If $3\in C(u_1)\cap C(u_2)$, one sees that $1\not\in S_u\setminus C(u_3)$,
then $uv\to \alpha\in T_1\cap R_1$, and $(u u_1, u u_2, u u_3, u u_4)\to (T_1\setminus \{\alpha\}, 1, 2, 3)$.
Otherwise, $3\not\in C(u_1)\cap C(u_2)$.

* $7\in C(u_3)$.
It follows that $W_3 = \{1, 3, 4, 5, 7\}$, $W_4 = \{4, 6, 2, 5/1/7\}$.
Together with $w_1 + w_3\le 9$ and $3, 5\in C(u_1)\cup C(u_3)$,
we have $W_1 = \{1, 2, 3, 6\}$, $W_4 = \{4, 6, 2, 5\}$, and then $3\in C(u_2)$.
Thus, are done.

* $7\not\in C(u_3)$. Then $5\in C(u_1)\cap C(u_4)$.
It follows from <ref>(2.2) that $7\in C(u_1)$ or $H$ contains a $(7, 2/4)_{(u_1, u)}$-path.
If $7\not\in C(u_4)$ and $H$ contains no $(7, i)_{(u_4, u_i)}$-path for each $i\in \{1, 2\}$,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (7, T_3, T_1)$.
Otherwise, $7\in C(u_4)$ or $H$ contains a $(7, 1/2)_{(u_4, u)}$-path.
Hence, $7\in C(u_1)\cup C(u_4)$ and $W_1\cup W_4 = \{1, 2, 5, 6\}\cup \{4, 5, 6\}\cup \{3, 7\}$.
We will have $7\in C(u_2)$ no matter on whether or not $7\in C(u_1)$, a contradiction.

* $2\not\in C(u_1)$ and $2\in C(u_3)\cap C(u_4)$.
One sees that $d(u_3) + d(u_4)\ge t_{uv} - t_1 + t_3 + |\{1, 4, 6\}| + |\{4, 6, 2, 5/1/7/3\}| = \Delta + 5 + t_3 - t_1\ge \Delta + 7 - t_1$.
If $T_1\subseteq C(u_4)$, then $W_1 = \{1, 4, 6\}$, $3\in C(u_4)$,
and it follows from <ref> that $5\in C(u_4)$, a contradiction.
Otherwise, $T_1\setminus C(u_4)$ and it follows from <ref>(2) that $4\in C(u_3)$ and $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path.
Hence, $W_3 = [1, 5]$, $W_4 = \{4, 6, 2, 5/1/3/7\}$, $T_1\cap C(u_4) = \emptyset$, and then $W_4 = \{4, 6, 2, 5/1/7\}$, $3\in C(u_1)$.
Thus, $w_1 + w_4\ge |\{1, 6, 4, 3, 7\}| + 5 = 10$,  a contradiction.

For (4), if mult$_{S_u}(7) + $mult$_{S_u\setminus C(u_3)}(5) = 2$, then it follows from (1) and (2) naturally that $6\in C(u_3)$ and $H$ contains a $(6, T_3)_{(u_3, u_4)}$-path.

For (5), it $C(u_4) = T_{uv}\cup \{4, 6\}$, it follows from <ref>(2) that mult$_{S_u}(4)\ge 2$,
from Corollary <ref>(7.2) that $2\in C(u_1)\cap C(u_3)$ and $T_1\cup T_3\subseteq C(u_2)$,
thus, $d(u_1) + d(u_2) + d(u_3)\ge |\{1, 2\}| + |\{2\}| + |\{3, 5, 2\}| + |\{5, 7, 4, 4, 6, 6\}| + 2t_{u v} = 2\Delta + 8$.
Otherwise, $T_4\ne \emptyset$.

$T_1\setminus C(u_4)\ne \emptyset$ and $T_3\setminus C(u_2)\ne \emptyset$.

One sees from <ref>(1) that $5\in S_u\setminus C(u_3)$, $7\in S_u$,
from <ref>(3) that mult$_{S_u}(6)\ge 2$,
and from Corollary <ref>(7.1) that $2, 4\in C(u_1)$, or $1\in S_u$,
and $2, 4\in C(u_3)$, or $3\in S_u$.
Then $|[1, 6]| + |\{5, 6, 7, 7\}| + 2t_{u v} = 2\Delta + 6$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |[1, 6]| + |\{5, 6, 7, 7\}| + |\{2, 4\}/\{a_1, 1\}| + |\{2, 4\}/\{a_3, 3\}| + 2t_{u v} + t_0 = 2\Delta + 10 + t_0$.
It follows that if $T_1\cup T_3\subseteq T_0$, then $t_1 = 1$, $t_3 = 2$.
One sees from <ref>(4) that if $T_1\cup T_3\subseteq C(u_4)$, then $C_{13}\ne \emptyset$.
We discuss the following two subcases.

Case 1. $T_1\cup T_3\subseteq C(u_2)$. Then $t_0\ge t_3$.
We distinguish whether or not $T_1\subseteq C(u_4)$:

Case 1.1. $T_1\setminus C(u_4)\ne \emptyset$.
Then $2\in C(u_1)$,
and it follows from Corollary <ref>(5.1) that $1\in C(u_2)$ or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path,
from Corollary <ref>(5.2) that mult$_{S_u}(4)\ge 2$.
One sees from  Corollary <ref>(7.1) that $2\in C(u_3)$ or $3\in S_u$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 6 + |\{4, 4, 2, 1, 3/2\}| + t_0\ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $\Delta = 7$, $t_3 = 2$, $T_1\cap C(u_4) = \emptyset$,
mult$_{S_u}(4) = 2$, mult$_{S_u}(i) = 1$, $i = 1, 5, 7$,
and from Lemma <ref>(3) that $6\in C(u_3)$.
One sees that $W_3 = \{3, 5, 6, 2/4\}$.
It follows from Corollary <ref>(7.1) that $3\in S_u$ and $2\not\in S_u\setminus C(u_1)$.

* $1\in C(u_2)$.
One sees that $1, 2\not\in C(u_4)$.
It follows from Corollary <ref>(5.1) and <ref>(1) that $3, 5\in C(u_4)$,
and then from <ref> that $4\in C(u_1)$.
Then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (T_1, 5, T_3)$
[Or, by <ref>(1), we are done. ].

* $1\in C(u_4)$.
It follows from Corollary <ref>(5.1) that $4\in C(u_2)$ and $H$ contains a $(1, 4)_{(u_2, u_4)}$-path,
and from <ref> and <ref> that for each $j\in \{3, 5\}$, $j\in C(u_1)$, or $H$ contains a $(2, j)_{(u_1, u_2)}$-path.
Together with Corollary <ref>(5.1) and <ref>(1), we have $3, 5\in C(u_1)$.
Let $(u u_2, u u_3)\to (5, 2)$.
If $H$ contains no $(5, j)_{(u_1, v_1)}$-path for some $j\in T_1$,
then $(u u_1, u v)\overset{\divideontimes}\to (j, T_3)$.
Otherwise, $H$ contains a $(5, T_1)_{(u_1, u_2)}$-path and $uv\overset{\divideontimes}\to T_1$.

Case 1.2. $T_1\subseteq C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0\ge 2\Delta + 10 + t_1 + t_3\ge 2\Delta + 10 + 1 + 2 = 2\Delta + 13$.
It follows that $\Delta = 7$, $t_1 = 1$, $t_3 = 2$, $c_{13} = 2$, $C_{13}\cap (C(u_2)\cup C(u_4)) = \emptyset$,
$\sum_{i\in [1, 4]} $mult$_{S_u}(i) = 4$, mult$_{S_u}(i) = 1$, $i = 5, 7$,
and from Lemma <ref>(3) that $6\in C(u_3)$.
One sees that $W_3 = \{3, 5, 6, 2/4\}$ and then $3\in S_u$.
It follows from Corollary <ref>(4.2) that there exists a $\gamma\in \{2, 4\}$ such that mult$_{S_u}(\gamma)\ge 2$.

* $2, 4\in C(u_1)$.
It follows that $1\not\in S_u$ and from Corollary <ref>(5.1) that $3\in C(u_{a_3})$.
Thus, one sees from Corollary <ref>(3) that mult$_{S_u}(j)\ge 2$, $j\in \{2, 4\}\setminus \{a_2\}$,
a contradiction.

* $\{2, 4\}\setminus C(u_1)\ne \emptyset$.
Then $a_1 = a_3 = \gamma$, $4\not\in C(u_2)$, $2\not\in C(u_4)$,
and from Corollary <ref>(3) that $1/3\in C(u_\beta)$, where $\{\gamma, \beta\} = \{2, 4\}$.
One sees that $1\not\in C(u_3)$ and $\beta\not\in C(u_\gamma)$.
It follows from Corollary <ref>(5.1) that $1\in C(u_\gamma)$ and then $3\in C(u_\beta)$.
Then $(u u_\beta, u u_1, u v)\overset{\divideontimes}\to (1, \beta, T_3)$.

Case 2. $(T_1\cup T_3)\setminus C(u_2)\ne \emptyset$ and $T_1\cup T_3\subseteq C(u_4)$.
Then $C_{13}\ne \emptyset$, $w_1 + w_3\le 9 - c_{13}$,
and for each $i\in \{1, 3\}$ where $T_i\setminus C(u_2)\ne \emptyset$,
$4\in C(u_i)$, and $H$ contains a $(4, T_i)_{(u_i, u_4)}$-path,
and from Corollary <ref>(5.2) and (3) that $2\in B_1\cup B_3$,
for each $j\in \{1, 3\}$, $2\in C(u_j)$, or $H$ contains a $(2, 4)_{(u_j, u_4)}$-path.
One sees clearly that mult$_{S_u}(4)\ge 1$ and mult$_{S_u}(2)\ge 2$.

Case 2.1. $T_3\subseteq C(u_2)$ and $T_1\setminus C(u_2)\ne \emptyset$.
Then $4\in C(u_1)$ and follows from  Corollary <ref>(5.1) that $1\in C(u_4)$ or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path,
from Corollary <ref>(7.1) that $4\in C(u_3)$ or $3\in S_u$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 6 + |\{2, 2, 4, 1, 4/3\}| + t_0\ge 2\Delta + 11 + t_3\ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $t_3 = 2$,  mult$_{S_u}(2) = 2$, mult$_{S_u}(i) = 1$, $i = 5, 7$,
and from Lemma <ref>(3) that $6\in C(u_3)$.
Since $W_3 = \{3, 5, 6, 2/4\}$, we have $3\in S_u$, $4\not\in S_u\setminus C(u_1)$,
$W_3 = \{3, 5, 6, 2\}$ and from Corollary <ref>(5.1) that $3\in C(u_2)$ or $H$ contains a $(1, 3)_{(u_1, u_2)}$-path.
Since $1\in C(u_4)$ or $H$ contains a $(1, 2)_{(u_2, u_4)}$-path, $1\in C(u_2)$, it follows from <ref> that $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path.

* $1\in C(u_2)$.
One sees that $2\in C(u_4)$, and $3\in C(u_1)$ or $H$ contains a $(4, 3)_{(u_1, u_4)}$-path.
Hence, $3\in C(u_1)$.
Then $uu_3\to \alpha\in T_3$,
and $(u u_1, u u_2, u u_4, u v) \overset{\divideontimes} \to (2, 4, 1, T_3\setminus \{j\})$.

* $1\in C(u_4)$.
One sees that $1, 4\not\in C(u_2)$.
Hence, $3\in C(u_2)$ and $2\in C(u_1)$.
Then $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 4, 1, 3)$ reduces the proof to (2.1.2.).

Case 2.2. $T_1\subseteq C(u_2)$ and $T_3\setminus C(u_2)\ne \emptyset$.
Then $4\in C(u_3)$ and follows from Corollary <ref>(5.1) that $3\in C(u_4)$ or $H$ contains a $(3, 1/2)_{(u_4, u)}$-path,
from Corollary <ref>(7.1) that $4\in C(u_1)$ or $1\in S_u$,
from <ref>(1) that $5\in C(u_4)$, or $H$ contains a $(5, 1/2)_{(u_4, u)}$-path.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 6 + |\{2, 2, 4, 3, 4/1\}| + t_0\ge 2\Delta + 11 + t_1\ge 2\Delta + 11 + 1 = 2\Delta + 12$ and $|C_{13}\cap T_0|\le 1$.

When $3, 5\in C(u_1)\cup C(u_4)$,
one sees that $d(u_1) + d(u_3) + d(u_4)\ge 2t_{uv} + |\{1\}| + |\{2\}| + |\{3, 5, 4\}| + |\{4, 6\} | + |\{2, 2, 3, 5, 6\}| = 2\Delta + 7$.
It follows that $2\in C(u_1)$ and $1, 7\not\in C(u_1)\cup C(u_3)\cup C(u_4)$.
If $2\not\in C(u_4)$, one sees that $W_4 = [3, 6]$, $W_3 = [2, 5]$ and $W_1 = \{1, 2, 6\}$,
then $(vu_3, uv)\overset{\divideontimes} \to (T_3, 5)$.
Otherwise, $2\in C(u_4)\setminus C(u_3)$, $2\in B_1$, and from <ref>(1) that $R_1\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $H$ contains a $(2, 7)_{(u_2, u_4)}$-path.
Since $u u_1\to 7$ and $(u u_1, u u_3)\to (7, 1)$ reduce to an acyclic edge $(\Delta + 5)$-coloring for $H$,
by <ref>, we are done.

In the other case, $\{3, 5\}\setminus (C(u_1)\cup C(u_4))\ne \emptyset$.
It follows that $2\in C(u_4)$ and $H$ contains a $(2, \{3, 5\}\setminus (C(u_1)\cup C(u_4)))_{(u_2, u_4)}$-path.
If $H$ contains no $(2, j)_{(u_2, u_4)}$-path for some $j\in \{3, 5\}\setminus C(u_4)$,
one sees that $1\in C(u_4)$, $H$ contains a $(1, j)_{(u_1, u_2)}$-path,
and then $W_1 = \{1, 2/4, j\}$, $W_4 = \{4, 6, 2, 1\}$.
Then $(u u_1, u u_4, u u_3)\to (\{3, 5\}\setminus \{j\}, j, T_3\cap T_2)$ reduces the proof to (2.4.1.) or (2.4.2.).
Otherwise, $H$ contains a $(2, \{3, 5\}\setminus C(u_4))_{(u_2, u_4)}$-path.

* $1\not\in S_u$. One sees that $2, 4\in C(u_1)$.
If $5\not\in C(u_1)\cup C(u_4)$,
then $(u u_1, u u_4, u u_3)\to (5, 1, T_3\cap T_2)$ reduces the proof to (2.4.1.).
Otherwise, $5\in C(u_1)\cup C(u_4)$.
It follows that $H$ contains a $(2, 3)_{(u_2, u_4)}$-path, $W_3 = [3, 6]$, and then $2\in B_1$, $R_1\ne \emptyset$.
Then, by <ref>, we are done.

* $1\in S_u$. One sees that if $4\not\in S_u\setminus C(u_3)$,
then for each $j\in C_{13}$, $H$ contains a $(j, i)_{(u_4, u_i)}$-path for some $i\in [1, 3]$,
and that if $5\not\in C(u_1)\cup C(u_4)$, then it follows from <ref> that $5/3\in C(u_1)$.

* $4\not\in S_4\setminus C(u_3)$.
If $W_4 = \{4, 6, 2\}$, one sees that $C_{13}\subseteq C(u_2)$, $3, 5\in C(u_2)$, and $5/3\in C(u_1)$,
then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 6 + |\{2, 2, 4, 3, 1, 5/3\}| + t_0\ge 2\Delta + 12 + t_1 + c_{13}\ge 2\Delta + 12 + 1 + 1 = 2\Delta + 14$.
Otherwise, $w_4 = 4$, $c_{13}\ge 2$, and then $1/3\in C(u_4)$.
It follows that $W_1 = \{1, 2, 3/5\}$, $W_3 = [3, 6]$, and $1, 5\in C(u_2)$.
Since $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{3/5\}| = 2\Delta + 13$,
$W_4 = \{4, 6, 2, 3\}$.
It follows from Corollary <ref>(5.1) that $H$ contains a $(2, 7)_{(u_2, u_4)}$-path.
Since $u u_3\to 7$ and $(u u_1, u u_3)\to (7, 1)$ reduce to an acyclic edge $(\Delta + 5)$-coloring for $H$, by <ref>, we are done.

* $4\in C(u_1)\cup C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{4\}| = 2\Delta + 13$.
It follows that $t_0 = t_1 = 1$, mult$_{S_u}(i) = 1$, $i = 1, 3, 5, 7$,
mult$_{S_u}(j) = 2$, $j = 2, 4, 6$,
and from Lemma <ref>(4) that $6\in C(u_3)$ and $H$ contains a $(6, T_3)_{(u_3, u_4)}$-path.
If $5\not\in C(u_4)$, then $5\in C(u_2)$, $3\in C(u_2)\cup C(u_4)$, and $5/3\in C(u_1)$, a contradiction. Otherwise, $W_4 = \{4, 6, 2, 5\}$, $W_3 = [3, 6]$, and $H$ contains a $(2, 3)_{(u_2, u_4)}$-path.
Then by <ref>, we are done.

Case 2.3. $T_3\setminus C(u_2)\ne \emptyset$ and $T_1\setminus C(u_2)\ne \emptyset$.
Then $4\in C(u_1)\cap C(u_3)$, $H$ contains a $(4, T_1\cup T_3)_{(u_4, u)}$-path,
and for each $j\in \{1, 3\}$, $j\in C(u_4)$, or $H$ contains a $(j, i)_{(u_4, u)}$-path.
When $1, 3\not\in S_u\setminus C(u_2)$,
it follows that $2\in C(u_4)$, $H$ contains a $(2, \{1, 3\})_{(u_2, u_4)}$-path,
then by <ref>, we are done.
In the other case, $1/3\in S_u\setminus C(u_2)$.
One sees that $d(u_1) + d(u_3) + d(u_4)\ge 2t_{u v} + |\{1, 4\}| + |[3, 5]| + |\{4, 6\}| + |\{2, 2, 6, 1/3\}| = 2\Delta + 7$.
It follows that $d(u_i) = \Delta = 7$, $i = 1, 3, 4$, $d(u_2)\le 6$, $|\{1, 3\}\cap S_u\setminus C(u_2)|\le 1$,
and then $5, 6, 7\in C(u_2)$.
Further, $2\in C(u_4)$, $6\in C(u_3)$, and $H$ contains a $(2, \{1, 3\}\setminus (S_u\setminus C(u_2)))_{(u_2, u_4)}$-path.

* $1\in C(u_3)\cup C(u_4)$.
It follows that $W_3\cup W_4 = [3, 6]\cup \{2, 4, 6\}\cup \{1\}$, $W_1 = \{1, 2, 4\}$, and $H$ contains a $(2, 3)_{(u_2, u_4)}$-path.
Further, it follows from Corollary <ref>(5.1) that $H$ contains a $(2, 7)_{(u_2, u_4)}$-path.
Since $u u_1\to 7$ and $(u u_1, u u_3)\to (3, 7)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$, by <ref>, we are done.

* $1\not\in C(u_3)\cup C(u_4)$.
Then $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
It follows from <ref> that $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path.
If $2\in C(u_3)$, since $W_1 = \{1, 3, 4\}$, $W_4 = \{4, 6, 2\}$, and $H$ contains a $(3, 6)_{(u_1, u_3)}$-path, then $(v u_4, u v)\overset{\divideontimes}\to (T_4, 6)$.
Otherwise, $2\not\in C(u_3)$.
It follows from Corollary <ref>(5.1) that $H$ contains a $(2, 7)_{(u_2, u_4)}$-path.
Since $u u_3\to 7$ and $(u u_1, u u_3)\to (7, 1)$ reduce to an acyclic edge $(\Delta + 5)$-coloring for $H$, by <ref>, we are done.

Hence, $T_1\setminus C(u_4)\ne \emptyset$, $T_1\subseteq C(u_2)$, and $T_3\setminus C(u_2)\ne \emptyset$.

It follows from Corollary <ref>(5.1) that $1, 3\in S_u$,
from Corollary <ref>(5.2) and (3) that $i\in B_1\cup B_3$, mult$_{S_u}(i)\ge 2$, $i\in \{2, 4\}$,
and $4\in C(u_3)$, $H$ contains a $(4, T_3)_{(u_3, u_4)}$-path,
$2\in C(u_1)$, $H$ contains a $(2, T_1)_{(u_1, u_2)}$-path.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + \sum_{i\in [1, 4]}$ mult$_{S_u}(i) + |\{6, 6, 5, 7\}| + 2t_{u v} + t_0 = 6 + 2 + 4 + 4 + 2\Delta - 4 + t_0 = 2\Delta + 12 + t_0$.
One sees that $2\le $ mult$_{S_u}(1) + $ mult$_{S_u}(3) \le 3$.
One sees that $1\in C(u_2)$ or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path,
and that $3\in C(u_4)$ or $H$ contains a $(3, 1/2)_{(u_4, u)}$-path.
We distinguish the following two subcases on whether mult$_{S_u}(1) + $ mult$_{S_u}(3) = 2$ or not.

Case 1. mult$_{S_u}(1) + $ mult$_{S_u}(3) = 2$.

Case 1.1.  $1\in C(u_3)$.
Then $3\in C(u_2)$ and $2\in C(u_4)$.
If $2\not\in C(u_3)$, since $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 1, 2, 3)$ reduces to acyclic edge $(\Delta + 5)$-coloring for $H$,
then by <ref>, we are done.
Otherwise, $2\in C(u_3)$, mult$_{S_u}(5) = 1$ and by Lemma <ref>(4), $6\in C(u_3)$.
Hence, $W_3 = [1, 6]$, $W_4 = \{2, 4, 6\}$, and by <ref>(1), $5\in C(u_2)$ and $H$ contains a $(2, 5)_{(u_2, u_4)}$-path.
Then $(u u_1, u v)\overset{\divideontimes}\to (5, T_3)$[Or, contradiction to <ref>. ].

Case 1.2.  $1\in C(u_3)$. Then $4\in C(u_2)$ and $H$ contains a $(4, 1)_{(u_2, u_4)}$-path.
It follows from <ref> that $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path.
Hence, $3\in C(u_1)\cup C(u_2)$.
If $3\in C(u_4)$, it follows that $4\in C(u_1)$ and $H$ contains a $(4, 3)_{(u_1, u_4)}$-path,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (T_1\cap T_4, 3, 1, T_3\cap T_2)$.

Otherwise, $3\in C(u_1)$.
If $2\not\in C(u_4)$, then $(u u_1, u u_2, u u_4, u v)\to (T_1\cap T_4, 1, 2, T_3)$.
If $2\not\in C(u_3)$, since $(u u_1, u u_2, u u_3, u u_4)\to (T_1\cap T_4, 1, 2, 3)$ reduces to acyclic edge $(\Delta + 5)$-coloring for $H$,
then by <ref>, we are done.
Otherwise, $2\in C(u_3)\cap C(u_4)$ and $4\not\in C(u_1)$.
Since $(u u_1, u u_2, u u_3, u u_4)\to (4, 1, T_3, 3)$ reduces to acyclic edge $(\Delta + 5)$-coloring for $H$,
then by <ref>, we are done.

Case 1.3.  $1\in C(u_2)$.
One sees that $3\in C(u_4)$ or $H$ contains a $(2, 3)_{(u_2, u_4)}$-path.
When $3\in C(u_2)$, it follows that $2\in C(u_4)$, $H$ contains a $(2, 3)_{(u_2, u_4)}$-path,
and from <ref> that $2\in C(u_3)$ and $H$ contains a $(1, 2)_{(u_2, u_3)}$-path.
Then $(u u_1, u u_4, u u_3, u v)\overset{\divideontimes}\to (3, 1, T_3\cap T_2, T_1\cap T_4)$
[In fact, $T_3\cap C(u_2) = \emptyset$, $T_1\cap C(u_4) = \emptyset$,
i.e., $T_3\cap T_2 = T_3$, $T_1\cap T_4 = T_1$. ].
In the other case, $3\in C(u_4)$.
If $H$ contains no $(4, 3)_{(u_2, u_4)}$-path,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (T_1\cap T_4, 3, 1, T_3\cap T_2)$.
Otherwise, $H$ contains a $(4, 3)_{(u_2, u_4)}$-path.
It follows from <ref> that $2\in C(u_3)$ and $H$ contains a $(1, 2)_{(u_2, u_3)}$-path.
If $2\not\in C(u_4)$,
then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1\cap T_4, 3, 1, 2, T_3\cap T_2)$.
Otherwise, $2\in C(u_4)$, mult$_{S_u}(5) = 1$ and by Lemma <ref>(4), $6\in C(u_3)$.
Hence, $W_3 = [2, 6]$, $W_4 = \{2, 3, 4, 6\}$, and by <ref>(1), $5\in C(u_2)$ and $H$ contains a $(2, 5)_{(u_2, u_4)}$-path.
Then $(u u_1, u v)\overset{\divideontimes}\to (5, T_3)$[Or, contradiction to <ref>. ].

Case 2. mult$_{S_u}(1) + $ mult$_{S_u}(3) = 3$.
Then mult$_{S_u}(i) = 2$, $i = 2, 4, 6$, mult$_{S_u}(j) = 1$, $j = 5, 7$,
and $t_0 = 0$, i.e., $T_3\cap C(u_2) = \emptyset$, $T_1\cap C(u_4) = \emptyset$.
If $7\not\in C(u_4)\cup C(u_3)$ and $H$ contains no $(7, 1/2)_{(u_4, u)}$-path,
then $(u u_4, u u_3, u v)\overset{\divideontimes}\to (7, T_3, T_1)$.
Otherwise, $7\in C(u_3)\cup C(u_4)$, or $H$ contains a $(7, 1/2)_{(u_4, u)}$-path.
We consider the following three scenarios due to $5\in C(u_i)$, $i = 1, 2, 4$, respectively.

Case 2.1. $5\in C(u_2)$.
It follows from <ref>(1) that $2\in C(u_4)$, $H$ contains a $(2, 5)_{(u_2, u_4)}$-path,
from <ref> that $3\in C(u_1)$,
and from <ref> that $1\in C(u_4)$ or $H$ contains a $(1, 4)_{(u_3, u_4)}$-path.
If $3\not\in C(u_2)\cup C(u_4)$, then $(u u_1, u u_3, u u_4)\to (5, T_3, 3)$ reduces the proof to (2.4.2.).
Otherwise, $3\in C(u_2)\cup C(u_4)$ and mult$_{S_u}(1) = 1$.
Recall that $1\in C(u_2)$ or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path.
It follows that $1\in C(u_3)$ and $3\in C(u_2)$.
One sees that $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 1, 2, 3)$ reduces to acyclic edge $(\Delta + 5)$-coloring for $H$,
then by <ref>, we are done.

Case 2.2. $5\in C(u_4)$.
It follows from <ref> and <ref>(1) that $H$ contains a $(5, \alpha)_{(u_1, u_\alpha)}$-path and a $(5, \beta)_{(u_2, u_\beta)}$-path, where $\{\alpha, \beta\} = \{3, 4\}$, and $\alpha\in C(u_1)$, $\beta\in C(u_2)$.

(2.2.1.) $\alpha = 4$, $\beta = 3$.
It follows from <ref>(2) that $1\in C(u_4)$ or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path.
If $1\not\in C(u_3)$ and $H$ contains no $(1, 4/5)_{(u_3, u)}$-path,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (T_1, 5, 1, T_3)$.
Otherwise, $1\in C(u_3)$ or $H$ contains a $(1, 4/5)_{(u_3, u)}$-path.

When $1\not\in C(u_3)\cup C(u_4)$,
it follows that $2\in C(u_4)$, $H$ contains a $(1, 2)_{(u_2, u_4)}$-path and a $(1, 5)_{(u_2, v_1)}$-path,
then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 5, 2, 1, T_3)$.
In the other case, $1\in C(u_3)\cup C(u_4)$.

One sees clearly that $3\not\in C(u_3)\cup C(u_4)$.
If $2\not\in C(u_3)$, then $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 5, 2, 3)$ reduces the proof to (2.4.2.).
Otherwise, $2\in C(u_3)$ and then $1\in C(u_4)$, $3\in C(u_1)$.
Thus, $1\not\in C(u_2)\cup C(u_3)$ and $H$ contains a $(1, 4)_{(u_2, u_4)}$-path and a $(1, 4)_{(u_3, u_4)}$-path, a contradiction.

(2.2.2.) $\alpha = 3$, $\beta = 4$.
It follows from <ref> that $1\in C(u_3)$ or $H$ contains a $(1, 2/4)_{(u_3, u)}$-path.
If $1\not\in C(u_4)$, and $H$ contains no $(1, 3/5)_{(u_4, u)}$-path,
then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (T_1, 5, 1, T_3)$.
Otherwise, $1\in C(u_4)$ or $H$ contains a $(1, 3/5)_{(u_4, u)}$-path.

When $1\not\in C(u_3)\cup C(u_4)$,
it follows that $2\in C(u_3)$, $H$ contains a $(1, 2)_{(u_2, u_3)}$-path and a $(1, 5)_{(u_2, u_4)}$-path,
then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 5, 1, 2, T_3)$.
In the other case, $1\in C(u_3)\cup C(u_4)$.
One sees clearly that $3\not\in C(u_3)\cup C(u_4)$.
If $H$ contains no $(2, 3)_{(u_2, u_4)}$-path,
then $(u u_1, u u_3, u u_4)\to (5, T_3, 3)$ reduces the reduces the proof to (2.4.2.).
Otherwise, $2\in C(u_4)$, $3\in C(u_2)$, and  mult$_{S_u}(1) = 1$.
Hence, $1\in C(u_4)$.
Thus, $H$ contains a $(1, 4)_{(u_2, u_4)}$-path and a $(1, 4)_{(u_3, u_4)}$-path, a contradiction.

Case 2.3. $5\in C(u_1)$. One sees that $1\in C(u_4)$ and $H$ contains a $(1, 5)_{(u_1, u_4)}$-path.
When $3\not\in C(u_1)\cup C(u_4)$,
one sees that $2\in C(u_4)$ and $H$ contains a $(2, 3)_{(u_2, u_4)}$-path,
then $(u u_1, u u_3, u u_4)\to (3, T_3, 5)$ reduces the proof to (2.4.1.).
In the other case, $3\in C(u_1)\cup C(u_4)$.

(2.3.1.) $3\in C(u_4)$.
Then $1\not\in C(u_3)$.
If $1\not\in C(u_2)$, one sees that $4\in C(u_2)$,
then $u u_3\to \alpha$,
and $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (4, 1, 5, T_3\setminus \{\alpha\})$.
Otherwise, $1\in C(u_2)$ and then $3\not\in C(u_1)\cup C(u_2)$.
If $2\not\in C(u_4)$, then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 3, 1, 2, T_3)$.
Otherwise, $2\in C(u_4)$ and $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 3, 2, 5, T_3)$.

(2.3.2.) $3\not\in C(u_4)$ and $3\in C(u_1)$.
When $1\not\in C(u_3)\cup C(u_4)$,
one sees that $4\in C(u_2)$, then $u u_3\to \alpha$,
and $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (4, 1, 5, T_3\setminus \{\alpha\})$.
In the other case, $1\in C(u_3)\cup C(u_4)$ and  mult$_{S_u}(3) = 1$.
If $1\in C(u_3)$, one sees that $2\in C(u_4)$, $2\in B_1$ and $R_1\ne \emptyset$,
then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\\to (4, 1, 2, 3, T_1\cap R_1)$.
Otherwise, $1\not\in C(u_3)$ and $1\in C(u_2)$.

If $2\in C(u_3)$, one sees that $2\not\in C(u_4)$, $4, 7\not\in C(u_1)$, and then $4\in B_3$,
it follows that $7\in C(u_4)$,
then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (4, 7, 2, T_3)$.
Otherwise, $2\in C(u_4)\setminus C(u_3)$ and then $2\in B_1$, $(T_1\cup C_{13})\cap R_1\ne \emptyset$.
If $H$ contains no $(7, j)_{(v_1, v_7)}$-path for some $j\in (T_1\cup C_{13})\cap R_1$,
then $(v v_1, v u_3, u v)\overset{\divideontimes}\to (j, 1, T_3)$.
Otherwise, it follows from Lemma <ref>(1.2) that $7\in C(v_1)$,
and $H$ contains a $(7, (T_1\cup C_{13}))_{(v_1, v_7)}$-path.
One sees that $H$ contains no $(j, i)_{(u_4, v)}$-path for each $j\in (T_1\cup C_{13})\cap R_1$.
It follows from Corollary <ref>(4) that $6\in B_1\cup B_3$.
One sees that if $C(v_3) = \{3, 6\}\cup T_{u v}$, then $(v v_3, u v)\overset{\divideontimes}\to (4, T_1)$,
and that if $4\in C(v_3)$, then from Corollary <ref>(6), $(T_3\cup C_{13})\cap R_3\ne \emptyset$.
Note that $\{4, 6\}\setminus C(u_1)\ne \emptyset$ and $\{4, 6\}\cap B_3\ne \emptyset$.
Hence, $(T_3\cup C_{13})\cap R_3\ne \emptyset$ and it follows from Lemma <ref>(1.2) that $7\in C(v_3)$, and $H$ contains a $(7, (T_3\cup C_{13})\cap R_3)_{(v_3, v_7)}$-path.
One sees that if $2\in C(v_3)$, since $2, 3, 4/6, 7\in C(v_3)$ and $c_{13}\le 1$,
then $R_3\ge 2$, and $R_3\cap T_3\ne \emptyset$.

* $7\in C(u_1)$. Then $(u u_3, u v)\overset{\divideontimes}\to (7, T_1\cap R_1)$.

* $7\in C(u_4)$.
Then $(u u_1, u u_2, u u_3, u u_4)\overset{\divideontimes}\to (T_1, 7, 1, 3, (2\cup T_3)\setminus C(v_3))$.

* $7\in C(u_3)$. One sees that $C_{13} = \emptyset$ and $R_1\subseteq T_1$.
Then $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (7, 1, 3, T_1\cap R_1)$.

* $7\in C(u_2)$.
Then $H$ contains a $(2, 7)_{(u_2, u_4)}$-path.
One sees that $u u_1\to 7$ or $u u_3\to 7$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$.
Hence, $T_1\cup T_3\subseteq C(v_7)$ and then $T_1\subseteq C(v_1)$, $T_3\subseteq C(v_3)$.
Thus, $H$ contains a $(7, C_{13})_{(v_1, v_7)}$-path and a $(7, C_{13})_{(v_3, v_7)}$-path, a contradiction.

This finishes the inductive step for the case (2.4.3.).

Case 2.5. $\{i_1, i_2\} = \{1, 2\}$ and assume w.l.o.g. $c(v u_3) = 5$.
Assume that $5\not\in C(u_4)$.
Let $u u_4\to 5$.
If $5\not\in C(u_1)$, $H$ contains no $(5, 2)_{(u_2, u_4)}$-path,
and $H$ contains no $(3, j)_{(u_1, u_3)}$-path for some $j\in T_1$,
then $u u_1\to j$.
If $H$ contains no $(5, 1)_{(u_4, u_1)}$-path, $4\not\in C(u_2)$,
and $H$ contains no $(4, i)_{(u_2, u_i)}$-path for each $i\in \{1, 3\}$,
then $u u_2\to 4$.
If $H$ contains no $(5, 2)_{(u_4, u_2)}$-path, $4\not\in C(u_1)$,
and $H$ contains no $(4, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $u u_1\to 4$.
Assume that $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$.
If $4\not\in B_1\cup B_2$, then $u v\overset{\divideontimes}\to 4$.
If for some $i\in [1, 2]$, $4\not\in C(u_i)$ and $H$ contains no $(3, 4)_{(u_i, u_3)}$-path,
then $uu_i\to 4$.
These reduce the proof to Case (2.2.1.) or Case (2.4.1.).
Hence, we proceed with the following proposition, or otherwise we are done:

Assume that $5\not\in C(u_4)$. Then

if $5\not\in C(u_1)$, $H$ contains no $(5, 2)_{(u_2, u_4)}$-path,
then $3\in C(u_1)$, $H$ contains a $(3, T_1)_{(u_1, u_3)}$-path;

if $5\not\in C(u_2)$, $H$ contains no $(5, 1)_{(u_1, u_4)}$-path,
then $3\in C(u_2)$, $H$ contains a $(3, T_2)_{(u_2, u_3)}$-path;

if $H$ contains no $(5, 1)_{(u_4, u_1)}$-path,
then $4\in C(u_2)$ or $H$ contains a $(4, 1/3)_{(u_2, u)}$-path;

if $H$ contains no $(5, 2)_{(u_4, u_2)}$-path,
then $4\in C(u_1)$ or $H$ contains a $(4, 2/3)_{(u_1, u)}$-path;

if $H$ contains no $(5, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then $4\in B_1\cup B_2$, mult$_{S_u}(4)\ge 2$, and for each $i\in[1, 2]$, $4\in C(u_i)$,
or $H$ contains a $(3, 4)_{(u_i, u_3)}$-path, $3\in C(u_i)$, $4\in C(u_3)$.

For each $j\in [6, 7]\setminus C(u_4)$, let $uu_4\to j$.
If $j\not\in C(u_1)$, $H$ contains no $(j, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
and $H$ contains no $(3, l)_{(u_1, u_3)}$-path for some $l\in T_1$,
then $u u_1\to l$.
If $H$ contains no $(j, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
$4\in C(u_1)$ and $H$ contains no $(4, l)_{(u_1, u_l)}$-path for each $l\in [2, 3]$,
then $u u_1\to 4$.
These reduce the proof to Case (2.2.2.) or Case (2.4.3.).
Hence, assume the following proposition, or otherwise we are done:

For each $j\in [6, 7]\setminus C(u_4)$, then

if $j\not\in C(u_1)$ and $H$ contains no $(j, i)_{(u_i, u_4)}$-path for each $i\in [2, 3]$,
then $3\in C(u_1)$ and $H$ contains a $(3, T_1)_{(u_1, u_3)}$-path;

if $j\not\in C(u_2)$ and $H$ contains no $(j, i)_{(u_i, u_4)}$-path for each $i\in \{1, 3\}$,
then $3\in C(u_2)$ and $H$ contains a $(3, T_2)_{(u_2, u_3)}$-path;

For each $j\in [6, 7]\setminus C(u_3)$, let $u u_3\to j$.
If $j\not\in C(u_1)$, $H$ contains no $(j, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
and $H$ contains no $(4, l)_{(u_1, u_4)}$-path for some $l\in T_1$,
then $u u_1\to l$ reduces the proof to Case (2.4.).
Hence, we proceed with the following proposition, or otherwise we are done:

For each $j\in [6, 7]\setminus C(u_3)$, then

if $j\not\in C(u_1)$ and $H$ contains no $(j, i)_{(u_i, u_3)}$-path for each $i\in \{2, 4\}$,
then $4\in C(u_1)$ and $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path;

if $j\not\in C(u_2)$ and $H$ contains no $(j, i)_{(u_i, u_3)}$-path for each $i\in \{1, 4\}$,
then $4\in C(u_2)$ and $H$ contains a $(4, T_2)_{(u_2, u_4)}$-path;

Assume that $j\in T_4\cup [6, 7]\cup C(u_4)$ and $H$ contains no $(j, 3)_{(u_4, u_3)}$-path.
If $H$ contains no $(j, 2)_{(u_4, u_2)}$-path,
$4\in C(u_1)$ and $H$ contains no $(4, l)_{(u_1, u_l)}$-path for each $l\in [2, 3]$,
then $u u_1\to 4$ reduces the proof to Case (2.2.2.) or Case (2.4.3.),
or reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
If $H$ contains no $(j, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then if $4\not\in B_1\cup B_2$, $uv\overset{\divideontimes}\to 4$;
and if for some $i\in [1, 2]$, $4\not\in C(u_i)$ and $H$ contains no $(3, 4)_{(u_i, u_3)}$-path,
then $uu_i\to 4$ and we are done.

Hence, assume the following proposition, or otherwise we are done:

Assume $j\in (T_4\cup [6, 7])\setminus C(u_4)$, $H$ contains no $(j, 3)_{(u_4, u_3)}$-path. Then

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

if $H$ contains no $(j, 1)_{(u_4, u_1)}$-path,
then $4\in C(u_2)$ or $H$ contains a $(4, 1/3)_{(u_2, u)}$-path.

if $H$ contains no $(j, i)_{(u_4, u_i)}$-path for each $i\in [1, 2]$,
then $4\in B_1\cup B_2$, mult$_{S_u}(4)\ge 2$,
and for each $i\in[1, 2]$, $4\in C(u_i)$,
or $H$ contains a $(3, 4)_{(u_i, u_3)}$-path, $3\in C(u_i)$, $4\in C(u_3)$.

Assume $j\in T_3\cup [6, 7]\cup C(u_3)$ and $H$ contains no $(j, 4)_{(u_4, u_3)}$-path.
If $H$ contains no $(j, 2)_{(u_3, u_2)}$-path,
$3\in C(u_1)$ and $H$ contains no $(3, l)_{(u_1, u_l)}$-path for each $l\in \{2, 4\}$,
then $u u_1\to 3$ reduces the proof to Case (2.4.2.) or Case (2.4.3.),
or reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
If $H$ contains no $(j, i)_{(u_3, u_i)}$-path for each $i\in [1, 2]$,
then if $3\not\in B_1\cup B_2$, $u v\overset{\divideontimes}\to 3$;
and if for some $i\in [1, 2]$, $3\not\in C(u_i)$ and $H$ contains no $(3, 4)_{(u_i, u_4)}$-path,
then $u u_i\to 3$ and we are done.
Hence, we proceed with the following proposition, or otherwise we are done:

Assume $j\in (T_3\cup [6, 7])\setminus C(u_3)$, $H$ contains no $(j, 4)_{(u_4, u_3)}$-path. Then

if $H$ contains no $(j, 2)_{(u_3, u_2)}$-path,
then $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path;

if $H$ contains no $(j, 1)_{(u_3, u_1)}$-path,
then $3\in C(u_2)$ or $H$ contains a $(3, 1/4)_{(u_2, u)}$-path.

if $H$ contains no $(j, i)_{(u_3, u_i)}$-path for each $i\in [1, 2]$,
then $3\in B_1\cup B_2$, mult$_{S_u}(3)\ge 2$,
and for each $i\in[1, 2]$, $3\in C(u_i)$,
or $H$ contains a $(3, 4)_{(u_i, u_4)}$-path, $4\in C(u_i)$, $3\in C(u_4)$.

By <ref>(1–2),
we proceed with the following proposition, or otherwise we are done:

If $4\not\in C(u_2)$ and $H$ contains no $(4, l)_{(u_2, u_l)}$-path for each $l\in \{1, 3\}$,
then $6, 7\in C(u_3)\cup C(u_4)$,
or $H$ contains a $(1, [6, 7]\setminus (C(u_3)\cup C(u_4)))_{(u_1, u_4)}$-path, $1\in C(u_4)$;

If $4\not\in C(u_1)$ and $H$ contains no $(4, l)_{(u_1, u_l)}$-path for each $l\in \{2, 3\}$,
then $6, 7\in C(u_3)\cup C(u_4)$,
or $H$ contains a $(2, [6, 7]\setminus (C(u_3)\cup C(u_4)))_{(u_2, u_4)}$-path, $2\in C(u_4)$.

By <ref>(1–2),
we proceed with the following proposition, or otherwise we are done:

If $3\not\in C(u_2)$ and $H$ contains no $(3, l)_{(u_2, u_l)}$-path for each $l\in \{1, 4\}$,
then $6, 7\in C(u_3)\cup C(u_4)$,
or $H$ contains a $(1, [6, 7]\setminus (C(u_3)\cup C(u_4)))_{(u_1, u_3)}$-path, $1\in C(u_3)$;

If $3\not\in C(u_1)$ and $H$ contains no $(3, l)_{(u_1, u_l)}$-path for each $l\in \{2, 4\}$,
then $6, 7\in C(u_3)\cup C(u_4)$,
or $H$ contains a $(2, [6, 7]\setminus (C(u_3)\cup C(u_4)))_{(u_2, u_3)}$-path, $2\in C(u_3)$.

If $2\not\in C(u_3)$, $H$ contains no $(1, 2)_{(u_1, u_3)}$-path,
and $3\not\in C(u_2)$, $H$ contains no $(3, i)_{(u_2, u_i)}$-path for each $i\in \{1, 4\}$,
then $(u u_3, u u_2)\to (2, 3)$ reduces the proof to Case (2.4.).
If $1\not\in C(u_3)$, $H$ contains no $(1, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$,
and $3\not\in C(u_1)$, $H$ contains no $(3, i)_{(u_1, u_i)}$-path for each $i\in \{2, 4\}$,
then $(u u_3, u u_1)\to (1, 3)$ reduces the proof to Case (2.4.).
Hence, we proceed with the following proposition, or otherwise we are done:

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

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

If $1\not\in C(u_4)$, $H$ contains no $(1, i)_{(u_4, u_i)}$-path for each $i\in [2, 3]$,
and $4\not\in C(u_1)$, $H$ contains no $(4, i)_{(u_1, u_i)}$-path for each $i\in [2, 3]$,
then $(u u_4, u u_1)\to (1, 4)$.
If $2\not\in C(u_4)$, $H$ contains no $(2, i)_{(u_4, u_i)}$-path for each $i\in \{1, 3\}$,
and $4\not\in C(u_2)$, $H$ contains no $(4, i)_{(u_1, u_i)}$-path for each $i\in \{1, 3\}$,
then $(u u_4, u u_2)\to (2, 4)$.
These reduce the proof to Case (2.2.2.) or Case (2.4.3.).
Hence, we proceed with the following proposition, or otherwise we are done:

If $1\not\in C(u_4)$, $H$ contains no $(1, i)_{(u_4, u_i)}$-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 $2\not\in C(u_4)$, $H$ contains no $(2, i)_{(u_4, u_i)}$-path for each $i\in \{1, 3\}$,
then $4\in C(u_2)$, or $H$ contains a $(4, i)_{(u_2, u_i)}$-path for some $i\in \{1, 3\}$.

If $3\not\in C(v_1)$ and $H$ contains no $(3, i)_{(v_1, v_i)}$-path for each $i\in C(v)$,
then $v v_1\to 3$ reduces the proof to Case (2.4.).
Hence, we proceed with the following proposition:

$3\in C(v_1)$, or $H$ contains a $(3, 2/5/6/7)_{(v_1, v)}$-path.

$3\in C(v_2)$, or $H$ contains a $(3, 1/5/6/7)_{(v_2, v)}$-path.

(2.5.1) $c(vu_4) = 2$.
Let $C(u_1) = W_1\cup T_2\cup C_{12}$, $C(u_2) = W_2\cup T_1\cup C_{12}$,
where $T_{u v} = T_1\cup T_2\cup C_{12}$.
Then $T_1\subseteq B_2\subseteq C(u_2)\cap C(u_4)$, $T_2\subseteq B_1\subseteq C(u_1)$.

By <ref>(2), we proceed with the following proposition:

$3\in C(u_4)$, or $H$ contains a $(3, i)_{(u_4, v_i)}$-path for some $i\in \{1, 6, 7\}$.

$3/1/6/7\in C(u_4)$, $T_4\ne \emptyset$, and there exists a $b_2\in [5, 7]\cap C(u_4)$.
such that $H$ contains a $(b_2, j)_{(u_4, v)}$-path for some $j\in T_4$.

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

Assume that $5\not\in C(u_1)$ and $H$ contains no $(5, 4)_{(u_1, u_4)}$-path.

$H$ contains a $(5, 2)_{(u_1, u_2)}$-path.

If $1\not\in C(u_2)$, $H$ contains no $(1, i)_{(u_2, u_i)}$-path for each $i\in \{3, 4\}$,
then $H$ contains a $(5, B_1)_{(u_1, u_3)}$-path, a $(5, T_4)_{(u_1, u_3)}$-path, and $T_4\subseteq C(u_3)$

Assume that $2\not\in C(u_1)$ and $H$ contains no $(2, 3)_{(u_1, u_3)}$-path.
Let $u u_1\to 2$.
If $5\not\in C(u_2)$ and $H$ contains no $(4, 5)_{(u_2, u_4)}$-path,
then $u u_2\to 5$ reduces the proof to the Case 2.3.

Together with Corollary <ref>(2.1),
we proceed with the following proposition, or otherwise we are done:

Assume that $2\not\in C(u_1)$ and $H$ contains no $(2, 3)_{(u_1, u_3)}$-path.

$5\in C(u_2)$ or $H$ contains a $(4, 5)_{(u_2, u_4)}$-path.

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

Assume that $5\not\in C(u_2)$ and $H$ contains no $(5, i)_{(u_2, u_i)}$-path for each $i\in \{1, 4\}$.
One sees that the same argument applies if $uu_2\to 5$.
Hence, we proceed with the following proposition:

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 $(5, T_{uv}\setminus B_1)_{(u_2, u_3)}$-path, a $(5, T_1)_{(u_2, u_3)}$-path,
and $T_1\subseteq T_{u v}\setminus B_1\subseteq C(u_3)$;

$T_3\ne \emptyset$ and there exists a $b_5\in \{2, 6, 7\}\cap C(u_3)$ such that $H$ contains a $(b_5, j)_{(u_3, v)}$-path for some $j\in T_3$.

If $H$ contains no $(4, i)_{(u_4, v)}$-path for each $i\in \{1\}\cup [5, 7]$,
and $H$ contains no $(j, 3)_{(u_4, u_3)}$-path for some $j\in T_4$,
then $(v u_4, u u_4, u v)\overset{\divideontimes}\to (4, j, T_1)$.
Hence, we proceed with the following proposition:

If $H$ contains no $(4, i)_{(u_4, v)}$-path for each $i\in \{1\}\cup [5, 7]$,
then $H$ contains a $(3, T_4)_{(u_4, u_3)}$-path.

If $3\not\in C(u_4)$, or $T_2\setminus C(u_4)\ne \emptyset$, or $T_4\setminus C(u_3)\ne \emptyset$,
then $H$ contains a $(4, i)_{(u_4, v)}$-path for some $i\in \{1\}\cup [5, 7]$.

One sees that $t_1 = \Delta - 2 - (d(u_1) - w_1) = \Delta - d(u_1) + w_1 - 2\ge w_1 - 2$,
and $t_2 = \Delta - 2 - (d(u_2) - w_2) = \Delta - d(u_2) + w_2 - 2 \ge w_2 - 2$.

$w_2\ge 3$ and $t_2 = \Delta - d(u_2) + w_2 - 2 \ge w_2 - 2 \ge 1$.

Assume that $w_2\le 2$.
One sees from <ref>(1) that $5/2/4\in C(u_1)$,
and if $5\not\in C(u_1)$ and $H$ contains no $(5, 2)_{(u_1, u_2)}$-path,
then $H$ contains a $(5, 4)_{(u_1, u_4)}$-path.

Condition (C1) states that $|W_3\cap [1, 5]| + |W_4\cap [1, 5]| + |W_1\cap [1, 5]|\ge 10$,
$6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$,
and $T_1\subseteq C(u_3)\cap C(u_4)$, $T_{uv}\subseteq C(u_3)\cup C(u_4)$.
Assume that (C1) holds.
Let $\kappa_{u_1} = |[6, 7]\cap C(u_1)|$.
One sees that $t_1\ge w_1 - 2\ge \kappa_{u_1} + |W_1\cap [1, 5]| - 2$.
Thus, $d(u_3) + d(u_4)\ge |W_3\cap [1, 5]| + |W_4\cap [1, 5]| + 2 - \kappa_{u_1} + t_{uv} + t_1\ge |W_3\cap [1, 5]| + |W_4\cap [1, 5]| + 2 - \kappa_{u_1} + t_{uv} + \kappa_{u_1} + |W_1\cap [1, 5]| - 2 \ge \Delta + 10 - 2 = \Delta + 8$,
and none of ($A_{8.1}$)–($A_{8.4}$) holds.
Hence, we proceed with the following proposition:

If (C1) holds, then we are done.

When $1, 3, 4, 5\not\in C(u_2)$,

it follows from <ref>(1) that $5/4\in C(u_1)$,
from <ref> that $2/3\in C(u_1)$,
and from <ref>(1) that $T_1\subseteq C(u_3)$.
One sees that $T_4\subseteq B_1$.
It follows from <ref>(3) that $3\in C(u_4)$,

$T_4\subseteq C(u_3)$, i.e., $T_{uv}\subseteq C(u_3)\cup C(u_4)$,
from <ref>(3) that $5/1\in C(u_4)$,
from <ref>(2) that $2/1\in C(u_3)$,
and from <ref>(1) that $6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$.
One sees that (C1) holds.
Thus, by <ref>, we are done.

In the other case, $5/1/3/4\in C(u_2)$.
We consider below the four cases, respectively.

Case 1. $W_2 = \{2, 5\}$.
It follows from <ref>(2) that $2/3\in C(u_1)$,
from <ref>(1) that $5/4/2\in C(u_1)$,
from <ref> that $T_1\subseteq C(u_3)$,
and from <ref>(3) that $3\in C(u_4)$,

$T_4\subseteq C(u_3)$, i.e., $T_{uv}\subseteq C(u_3)\cup C(u_4)$.
It follows from <ref>(3) that $5/1\in C(u_4)$,
from <ref>(2) that $2/1\in C(u_3)$,
and from <ref>(1) that $6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$.
It follows from <ref> that mult$_{C(u_3)}(1) + $mult$_{C(u_3)}(2)= 1$, $4\not\in C(u_3)$,
$[2, 5]\cap C(u_1) = \{2\}$,
and then from <ref>(1) that $1\in C(u_3)$.

Then $2\not\in C(u_3)$, and $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (3, 1, 2, T_4)$.

Case 2. $W_2 = \{2, 1\}$.
It follows from <ref>(1) that $5/4\in C(u_1)$,
from <ref>(1) that $2/3\in C(u_1)$,
from <ref> that $T_1\subseteq C(u_3)$,
and from <ref>(3) that $3\in C(u_4)$,

$T_4\subseteq C(u_3)$, i.e., $T_{uv}\subseteq C(u_3)\cup C(u_4)$.
It follows from <ref>(5) that $5/1\in C(u_4)$,
from <ref>(3) that $6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$.
If $1/2\in C(u_3)$, then (C1) holds and by <ref>, we are done.
Otherwise, $1, 2\not\in C(u_3)$.
It follows from <ref>(2) that $H$ contains a $(1, 3)_{(u_1, u_2)}$-path.

If there exists a $j\in [5, 7]\setminus C(u_1)$,
then $(u u_2, u u_3, u v)\overset{\divideontimes}\to (j, 2, 3)$.
Otherwise, $[5, 7]\subseteq C(u_1)$ and $t_1\ge 3$.
Since $d(u_3) + d(u_4)\ge  |\{3, 5\}| + |\{2, 4, 3, 5/1\}| + t_{uv} + t_1 = \Delta + 4 + t_1$,
we have $t_1 = 3$, $W_3 = \{3, 5\}$ and $W_1 = \{1, 3\}\cup [5, 7]$.
Then by <ref>(1), we are done.

Case 3. $W_2 = \{2, 3\}$.
It follows from <ref>(1) that $5/4\in C(u_1)$,
from <ref>(1) that $2/3\in C(u_1)$,
from <ref>(2) that $5/6/7\in C(u_4)$,
and from <ref>(1) that $T_1\subseteq T_{uv}\setminus B_1\subseteq C(u_3)$,
from <ref>(2) that $2/6/7\in C(u_3)$.
We distinguish whether or not $H$ contains a $(3, 4)_{(u_2, u_3)}$-path.

Case 3.1. $H$ contains no $(3, 4)_{(u_2, u_3)}$-path.
It follows from <ref>(3) that $3\in C(u_4)$,
$T_4\subseteq C(u_3)$, i.e., $T_{u v}\subseteq C(u_3)\cup C(u_4)$,
from <ref>(5) that $5/1\in C(u_4)$,
from <ref>(2) that $6, 7\in C(u_1)\cup C(u_3)\cup C(u_4)$,
and if $1\not\in C(u_4)$, then $6, 7\in C(u_3)\cup C(u_4)$.
One sees that $|\{3, 5\}| + |[2, 4]| + t_{uv} + t_1 = \Delta + 3 + t_1$.
If $1\not\in C(u_4)$, one sees that $5\in C(u_4)$, $6, 7\in C(u_3)\cup C(u_4)$,
and $d(u_3) + d(u_4)\ge \Delta + 3 + t_1 + |[5, 7]| = \Delta + 6 + t_1$,
then $t_1 = 1$, $1, 2\not\in C(u_3)$, $W_1 = \{1, 2, 4/5\}$,
and by Corollary <ref>(7.2), we are done.
Otherwise, $1\in C(u_4)$,
and $d(u_3) + d(u_4)\ge \Delta + 3 + t_1 + |\{2/6/7\}| + |\{1, 5/6/7\}| = \Delta + 6 + t_1$.
Then $t_1 = 1$, $6, 7\not\in C(u_1)$, $1, 2\not\in C(u_3)$, and $5\not\in C(u_4)$.
Hence, $W_1 = \{1, 2, 5\}$, a contradiction.

Case 3.2. $H$ contains a $(3, 4)_{(u_2, u_3)}$-path and $4\in C(u_3)$.
When there exists a $8\in T_{uv}\setminus C(u_3)\cup C(u_4)$,
then assume w.l.o.g. $H$ contains a $(6, 8)_{(u_3, v_6)}$-path, a $(7, 8)_{(u_4, v_7)}$-path,
and $6\in C(u_3)$, $7\in C(u_4)$.
It follows from <ref>(3) that $4\in B_1$.
If $3\not\in B_1$, then $(u u_3, u u_2, u v)\overset{\divideontimes}\to (8, 5, 3)$.
Otherwise, $3\in B_1$ and then $3, 4\in C(u_1)$.
If $7\not\in C(u_1)\cup C(u_3)$,
then $(u u_2, u u_3, u u_4)\to (4, 7, 8)$ reduces the proof to (2.4.3.).
Otherwise, $7\in C(u_1)\cup C(u_3)$.
Since $[3, 6]\subseteq C(u_3)$, we have $t_3\ge 2$ and assume $9\in T_3$.
One sees from $T_2\subseteq C(u_3)$ and $T_{u v}\setminus B_1\subseteq C(u_3)$ that $9\in C(u_2)\cap B_1$.
Assume that $6\not\in C(u_1)\cup C(u_4)$ and $9\not\in C(u_4)$.
Let $(u u_1, u u_2)\to (6, 5)$.
If $H$ contains no $(6, 9)_{(u_1, v_6)}$-path,
then $(u u_3, u v)\overset{\divideontimes}\to (8, 9)$.
Otherwise, $H$ contains a $(6, 9)_{(u_1, v_6)}$-path and $(u u_3, u v)\overset{\divideontimes}\to (9, 8)$.
Hence, $6\in C(u_1)\cup C(u_4)$ or $9\in C(u_1)\cap C(u_2)\cap C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3, 4\}| + |[2, 3]| +  |\{3, 5, 4, 6\}| + |\{2, 4, 7\}| + |\{2, 5, 7, 6/9\}| + 2t_{u v} + t_0 \ge 2\Delta + 12 + t_1\ge 2\Delta + 12 + 1 = 2\Delta + 13$.
It follows that $t_1 = 1$, $1\not\in S_u$, and $3\not\in C(u_4)$.
Then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, 5, 1, 3)$.

In the other case, $T_{u v}\subseteq C(u_3)\cup C(u_4)$.
One sees that $d(u_3) + d(u_4)\ge |\{3, 5, 4, 2/6/7\}| + |\{2, 4, 5/6/7\}| + t_{u v} + t_1 = \Delta + 5 + t_1$.
It follows that $t_1\le 2$ and $1\not\in C(u_3)\cap C(u_4)$.

* $1\in C(u_3)\cup C(u_4)$.
Then $t_1 = 1$, $3\not\in C(u_4)$,  $\sum_{i\in \{2, 6, 7\}}$ mult$_{C(u_3)}(i) = 1$,
$\sum_{i\in [5, 7]}$ mult$_{C(u_4)}(i) = 1$,
and $W_1 = \{1, 2/3, 4/5\}$.
When $1\in C(u_3)\setminus C(u_4)$,
it follows from Corollary <ref>(7.2) that $3\in C(u_1)$ and then $2\in C(u_3)$.
If $4\not\in C(u_1)$, then $(u u_3, u u_1)\to ([6, 7]\setminus C(u_4), T_1)$ reduces the proof to (2.4.2.).
Otherwise, $W_1 = \{1, 3, 4\}$ and $5\in C(u_4)$.
Then $v u_4\to 3$ reduces the proof to (2.4.1.).
In the other case, $1\in C(u_4)\setminus C(u_3)$.
If $H$ contains no $(3, j)_{(u_1, u_4)}$-path,
then $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (j, 1, 3, T_4)$.
Otherwise, $H$ contains a $(3, T_1)_{(u_1, u_4)}$-path and then contains a $(4, T_1)_{(u_1, u_4)}$-path.
Hence, $W_1 = \{1, 3, 4\}$, $2\in C(u_3)$, $5\in C(u_4)$, $6, 7\not\in C(u_3)\cup C(u_4)$,
and from <ref> that $3\in B_1$.
Thus, $v u_4\to 3$ reduces the proof to (2.4.1.).

* $1\not\in C(u_3)\cup C(u_4)$.
It follows from Corollary <ref>(7.2) that $3, 4\in C(u_1)$.
When $2\in C(u_1)$, then $t_1 = 2$, $5\in C(u_4)$, $W_4 = \{2, 4, 5\}$,
and $v u_4\to 3$ reduces the proof to (2.4.1.).
Hence, $2\not\in C(u_1)$.
If $3\not\in C(u_4)$, then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, 5, 1, 3)$.
Otherwise, $3\in C(u_4)$.
Then $t_1 = 1$, $W_1 = \{1, 3, 4\}$.
If $H$ contains no $(5, j)_{(u_2, u_3)}$-path for some $j\in T_4$,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, 5, 1, j)$.
Otherwise, $H$ contains a $(5, T_4)_{(u_2, u_3)}$-path.
Then $(u u_1, u u_4, u v)\overset{\divideontimes}\to (5, 1, T_4)$.

Case 4. $W_2 = \{2, 4\}$.
If $H$ contains no $(4, 5)_{(u_2, u_4)}$-path, then $u u_2\to 5$ reduces the proof to the above Case 3.
Otherwise, $H$ contains a $(4, 5)_{(u_2, u_4)}$-path, $5\in C(u_4)$,
and follows from <ref>(1) that $5\in C(u_1)$.
When $W_1 = \{1, 5\}$,
it follows from <ref>(2) that $H$ contains a $(4, 1)_{(u_2, u_4)}$-path, $1\in C(u_4)$,
and from <ref>(1) that $1\in C(u_3)$.
It follows from <ref>(3) that $3\in C(u_4)$, $T_4\subseteq C(u_3)$ and $6, 7\in C(u_3)\cup C(u_4)$.
Thus, $d(u_3) + d(u_4)\ge |\{1, 3, 5\}| + |[1, 5]| + |[6, 7]| + t_{u v} = \Delta + 8$.

In the other case, $W_1\ge 3$ and $t_1\ge 1$.
One sees that $H$ contains a $(3/4, T_1)_{(u_1, u)}$-path.

Case 4.1. $3\not\in C(u_1)$.
Then $4\in C(u_1)$ and follows from Corollary <ref>(5.1) that $1, 6, 7\in S_u$,
and if $1\not\in C(u_4)$, then $H$ contains a $(1, 3)_{(u_3, u_4)}$-path,
from Corollary <ref>(5.1) and (4) that $T_1\subseteq C(u_1)$.

* There exists a $8\in T_{u v}\setminus (C(u_3)\cup C(u_4))$.
It follows from <ref>(3) that $2\in C(u_3)$,
from <ref>(2) that $H$ contains a $(4, 3)_{(u_2, u_4)}$-path, $3\in C(u_4)$,
and from <ref>(2) that $H$ contains a $(6, 8)_{(u_4, v_6)}$-path, $6\in C(u_4)$.
If $6\not\in C(u_1)\cup C(u_3)$, one sees from <ref>(3) that $H$ contains a $(4, 6)_{(u_3, u_4)}$-path, then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $6\in C(u_1)\cup C(u_3)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 4, 5\}| + |\{2, 4\}| + |\{3, 5, 2\}| + |[2, 6]| + |\{1, 6, 7\}| + 2t_{u v} + t_0\ge 2\Delta 2\Delta + 12 + t_1\ge 2\Delta + 13$.
It follows that $t_1 = 1$, $4\not\in C(u_3)$, $W_1 = \{1, 4, 5\}$,
and from <ref>(2) that $1\in C(u_4)\setminus C(u_3)$.
Then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (3, 1, 8)$.

* $T_{u v}\subseteq C(u_3)\cup C(u_4)$.
Let $\kappa_{u_1} = |[6, 7]\cap C(u_1)|$.
One sees that $t_1\ge w_1 - 2\ge \kappa_{u_1} + 3 - 2 = \kappa_{u_1} + 1$.
Then $d(u_3) + d(u_4)\ge |\{3, 5\}| + |\{2, 4, 5\}| + |\{1\}| + 2 - \kappa_{u_1} + t_{u v} + t_1\ge 8 - \kappa_{u_1} + t_{u v} + \kappa_{u_1} + 1 = \Delta + 7$.
It follows that $3\not\in C(u_4)$ and then $1\in C(u_4)\setminus C(u_3)$.
Then $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, 3, T_4)$.

Case 4.2. $3\in C(u_1)$.
When $4\not\in B_1$, it follows from <ref>(3) that $3\in C(u_4)$, $T_4\subseteq C(u_3)$,
and $6, 7\in C(u_3)\cup C(u_4)\cup C(u_1)$.
Let $\kappa_{u_1} = |[6, 7]\cap C(u_1)|$.
One sees that $t_1\ge w_1 - 2\ge \kappa_{u_1} + 3 - 2 = \kappa_{u_1} + 1$.
Then $d(u_3) + d(u_4)\ge |\{3, 5\}| + |[2, 5]| + |\{1\}| + 2 - \kappa_{u_1} + t_{u v} + t_1\ge 9 - \kappa_{u_1} + t_{u v} + \kappa_{u_1} + 1 = \Delta + 8$.

In the other case, $4\in B_1\subseteq C(u_1)$ and then $t_1\ge 2$.
One sees that $T_4\subseteq C(u_1)\cap C(u_2)$.

(4.2.1.) $T_1\setminus C(u_3)\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $1, 6, 7\in S_u$,
from Corollary <ref>(5.2) and (3) that $3\in B_1$ and $H$ contains a $(4, 3)_{(u_2, u_4)}$-path, $3\in C(u_4)$.
If $1\in C(u_4)$, ones sees that $d(u_4)\ge |[1, 5]| + t_1\ge 7$,
it follows that $t_1 = 2$, $W_4 = [1, 5]$, $W_1 = \{1, 3, 4, 5\}$,
and from Lemma <ref>(1.2) that $T_4\subseteq C(u_3)$,
then $C(u_3) = \{3, 5, 6, 7\}\cup T_4$,
and $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 4, 6, T_4)$.
Otherwise, $1\not\in C(u_4)$.
It follows from <ref>(2) that $2\in C(u_1)$ or $H$ contains a $(2, 3)_{(u_1, u_3)}$-path.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3, 4, 5\}| + |\{2, 4\}| + |\{3, 5, 1\}| + |[2, 5]| + |\{6, 7, 2\}| + 2t_{u v} + t_0 = 2\Delta + 12 + t_0$.
Since $t_4\ge 2$, we have $T_4\setminus C(u_3)\ne \emptyset$.
Assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
ones sees that $d(u_4)\ge |[2, 6]| + t_1\ge 7$,
it follows that $t_1 = 2$, $C(u_4) = [2, 6]\cup T_1$, $T_{uv} = T_1\cup T_4$,
$W_1 = \{1, 3, 4, 5\}$, and $1, 2, 7\in C(u_3)$.
If $H$ contains no $(4, 6)_{(u_4, v_6)}$-path,
then $(v u_4, u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (4, T_1\cap T_3, 1, 2, T_4)$.
Otherwise, $H$ contains a $(4, 6)_{(u_4, v_6)}$-path and $4\in C(v_6)$.
If $6\not\in C(u_3)$, then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $6\in C(u_3)$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + t_0 + |\{6\}|\ge 2\Delta + 13$.
It follows that $t_0 = 0$ and $H$ contains a $(6, T_4)_{(u_4, v_6)}$-path.
One sees that the same argument applies if $u u_2\to 6$, and thus $T_1\subseteq C(v_6)$.
Hence, $C(v_1) = \{4, 6\}\cup T_{uv}$.
Then $(v v_6, u v)\overset{\divideontimes}\to (3, 6)$.

(4.2.2.) $T_1\subseteq C(u_3)$.
One sees that $|\{1, 3, 4, 5\}| + |\{2, 4\}| + |\{3, 5\}| + |\{2, 4, 5\}| + 2t_{u v} + t_0 = 2\Delta + 7 + t_0$.
We distinguish whether or not $T_4\setminus C(u_3) \ne \emptyset$.

* $T_4\setminus C(u_3)\ne \emptyset$.
Assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
It follows from <ref>(2) that $3\in C(u_4)$.
Ones sees that $d(u_4)\ge |[2, 6]| + t_1\ge 7$.
It follows that $t_1 = 2$, $C(u_4) = [2, 6]\cup T_1$, $T_{u v} = T_1\cup T_4$,
$W_1 = \{1, 3, 4, 5\}$,
and from <ref>(2) that $2\in C(u_3)$.
Assume that $6\not\in C(u_3)$.
If $H$ contains no $(4, 6)_{(u_1, u_4)}$-path, then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $H$ contains a $(4, 6)_{(u_1, u_4)}$-path and $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, 1, 6, 8)$.
Hence, $6\in C(u_3)$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 7 + t_0 + |\{3, 6, 6, 2\}| = 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1\ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $1, 4, 7\not\in C(u_3)$,
and from Corollary <ref>(7.2) that $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path.
Then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, 4, 7, 8)$.

* $T_4\subseteq C(u_3)$.
One sees from <ref>(2) that $1\in C(u_4)$ or $2\in C(u_1)\cup C(u_3)$,
and from <ref>(1) that $1/3/6/7\in C(u_4)$, $t_4\ge 2$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 7 + t_0 + |\{1/2\}|\ge 2\Delta + 8 + t_1 + t_4\ge 2\Delta + 8 +2 + 2 = 2\Delta + 12$.
Hence, assume w.l.o.g. $6\not\in S_u$,
and it follows from <ref>(3) that $3\in C(u_4)$.
If $1\not\in C(u_3)$, then $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, 6, T_4)$.
Otherwise, $1\in C(u_3)$ and then $1\not\in C(u_4)$, $2\not\in C(u_1)\cup C(u_3)$.
Thus, $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 1, T_4)$
[Or, by <ref>(2), we are done.  ].

Hence, $w_2\ge 3$ and $t_2 = \Delta - d(u_2) + w_2 - 2 \ge w_2 - 2 \ge 1$.

$w_1\ge 3$ and $t_1 = \Delta - d(u_1) + w_1 - 2 \ge w_1 - 2 \ge 1$.

Assume $w_1\le 2$.
One sees from <ref>(1) that $5/2/4\in C(u_1)$.
Then $W_1 = \{1, 5/2/4\}$.
We consider below the three cases, respectively.

Case 1. $W_1 = \{1, 5/2\}$.
It follows from <ref>(3) that $T_2\subseteq C(u_4)$, $3\in C(u_4)$, $T_4\subseteq C(u_3)$,
from <ref>(3) and <ref>(3) that $6, 7\in C(u_3)\cup C(u_4)$.
One sees that $|\{3, 5\}| + |[2, 4]| + |\{6, 7\}| + t_{u v} + t_2 = \Delta + 5 + t_2$.
When $5\in C(u_1)$, it follows from <ref>(1) and <ref>(1) that $1\in C(u_3)\cup C(u_4)$.
One sees that the same argument applies if $uu_1\to 6/7$.
Hence, $H$ contains a $(\{1, 6, 7\}, T_4)_{(u_1, u)}$-path.
It follows from <ref>(2) that $5\in C(u_4)$.
Then $d(u_3) + d(u_4)\ge \Delta + 5 + t_2 + |\{1, 5\}| = \Delta + 7 + t_2\ge\Delta + 7 + 1 = \Delta + 8$.

In the other case, $W_2 = \{1, 2\}$.
Then $H$ contains a $(2, 5)_{(u_1, u_2)}$-path.
It follows from <ref>(4) that $5\in C(u_4)$.
Then $d(u_3) + d(u_4)\ge \Delta + 5 + t_2 + |\{5\}| = \Delta + 6 + t_2\ge \Delta + 6 +1 = \Delta + 7$.
It follows that $t_2 = 1$, $1\not\in S_u$, $2\not\in C(u_3)$,
and from <ref>(1) that $4\in C(u_2)$.
Hence, $W_2 = \{2, 4, 5\}$ and $H$ contains a $(4, T_2)_{(u_2, u_4)}$-path.
Then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (3, T_2, 4, 1, T_4)$.

Case 2. $W_1 = \{1, 4\}$.
Then $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
It follows from <ref>(1) that $5\in C(u_2)$,
from <ref>(2) that $1\in C(u_2)$, or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path.

Case 2.1. $T_2\setminus C(u_3)\ne\emptyset$.
Then $4\in C(u_2)$ and $T_2\subseteq C(u_4)$.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$,
and from <ref>(3) that $3\in C(u_2)\cap C(u_4)$.
If $1\not\in C(u_2)\cup C(u_4)$,
one sees that $H$ contains a $(1, 3)_{(u_2,u_3)}$-path,
then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (2, T_2\cap T_3, 1, T_4)$.
Otherwise, $1\in C(u_2)\cup C(u_4)$.
Since $d(u_2) + d(u_4)\ge t_{uv} + 2\times |[2, 5]| + |\{1\}| = \Delta + 7$,
it follows that $d(u_2) = \Delta$, $W_1\cup W_4 = [2, 5]\cup \{1\}$, $6, 7\in C(u_3)$, $d(u_4) = 7$,
and then $C(u_1) = \{1, 4\}\cup T_2\cup T_4$,
and from <ref>(2) that $T_4\subseteq C(u_3)$.
One sees from $d(u_1) = \Delta$ that $\Delta = 7$ and $d(u_3)\le 6$.
Since $d(u_3)\ge |\{3, 5, 6, 7\}\cup T_4| = 4 + t_4$,
we have $t_4 = 2$ and $W_1 = [1, 5]$, $W_4 = [2, 5]$, $W_3 = \{3, 5, 6, 7\}\cup T_4$.
Then $(v u_4, u u_4, u u_2)\to (4, 2, T_2)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.

Case 2.2. $T_2\subseteq C(u_3)$.
When $3\not\in C(u_2)$,
then $4\in C(u_2)$, $T_2\subseteq C(u_4)$,
and $1\in C(u_2)$, or $H$ contains a $(1, 4)_{(u_2, u_4)}$-path.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$,
from <ref>(2) that $2\in C(u_3)$ or $3\in C(u_4)$.
Let $\kappa_{u_2} = |\{1, 6, 7\}\cap C(u_2)|$.
If $T_4\subseteq C(u_3)$,
then $d(u_3) + d(u_4)\ge |\{3, 5\}| + |\{2, 4, 5\}| + |\{2/3\}| + 3 - \kappa_{u_2} + t_{uv} + t_0 \ge \Delta + 7 - \kappa_{u_2} + t_2\ge \Delta + 7 - \kappa_{u_2} + \kappa_{u_2} + 3 - 2 = \Delta + 8$.
Otherwise, $T_4\setminus C(u_3)\ne \emptyset$.
Assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
It follows from <ref>(3) that $2\in C(u_3)$, $4\in B_1$,
and from <ref>(2) that $3\in C(u_4)$.
Recall that $1\in C(u_2)\cup C(u_4)$.
One sees that $d(u_2) + d(u_4)\ge |\{2, 4, 5\}| + |[2, 6]| + |\{1\}| + t_{uv} = \Delta + 7$.
It follows that $d(u_2) = \Delta$, $C(u_1)= \{1, 4\}\cup T_2\cup T_4 = \{1, 4\}\cup T_{uv}$, $d(u_4) = 7$,
and $7\not\in C(u_4)$.

If $H$ contains no $(4, 6)_{(u_1, u_4)}$-path, then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $H$ contains a $(4, 6)_{(u_1, u_4)}$-path,
and $(v u_4, u u_4)\to (4, 8)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.

In the other case, $3\in C(u_2)$.
One sees that $|\{1, 4\}| + |\{2, 3, 5\}| + |\{3, 5\}| + |\{2, 4, 5\}| + 2t_{u v} = 2\Delta + 6$.

(2.2.1.) $T_2\setminus C(u_4)\ne \emptyset$.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$,
Corollary <ref>(5.2) and (3) that $4\in B_1$ and $4\in C(u_2)\cup C(u_3)$,
and from Corollary <ref>(7.2) that $2\in C(u_3)$.
If $H$ contains no $(4, i)_{(u_4, v)}$-path for each $i\in [6, 7]$,
then $(v u_4, u u_4)\to (4, T_2\cap T_4)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
Otherwise, assume w.l.o.g. $6\in C(u_4)$ and $H$ contains a $(4, 6)_{(u_4, v)}$-path.
One sees that the same argument applies if let $u u_1\to 6$ and thus $H$ contains a $(6, T_4)_{(u_1, v_6)}$-path.
If $1\not\in C(u_2)\cup C(u_3)$,
one sees that $H$ contains a $(1, 4)_{(u_2, u_4)}$-path,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, T_2\cap T_4, 1, T_4)$.
If $6\not\in C(u_2)\cup C(u_3)$,
then $(u u_2, u u_3)\to (T_2\cap T_4, 6)$ reduces the proof to (2.4.3.).
Otherwise, $1, 6\in C(u_2)\cup C(u_3)$.
Hence, $d(u_2) + d(u_3) \ge t_{u v} + 2\times |\{2, 3, 5\}| + |\{1, 4, 6\}| = \Delta + 7$.
It follows that $d(u_2) = \Delta$, $d(u_3) = 7$, and $7\in C(u_4)\setminus (C(u_2)\cup C(u_3))$.
If $3\not\in C(u_4)$, then $(u u_2, u u_3, u u_4)\to (T_2\cap T_4, 7, 3)$ reduces the proof to (2.4.3.).
Otherwise, $3\in C(u_4)$.
Recall that $4\in C(v_1)$.
It follows from <ref>(1) that $R_1\ne \emptyset$.
Assume w.l.o.g. $8\in R_1$.
Then $8\in C(u_2)\cap C(u_4)$ and $C(u_4) = [2, 8]$.
It follows from <ref>(2) that $4\in C(u_3)$, $H$ contains a $(4, 7)_{(u_3, u_4)}$-path.
One sees clearly that $T_4\setminus C(u_3)\ne \emptyset$.
Assume w.l.o.g. $9\in T_3\cap T_4$ and then $H$ contains a $(7, 9)_{(u_4, v_7)}$-path.
Then $(u u_1, u v)\overset{\divideontimes}\to (7, 9)$.

(2.2.2.) $T_2\subseteq C(u_4)$.
Recall that $1\in S_u$.
One sees that the same argument applies if $i\not\in S_u$ and $u u_1\to i$, $i\in [6, 7]$,
and thus $6, 7\in S_u$.

* $4\not\in C(u_2)$.
It follows from <ref>(2) that $1\in C(u_2)\cup C(u_3)$,
from Corollary <ref>(7.2) that $2\in C(u_3)$.
Assume that $6\not\in C(u_2)\cup C(u_3)$.
It follows from <ref>(2) that $H$ contains a $(4, 6)_{(u_3, u_4)}$-path.
If $H$ contains no $(6, j)_{(u_4, v)}$-path for some $j\in T_4$,
then $(u u_1, u v)\overset{\divideontimes}\to (6, j)$.
Otherwise, $H$ contains a $(6, T_4)_{(u_1, v)}$-path.
Then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 6, T_4)$.
Hence, $6, 7\in C(u_2)\cup C(u_3)$,
and $d(u_2) + d(u_3)\ge 2\times |\{2, 3, 5\}| + |\{1, 6, 7\}| + t_{uv} = \Delta + 7$.
It follows that $4\not\in C(u_2)\cup C(u_3)$,
and from <ref>(3) that $3\in C(u_4)$ and $T_4\subseteq C(u_3)$.
Let $\kappa_{u_2} = |\{1, 6, 7\}\cap C(u_2)|$.
One sees that $t_2\ge  \kappa_{u_2} + |\{2, 3, 5\}| - 2 = \kappa_{u_2} + 1$.
Thus, $d(u_3) + d(u_4)\ge |\{2, 3, 5\}| + |[2, 5]| + 3 - \kappa_{u_2} + t_{uv} + t_0 \ge \Delta + 8 - \kappa_{u_2} + t_2\ge \Delta + 8 - \kappa_{u_2} + \kappa_{u_2} + 1 = \Delta + 9$.

* $4\in C(u_2)$.
Let $\kappa_{u_2} = |\{1, 6, 7\}\cap C(u_2)|$.
One sees that $t_2\ge \kappa_{u_2} + |[2, 5]| - 2 = \kappa_{u_2} + 2$.
If $T_4\subseteq C(u_3)$,
then $d(u_3) + d(u_4)\ge |\{3, 5\}| + |\{2, 4, 5\}| + 3 - \kappa_{u_2} + t_{u v} + t_0 \ge \Delta + 6 - \kappa_{u_2} + t_2\ge \Delta + 6 - \kappa_{u_2} + \kappa_{u_2} + 2 = \Delta + 8$.
Otherwise, $T_4\setminus C(u_3)\ne \emptyset$.
Assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
It follows from <ref>(3) that $4\in B_1$.
If $H$ contains no $(4, 6)_{(u_1, u_4)}$-path,
then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $H$ contains a $(4, 6)_{(u_1, u_4)}$-path.
If $H$ contains no $(4, 7)_{(u_4, v)}$-path,
then $(v u_4, u u_4)\to (4, 8)$ reduces to an acyclic edge $(\Delta + 5)$-coloring for $H$ such that $|C(u)\cap C(v)| = 1$.
Otherwise, $7\in C(u_4)$ and $H$ contains a $(4, 7)_{(u_4, v)}$-path.
It follows that $t_2 = 2$, $C(u_4) = \{2\}\cup [4, 7]\cup T_2$, and $W_2 = [2, 5]$, $1\in C(u_3)$.
If $H$ contains no $(7, 8)_{(u_1, v)}$-path,
then $(u u_1, u v)\overset{\divideontimes}\to (7, 8)$.
Otherwise, $H$ contains a $(7, 8)_{(u_1, v)}$-path.
If $7\not\in C(u_3)$, then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 7, 8)$.
Otherwise, $7\in C(u_3)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 4\}| + |[2, 5]| + |\{3, 5, 1, 7\}| + |\{2\}\cup [4, 7]| + 2t_{u v} + t_0 = 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_2 = 2\Delta + 13$.
It follows that $6\not\in C(u_2)\cup C(u_3)$.
Then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 6, 8)$.

Hence, $w_1\ge 3$ and $t_1 = \Delta - d(u_1) + w_1 - 2 \ge w_1 - 2 \ge 1$.

Assume that $H$ contains a $(6, 8)_{(u_4, v_6)}$-path, $8\in T_4$.
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 $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $6\in C(u_1)$ or $H$ contains a $(6, 2/3/4)_{(u_1, u)}$-path.
If $6\not\in S_u\setminus C(u_4)$ and $2\not\in C(u_1)$, $H$ contains no $(2, 3)_{(u_1, u_3)}$-path,
one sees that $H$ contains a $(4, 6)_{(u_1, u_4)}$-path,
then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 6, 8)$.
Hence, together with <ref>, <ref>(3),
we proceed with the following proposition, or otherwise we are done:

Assume that $H$ contains a $(6, 8)_{(u_4, v_6)}$-path, $8\in T_4$. Then

$6\in C(u_1)$, or $H$ contains a $(6, 2/3/4)_{(u_1, u)}$-path;

if $6\not\in S_u\setminus C(u_4)$, then

$4\in C(u_2)\cap C(u_1)$,
and $H$ contains a $(4, T_2)_{(u_2, u_4)}$-path, a $(4, T_1)_{(u_1, u_4)}$-path;

$3\in B_1\cup B_2$, mult$_{S_u}(3)\ge 2$, and for each $i\in[1, 2]$, $3\in C(u_i)$, or $H$ contains a $(3, 4)_{(u_i, u_4)}$-path, $4\in C(u_i)$, $3\in C(u_4)$.

$2\in C(u_1)$, or $H$ contains a $(2, 3)_{(u_1, u_3)}$-path.

Assume that $H$ contains a $(7, 8)_{(u_3, v_6)}$-path, $8\in T_4$.
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 $(u u_1, u v)\overset{\divideontimes}\to (7, 8)$.
Otherwise, $7\in C(u_1)$ or $H$ contains a $(7, 2/3/4)_{(u_1, u)}$-path.
If $7\not\in S_u\setminus C(u_3)$ and $2\not\in C(u_1)$, $H$ contains no $(2, 3)_{(u_1, u_3)}$-path,
one sees that $H$ contains a $(3, 7)_{(u_1, u_3)}$-path,
then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 7, 8)$.
Hence, together with <ref>, <ref>(3),
we proceed with the following proposition, or otherwise we are done:

Assume that $H$ contains a $(7, 8)_{(u_3, v_7)}$-path, $8\in T_4$. Then

$7\in C(u_1)$, or $H$ contains a $(7, 2/3/4)_{(u_1, u)}$-path;

if $7\not\in S_u\setminus C(u_3)$, then

$3\in C(u_1)\cap C(u_2)$,
and $H$ contains a $(3, T_1)_{(u_1, u_3)}$-path, a $(3, T_2)_{(u_2, u_3)}$-path;

$4\in B_1$, mult$_{S_u}(4)\ge 2$, and for each $i\in[1, 2]$, $4\in C(u_i)$, or $H$ contains a $(3, 4)_{(u_i, u_3)}$-path, $3\in C(u_i)$, $4\in C(u_3)$.

$2\in C(u_1)$, or $H$ contains a $(2, 3)_{(u_1, u_3)}$-path.

By <ref>(2.1),
we proceed with the following proposition, or otherwise we are done:

Assume that $T_2\setminus C(u_4)\ne \emptyset$. Then

for each $i\in [6, 7]$, if $H$ contains a $(i, j)_{(u_4, v_i)}$-path for some $j\in T_4$,
then $i\in S_u\setminus C(u_4)$, i.e., mult$_{S_u}(i)\ge 2$.

if mult$_{S_u}(6)\le 1$ and mult$_{S_u}(7)\le 1$,
then $H$ contains no $(i, j)_{(u_4, v)}$-path, $i = 6, 7$, $j\in T_4$,
and $5\in C(u_4)$, $H$ contains a $(5, T_4)_{(u_3, u_4)}$-path.

Assume that $H$ contains a $(4, i)_{(u_4, v_i)}$-path and a $(i, j)_{(u_4, v_i)}$-path for some $j\in T_4$, $i\in [6, 7]\cap C(u_4)$.
If $i\not\in C(u_1)$, and $H$ contains no $(i, l)_{(u_1, u_l)}$-path for each $l\in [2, 3]$,
then $(u u_1, u v)\overset{\divideontimes}\to (i, j)$.
Hence, we proceed with the following proposition, or otherwise we are done:

If $H$ contains a $(4, i)_{(u_4, v_i)}$-path and a $(i, j)_{(u_4, v_i)}$-path for some $j\in T_4$, $i\in [6, 7]\cap C(u_4)$, then $i\in S_u\setminus C(u_4)$,
and if $i\not\in C(u_1)$, then $H$ contains a $(i, 2/3)_{(u_1, u)}$-path.

If $T_2\setminus C(u_4)\ne \emptyset$ and $T_1\setminus C(u_3)\ne \emptyset$,
then we are done.

Then $4\in C(u_1)$, $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path,
and $3\in C(u_2)$, $H$ contains a $(3, T_2)_{(u_2, u_3)}$-path, $T_2\subseteq C(u_3)$.
It follows from Corollary <ref>(5.1) that $6, 7\in S_u$,
$1\in C(u_4)$ or $H$ contains a $(1, 3/2)_{(u_4, u)}$-path,
and $2\in C(u_3)$ or $H$ contains a $(1, 2)_{(u_1, u_3)}$-path,
from Corollary <ref>(5.2) and (3) that $4\in B_1$, $3\in B_1\cup B_2$,
and $4\in C(u_2)\cup C(u_3)$, $3\in C(u_1)\cup C(u_4)$.
It follows from <ref>(1) and <ref>(1) that $5\in C(u_1)\cup C(u_4)$,
from <ref>(1) and <ref>(1) that $5\in C(u_1)\cup C(u_2)$.
We distinguish whether or not mult$_{S_u}(5) = 1$.

Case 1. mult$_{S_u}(5) = 1$.
It follows that $5\in C(u_1)\setminus (C(u_2)\cup C(u_4))$,
and from <ref>(1) that $H$ contains a $(1, 5)_{(u_1, u_2)}$-path, $1\in C(u_2)$.
If $H$ contains no $(5, j)_{(u_1, u_4)}$-path for some $j\in T_1\cap T_3$,
then $(u u_4, u u_1, u v)\overset{\divideontimes}\to (5, j, T_4)$.
Otherwise, $H$ contains a $(5, T_1\cap T_3)_{(u_1, u_4)}$-path.
If $1\not\in C(u_4)$ and $H$ contains no $(1, 3)_{(u_3, u_4)}$-path,
then $(u u_1, u u_2, u u_4)\to (T_1\cap T_3, 5, 1)$ reduces the proof to (2.4.2.).
Otherwise, $1\in C(u_4)$ or $H$ contains a $(1, 3)_{(u_3, u_4)}$-path.
Hence, $1\in C(u_3)\cup C(u_4)$ and mult$_{S_u}(1)\ge 2$.
Since $5\not\in C(u_4)$, it follows from <ref>(2) that for some $j\in T_2\cap T_4$,
$H$ contains a $(6, j)_{(u_4, v_6)}$-path and $6\in C(u_4)$, say $j = 8$.

It follows from <ref>(1) that $6\in S_u\setminus C(u_4)$ and mult$_{S_u}(6)\ge 2$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + 2\times |\{1, 6, 3, 4\}| + |\{2, 5, 7\}| + 2t_{u v} + t_0 = 2\Delta + 13 + t_0$.
It follows that $t_0 = 0$, $\Delta = 7$,
and mult$_{S_u}(i) = 2$, $i = 1, 3, 4, 5$,
mult$_{S_u}(j) = 1$, $j = 2, 7$.

* $1\in C(u_4)\setminus C(u_3)$.
It follows that $2\in C(u_3)\setminus C(u_1)$ and from <ref>(1) that $H$ contains a $(2, 3)_{(u_1, u_3)}$-path, $3\in C(u_1)\setminus C(u_4)$.
Then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (2, 5, 1, 3, T_1\cap T_4)$.

* $1\in C(u_3)\setminus C(u_4)$.
It follows that $3\in C(u_4)\setminus C(u_1)$,
and from <ref>(1) that $2\in C(u_1)\setminus C(u_3)$.
One sees that $H$ contains a $(1, 2)_{(u_1, u_3)}$-path.
Then $u u_2\to \alpha\in T_2\cap T_4$,
and $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (3, 2, 1, T_4\setminus \{\alpha\})$.

Case 2. mult$_{S_u}(5)\ge 2$.
Condition (C2) states that mult$_{S_u}(1) = 1$, mult$_{S_u\setminus C(u_4)}(2) = 1$,
and mult$_{S_u}(i) = 2$, $i = 3, 4$.
When (C2) holds, if $2\in C(u_1)$,
one sees that $1\in C(u_3)$ and $3\in C(u_4)$,
then $uu_2\to \alpha\in T_2\cap T_4$,
and $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (3, 2, 1, T_4\setminus \{\alpha\})$.
Otherwise, $2\in C(u_3)\setminus C(u_1)$.
Then $1\in C(u_4)$ or $H$ contains a $(1, 3)_{(u_3, u_4)}$-path.
It follows from <ref> (2) that $1\in C(u_4)$,
and $3\in C(u_1)$ or $4\in C(u_2)$.
Let $(u u_1, u u_2, u u_3)\to (2, T_2\cap T_4, 1)$.
If $4\not\in C(u_3)$, then reduces the proof to (2.4.2.).
Otherwise, $4\in C(u_3)$ and $3\in C(u_1)$.
Then $u u_4\to 3$ reduces the proof to (2.4.2.).

In the other case, $\sigma_{[1, 4]} = \sum_{i\in \{1, 3, 4\}}$ mult$_{S_u}(1) + $mult$_{S_u\setminus C(u_4)}(2)\ge 7$. Let $\sigma_{[5, 7]} = \sum_{i\in [5, 7]}$ mult$_{S_u}(i)$.
Recall that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + 2t_{u v} + t_0 = 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0$.
If $\sigma_{[5, 7]}\ge 5$, one sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + 5 + t_0 = 2\Delta + 7 + \sigma_{[1, 4]} + t_0$, then $\sigma_{[1, 4]}\le 6$ and (C2) holds.
Otherwise, $\sigma_{[5, 7]}\le 4$.
It follows that mult$_{S_u}(5) = 2$, mult$_{S_u}(i) = 1$, $i = 6, 7$,
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + 4 + t_0 \ge 2\Delta + 6 + 7 + t_0 = 2\Delta + 13 + t_0\ge 2\Delta + 13$.
Hence, $\Delta = 7$, $\sigma_{[1, 4]} = 7$, $t_0 = 0$, and $T_2\cap C(u_4) = T_1\cap C(u_3) = \emptyset$.
One sees from <ref>(2) that $5\in C(u_4)$, $H$ contains a $(5, T_4)_{(u_3, u_4)}$-path.
Since $t_0 = 0$ and $T_4\subseteq C(u_3)$,
we have $C_{12} = \emptyset$, i.e., $T_{uv} = T_1\cup T_2$.

When $6\in C(u_3)\cup C(u_4)$, one sees that $6\not\in B_1\cup B_2$,
and it follows from <ref>(2) that if $6\in C(u_4)$, then $H$ contains a $(4, 6)_{(u_3, u_4)}$-path;
from <ref>(1) that if $6\in C(u_3)$, then $H$ contains a $(3, 6)_{(u_3, u_4)}$-path.
Since the same argument applies if $u u_1\to 6$ or $u u_2\to 6$, $C(v_6) = \{4, 6\}$ $\cup T_1\cup T_2$.
Then $(v v_6, u v)\overset{\divideontimes}\to (3, 6)$.
In the other case, $6, 7\not\in C(u_3)\cup C(u_4)$.
Recall that $4\in B_1 \subseteq C(v_1)$.

One sees from <ref>(2) that $H$ contains a $(5, 4)_{(u_3, u_4)}$-path.
It follows from <ref>(1) that if $5\not\in C(u_1)$, then $H$ contains a $(2, 5)_{(u_1, u_2)}$-path,
and from <ref>(1) that if $5\not\in C(u_2)$, then $H$ contains a $(1, 5)_{(u_1, u_2)}$-path,
i.e., $5\not\in B_1\cup B_2$.
One sees from $\{1, 4\}\cup T_2\subseteq C(v_1)$ and <ref>(1) that $T_1\setminus C(v_1)\ne \emptyset$.
Then $(v u_3, u v)\overset{\divideontimes}\to (T_1\cap R_1, 5)$.

Recall that $t_1 + t_2 = \Delta - d(u_1) + \Delta - d(u_2) + w_1 + w_2 - 4\ge w_1 + w_2 - 4$.
It follows that $w_1 + w_2\le 4 + t_1 + t_2$,
and $t_1 + t_2 = w_1 + w_2 - 4$ if and only if $d(u_1) = d(u_2) = \Delta$.

If $t_1 + t_2\ge 4$, then $T_2\setminus C(u_4)\ne \emptyset$ and $T_1\setminus C(u_3)\ne \emptyset$.

Assume w.l.o.g. $[8, 11]\subseteq T_1\cup T_2$.
It follows that $w_1 + w_2\le 8$ and if $w_1 + w_2 = 8$,
then $d(u_1) = d(u_2) = \Delta = 7$, and $d(u_3)\le 6$.
We consider the following two cases, respectively.

Case 1. $T_2\subseteq C(u_4)$.
One sees from <ref>(2) that $5/6/7\in C(u_4)$.
It follows that $t_1 + t_2 = 4$ and $C(u_4) = \{4, 2, 5/6/7\}\cup [8, 11]$, $C_{12} = T_4 = T_{u v}\setminus [8, 11]$.
Further, it follows from <ref>(1) that $6\in C(u_4)$, $H$ contains a $(3, 6)_{(u_4, v_6)}$-path,
from Lemma <ref>(1.2) that $H$ contains a $(6, C_{12})_{(u_4, v_6)}$-path,
and from <ref>(2) that $H$ contains a $(6, 4)_{(u_4, v_6)}$-path.
Hence, it follows from <ref> that $6\in S_u\setminus C(u_4)$.
One sees from <ref>(3) that $4\in B_1$, $4\in C(u_2)\cup C(u_3)$,
and from <ref>(1) that $5\in C(u_1)$, or $H$ contains a $(2, 5)_{(u_1, u_2)}$-path.

Case 1.1. $H$ contains no $(1, 2)_{(u_2, v)}$-path.
It follows from Corollary <ref>(7.2) that $3\in C(u_1)$ and $H$ contains a $(3, T_1)_{(u_1, u_3)}$-path, $T_1\subseteq C(u_3)$.
If $1\not\in S_u$, then $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, 3, T_4)$.
Since the same argument applies if $7\not\in S_u$ and $uu_1\to 7$, we have $1, 7\in S_u$.
One sees that $6 + |\{3, 4, 4\}| + |\{6\}| + |\{6, 7, 1, 5\}| + 2t_{u v} + t_0 = 2\Delta + 10 + t_0$.
It follows that $T_2\setminus C(u_3)\ne \emptyset$,
and then from Corollary <ref>(5.2) that $3\in C(u_2)$.
One sees from $1, 3, 4, 5/2\in C(u_1)$ that $t_1\ge 2$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0 + |\{3\}|\ge 2\Delta + 11 + t_1\ge 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $t_1 = 2$, $2\not\in S_u\setminus C(u_4)$, $C_{12}\cap C(u_3) = \emptyset$,
and then $W_1 = \{1, 3, 4, 5\}$.
Thus, $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 5, C_{12})$.

Case 1.2. $H$ contains a $(1, 2)_{(u_2, v)}$-path.
Then $1\in C(u_2)$.
If $1\not\in C(u_3)$ and $3\not\in C(u_1)$,
then $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, 3, T_4)$.
Otherwise, $1\in C(u_3)$ or $3\in C(u_1)$.
When $T_1\cup T_2\subseteq C(u_3)$,
one sees that $6 + |\{1, 4, 4, 6, 5, 6, 1/3\}| + 2t_{uv} + t_0 = 2\Delta + 9 + t_0\ge 2\Delta + 9 + t_1 + t_2 = 2\Delta + 13$.
It follows that $7\not\in S_u$ and mult$_{S_u}(3)\le 1$.
Thus, by <ref>, we are done.

In the other case, $(T_1\cup T_2)\setminus C(u_3)\ne \emptyset$.
It follows from Corollary <ref>(5.2) and (3) that $3\in B_1$ and $3\in C(u_2)$,
and from <ref> that $7\in C(u_1)\cup C(u_2)$.
Together with $5\in C(u_1)\cup C(u_2)$, we have $W_1\cup W_2 = [1, 3]\cup \{1, 3, 4, 5\}\cup \{7\}$,
$4, 6\in C(u_3)$.
It follows from <ref>(1) that $2\in C(u_3)$.
One sees that $6 + |\{1, 4, 4, 3, 3, 6, 6, 5, 7, 2\}| + 2t_{u v} + t_0\ge 2\Delta + 12 + t_2\ge 2\Delta + 12 + 1 = 2\Delta + 13$.
It follows that $t_2 = 1$, $7\in C(u_1)$, $T_1\cap C(u_3) = \emptyset$, and $1\not\in C(u_3)$.
Thus, $(u u_2, u u_3, u v)\overset{\divideontimes}\to (T_2, 7, T_1)$
[Or, by <ref>(2), we are done. ].

Case 2. $T_2\setminus C(u_4)\ne \emptyset$ and $T_2\cup T_1 \subseteq C(u_3)$.
Then $3\in C(u_2)$ and $H$ contains a $(3, T_2)_{(u_2, u_3)}$-path.
It follows from Corollary <ref>(5.2) and (3) that $4\in B_1$, and $4\in C(u_2)\cup C(u_3)$,
and from Corollary <ref>(7.2) that $2\in C(u_3)$ or $H$ contains a $(1, 2)_{(u_1, u_3)}$-path.
Hence, $t_1 + t_2 = 4$, $C(u_3) = \{3, 5, 1/2\}\cup T_1\cup T_2$, $4\in C(u_1)\cap C(u_2)$,
and from <ref>(2) that $6, 7\in C(u_1)\cup C(u_2)$.

If $C(u_3) = \{3, 5, 1\}\cup T_1\cup T_2$, one sees that $2\in C(u_1)$ and $5\not\in C(u_1)\cup C(u_2)$,
then $(v u_3, u u_3, u u_2)\to (4, 5, T_2\cap T_4)$ reduces the proof to (2.4.1.).
Otherwise, $C(u_4) = \{3, 5, 2\}\cup T_1\cup T_2$.
It follows from <ref>(2) that $6, 7\in C(u_2)$.
By Lemma <ref> and Lemma <ref>, $W_1 = [3, 4]\cup [6, 7]$, $W_2 = \{1, 4, a_2\}$,
and $d(u_1) = d(u_2) = \Delta$.
It follows that $\Delta = 7$ and together with $d(u_3) = 7$, $d(u_4)\le 6$.
If $2\not\in C(u_1)$, then $(u u_1, u u_2, u u_3)\to (2, T_2\cap T_4, 1)$ reduces the proof to (2.4.2.).
Otherwise, $2\in C(u_1)$ and $W_1 = \{1, 4, 2\}$.
It follows from Corollary <ref>(5.1) that $1\in C(u_4)$,
from Corollary <ref>(4) that $12\in C(u_4)$,
and from <ref>(1) that $5\in C(u_4)$, $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
Hence, $C(u_4) = \{2, 4, 5, 1, 12\}\cup T_1$,
and by <ref>(2), we are done.

Hence, $T_2\setminus C(u_4)\ne \emptyset$.
One sees from Corollary <ref>(5) that $T_2\subseteq C(u_3)$.
Then $T_1\setminus C(u_3)\ne \emptyset$.

By Lemma <ref> and Lemma <ref>, hereafter we assume that $t_1 + t_2\le 3$,
and $T_1\cup T_2\subseteq C(u_i)$ for some $i\in [3, 4]$.
Together with Lemma <ref> and Lemma <ref>, $6\le w_1 + w_2\le 4 + t_1 + t_2 \le 7$.
It follows that $w_1 = w_2 = 3$, or $\{w_1, w_2\} = \{3, 4\}$.
One sees from <ref>(1) that $5\in S_u\setminus C(u_3)$.

Assume that $6\not\in S_u$.
It follows from Corollary <ref>(5.1) that $3, 4\in C(u_1)\cap C(u_2)$ and $T_1\cup T_2\subseteq C(u_3)\cap C(u_4)$.
If $5\not\in C(u_1)$,
one sees from $w_1 + w_2\le 7$ that $H$ contains no $(2, 5)_{(u_1, u_2)}$-path and $H$ contains a $(4, 5)_{(u_1, u_4)}$-path,
then $(u u_1, u u_4, u v)\overset{\divideontimes}\to (5, 6, 4)$.
Otherwise, $W_1 = \{1\}\cup [4, 5]$ and $W_2 = [2, 4]$.
It follows from <ref>(1) that $H$ contains a $(5, 4)_{(u_2, u_4)}$-path,
or a $(3, 2)_{(u_1, u_3)}$-path.
If $H$ contains no $(5, 4)_{(u_2, u_4)}$-path,
one sees that $H$ contains a $(3, 2)_{(u_1, u_3)}$-path,
then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, 5, 6, 3)$.
Otherwise, $H$ contains a $(5, 4)_{(u_2, u_4)}$-path.
If $2\not\in C(u_3)$, then $(u u_1, u u_2, u u_4, u v)\overset{\divideontimes}\to (2, 5, 6, 4)$.
Otherwise, $2\in C(u_3)$.
One sees that $|\{1, 3, 4, 5\}| + |[2, 4]| + |\{3, 5, 2\}| + |\{2, 4, 5\}| + 2t_{uv} + t_0  = 2\Delta + 9 + t_0$.
If $C_{12}\subseteq C(u_3)\cup C(u_4)$, then $t_0\ge t_{u v} = \Delta - 2$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + \Delta - 2 = 3\Delta + 7\ge 2\Delta + 14$.
Otherwise, $C_{12}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$.
It follows from <ref>(2) that $7\in C(u_4)$ and $H$ contains a $(7, j)_{(u_4, v_7)}$-path for some $j\in T_4$.
Hence, $C(u_4) = \{2, 4, 5, 7\}\cup T_1\cup T_2$.
It follows from <ref>(2) that $H$ contains a $(4, 7)_{(u_4, v_7)}$-path,
and from <ref> that $7\in S_u\setminus C(u_4)$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + t_0 + |\{7, 7\}| = 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1 + t_2 = 2\Delta + 14$.
Hence, we proceed with the following proposition, or otherwise we are done:

$5\in S_u\setminus C(u_3)$ and $6, 7\in S_u$.

If $1\not\in S_u$ and $3\not\in C(u_4)$,
one sees from Corollary <ref>(7.2) that $3, 4\in C(u_1)$ and $H$ contains a $(3, T_1)_{(u_1, u_3)}$-path, a $(4, T_1)_{(u_1, u_4)}$-path,
then $(u u_1, u u_3, u u_4, u v)\to (T_1, 1, 3, T_4)$.
Hence, we proceed with the following proposition, or otherwise we are done:

If $1\not\in S_u$, then $3, 4\in C(u_1)$, $T_1\subseteq C(u_3)\cap C(u_4)$, and $3\in C(u_4)$.

It follows that $6, 7\in S_u$, $5\in S_u\setminus C(u_3)$, and $1\in S_u$ or $3\in C(u_4)$.
One sees that $c_{12} = t_{uv} - t_1 - t_2 = \Delta - 2 - (t_1 + t_2)\ge \Delta - 2- 3 = \Delta - 5\ge 1$.

Assume that $8\in C_{12}\setminus (C(u_3)\cup C(u_4))$.
It follows from <ref>(2) that $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
If  mult$_{S_u\setminus C(u_3)}(5) = 1$, then it follows from mult$_{S_u}(5) = 1$ that $7\in C(u_3)$ and $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.
Hence, we proceed with the following proposition:

Assume that $8\in C_{12}\setminus (C(u_3)\cup C(u_4))$.
Then $6\in C(u_4)$, $H$ contains a $(6, 8)_{(u_3, u_4)}$-path,
and if mult$_{S_u\setminus C(u_3)}(5) = 1$, then $7\in C(u_3)$ and $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.

If $T_2\setminus C(u_4)\ne \emptyset$, then we are done.

Then $3\in C(u_2)$ and $H$ contains a $(3, T_2)_{(u_2, u_3)}$-path.
It follows from Corollary <ref>(5.1) that $2\in C(u_3)$ or $H$ contains a $(1, 2)_{(u_1, u_2)}$-path,
from Corollary <ref>(5.2) and (3) that $4\in B_1$ and $4\in C(u_2)\cup C(u_3)$.
Let $\sigma_{13} = $ mult$_{S_u}(1) + $mult$_{C(u_1)\cup C(u_4)}(3)$.

When $\sigma_{13}\ge 2$,
one sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 4\}| + |\{2, 3\}| + |\{3, 5\}| + |\{2, 4\}| + |\{2, 4, 5, 6, 7\}| + \sigma_{13} + 2t_{u v} + t_0 = 2\Delta + 9 + \sigma_{13} + t_0
\ge 2\Delta + 9 + 2 + t_1\ge 2\Delta + 11 + 1 = 2\Delta + 12$.
If $C_{12}\subseteq C(u_3)\cup C(u_4)$,
then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + \Delta - 5 = 3\Delta + 7\ge 2\Delta + 14$.
Otherwise, $C_{12}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$.
Assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
It follows from <ref>(1) that $6\in S_u\setminus C(u_4)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + \sigma_{13} + t_0 + |\{6\}|\ge 2\Delta + 10 + \sigma_{13} + t_1\ge 2\Delta + 10 + 2 + 1 = 2\Delta + 13$.
It follows that $\Delta = 7$, $t_1 = 1$, $\sigma_{13} = 2$, $T_2\cap C(u_4)= \emptyset$,
$C_{12}\cap (C(u_3)\cup C(u_4)) = \emptyset$,
and mult$_{S_u}(i) = 1$, $i = 2, 5, 7$, mult$_{S_u}(j) = 2$, $j = 4, 6$, holds.
It follows from <ref> that $7\in C(u_3)$, $H$ contains a $(7, 8)_{(u_3, v_7)}$-path,
and from <ref>(2.1) that $3\in C(u_1)$.
Hence, $W_1 = \{1, 3, 4\}$.
It follows from <ref>(1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path, $5\in C(u_4)$,
from <ref>(1) that $2\in C(u_3)$,
and from <ref>(2) that $H$ contains a $(4, 6)_{(u_1, u_4)}$-path.
If $6\not\in C(u_3)$, then $(u u_1, u v)\overset{\divideontimes}\to (6, 8)$.
Otherwise, $6\in C(u_3)$ and $C(u_3) = [2, 3]\cup [5, 7]\cup T_1\cup T_2$, $4\in C(u_2)$.
Since $d(u_2) + d(u_3)\ge |[2, 4]| + |[2, 3]| + |[5, 7]| + t_{uv} + t_1 = \Delta + 6 + t_1 \ge \Delta + 6 + 1= \Delta + 7$, we have $1\not\in C(u_2)\cup C(u_3)$.
Then $(u u_1, u u_2, u u_3, u v)\overset{\divideontimes}\to (2, T_2, 1, T_4)$.
In the other case, $\sigma_{13}\le 1$.

One sees from <ref> that $1\in S_u$ or $3\in C(u_4)$.
Hence, $3\not\in C(u_1)$.
It follows from Corollary <ref>(5.1) that $1\in C(u_4)$ or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path.
One sees from $1\in S_u$ that $3\not\in C(u_4)$,
and $1\in C(u_4)$ or $H$ contains a $(1, 2)_{(u_2, u_4)}$-path.
Let $(uu_3, uu_4)\to (1, 3)$.
If $1\in C(u_4)$, then $(u u_1, u u_2)\to (2, T_2\cap T_4)$;
if $1\in C(u_2)$, then $u u_1\to T_1$.
These reduce the proof to (2.4.2.).

Now we assume that $T_2\subseteq C(u_4)$.
One sees from <ref> that $5\in S_u\setminus C(u_3)$ and $6, 7\in S_u$.

When $1\not\in S_u$,
it follows from <ref> that $3, 4\in C(u_1)$, $T_1\subseteq C(u_3)$, $3\in C(u_4)$,
and from <ref>(2) that $2\in S_u\setminus C(u_4)$.
If $2, 5\not\in C(u_1)$, one sees from <ref>(1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path,
then $(u u_1, u u_4, u v)\overset{\divideontimes}\to (5, 1, 4)$.
Otherwise, $5/2\in C(u_1)$.
It follows that $t_1 = 2$ and $t_2 = 1$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3, 4\}| + |\{2, a_2\}| + |\{3, 5\}| + |\{2, 4, 3\}| + |\{2, 5, 6, 7\}| + 2t_{uv} + t_0 = 2\Delta + 10 + t_0$.
If $C_{12}\subseteq C(u_3)\cup C(u_4)$, then $t_0\ge c_{12} + t_1 = t_{u v} - t_2 = \Delta - 3$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0\ge 2\Delta + 10 + \Delta - 3 = 3\Delta + 7\ge 2\Delta + 14$.
Otherwise, assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
Hence, $C(u_4) = [2, 4]\cup \{6\}\cup T_1\cup T_2$.
It follows that $4\in B_1$, mult$_{S_u}(4)\ge 2$,
from <ref>(2) that $H$ contains a $(4, 6)_{(u_4, v_6)}$-path,
and from <ref> that $6\in S_u\setminus C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{6\}| + t_0\ge 2\Delta + 11 + t_1 = 2\Delta + 13$.
It follows that mult$_{S_u}(i) = 1$, $i = 5, 7$, $a_2 = 4$, $3\not\in C(u_2)$,
and follows from <ref> that $7\in C(u_4)$,
from <ref>(2.1) that $H$ contains a $(3, 7)_{(u_1, u_3)}$-path.
Thus, by <ref>(2), we are done.
In the other case, $1\in S_u$.
Recall that $\sum_{x\in N(u)\setminus \{v\}}d(x)\le 3\Delta + 6$ for ($A_{8.1}$)–($A_{8.4}$).

Let $\sigma_{[1, 4]} = \sum_{i\in \{1, 3, 4\}}$ mult$_{S_u}(i) + $ mult$_{C(u_1)\cup C(u_3)}(2)$,
$\sigma_{[5, 7]} = \sum_{i\in [6, 7]}$ mult$_{S_u}(i) + $ mult$_{S_u\setminus C(u_3)}(5)$.

Assume that $\sigma_{[1, 4]}\le 3$.
One sees that if $(T_1\cup T_2)\setminus C(u_3)\ne \emptyset$, then $4\in C(u_1)\cup C(u_2)$ and mult$_{S_u}(3)\ge 2$, $\sigma_{[1, 4]}\ge |\{4, 3, 3, 1\}| = 4$,
and that if $|C(u_1)\cap [3, 4]|\ge 2$, or $|C(u_2)\cap [3, 4]|\ge 2$, then $\sigma_{[1, 4]}\ge 4$.
Then $|C(u_1)\cap [3, 4]| = |C(u_2)\cap [3, 4]| = 1$.
One sees that if $a_2 = 3$, then $2\in C(u_1)\cup C(u_3)$, and $\sigma_{[1, 4]}\ge |\{a_1, a_2, 1, 2\}| = 4$.
Hence, $a_2 = 4$.
It follows from <ref>(2) $1\in C(u_2)\cup C(u_4)$,

and from Corollary <ref>(7.2) that $a_1 = 4$.
Thus, $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, 3, T_4)$.
Hence, we proceed with the following proposition, or otherwise we are done:

$1\in S_u$ and $\sigma_{[1, 4]}\ge 4$, $\sigma_{[5, 7]}\ge 3$.

Recall that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + 2t_{uv} + t_0 = 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0$.

If $H$ contains a $(6, 8)_{(u_4, v_6)}$-path for $8\in T_4$, then $6\in S_u\setminus C(u_4)$.

Assume that $6\not\in S_u\setminus C(u_4)$.
One sees from <ref>(1) that $H$ contains a $(4, 6)_{(u_1, u_4)}$-path,
from <ref>(2) that $4\in C(u_1)\cap C(u_2)$,
$3\in C(u_1)\cup C(u_2)$, mult$_{C(u_1)\cup C(u_2)\cup C(u_4)}(3)\ge 2$,
and $2\in C(u_1)$ or $H$ contains a $(2, 3)_{(u_1, u_3)}$-path.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + 2t_{u v} + t_0\ge 6 + |\{4, 4, 3, 3, 2, 1\}| + |\{5, 6, 7\}| + 2t_{u v} + t_0 = 2\Delta + 11 + t_0$,
and $d(u_1) + d(u_2) + d(u_4)\ge 4 + |\{3, 3, 4, 4, 5, 6\}| + 2t_{u v} = 2\Delta + 6$.
It follows that mult$_{C(u_1)}(2) + $mult$_{S_u\setminus C(u_3))}(5) + $mult$_{S_u\setminus C(u_3))}(7)\le 2$.

Case 1. $H$ contains a $(4, j)_{(u_4, v)}$-path for some $j\in [5, 7]$.

Case 1.1. $5\in C(u_4)$ and $H$ contains a $(4, 5)_{(u_3, u_4)}$-path.
It follows from <ref>(1) that $5\in C(u_1)$,
or $H$ contains a $(2, 5)_{(u_1, u_2)}$-path, $2\in C(u_2)$.
Hence, $5\in C(u_1)$, $d(u_1) + d(u_2) + d(u_4)\ge = 2\Delta + 7$,
and then $2\not\in C(u_1)$, $3\in C(u_1)$.
It follows that $W_1 = \{1\}\cup [3, 5]$, $W_2 = [2, 4]$, $C(u_4) = \{2\}\cup [4, 6]\cup T_1\cup T_2$, $C(u_3) = [1, 5]\cup \{7\}$.
Thus, $(u u_1, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, T_4)$
[Or, by Corollary <ref>(7.2), we are done].

Case 2.1. $H$ contains a $(4, 7)_{(u_4, v)}$-path.
Then $d(u_1) + d(u_2) + d(u_4)\ge 2\Delta + 6 + |\{7\}| = 2\Delta + 7$.
It follows that $d(u_3)\le 6$, $2\not\in C(u_1)$, $1, 2\in C(u_4)$, and then $3\in C(u_1)$.
One sees that $5\in C(u_1)$ or $H$ contains a $(4, 5)_{(u_1, u_4)}$-path,
and $d(u_1) + d(u_4)\ge t_{u v} + t_2 + |\{1, 3, 4\}| + |\{2, 4, 6, 7\}| + |\{5\}| = \Delta + 6 + t_2\ge \Delta + 6 + 1 = \Delta + 7$.
It follows that $t_2 = 1$, $3\not\in C(u_4)$, and $W_2 = [2, 4]$,
from <ref>(1) that $T_1\subseteq C(u_3)$.
If $T_2\setminus C(u_3)\ne \emptyset$,
one sees from <ref>(2) that if $5\not\in C(u_4)$, then $T_2\subseteq C(u_3)$,
then $5\in C(u_4)$, and $(u u_1, u u_2, u u_4)\to (5, T_2\cap T_3, 1)$ reduces the proof to (2.4.2.).
Otherwise, $T_2\subseteq C(u_3)$.
Thus, $d(u_3)\ge |\{5\}\cup [1, 3]\cup T_1\cup T_2|\ge 7$.

Case 2. $H$ contains no $(4, i)_{(u_4, v_i)}$-path for each $i\in [5, 7]$.

One sees from <ref>(2) that $3\in C(u_4)$ and $T_4\subseteq C(u_3)$, i.e., $C_{12}\subseteq C(u_3)\cup C(u_4)$, and $t_0\ge c_{12} = \Delta - 2 - t_1 - t_2\ge \Delta - 2 - 3 = \Delta - 5\ge 1$.
Recall that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + t_0\ge 2\Delta + 11 + c_{12}\ge 2\Delta + 11 + 1 = 2\Delta + 12$.
We have $c_{12}\le t_0\le 2$ and $\Delta \le 7$.
When $w_4\ge 5$, one sees that $d(u_4)\ge w_4 + t_1 + t_2\ge 5 + 1 + 1 = 7$,
it follows that $\Delta = 7$, $t_1 = t_2 = 1$, and thus $t_0\ge c_{12} = \Delta - 4\ge 3$.
In the other case, $W_4 = [2, 4]\cup \{6\}$.

Case 2.1. $2\in C(u_1)$.
Then $d(u_1) + d(u_2) + d(u_4)\ge 4 + |\{3, 3, 4, 4, 5, 6, 2\}| + 2t_{uv} = 2\Delta + 7$.
It follows that $d(u_i) = \Delta = 7$, $i = 1, 2, 4$, $d(u_3)\le 6$,
and $\{1, 7, 3, 5\}\subseteq C_{12}\subseteq C(u_3)$, mult$_{S_u\setminus C(u_3)}(5) = 1$.
One sees from $w_1 + w_2\ge |\{1, 2\}| + |\{4, 4, 3, 5, 2\}| = 7$ that $t_1 + t_2 = 3$ and then $c_{12} = 2$.
Hence, $C(u_3) = \{1, 7, 3, 5\}\cup C_{12}$.
Since mult$_{S_u\setminus C(u_3)}(5) = 1$, it follows from Corollary <ref>(4) that $H$ contains a $(7, T_2)_{(u_3, v)}$-path.
If $H$ contains no $(3, 7)_{(u_1, u_3)}$-path, then $(u u_1, u v)\to (7, T_2)$.
Otherwise, $3\in C(u_1)\setminus C(u_2)$ and $H$ contains a $(3, 7)_{(u_1, u_3)}$-path.
Thus, by <ref>(2), we are done.

Case 2.2. $2\not\in C(u_1)$.  Then $3, 5\in C(u_1)$ and $W_1 = \{1\}\cup [3, 5]$.
It follows that $t_1 = 2$, $t_2 = 1$, $C(u_4) = \{2, 3, 4, 6\}\cup T_1\cup T_2$,
and $d(u_i) = \Delta = 7$, $i = 1, 2, 4$, $d(u_3)\le 6$.
Hence $W_1 = \{2, 4, \alpha\}$, $C(u_3) = \{2, 3, 5, \beta\}\cup C_{12}$, where $\{\alpha, \beta\} = \{1, 7\}$.
One sees that $5, 3\not\in C(u_2)$ and $1, 5\not\in C(u_4)$.
Thus, by <ref>(2), we are done.

One sees that if $T_1\cup T_2\subseteq C(u_3)$ and $C_{12}\subseteq C(u_3)\cup C(u_4)$,
then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0\ge 2\Delta + 2 + 4 + 3 + t_{u v}\ge 2\Delta + 9 + \Delta - 2 = 3\Delta + 7$.
Hence, if $T_1\cup T_2\subseteq C(u_3)$, then $C_{12}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$,
and if $C_{12}\subseteq C(u_3)\cup C(u_4)$, then $(T_1\cup T_2)\setminus C(u_3) \ne \emptyset$.

If $H$ contains a $(7, 8)_{(u_4, v_7)}$-path, where $7\in T_1\cup T_2\cup T_4$,
then $7\in S_u\setminus C(u_3)$.

Assume that $7\not\in S_u\setminus C(u_3)$.
One sees from <ref>(1) that $H$ contains a $(3, 7)_{(u_i, u_3)}$-path for some $i\in [1, 2]$,
from <ref>(2) that $3\in C(u_1)\cap C(u_2)$, $T_1\cup T_2\subseteq C(u_3)$,
$4\in B_1$, mult$_{S_u}(4)\ge 2$.

Hence, it suffices to assume that $8\in T_4$ and $3\in C(u_1)$, $H$ contains a $(3, 7)_{(u_1, u_3)}$-path.
One sees from <ref>(2) that $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
It follows from Lemma <ref> that $6\in S_u\setminus C(u_4)$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + 2t_{uv} + t_0
\ge 6 + |\{4, 4, 3, 3, 1\}| + |\{5, 6, 6, 7\}| + 2t_{u v} + t_0 = 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1 + t_2\ge 2\Delta + 11 + 1 + 1 = 2\Delta + 13$.
It follows that $t_1 = t_2 = 1$, $W_1 = \{1, 3, 4\}$,
$2\not\in S_u\setminus C(u_4)$, and mult$_{S_u}(1) = $ mult$_{S_u\setminus C(u_3)}(5) = 1$.
Hence, $5\in C(u_4)$ and $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
Thus, $(u u_1, u u_2)\to (2, 5)$ reduces the proof to (2.3.)
[Or, by <ref>(1), we are done].

If $6\not\in C(u_1)\cup C(u_2)$ and $H$ contains no $(6, i)_{(u_i, u_j)}$-path,
$i = 1, 2$, $j = 3, 4$,
one sees that the same argument applies if $uu_1\to 6$,
and it follows from <ref>(2) that $2\in C(u_1)$ or $H$ contains a $(3, 2)_{(u_1, u_3)}$-path.
Otherwise, we proceed with the following proposition, or otherwise we are done:

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

Assume $6\not\in S_u\setminus C(u_4)$ and $H$ contains no $(4, 6)_{(u_3, u_4)}$-path.
One sees from <ref> that $4\in C(u_1)\cap C(u_2)$,
and from <ref>(3) that mult$_{S_u}(3)\ge 2$,
then $\sigma_{[1, 4]}\ge |\{4, 4, 3, 3, 1\}|\ge 5$.
Together with <ref>, we proceed with the following proposition, or otherwise we are done:

Assume $6\not\in S_u\setminus C(u_3)$ and $H$ contains no $(4, 6)_{(u_3, u_4)}$-path. Then

$4\in C(u_1)\cap C(u_2)$, mult$_{S_u}(3)\ge 2$, and $\sigma_{[1, 4]}\ge |\{4, 4, 3, 3, 1\}|\ge 5$;

if $H$ contains a $(4, 6)_{(u_4, v)}$-path, then $\sigma_{[1, 4]}\ge 6$,
and $2\in C(u_1)$ or $H$ contains a $(3, 2)_{(u_1, u)}$-path.

Assume $7\not\in S_u\setminus C(u_3)$ and $H$ contains no $(3, 7)_{(u_3, u_4)}$-path.
One sees from <ref> that $3\in C(u_1)\cap C(u_2)$, $T_1\cup T_2\subseteq C(u_3)$,
and from <ref>(3) that $4\in B_1$, mult$_{S_u}(4)\ge 2$,
then $\sigma_{[1, 4]}\ge |\{4, 4, 3, 3, 1\}|\ge 5$.
Together with <ref>, we proceed with the following proposition, or otherwise we are done:

Assume $7\not\in S_u\setminus C(u_3)$ and $H$ contains no $(3, 7)_{(u_3, u_4)}$-path. Then

$3\in C(u_1)\cap C(u_2)$, $T_1\cup T_2\subseteq C(u_3)$,
$4\in B_1$, mult$_{S_u}(4)\ge 2$, and $\sigma_{[1, 4]}\ge |\{4, 4, 3, 3, 1\}|\ge 5$;

if $H$ contains a $(3, 7)_{(u_3, v)}$-path, then $\sigma_{[1, 4]}\ge 6$,
and $2\in C(u_1)$ or $H$ contains a $(3, 2)_{(u_1, u)}$-path.

Assume that $8\in C_{12}\setminus (C(u_3)\cup C(u_4))$, and $H$ contains a $(6, 8)_{(u_3, u_4)}$-path.
It follows from Lemma <ref> that $6\in S_u\setminus C(u_4)$.
One sees from Corollary <ref>(4) that if $H$ contains no $(7, 8)_{(u_3, v)}$-path,
then mult$_{S_u\setminus C(u_3)}(5)\ge 2$.
If $H$ contains a $(7, 8)_{(u_3, v)}$-path,
then it follows Lemma <ref> that $7\in S_u\setminus C(u_3)$.
Hence, $\sigma_{[5, 7]}\ge |\{5, 6, 6, 7, 5/7\}|\ge 5$,
and we proceed with the following proposition:

If $C_{12}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$, then $\sigma_{[5, 7]}\ge |\{5, 6, 6, 7, 5/7\}|\ge 5$.

If $T_2\cup T_1\subseteq C(u_4)\cap C(u_3)$, then we are done.

One sees that if $C_{12}\subseteq C(u_3)\cup C(u_4)$,
then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0\ge 2\Delta + 2 + 4 + 3 + t_{u v}\ge 2\Delta + 9 + \Delta - 2 = 3\Delta + 7$.
Hence, $C_{12}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$, say $8\not\in C(u_3)\cup C(u_4)$.
It follows from <ref> that $4\in B_1$ and $4\in C(u_2)\cup C(u_3)$,
from <ref>(2) that $6\in C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path,
from <ref>(2) that $H$ contains a $(4, 5/6/7)_{(u_4, v)}$-path.
One sees from <ref> that $\sigma_{[5, 7]}\ge 5$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0\ge 2\Delta + 2 + 4 + 5 + t_0 = 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1 + t_2 = 2\Delta + 13$.
It follows that $\sigma_{[1, 4]} = 4$, $t_1 = t_2 = 1$, $\sigma_{[5, 7]} = 5$,
and mult$_{S_u}(6) = 2$.
One sees from <ref> that if mult$_{S_u}(7) = 1$ and $H$ contains a $(4, 7)_{(u_4, v_7)}$-path,
then $\sigma_{[1, 4]} \ge 5$ and we are done.
Otherwise, mult$_{S_u}(7) \ge 2$ or $H$ contains no $(4, 7)_{(u_4, v_7)}$-path.

When $W_1 = \{1, 4, 5\}$, it follows from <ref>(1) that $H$ contains a $(4, 6)_{(u_1, u_4)}$-path,
and from <ref>(1) that $5\in C(u_2)$ or $H$ contains a $(4, 5)_{(u_2, u_4)}$-path.
Hence, mult$_{S_u}(7) = 1$ and $H$ contains a $(4, 7)_{(u_4, v_7)}$-path, a contradiction.
In the other case, $5\not\in C(u_1)$.

When $H$ contains no $(7, 8)_{(u_3, v_7)}$-path,
it follows from Corollary <ref>(4) that $5\in B_2\subseteq C(u_2)\cap C(u_4)$, $W_2 = \{2, a_2, 5\}$,
and $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
Then mult$_{S_u}(7) = 1$.
We further assume that $H$ contains a $(4, 6)_{(u_4, v_6)}$-path,
and it follows from <ref>(1) that $6\in C(u_1)$ or $H$ contains a $(3, 6)_{(u_1, u_3)}$-path.
If $H$ contains no $(2, 3)_{(u_1, u_2)}$-path, then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (2, 6, 8)$.
Otherwise, $3\in C(u_1)$ and $H$ contains a $(3, \{2, 6\})_{(u_1, u_3)}$-path.
Thus, $\sigma_{[1, 4]} \ge |\{3, 4, a_2, 2, 1\}| = 5$.

Now assume $H$ contains a $(7, 8)_{(u_3, v_7)}$-path.
It follows from Lemma <ref> that $7\in S_u\setminus C(u_3)$ and mult$_{S_u\setminus C(u_3)}(5) = 1$.
If $5\in C(u_2)$, since $W_1 = \{1, 2, 4\}$, $W_2 = \{2, 5, a_2\}$,
and it follows from <ref>(1) that $H$ contains a $(4, 6)_{(u_1, u_4)}$-path,
and then a $(4, 7)_{(u_4, v_7)}$-path,
then $(uu_1, uv)\overset{\divideontimes}\to (7, 8)$
[Or, by <ref>(1), we are done. ].
Otherwise, $5\in C(u_4)$ and $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
It follows from <ref>(1) that $2\in C(u_1)$ or $H$ contains a $(3, 2)_{(u_1, u_3)}$-path.
Since $\sigma_{[1, 4]} = 4$, $W_1 = \{1, 4, 2\}$.
If $H$ contains no $(4, 6)_{(u_1, u_4)}$-path,
then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (5, 6, 8)$.
Otherwise, $H$ contains a $(4, 6)_{(u_1, u_4)}$-path, a $(4, 7)_{(u_1, u_4)}$-path,
and it follows from <ref>(1) that $W_2 = \{2, 4, 7\}$.
Thus, $(u u_1, u u_4, u v)\overset{\divideontimes}\to (T_1, 1, T_4)$
[Or, by Corollary <ref>(7.2), we are done. ].

Hence, $(T_1\cup T_2)\setminus C(u_3)\ne \emptyset$.
It follows from Corollary <ref>(5.2) and (3) that $3\in B_1\cup B_2$,
and for each $i\in [1, 2]$, $3\in C(u_i)$ or $H$ contains a $(4, 3)_{(u_i, u_4)}$-path.
And for each $j\in [1, 2]$ such that $T_j\setminus C(u_3)\ne \emptyset$, $4\in C(u_j)$ and $H$ contains a $(4, T_j)_{(u_j, u_4)}$-path.

Assume that $2\not\in C(u_1)$ and $H$ contains no $(2, 3)_{(u_1, u_3)}$-path.
One sees from <ref>(1) that $5\in C(u_2)$ or $H$ contains a $(4, 5)_{(u_2, u_4)}$-path,
from <ref>(2) that $1\in C(u_2)$ or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path,
and from <ref>(1) that $5\in C(u_1)$ or $H$ contains a $(4, 5)_(u_1, u_4)$-path.
It follows that mult$_{S_u\setminus C(u_3)}(5)\ge 2$.
If $6\not\in C(u_1)\cup C(u_2)\cup C(u_4)$,
then by <ref>, we are done.
Otherwise, $6, 7\in C(u_1)\cup C(u_2)\cup C(u_4)$.
One sees that $d(u_1) + d(u_2) + d(u_4)\ge 4 + |\{4, 3, 3, 5, 5, 6, 7\}| + 2t_{uv} = 2\Delta + 7$.
It follows that mult$_{C(u_1)\cup C(u_2)}(4) = 1$, mult$_{S_u)}(3) = 2$,
$3\in C(u_2)$ and $H$ contains a $(1, 3)_{(u_1, u_3)}$-path,
and from Corollary <ref>(7.2) that $3\in C(u_1)$, $T_1\subseteq C(u_3)$.
One sees from <ref>(3) that $4\in B_1$.
Thus, $T_2\subseteq C(u_3)$, a contradiction.
In the other case, we proceed with the following proposition, or otherwise we are done:

$2\in C(u_1)$ or $H$ contains a $(2, 3)_{(u_1, u_3)}$-path.

Assume that $C_{12}\subseteq C(u_3)\cup C(u_4)$, i.e., $T_4\subseteq C(u_3)$.
Then $t_0\ge c_{12} = t_{u v} - (t_1 + t_2)\ge t_{uv} - 3 = \Delta - 5$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0 \ge 2\Delta + 2 + |\{4, 3, 3, 1, 2\}| + |[5, 7]| + t_0 = 2\Delta + 10 + t_0$.
When mult$_{S_u}(4)\ge 2$, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + |\{4\}| + t_0 \ge 2\Delta + 11 + c_{12} \ge 2\Delta + 11 + \Delta - 5 = 3\Delta + 6$.
Hence, $t_0 = c_{12} = \Delta - 5$, $(T_1\cup T_2)\cap C(u_3) = \emptyset$,
and mult$_{S_u}(i) =  $mult$_{S_u\setminus C(u_3)}(5) = 1$, $i = 6, 7$.
It follows from Lemma <ref> and Lemma <ref> that $5\in C(u_4)$,
and from <ref>(1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
Thus, $(uu_2, uv)\to (5, T_1)$
[Or, by <ref>(1), we are done. ].
In the other case, mult$_{S_u}(4) = 1$.
It follows that $3\in C(u_4)$ and $T_i\subseteq C(u_i)$ for some $i\in [1, 2]$ such that $4\not\in C(u_i)$.
One sees from <ref>(2) that $5/6/7\in C(u_4)$.
Hence, $t_0\ge c_{12} + t_i\ge t_4 + t_i\ge 2 + 1 = 3$,
and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0 \ge 2\Delta + 10 + 3 = 2\Delta + 13$.
It follows that $t_0 = 3$, $c_{12} = t_4 = 2$, $t_i = 1$,
and mult$_{S_u}(i) =  $mult$_{S_u\setminus C(u_3)}(5) = 1$, $i = 6, 7$.
It follows from Lemma <ref> and Lemma <ref> that $5\in C(u_4)$,
and from <ref>(1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path, $4\in C(u_1)$.
Thus, it follows from <ref>(1) that $T_1\subseteq C(u_3)$.
Since $4\not\in C(u_2)$, $T_2\subseteq C(u_3)$, a contradiction.
Hence, we proceed with the following proposition, or otherwise we are done:

$C_{12}\setminus (C(u_3)\cup C(u_4))\ne \emptyset$.

Assume w.l.o.g. $8\not\in C(u_3)\cup C(u_4)$ and $H$ contains a $(6, 8)_{(u_4, v_6)}$-path.
It follows from Lemma <ref> that $6\in S_u\setminus C(u_4)$,
from <ref> that $\sigma_{[5, 7]}\ge |\{5, 6, 6, 7, 5/7\}|\ge 5$,
and from <ref>(3) that $4\in B_1$ and mult$_{S_u}(4)\ge 2$,
from <ref>(2) that $H$ contains a $(4, 5/6/7)_{(u_4, v)}$-path.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 2 + \sigma_{[1, 4]} + \sigma_{[5, 7]} + t_0\ge 2\Delta + 2 + |\{4, 4, 3, 3, 1, 2\}| + |\{5, 6, 6, 7, 5/7\}| + t_0 = 2\Delta + 13 + t_0$.
It follows that $T_0 = \emptyset$, $4\in C(u_1)\cap C(u_2)$,
and $H$ contains a $(4, T_i)_{(u_, u_4)}$-path, $i = 1, 2$.

\[
S'_u = \biguplus_{i\in \{1, 2, 4\}}(C(u_i)\setminus \{c(u u_i)\}).
\]

One sees from $d(u_1) + d(u_2) + d(u_4)\ge 4 + |\{4, 4, 3, 3, 5, 6, 5/7\}| + 2t_{uv} = 2\Delta + 7$.
Hence, $d(u_i) = 7$, $i = 1, 2, 4$, and $d(u_3)\le 6$, $1, 2\in C(u_3)$.
It follows from Corollary <ref>(7.2) that $3\in C(u_4)$, $H$ contains a $(1, 3)_{(u_3, u_4)}$-path,
and from <ref> that $3\in C(u_1)$.

One sees from <ref> that $5\in C(u_2)$ or $H$ contains a $(4, 5)_{(u_2, u_4)}$-path.
If $5\not\in C(u_1)$, and $H$ contains no $(4, 5)_{(u_1, u_4)}$-path,
then $(u u_1, u u_2, u v)\overset{\divideontimes}\to (5, 1, T_2)$.
Otherwise, $5\in C(u_1)$, or $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
It follows that mult$_{S'_u}(5) = 2$.
One sees
Then $d(u_1) + d(u_4)\ge t_{u v} + t_2 + |\{1, 4, 3\}| + |\{2, 3, 4, 6\}| + |\{5\}| = \Delta + 6 + t_2\ge \Delta + 6 + 1 = \Delta + 7$.
It follows that $t_2 = 1$, $W_2 = \{2, 4, 5\}$, $W_1 \biguplus W_4 = \{1, 4, 3\}\cup \{2, 3, 4, 6\}\cup \{5\}$,
and $C(u_3) = [1, 3]\cup [5, 7]$.
One sees that $5\in C(u_4)$ or $H$ contains a $(4, 5)_{(u_1, u_4)}$-path.
It follows that $H$ contains a $(4, 6)_{(u_4, v)}$-path.
If $H$ contains no $(6, j)_{(u_2, v_6)}$-path, then $(u u_2, u v)\overset{\divideontimes}\to (6, j)$.
Otherwise, $H$ contains a $(6, T_1)_{(u_2, v_6)}$-path.
Since $\{1, 3, 4\}\cup T_2\cup T_4\subseteq C(v_1)$, $T_1\setminus C(v_1)\ne \emptyset$.
It follows from Lemma <ref>(1.2) that $H$ contains a $(7, T_1\cap R_1)_{(u_3, v_1)}$-path,
and from Lemma <ref> that $7\in S'_u$, a contradiction.

This finishes the inductive step for the Case 2.5.1.

(2.5.2.) $c(vu_4) = 6$.
One sees from <ref>(1)–(2) that for each $j\in [3, 4]$,
$j\in C(v_i)$ or $H$ contains a $(j, l)_{(v_1, v_l)}$-path for some $l\in C(v)\setminus \{i\}$, $i\in [1, 2]$.
Hence, together with Corollary <ref>(2.1),
we have the following proposition, or otherwise we are done:

$R_1\cup R_2\ne \emptyset$.

If $R_i = \emptyset$, $i\in [1, 2]$,
then $C(v_i) = \{i, 7\}\cup T_{uv}$ and $H$ contains a $(7, [3, 4])_{(v_1, v_7)}$-path.

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

By <ref> and <ref>,
we have the following proposition, or otherwise we are done:

Assume that $6\not\in C(u_3)$. Then

if $6\not\in C(u_1)$, $H$ contains no $(6, 2)_{(u_2, u_3)}$-path,
then $4\in C(u_1)$, $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path;

if $6\not\in C(u_2)$, $H$ contains no $(6, 1)_{(u_1, u_3)}$-path,
then $4\in C(u_2)$, $H$ contains a $(4, T_2)_{(u_2, u_4)}$-path;

if $H$ contains no $(6, 1)_{(u_3, u_1)}$-path,
then $3\in C(u_2)$ or $H$ contains a $(3, 1/4)_{(u_2, u)}$-path;

if $H$ contains no $(6, 2)_{(u_3, u_2)}$-path,
then $3\in C(u_1)$ or $H$ contains a $(3, 2/4)_{(u_1, u)}$-path;

if $H$ contains no $(6, i)_{(u_3, u_i)}$-path for each $i\in [1, 2]$,
then $3\in B_1\cup B_2$, mult$_{S_u}(3)\ge 2$, and for each $i\in[1, 2]$, $3\in C(u_i)$,
or $H$ contains a $(3, 4)_{(u_i, u_4)}$-path, $4\in C(u_i)$, $3\in C(u_4)$.

Assume that $5\not\in C(u_1)$ and $H$ contains no $(5, 4)_{(u_1, u_4)}$-path.
If $H$ contains no $(5, 2)_{(u_2, u_1)}$-path,
then $u u_1\to 5$ reduces the proof to the Case (2.5.1.).
If $1\not\in C(u_2)$, $H$ contains no $(1, i)_{(u_2, u_i)}$-path for each $i\in \{3, 4\}$,
and $(u u_1, u u_2)\to (5, 1)$ reduces the proof to the Case (2.5.1.).
Hence, and by symmetry, we proceed with the following propositions, or otherwise we are done:

If $5\not\in C(u_1)$ and $H$ contains no $(5, 4)_{(u_1, u_4)}$-path, then

$H$ contains a $(5, 2)_{(u_1, u_2)}$-path.

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

If $5\not\in C(u_2)$ and $H$ contains no $(5, 4)_{(u_2, u_4)}$-path, then

$H$ contains a $(5, 1)_{(u_1, u_2)}$-path.

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

If $6\not\in C(u_1)$ and $H$ contains no $(6, 3)_{(u_1, u_3)}$-path, then

$H$ contains a $(6, 2)_{(u_1, u_2)}$-path.

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

If $6\not\in C(u_2)$ and $H$ contains no $(6, 3)_{(u_2, u_3)}$-path, then

$H$ contains a $(6, 1)_{(u_1, u_2)}$-path.

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

By <ref>(1.1), (2.1), and <ref>(1.1), (2.1),
we have the following proposition:

$5/2/4, 6/2/3\in C(u_1)$, $5/1/4, 6/1/3\in C(u_2)$, and $5, 6\in C(u_1)\cup C(u_2)$.

For each $i\in [1, 2]$, $w_i\ge 3$, $t_i\ge 1$,
and for each $j\in T_i$, there exists a $a_i\in [3, 4]\cap C(u_i)$ such that $H$ contains a $(a_i, j)_{(u_i, u_{a_i})}$-path.

Assume that $w_1\le 2$.
It follows from <ref> that $W_1 = [1, 2]$, $5, 6\in C(u_2)$,
and then $C(u_1) = [1, 2]\cup T_{uv}$, there exists a $a_2\in [3, 4]\cap C(u_2)$,
and from <ref>(1.1) that $H$ contains a $(2, 5)_{(u_1, u_2)}$-path,
from <ref>(1.1) that $H$ contains a $(2, 6)_{(u_1, u_2)}$-path.

One sees clearly that $3, 4\not\in C(u_1)$.
It follows from <ref>(5) that $5\in C(u_4)$,
and <ref>(5) that $6\in C(u_3)$.
Hence, $T_3\ne \emptyset$ and $T_4\ne \emptyset$.
It follows from <ref>(3) and
Corollary <ref>(4) that $T_2\subseteq C(u_4)$, for each $j\in T_4$,
there exists a $a_4\in [1, 3]\cap C(u_4)$, $b_6\in \{1, 2, 5, 7\}\setminus \{a_4\}$ such that $H$ contains a $(a_4, j)_{(u_4, u)}$-path, a $(b_6, j)_{(u_4, v)}$-path,
and from <ref>(3) and
Corollary <ref>(4) that $T_2\subseteq C(u_3)$, for each $j\in T_3$,
there exists a $a_3\in \{1, 2, 4\}\cap C(u_3)$, $b_5\in \{1, 2, 6, 7\}\setminus \{a_3\}$ such that $H$ contains a $(a_3, j)_{(u_3, u)}$-path, a $(b_5, j)_{(u_3, v)}$-path.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |[1, 2]| + |\{2, 5, 6, a_2\}| + |\{3, 5, 6, a_3, b_5\}| + |\{4, 5, 6, a_4, b_6\}| + 2t_{uv} + t_0\ge 2\Delta + 12 + 2 = 2\Delta + 14$, a contradiction.

Assume that $1\not\in S_u$.
It follows from Corollary <ref>(7.2) that $3, 4\in C(u_1)$, $T_1\subseteq C(u_3)\cap C(u_4)$,
from <ref>(3) that $2\in C(u_1)$ or $H$ contains a $(2, i)_{(u_2, u_i)}$-path,
from <ref>(1.1) that $5\in C(u_1)$ or $H$ contains a $(4, 5)_{(u_1, u_4)}$-path,
from <ref>(2.1) that $5\in C(u_2)$ or $H$ contains a $(4, 5)_{(u_2, u_4)}$-path,
and from <ref>(1.1) that $6\in C(u_1)$ or $H$ contains a $(3, 6)_{(u_1, u_3)}$-path,
from <ref>(2.1) that $6\in C(u_2)$ or $H$ contains a $(3, 6)_{(u_2, u_3)}$-path.
It follows that mult$_{S_u\setminus C(u_3)}(5)\ge 2$ and mult$_{S_u\setminus C(u_4)}(6)\ge 2$.
Since the same argument applies if $7\not\in S_u$ and $uu_2\to 7$, $7\in S_u$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3, 4\}| + |\{2, a_2\}| + |\{3, 5\}| + |\{4, 6\}| + |\{2, 7\}| + 2\times |\{5, 6\}| + 2t_{u v} + t_0 = 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1$.
One sees from $1, 3, 4, 2, a_2, 5, 6\in W_1\biguplus W_2$ that $t_1 + t_2\ge 3$.
It follows that $T_2\setminus T_0\ne \emptyset$.
Assume w.l.o.g. $T_2\setminus C(u_4)\ne \emptyset$.
Then $3\in C(u_2)$ and $T_2\subseteq C(u_3)$.
It follows from Corollary <ref>(5.1) that $2\in C(u_3)$,
or $H$ contains a $(2, i)_{(u_3, u_i)}$-path for each $i\in \{1, 4\}$,
and from Corollary <ref>(5.2) that mult$_{S_u}(4)\ge 2$.
One sees that if $2\not\in C(u_1)\cup C(u_4)$,
then $H$ contains a $(2, 4)_{(u_1, u_4)}$-path and a $(2, 4)_{(u_3, u_4)}$-path, a contradiction.
Hence, $2\in C(u_1)\cup C(u_3)$, and thus,
$d(u_1) + d(u_2) + d(u_3)\ge |\{1, 2, 3, 5\}| + 2\times |\{3, 4, 6\}| + |\{2, 5\}| + 2t_{u v} = 2\Delta + 8$.
Since the same argument applies if $7\not\in S_u$ and $u u_1\to 7$,
we proceed with the following proposition, or otherwise we are done:

$1, 2\in S_u$, and $7\in S_u$

For each $i\in [1, 2]$, $4\le w_i\le 5$, $t_i\ge 2$ and $\min\{w_1, w_2\} = 4$.

Assume that $w_1 = 3$.
By <ref>, assume w.l.o.g. $W_1 = \{1, 4, 6/2/3\}$.
Below we distinguish three cases where $W_1 = \{1, 4, i\}$, $i = 3, 6, 2$, respectively.

Case 1. $W_1 = \{1, 4, 3\}$.
It follows from <ref>(1.1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path, $5\in C(u_4)$,
from <ref>(1.1) that $H$ contains a $(6, 3)_{(u_1, u_3)}$-path, $6\in C(u_3)$,
and from <ref>(2.1) that $5\in C(u_2)$,
from <ref>(2.1) that $6\in C(u_2)$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 3, 4\}| + |\{2, 5, 6, a_2\}| + |\{3, 5, 6\}| + |\{4, 6, 5\}| + 2t_{u v} + t_0 = 2\Delta + 9 + t_0$.

Case 1.1. $1\not\in C(u_3)$ and $H$ contains no $(1, i)_{(u_3, u_i)}$-path for each $i\in \{2, 4\}$.
One sees from Corollary <ref>(7.2) that $T_1\subseteq C(u_4)$.
If $H$ contains no $(6, j)_{(u_1, u_4)}$-path for some $j\in T_2$,
then $(u u_1, u u_3, u v)\overset{\divideontimes}\to (6, 1, j)$.
Otherwise, $H$ contains a $(6, T_2)_{(u_1, u_4)}$-path and $T_2\subseteq C(u_4)$.
If there exists a $j\in (T_2\cup T_1)\setminus C(u_3)$,
then $(u u_1, u u_3)\to (6, j)$ reduces the proof to Case (2.5.1.).
Otherwise, $T_1\cup T_2\subseteq C(u_3)$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + t_0 + |\{1, 2, 7\}|\ge 2\Delta + 12 + t_1 + t_2\ge 2\Delta + 12 + 1 + 2 = 2\Delta + 15$,
and none of ($A_{8.1}$)–($A_{8.4}$) holds.

Case 1.2. $1\in C(u_3)$ or $H$ contains a $(1, 2/4)_{(u_3, u)}$-path,
and $1\in C(u_4)$ or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path.
It follows that there exists a $a_3\in \{1, 2, 4\}\cap C(u_3)$ and a $a_4\in \{1, 2, 3\}\cap C(u_4)$.
One sees clearly that $1\in C(u_3)\cup C(u_4)$, say $1\in C(u_3)$,
and if $1, 3\not\in C(u_4)$, i.e., $a_4 = 2$, then $1\in C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 9 + t_0 + |\{1\}| + |\{a_4\}| + |\{7, 2/1/3\}| = 2\Delta + 13 + t_0$.
It follows that $t_0 = 0$, $C(u_2)\cap [3, 4] = \{a_2\}$,
and $4\not\in C(u_3)$.
Then $T_2\subseteq C(u_{a_2})$,
and it follows from Corollary <ref>(5.2) that mult$_{S_u}(i)\ge 2$, where $\{a_2, i\} = [3, 4]$.
Hence, $a_2 = 4$, $a_4 = 3$, $1\not\in C(u_2)\cup C(u_4)$,
and $T_2\subseteq C(u_4)$, $T_2\cap C(u_3) = \emptyset$.
Thus, $(u u_1, u u_3, u u_4, u v)\overset{\divideontimes}\to (5, 4, 1, T_2)$.

Case 2. $W_1 = \{1, 4, 6\}$.
It follows from <ref>(1.1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path, $5\in C(u_4)$,
from <ref>(2.1) that $5\in C(u_2)$,
and from Corollary <ref>(5.1) that $1\in C(u_4)$, or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path.
When $6\not\in C(u_2)\cup C(u_3)$,
one sees from <ref>(2.1) that $1\in C(u_2)$ and $H$ contains a $(1, 6)_{(1, 2)}$-path.
Then it follows from <ref>(2.2) that $2\in C(u_4)$,
from <ref>(2) that $T_2\subseteq C(u_4)$,
and from <ref>(4) that $3\in C(u_4)$.
One sees from $1, 2, 5, a_2\in C(u_2)$ that $t_2\ge 2$.
Thus, $d(u_4)\ge |[2, 6]| + t_1 + t_2\ge 5 + 1 + 2 = 8$.

In the case, $6\in C(u_2)\cup C(u_3)$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 4, 6\}| + |\{2, 5, a_2\}| + |\{3, 5\}| + |\{4, 6, 5\}| + |\{1, 2, 6, 7\}| + 2t_{u v} + t_0 = 2\Delta + 11 + t_0$.
We distinguish whether or not $T_2\subseteq C(u_4)$:

Case 1.1. $T_2\setminus C(u_4)\ne \emptyset$.
Then $3\in C(u_2)$, $T_2\subseteq C(u_3)$,
and it follows from Corollary <ref>(5.2) that $4\in C(u_2)\cup C(u_3)$.
One sees that $T_1\subseteq C(u_3)$ or $3\in C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + |\{4\}| + |\{3/T_1\}|\ge 2\Delta + 13 + t_0\ge 2\Delta + 13$.
It follows that mult$_{S_u}(i) = 1$, $i = 1, 2$, mult$_{S_u}(4) = 2$.
One sees from Corollary <ref>(5.1) that $2\not\in C(u_4)$ or $H$ contains a $(2, 4)_{(u_3, u_4)}$-path.

It follows from <ref>(3) that $1\in C(u_2)$ or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path.
Hence, $1\in C(u_2)\cup C(u_4)$.

If $1\in C(u_2)$, one sees that $2\in C(u_4)$, $4\in C(u_3)$,
then $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 4, 2, 1)$ reduces the proof to Case 2.3.
Otherwise, $1\in C(u_4)$, one sees that $4\in C(u_2)$, $2\in C(u_3)$,
then $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 1, 2, 4)$ reduces the proof to Case 2.4.3.

Case 1.2. $T_2\subseteq C(u_4)$.
When $3\not\in C(u_4)$, it follows from Corollary <ref>(5.2) that $T_1\cup T_2\subseteq C(u_3)$,
then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + t_0\ge 2\Delta + 11 + t_1 + t_2\ge 2\Delta + 11 + 1 + 1 = 2\Delta + 13$.
It follows that $t_1 = t_2 = 1$, mult$_{S_u}(i) = 1$, $i = 1, 2$, and $4\not\in C(u_3)$, $3\not\in C(u_4)$.
One sees that $W_2 = \{2, 5, a_2\}$.
It follows from <ref> that $a_2 = 3$.
Hence, by Corollary <ref>(5.1), $1\in C(u_4)\setminus C(u_3)$ and $2\in C(u_3)\setminus C(u_4)$.
Thus, by <ref>, we are done.
In the other case, $3\in C(u_4)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 11 + t_0 + |\{3\}| = 2\Delta + 12 + t_0$.
It follows that $(T_1\cup T_2)\setminus C(u_3)\ne \emptyset$,
and from Corollary <ref>(5.2) and (3) that $3\in B_1$.
One sees that $4\in C(u_2)$ or $T_2\subseteq C(u_2)$.
Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 12 + |\{4/T_2\}| = 2\Delta + 13$.
It follows that mult$_{S_u}(2) = 1$, $4\not\in C(u_3)$,
and if $4\in C(u_2)$, then $H$ contains a $(4, T_2)_{(u_2, u_4)}$-path.
If $2\not\in C(u_4)$, then $(u u_2, u u_3, u u_4)\to (T_2, 4, 2)$ reduces the proof to Case 2.4.3.
Otherwise, $2\in C(u_4)$ and $C(u_4) = [2, 6]\cup T_1\cup T_2$.
One sees clearly that $W_2 = \{2, 3, 5\}$ and $C(u_2) = \{3, 5, 6, 1, 7\}\cup T_2$.
Thus, $(u u_2, u u_3, u v)\overset{\divideontimes}\to (T_2, 2, T_1)$
[Or, by Corollary <ref>(7.2), we are done. ].

Case 3. $W_1 = \{1, 4, 2\}$.
Then $T_1\subseteq C(u_1)$.
It follows from Corollary <ref>(5.1) that $1\in C(u_4)$,
or $H$ contains a $(1, 2/3)_{(u_4, u)}$-path,
and from <ref>(1.1) that $H$ contains a $(4, 5)_{(u_1, u_4)}$-path,
or a $(2, 5)_{(u_1, u_2)}$-path.
Together with <ref>(2.1) and <ref>(1), we have $5\in C(u_2)\cap C(u_4)$.
It follows from <ref>(1) that $H$ contains a $(2, 6)_{(u_1, u_2)}$-path,
and from <ref>(1.2) that $1\in C(u_2)$, or $H$ contains a $(1, 3/4)_{(u_2, u)}$-path.
Hence, $1\in C(u_2)\cup C(u_4)$.
One sees that $t_2\ge 2$ and $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, 2, 4\}| + |\{2, 5, 6, a_2\}| + |\{3, 5\}| + |\{4, 6, 5\}| + |\{1, 7\}| + 2t_{u v} + t_0 = 2\Delta + 10 + t_0$.

Further, if $3\in C(u_4)$, then $T_2\setminus C(u_4)\ne \emptyset$,
or else $d(u_2) + d(u_4)\ge t_{uv} + t_1 + w_1 + w_4\ge \Delta + 8$.
One sees from <ref>(5) that $6\in C(u_3)$ or $3\in C(u_4)$.

If $T_2\subseteq C(u_4)$, one sees that $3\not\in C(u_4)$, then $6\in C(u_3)$,
and it follows from Corollary <ref>(5.1) that $T_1\cup T_2\subseteq C(u_3)$,
thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0 + |\{6\}|\ge 2\Delta + 11 + t_1 + t_2\ge 2\Delta + 11 + 1 + 2 = 2\Delta + 14$.
Otherwise, $T_2\setminus C(u_4)\ne \emptyset$.
Then $3\in C(u_3)$, and it follows from Corollary <ref>(5.2) that $4\in C(u_2)\cup C(u_3)$,
from Corollary <ref>(5.1) that $2\in C(u_3)\cup C(u_4)$ or $1\in C(u_3)$,
and $3\in C(u_4)$ or $T_1\subseteq C(u_3)$.
Hence, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 2\Delta + 10 + t_0 + |\{4, 1/2, 3/T_1\}/|\ge 2\Delta + 13$.
It follows that $6\not\in C(u_3)$, $3\in C(u_4)$, and $t_0 = 0$, i.e., $T_1\cap C(u_3) = T_2\cap C(u_4) = \emptyset$,
mult$_{S_u}(1) + $mult$_{S_u}(2) = 3$, and mult$_{S_u\setminus C(u_4)}(4) = 2$.
If $1\in C(u_2)$, one sees that $2\in C(u_4)$, $4\in C(u_3)$,
then $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 4, 2, 1)$ reduces the proof to Case 2.3.
Otherwise, $1\in C(u_4)$ and $H$ contains a $(1, 4)_{(u_2, u_3)}$-path.
One sees that $4\in C(u_2)$, $2/1\in C(u_3)$.
Then $(u u_1, u u_2, u u_3, u u_4, u v)\overset{\divideontimes}\to (6, 1, 4, 2, T_2)$.

Hence, for each $i\in [1, 2]$, $w_i\ge 4$.
Together with $w_1 + w_2 + t_{uv}\le d(u_1) + c(u_2)\le \Delta + 7$,
$w_1 + w_2\le 9$.
It follows that $4\le w_1, w_2\le 5$ and $\min\{w_1, w_2\} = 4$.

Hereafter, for each $i\in [1, 2]$, $w_i\ge 4$ and $t_i\ge 2$.

For each $i\in [3, 4]$, $(T_1\cup T_2)\setminus C(u_i)\ne \emptyset$.

Assume that $T_1\cup T_2\subseteq C(u_4)$.
When $T_1\cup T_2\subseteq C(u_3)$,
$\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{1, a_1\}| + |\{2, a_2\}| + |\{3, 5\}| + |\{4, 6\}| + |\{5, 6, 1, 2, 7\}| + 2t_{u v} + t_0\ge 2\Delta + 9 + t_1 + t_2\ge 2\Delta + 9 + 2 + 2 = 2\Delta + 13$.
It follows that mult$_{S_u\setminus C(u_3)}(5) = $ mult$_{S_u\setminus C(u_4)}(6) = 1$, $t_0 = t_1 + t_2 = 2 + 2 = 4$,
and from Corollary <ref>(4) that $1/2/7/6\in C(u_3)$ and $1/2/7/5\in C(u_4)$.
Thus, $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge w_1 + w_2 + w_3 + w_4 + 2t_{uv} + t_0\ge 2\times 4 + 2\times 3 + 2t_{uv} + 4 = 2\Delta + 14$.
In the other case, $(T_1\cup T_2)\setminus C(u_3)\ne \emptyset$.
It follows that $4\in C(u_1)\cup C(u_2)$,
and from  Corollary <ref>(5.1) that mult$_{S_u}(3)\ge 2$.

One sees from Corollary <ref>(7.2) that if $1/2/3\not\in C(u_4)$,
then $T_1\cup T_2\subseteq C(u_3)$.
Hence, $1/2/3\in C(u_4)$, $C(u_4) = \{4, 6, 1/2/3\}\cup T_1\cup T_2$, and $t_1 = t_2 = 2$.
It follows that $w_i = 4$, $d(u_i) = \Delta$, $i = 1, 2$.

Case 1. $C(u_4) = \{4, 6, 3\}\cup T_1\cup T_2$.
It follows from <ref>(5) that mult$_{S_u}(4)\ge 2$,
from Corollary <ref>(4) that mult$_{S_u\setminus C(u_4)}(6)\ge 2$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + 2\times |\{3, 4, 6\}| + |\{5, 1, 2, 7\}| + 2t_{uv} + t_0\ge 2\Delta + 12 + t_0$.
It follows that $T_i\setminus C(u_3)\ne \emptyset$, $i = 1, 2$,
and from <ref>(1–2) that $5\in C(u_1)\cap C(u_2)$.
Hence, assume w.l.o.g. $W_1 = \{1, 4, 5, 6\}$, $W_2 = [2, 5]$, $C(u_3) = [1, 3]\cup [5, 7]$,
and from Corollary <ref>(5.1) that $H$ contains a $(1, 3)_{(u_3, u_4)}$-path.
Thus, by <ref>(3), we are done.

Case 2. $C(u_4) = \{4, 6, 1\}\cup T_1\cup T_2$.
Then $3\in C(u_1)\cap C(u_2)$.
When $H$ contains no $(3, j)_{(u_1, u_3)}$-path for some $j\in T_1$,
then $4\in C(u_1)$, and it follows from <ref>(1) that $5\in C(u_1)$,
from <ref>(1) that $7\in C(u_1)$,
thus, $W_1\ge |\{1, 3, 4, 5, 7\}| = 5$.
In the other case, $H$ contains a $(3, T_1)_{(u_1, u_3)}$-path.

One sees from Corollary <ref>(3–4) that  mult$_{S_u}(4)\ge 2$ or mult$_{S_u\setminus C(u_4)}(6)\ge 2$.

Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge 6 + |\{3, 3, 4, 6, 5, 1, 2, 7, 4/6\}| + 2t_{uv} + t_0\ge 2\Delta + 11 + t_1 = 2\Delta + 11 + 2 = 2\Delta + 13$.
It follows that $T_2\cap C(u_3) = \emptyset$ and then $4\in C(u_2)$.
One sees from Corollary <ref>(7.2) that $2\in C(u_1)$,
and from <ref>(2.1) that $5\in C(u_2)$,
from <ref>(2) that $7\in C(u_1)\cup C(u_2)$.
Thus, $w_1 + w_2\ge |[2, 5]| + |[1, 3]| + |[6, 7]| = 9 > 8$, a contradiction.

Otherwise, $(T_1\cup T_2)\setminus C(u_4)\ne \emptyset$,
and likewise, $(T_1\cup T_2)\setminus C(u_3)\ne \emptyset$.

Assuming w.l.o.g. $T_2\setminus C(u_4)\ne \emptyset$.
Then $3\in C(u_2)$, $H$ contains a $(3, T_2)_{(u_2, u_3)}$-path.
One sees from Corollary <ref>(5) that $T_2\subseteq C(u_3)$.
By Lemma <ref>, $T_1\setminus C(u_3)\ne \emptyset$ and then $T_1\subseteq C(u_4)$.
Then $4\in C(u_1)$, $H$ contains a $(4, T_1)_{(u_1, u_4)}$-path.
It follows from Corollary <ref>(5.2) and (3) that $3, 4\in B_1\cup B_2$ and mult$_{S_u}(i)\ge 2$, $i = 3, 4$.
One sees that $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge |\{6\}| + 2\times |\{3, 4\}| + |\{1, 2, 7, 5, 6\}| + 2t_{uv} + t_0 = 2\Delta + 11 + t_0$.

When mult$_{S_u\setminus C(u_3)}(5)\ge 2$ and mult$_{S_u\setminus C(u_4)}(6)\ge 2$,
then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge +  2\Delta + 11 + t_0 + |\{5, 6\}| = 2\Delta + 13 + t_0$.
It follows that mult$_{S_u}(i) = 1$, $i = 1, 2$, mult$_{S_u}(j) = 2$, $j = 1, 2$.
If $1, 2\in C(u_3)\cup C(u_4)$,
one sees from  Corollary <ref>(5.1) that $1\not\in C(u_3)$ or $H$ contains a $(1, 3)_{(u_3, u_4)}$-path,
and $2\not\in C(u_4)$ or $H$ contains a $(2, 4)_{(u_3, u_4)}$-path.
Thus, by <ref>(3), we are done.
Otherwise, assume w.l.o.g. $1\in C(u_2)$.
Then $2\in C(u_4)$, $4\in C(u_3)$,
and $(u u_1, u u_2, u u_3, u u_4)\to (T_1, 4, 2, 1)$ reduces the proof to Case 2.3.
In the other case, assume w.l.o.g. mult$_{S_u\setminus C(u_3)}(5) = 1$.

If $5\in C(u_2)$, one sees from <ref>(1.1) that $H$ contains a $(2, 5)_{(u_1, u_2)}$-path,
then by <ref>(1), we are done.
Otherwise, $5\in C(u_1)$,
and it follows from <ref>(2.1) that $1\in C(u_2)$, $H$ contains a $(1, 5)_{(u_1, u_2)}$-path.
If $H$ contains no $(5, j)_{(u_1, u_4)}$-path for some $j\in T_1\cap T_3$,
then $(u u_1, u u_4)\to (T_1\cap T_3, 5)$ reduces the proof to Case 2.4.1.
Otherwise, $H$ contains a $(5, T_1\cap T_3)_{(u_1, u_4)}$-path.
If $1\not\in C(u_4)$ and $H$ contains no $(1, 3)_{(u_3, u_4)}$-path,
then $(u u_1, u u_2, u u_4)\to (T_1\cap T_3, 5, 1)$ reduces the proof to Case 2.4.2.
Otherwise, $1\in C(u_4)$ or $H$ contains a $(1, 3)_{(u_3, u_4)}$-path.
Likewise, if mult$_{S_u\setminus C(u_4)}(6) = 1$, then $6\in C(u_2)$, $2\in C(u_1)$,
and $2\in C(u_3)$ or $H$ contains a $(2, 4)_{(u_3, u_4)}$-path.

Then $\sum_{x\in N(u)\setminus \{v\}}d(x)\ge +  2\Delta + 11 + t_0 + |\{1, 6/2\}| = 2\Delta + 13 + t_0$.
It follows that $t_0 = 0$, $(C_{12}\cup T_2)\cap C(u_4) = \emptyset$, $(C_{12}\cup T_1)\cap C(u_3) = \emptyset$,
and mult$_{S_u}(i) = 2$, $i = 3, 4$, and mult$_{S_u}(7) = 1$.

One sees that $5\not\in C(u_4)$ and $C_{12}\cap C(u_3) = \emptyset$.
Now we want to show that for some $j^*\in T_2\cup C_{12}\cup \{3\}$,
$H$ contains no $(i, j)_{(u_4, v)}$-path for each $i\in 1, 2, 7$.

Assume that $H$ contains a $(7, j)_{(u_4, v_7)}$-path for some $j\in T_2$.
If $H$ contains no $(4, 7)_{(u_1, u_4)}$-path,
then $(u u_1, u v)\overset{\divideontimes}\to (7, j)$.
Otherwise, $H$ contains a $(4, 7)_{(u_1, u_4)}$-path.
Thus, by <ref>(2), we are done.
Hence, $H$ contains no $(7, j)_{(u_4, v_7)}$-path for each $j\in T_2$.

Further, we assume that $H$ contains a $(2, T_2)_{(u_4, v_2)}$-path.
If $3\in C(v_2)$, one sees from <ref>(1) that $C_{12}\setminus C(v_2) \ne \emptyset$,
and it follows from Lemma <ref>(1.2) that $H$ contains a $(7, C_{12}\cap R_2)_{(v_2, v_7)}$-path.
Then $j^*\in C_{12}\cap R_2$.
Otherwise, $3\not\in C(v_2)$.
Since $3\in B_1$, $3\in C(u_1)\setminus C(u_4)$.
One sees from <ref>(2) that $H$ contains a $(3, 7)_{(v_2, v_7)}$-path.
Then $j^* = 3$.
Hence, by Corollary <ref>(4) or $vu_4\to 3$,
we have $6\in B_1\cup B_2$, and mult$_{S_u}(6) = 2$.
It follows that mult$_{S_u}(2) = 1$.

One sees from <ref>(2.2) that $2\in C(u_1)$ or $H$ contains a $(2, 3/4)_{(u_1, u)}$-path.
Together with $2\in C(u_3)$ or $H$ contains a $(2, 1/4)_{(u_3, u)}$-path, $2\in C(u_1)\cup C(u_3)$.
If $2\in C(u_2)$, one sees that $1\in C(u_3)$, $3\in C(u_4)$,
then $(u u_1, u u_2, u u_3, u u_4)\to (3, T_2, 2, 1)$ reduces the proof to Case 2.3.
Otherwise, $2\in C(u_3)$ and then $3\in C(u_1)$, $1\in C(u_4)$.
One sees from <ref>(1) that $7\in C(u_4)$,
and from <ref>(2) that $H$ contains a $(4, 7)_{(u_3, u_4)}$-path.
Then $(u u_1, u u_2)\to (7, 5)$ reduces the proof to Case (2.5.1.)
This finishes the inductive step for the Case (2.5.1.)

This finishes the inductive step for the case where $G$ contains the configuration ($A_6$)–($A_8$),
and thus completes the proof of Theorem <ref>.

§ CONCLUSIONS

Proving the acyclic edge coloring conjecture (AECC) on general graphs has a long way to go,
but on planar graphs it is very close to reach with recent efforts.
In particular, the AECC has been confirmed true for planar graphs without $i$-cycles, for each $i \in \{3, 4, 5, 6\}$,
and for planar graphs that containing no intersecting $3$-cycles.
In this paper, we deal with the general planar graph.
Prior to our work, their acyclic chromatic index was shown to be at most $\Delta + 6$,
and in this work we made another step that $a'(G) \le \Delta + 5$.
The main technique is a discharging method to show that at least one of the several specific local structures must exist in the graph,
followed by an induction on the number of edges in the graph.

The discharging method is classic;
different from the previous work, we do not always validate $\omega'(x)\ge 0$ for every element in $x\in V(H)\cup F(H)$ independently,
but sometimes collectively for subsets of elements.
For example, in Discharging-Lemma <ref> we validate $\sum_{x\in \{v\}\cup F_3(v)}\omega'(x)\le 0$,
and in Discharging-Lemma <ref>, when $u$ is a $4$-vertex in $V(H)$ adjacent to a vertex $v$,
we validate $\sum_{x\in \{u, v\}\cup (F_3(u)\cup F_3(v))}\omega'(x)\le 0$ if $d_H(v) = 4$,
and $\sum_{x\in \{u\}\cup F_3(u)}\omega'(x)\le 0$ if $d_H(v)\ge 5$.

Some notations defined and some properties unearthed in the paper are of importance during the proof.
For example, the notations of $T_i$, $W_i$, $C_{ij}$, $S_u$, $T_0$,
and if $C(u)\cap C(v) = \{1, 2\}$, then $C(u_1) = W_1\cup T_2\cup C_{12}$, $C(u_2) = W_2\cup T_1\cup C_{12}$,
and the characteristic stated in Lemma <ref>, Corollary <ref> and Corollary <ref>.

Surely, our detailed construction and proofs are built on top of several existing works,
and we believe our new design of the local structures and the associated discharging method can be helpful
for further lowering the upper bound of the acyclic chromatic number on planar graphs, or addressing the AECC eventually.
On the other hand, for general graphs, improvements from the approximation algorithm perspective,
that is, to reduce the coefficient for $\Delta$ in the upper bound (from the currently best $3.74$), are impactful pursuit.

[1]
N. Alon, C. Mcdiarmid, and B. Reed.
Acyclic coloring of graphs.
Random Structures & Algorithms, 2:277–288, 1991.

[2]
N. Alon, B. Sudakov, and A. Zaks.
Acyclic edge colorings of graphs.
Journal of Graph Theory, 37:157–167, 2001.

[3]
L. D. Andersen, E. Máčajová, and J. Mazák.
Optimal acyclic edge-coloring of cubic graphs.
Journal of Graph Theory, 71:353–364, 2012.

[4]
M. Basavaraju and L. S. Chandran.
Acyclic edge coloring of graphs with maximum degree $4$.
Journal of Graph Theory, 61:192–209, 2009.

[5]
M. Basavaraju, L. S. Chandran, N. Cohen, F. Havet, and T. Müller.
Acyclic edge-coloring of planar graphs.
SIAM Journal on Discrete Mathematics, 25:463–478, 2011.

[6]
L. Esperet and A. Parreau.
Acyclic edge-coloring using entropy compression.
European Journal of Combinatorics, 34:1019–1027, 2013.

[7]
J. Fiamcik.
The acyclic chromatic class of a graph.
Math. Slovaca, 28:139–145, 1978.

[8]
I. Giotis, L. Kirousis, K. I. Psaromiligkos, and D. M. Thilikos.
Acyclic edge coloring through the lovász local lemma.
Theoretical Computer Science, 665:40–50, 2017.

[9]
M. Molloy and B. Reed.
Further algorithmic aspects of the local lemma.

[10]
S. Ndreca, A. Procacci, and B. Scoppola.
Improved bounds on coloring of graphs.
European Journal of Combinatorics, 33:592–609, 2012.

[11]
Q. Shu, Y. Chen, S. Han, G. Lin, E. Miyano, and A. Zhang.
Acyclic edge coloring conjecture is true on planar graphs without
intersecting triangles.
In Proceedings of the 16th International Conference on Theory
and Applications of Models of Computation (TAMC 2020), LNCS 12337, pages
426–438, 2020.

[12]
Q. Shu, Y. Chen, S. Han, G. Lin, E. Miyano, and A. Zhang.
Acyclic edge coloring conjecture is true on planar graphs without
intersecting triangles.
Theoretical Computer Science, 882:77–108, 2021.

[13]
Q. Shu, W. Wang, and Y. Wang.
Acyclic edge coloring of planar graphs without $5$-cycles.
Discrete Applied Mathematics, 160:1211–1223, 2012.

[14]
Q. Shu, W. Wang, and Y. Wang.
Acyclic chromatic indices of planar graphs with girth at least $4$.
Journal of Graph Theory, 73:386–399, 2013.

[15]
Q. Shu, Y. Wang, Y. Ma, and W. Wang.
Acyclic edge coloring of $4$-regular graphs without 3-cycles.
Bulletin of the Malaysian Mathematical Sciences Society,
42:285–296, 2019.

[16]
S. Skulrattankulchai.
Acyclic colorings of subcubic graphs.
Information Processing Letters, 92:161–167, 2004.

[17]
V. G. Vizing.
On an estimate of the chromatic class of a $p$-graph.
Discret Analiz, 3:25–30, 1964.

[18]
T. Wang and Y. Zhang.
Further result on acyclic chromatic index of planar graphs.
Discrete Applied Mathematics, 201:228–247, 2016.

[19]
W. Wang, Y. Ma, Q. Shu, and Y. Wang.
Acyclic edge coloring of $4$-regular graphs (ii).
Bulletin of the Malaysian Mathematical Sciences Society,
42:2047–2054, 2019.

[20]
W. Wang, Q. Shu, and Y. Wang.
Acyclic edge coloring of planar graphs without $4$-cycles.
Journal of Combinatorial Optimization, 25:562–586, 2013.

[21]
W. Wang, Q. Shu, and Y. Wang.
A new upper bound on the acyclic chromatic indices of planar graphs.
European Journal of Combinatorics, 34:338–354, 2013.

[22]
Y. Wang, Q. Shu, J.-L. Wu, and W. Zhang.
Acyclic edge coloring of planar graphs without a $3$-cycle adjacent
to a $6$-cycle.
Journal of Combinatorial Optimization, 28:692–715, 2014.