes}\to (7, 2)$;
if $4\not\in B_1\cup B_3$, then $(u u_4, u v)\overset{\divideontimes}\to (7, 4)$.
Otherwise, $2, 4\in B_1\cup B_3$.

* $2\not\in C(u_3)\cup C(u_4)$. One sees that $2\in B_1$ and $H$ contains no $(2, 4)_{(u_3, u_4)}$-path.
If $H$ contains no $(7, j)_{(u_2, v_7)}$-path for some $j\in T_1$,
then $(u u_2, u u_3, uv)\overset{\divideontimes}\to (7, 2, j)$.
otherwise, $H$ contains a $(7, T_1)_{(u_2, v_7)}$-path and $(u u_3, u u_4, u v)\overset{\divideontimes}\to (7, 8, T_1)$.

* $4\not\in C(u_2)\cup C(u_3)$.
One sees that $4\in B_1$ and $H$ contains no $(2, 4)_{(u_2, u_3)}$-path.
If $H$ contains no $(7, j)_{(u_2, v_7)}$-path for some $j\in T_1$,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (7, 8, j)$.
otherwise, $H$ contains a $(7, T_1)_{(u_2, v_7)}$-path and $(u u_3, u u_4, u v)\overset{\divideontimes}\to (4, 7, T_1)$.

* $2\in C(u_3)\cup C(u_4)$ and $4\in C(u_2)\cup C(u_3)$.
If $1\not\in C(u_1)$ and $H$ contains no $(1, i)_{(u_3, u_i)}$-path,
then $(u u_1, u u_3, u v)\to (7, 1, 8)$.
Otherwise, $1\in C(u_1)$ or $H$ contains a $(1, 2/4)_{(u_3, u)}$-path.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |[1, 4]| + |\{2\}| + |\{3, 5, 7\}| + |\{4, 6\}| + |\{5, 6, 2, 4, 1\}| + 2t_{u v} + t_0\ge 2\Delta + 11 + t_1\ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $t_1 = 2$,  mult$_{S_u}(6) = 1$, $T_3\cap T_1 = \emptyset$ and then $4\in C(u_3)$.
One sees that $2\in C(u_3)\cup C(u_4)$ and $d(u_3) + d(u_4)\ge |\{3, 5, 7, 4\}| + |\{4, 6\}| + |\{2\}| + t_{uv} + t_1\ge \Delta + 5 + 2 = \Delta + 7$.
It follows that $1\in C(u_2)$, $H$ contains a $(1, 2)_{(u_2, u_3)}$-path and $2\in C(u_3)$.
One sees that $W_4 = \{4, 6\}$,
and $H$ contains no $(j, i)_{(u_4, v)}$-path for each $i\in C(v)$ and $j\in T_4$.
Thus, by Corollary <ref>(4), we are done.

For (3), assume that mult$_{S_u}(6) = 1$.
It follows from <ref>(4) that $T_4\ne \emptyset$, $1/3/5/7\in C(u_4)$,
and from <ref>(1) that $2\in B_1\cup B_3$,
and $2\in C(u_1)$ or $H$ contains a $(2, 4)_{(u_1, u_4)}$-path,
i.e., $2\in C(u_1)$ or $2\in C(u_3)\cap C(u_4)$.
By Corollary <ref>(4), $H$ contains a $(7, T_3)_{(u_3, v_7)}$-path,
or $5\in B_1\cap C(u_1)$, mult$_{S_u\setminus C(u_3)}(5)\ge 2$.
Hence, by (1) and (2), we have mult$_{S_u}(7) +  $mult$_{S_u\setminus C(u_3)}(5)\ge 3$.

When $6\in C(u_3)$, then $3\in C(u_1)$, $H$ contains a $(3, 6)_{(u_1, u_3)}$-path,
and it follows from <ref>(3) that $4\in C(u_1)$ and $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path.
We distinguish whether or not $2\in C(u_1)$:

* $2\in C(u_1)$.
One sees that $t_1\ge 2$ and $d(u_3) + d(u_4)\ge t_{uv} + t_1 + |\{3, 5, 6, 2/4\}| + |\{4, 6, 5/1/3/7\}|\ge\Delta + 5 + t_1\ge \Delta + 5 + 2 = \Delta + 7$.
It follows that $W_1 = [1, 4]$, $t_1 = 2$, $W_3 = \{3, 5, 6, a_3\}$, $W_4 = \{4, 6, 5/1/3/7\}$, $d(u_i) = \Delta = 7$, $i = 1, 3, 4$, $d(u_2)\le 6$,
and from <ref>(1.1–1.2) that $H$ contains a $(a_3, \{1, 7\})_{(u_3, u_{a_3})}$-path, $1, 7\in C(u_{a_3})$.
Hence, $a_3 = 2$, $\{1, 7\}\cup T_3\subseteq C(u_2)$, and $T_1\setminus C(u_2)\ne \emptyset$.
One sees from $5\not\in C(u_1)$ that $H$ contains a $(6, T_3)_{(u_3, u_4)}$-path.
Then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (6, T_3, T_1\cap T_2)$.

* $2\not\in C(u_1)$ and $2\in C(u_3)\cap C(u_4)$.
One sees that $t_1\ge 1$ and $d(u_3) + d(u_4)\ge t_{u v} + t_1 + |\{3, 5, 6, 2\}| + |\{4, 6, 2, 5/1/3/7\}|\ge\Delta + 6 + t_1\ge \Delta + 6 + 1 = \Delta + 7$.
It follows that $W_1 = \{1, 3, 4\}$, $W_3 = \{3, 5, 6, 2\}$, $W_4 = \{4, 6, 2, 5/1/3/7\}$,
$d(u_i) = \Delta = 7$, $i = 1, 3, 4$, $d(u_2)\le 6$, $t_1 = 1$, $t_3 = 2$, $c_{13} = 2$,
and from <ref>(1.1–1.2) that $H$ contains a $(2, \{1, 7\})_{(u_3, u_{a_3})}$-path, $1, 7\in C(u_2)$.
One sees that $1, 7\setminus C(u_4)\ne \emptyset$ and assume w.l.o.g. $1\not\in C(u_4)$[If $7\not\in C(u_4)$, then $u u_1\to 7$ reduces the proof to the case where $1\not\in C(u_4)$. ].
Then $(u u_1, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, T_3)$.

Hence, $6\in C(u_1)$, $1\in C(u_3)$, and $H$ contains a $(1, 6)_{(u_1, u_3)}$-path.
It follows from <ref> (2.1) that $3\in C(u_1)$, or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path.
Since $7\in C(u_3)$ or $5\in C(u_1)$,
$t_1 + t_3\ge w_1 + w_3 - 4 = |\{1, 6, 2\}| + |\{3, 5, 1, 2/4\}| + |\{5/7\}| - 4 = 4$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 6, 2\}| +|\{2\}| + |\{3, 5, 1, 2/4\}| + |\{4, 6\}|+  |\{3\}| + $mult$_{S_u}(7) +  $mult$_{S_u\setminus C(u_3)}(5) + 2t_{u v} \ge 11 + 3 + 2t_{u v} + t_0 = 2\Delta + 10 + t_0$.
It follows that $(T_3\cup T_1)\setminus T_0\ne \emptyset$.
We distinguish whether or not $2\in C(u_1)$:

* $2\in C(u_1)$.
When $H$ contains no $(4, j)_{(u_3, u_4)}$-path for some $j\in T_3$,
one sees from <ref>(2) that $T_1\subseteq T_0$ and then $T_3\setminus T_0\ne \emptyset$,
i.e., $T_3\setminus C(u_2)\ne \emptyset$,
thus, $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path, a contradiction.
In the other case, $4\in C(u_3)$ and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.

* $2\not\in C(u_3)\cup C(u_4)$.
It follows from Corollary <ref>(5.1) that $3\in C(u_4)$ or $H$ contains a $(3, 1)_{u_4, u_1}$-path,
from <ref> that $5\in C(u_4)$ or $H$ contains a $(5, 1)_{u_4, u_1}$-path,
and from <ref>(2) that $T_1\subseteq T_0$.
When $w_4 \ge 4$, it follows that $W_3 = \{1, 4, 3, 5\}$, $t_1 = 1$, $t_3 = 2$, $c_{13} = 2$,
and then $W_1 = \{1, 2, 6\}$.
Thus, $(v u_3, u v)\overset{\divideontimes}\to (T_3, 5)$
[a contradiction to Corollary <ref>(4). ].
In the other case, $w_4\le 3$.
Hence, $W_4 = \{1, 4, 6\}$, $3, 5\in C(u_1)$, and then $W_1 = \{1, 2, 3, 5, 6\}$, $W_3 = \{1, 3, 4, 5\}$. Thus, $(v u_3, u u_3, u v)\overset{\divideontimes}\to (T_3, 5, T_1)$
[a contradiction to Corollary <ref>(4). ].

* $2\in C(u_3)\cup C(u_4)$.
One sees that $w_3 + w_4\ge |\{1, 3, 4, 5\}| + |\{4, 6, 5/1/3/7\}| + |\{2\}| = 8$.
If $T_1\subseteq C(u_4)$, then $t_1 = 1$ and then $W_3 = \{1, 4, 3, 5\}$, $t_3 = 2$, $c_{13} = 2$,
and then $W_1 = \{1, 2, 6\}$.
Thus, $(v u_3, u v)\overset{\divideontimes}\to (T_3, 5)$
[a contradiction to Corollary <ref>(4). ].
Otherwise, $T_1\setminus C(u_4)\ne \emptyset$.
It follows from Corollary <ref>(5.2) and (3) that $4\in B_1\cup B_3$, $4\in C(u_1)\cup C(u_2)$,
and from <ref>(4) that $5/1/7\in C(u_4)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{2, 4\}| + t_0 = 2\Delta + 12 + t_0$.
Since $t_3\ge 2$, we have $T_3\setminus C(u_2)\ne \emptyset$,
and it follows from Corollary <ref>(5.1) that $3\in C(u_4)$ or $H$ contains a $(3, 1/2)_{(u_4, u)}$-path,
from <ref>(1) that $5\in C(u_4)$ or $H$ contains a $(5, 1/2)_{(u_4, u)}$-path.
If $5\not\in C(u_1)\cup C(u_4)$, one sees that $H$ contains a $(2, 5)_{(u_2, u_4)}$-path,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (5, 6, T_3)$.
Otherwise, together with <ref> (2.1), we have $3, 5\in C(u_1)\cup C(u_4)$.
If mult$_{S_u}(1) + $mult$_{S_u}(3) = 2$,
one sees from Corollary <ref>(5.1) that $3\in C(u_2)$,
then $(u u_1, u u_2, u u_3)\to (3, 1, 6)$, and $uv\overset{\divideontimes} \to 2$,
or $uv\overset{\divideontimes} \to T_1\cap R_1$
[Or, One sees that $(u u_1, u u_2, u u_3)\to (3, 1, 6)$ reduces to an acyclic edge $(\Delta + 5)$-coloring $c'$ for $H$ such that $C(u)\cap C(v) = \{1, 3\}$, $2\not\in C'(u)$,  then by <ref>, we are done. ].
Otherwise, mult$_{S_u}(1) + $mult$_{S_u}(3)\ge 3$.

It follows that $t_0 = 0$, mult$_{S_u}(7) + $mult$_{S_u\setminus C(u_3)}(5) = 3$,
mult$_{S_u}(i) = 2$, $i = 2, 4$,
and from <ref>(2) that if there exists a $j\in T_3$ such that $H$ contains no $(j, 7)_{(u_3, v)}$-path, then $5\in B_1$, $H$ contains a $(4, 5)_{(u_3, u_4)}$-path, and $5\in C(u_3)\cap C(u_4)$.

If $2\in C(u_3)$, one sees from $7\in C(u_3)$ or $5\in C(u_1)$ that $3, 4\not\in C(u_1)$ and from <ref>(2.1) that $H$ contains a $(2, 3)_{(u_1, u_2)}$-path,
then $(u u_1, u u_2, u u_3)\to (3, 1, 6)$, and $uv\overset{\divideontimes} \to 2$,
or $uv\overset{\divideontimes} \to T_1\cap R_1$
[Or, One sees that $(u u_1, u u_2, u u_3)\to (3, 1, 6)$ reduces to an acyclic edge $(\Delta + 5)$-coloring $c'$ for $H$ such that $C(u)\cap C(v) = \{1, 3\}$, $2\not\in C'(u)$,  then by <ref>, we are done. ].
Otherwise, $2\in C(u_4)\setminus C(u_3)$ and then $2\in B_1$, $T_1\cap R_1\ne \emptyset$.