© Wuhan University 2024

0 Introduction

In this paper, we consider only finite, undirected and simple graphs. Let $G=\left(V\left(G\right),E\left(G\right)\right)$ be a graph with vertex set $V\left(G\right)$ and edge set $E\left(G\right)$. For a set $X$, we use $\left|X\right|$ to denote the cardinality of $X$. For a subgraph $H$ of $G$ and a vertex $v\in V\left(G\right)$, we denote by $d_H\left(v\right)$ and $N_H\left(v\right)$ the degree and the neighborhood of $v$ in $H$, respectively. For a subgraph $H$, a nonempty subset $X$ of $V\left(G\right)$ and an integer $i\ge \mathrm{1}$, we write

$N_H\left(X\right)={\cup }_{x\in X}{N}_H\left(x\right),d_H\left(X\right)={\sum }_{x\in X}{d}_H\left(x\right)$

$N_i(X)=\{x\in V(G):|N(x)\bigcap X|=i\}$

and $N_{\ge i\mathrm{ }}(X)=\{x\in V(G):|N(x)\bigcap X|\ge i\}.$

If $H=G$, then for simplicity, we write $N\left(x\right)$, $N\left(X\right)$, $d\left(x\right)$ and $d\left(X\right)$ for $N_G\left(x\right)$, $N_G\left(X\right)$, $d_G\left(x\right)$ and $d_G\left(X\right)$, respectively.

The subgraph of $G$ induced by $X$ is denoted by ${〈X〉}_G$. We write $G-X$ for ${〈V(G)-X〉}_G$, and for a vertex $v$ of $G$, we write $G-v$ for $G-\left\{v\right\}$.

A spanning tree of a graph $G$ is a tree that contains all the vertices of $G$, respectively. A spanning tree is called a spanning $k$-ended tree if it contains at most $k$ leaves. Clearly, a Hamiltonian path can be regarded as a spanning 2-ended tree.

An independent set in a graph is a set of pairwise nonadjacent vertices. If $G$ has $k$ independent vertices, the minimum degree sum of $k$ independent vertices of $G$ is denoted by ${\sigma }_k\left(G\right)$, that is ${\sigma }_k(G)=\underset{S}{\mathrm{m}\mathrm{i}\mathrm{n}}\{\sum _{x\in S}d(x):S\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{p}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{e}\mathrm{t}\mathrm{o}\mathrm{f}G\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}k\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{e}\mathrm{s}\}$. Otherwise, we define ${\sigma }_k\left(G\right)=+\mathrm{\infty }$.

The graph constructed from two complete graphs $K_m$ and $K_n$ by identifying one vertex of $K_m$ with one vertex of $K_n$ is called a double complete graph, where $m,n\ge \mathrm{2}$. The common vertex of $K_m$ and $K_n$ is called the center, and other vertices are called non-central vertices. The complete bipartite graph with bipartition $\left(X,Y\right)$, where $\left|X\right|=m$ and $\left|Y\right|=n$, is denoted by $K_{m,n}$. If a graph $G$ contains no $K_{\mathrm{1,3}}$ as its induced subgraph, then we say $G$ is a claw-free graph (or $K_{\mathrm{1,3}}$-free graph).

Theorem 1^[1] Let $G$ be a connected claw-free graph. If $\delta \left(G\right)\ge \frac{\left|V\left(G\right)\right|-\mathrm{2}}{\mathrm{3}}$, then $G$ has a Hamiltonian path.

In view of Theorem 1, for a claw-free graph, Kano et al^[1] gave a degree sum condition to have a spanning tree with a bounded number of leaves, and got the following theorem.

Theorem 2^[2] Let $G$ be a connected claw-free graph and $k\ge \mathrm{2}$ be an integer. If ${\sigma }_{k+\mathrm{1}}\left(G\right)\ge \left|V\left(G\right)\right|-k$, then $G$ has a spanning $k$-ended tree.

1 Minimum Spanning $\mathit{t}$-Ended Systems

Let $\zeta$ be a set of vertices, paths or cycles in a graph $G$ such that any two distinct elements of $\zeta$ have no common vertices. Define a function $f$: $\zeta \to \left\{\mathrm{1,2}\right\}$ as follows:

$f\left(\beta \right)=\{\begin{array}{ll}\mathrm{1},& \beta \mathrm{i}\mathrm{s}{K}_{\mathrm{1}},K_{\mathrm{2}}\mathrm{o}\mathrm{r}\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{o}\mathrm{f}\zeta ;\\ \mathrm{2},& \beta \mathrm{i}\mathrm{s}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\zeta .\end{array}$

We call $\zeta$ a $t$-ended system of $G$ if ${\sum }_{\beta \in \zeta }f\left(\beta \right)\le t$. For a $t$-ended system $\zeta$ of $G$, let $V\left(\zeta \right)$ denote the union of the vertex sets of all the elements of $\zeta$. Moreover, we call $\zeta$ a spanning $t$-ended system of $G$ if $\zeta$ is a $t$-ended system and $V\left(\zeta \right)=V\left(G\right)$.

Lemma 1^[3] Let $t\ge \mathrm{2}$ be an integer and $G$ be a connected graph. If $G$ has a spanning $t$-ended system, then $G$ has a spanning $t$-ended tree.

Let $\zeta$ be a $t$-ended system in a graph. We define ${\zeta }_P=\left\{\beta \in \zeta :f\left(\beta \right)=\mathrm{2}\right\}$ and ${\zeta }_C=\left\{\beta \in \zeta :f\left(\beta \right)=\mathrm{1}\right\}$. For a path $P\in {\zeta }_P$, let $x_P,y_P$ be the end vertices of $P$. For an element $C\in {\zeta }_C$, choose one arbitrary vertex $x_C$ of $C$.

We define $\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)={\cup }_{\mathrm{P}\in {\mathrm{\zeta }}_{\mathrm{P}}}\left\{x_P,y_P\right\}$,$\mathrm{I}\mathrm{n}\mathrm{t}\left({\zeta }_P\right)=V\left({\zeta }_P\right)\backslash \mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)$, $\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_C\right)={\cup }_{\mathrm{C}\in {\mathrm{\zeta }}_{\mathrm{C}}}\left\{x_C\right\}$ and $\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)=\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)\bigcup$

$\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_C\right)$

Definition 1   Let $\zeta$ be a spanning$t$-ended system of a graph $G$. If there is no positive integer $s<t$ such that $G$ has a spanning $s$-ended system, then we call $\zeta$ a minimum spanning $t$-ended system of $G$.

From Definition 1, we can conclude that if a spanning $t$-ended system is minimum, then $\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)$ has exactly $t$ elements. Let $\zeta$ be a minimum spanning $t$-ended system of $G$ satisfying

$\left(\mathrm{C}\mathrm{1}\right)$ $\left|V\left({\zeta }_P\right)\right|$ is as large as possible;

$\left(\mathrm{C}\mathrm{2}\right)$ $\left|N_{\mathrm{2}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)\right|$ is as small as possible, subject to $\left(\mathrm{C}\mathrm{1}\right)$.

Lemma 2^[4] $\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)$ is an independent set with $t$ elements.

Fact 1^[4] If $a\in \mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)$, then $N_G\left(a\right)\subseteq \mathrm{I}\mathrm{n}\mathrm{t}\left({\zeta }_P\right)$.

Fact 2^[4] If $d\in V\left(C\right)$ for some $C\in {\zeta }_C$, then $N_G\left(d\right)\subseteq \mathrm{I}\mathrm{n}\mathrm{t}\left({\zeta }_P\right)\bigcup V\left(C\right)$.

Since $G$ is claw-free, by Facts 1 and 2, we conclude that

Fact 3 $N_{\mathrm{2}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)\subseteq \mathrm{I}\mathrm{n}\mathrm{t}\left({\zeta }_P\right)$ and $N_{\ge \mathrm{3}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)=\varphi$.

In the following, we take $P=v_{\mathrm{0}}{v}_{\mathrm{1}}\cdots v_p$ as an element of ${\zeta }_P$ and $X$ be a subset of $V\left(G\right)$. For two vertices $v_i,v_j$ of $V\left(P\right)$, we use $P\left(v_i,v_j\right)=v_i{v}_{i+\mathrm{1}}\cdots v_j$ and $\overline{P}\left(v_j,v_i\right)=v_j{v}_{j-\mathrm{1}}\cdots v_i$. Define

$N_P{\left(X\right)}^{+}=\{v_{i+\mathrm{1}}\in V\left(P\right):\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{e}\mathrm{x}x\in X\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}x{v}_i\in E\left(G\right)\}$

$N_P{\left(X\right)}^{-}=$

$\left\{v_{i-\mathrm{1}}\in V\left(P\right):\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{e}\mathrm{x}x\in X\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}x{v}_i\in E\left(G\right)\right\}$

If $X=\left\{x\right\}$, then we simplify $N_P{\left(\left\{x\right\}\right)}^{+}$ and $N_P{\left(\left\{x\right\}\right)}^{-}$ to $N_P{\left(x\right)}^{+}$ and $N_P{\left(x\right)}^{-}$, respectively.

Let $C$ be an element of ${\zeta }_C$ and $x$ be an arbitrary vertex of $C$, we use $x^{\mathrm{*}}$ to denote the neighbor of $x$ on $C$ if $C\ne K_{\mathrm{1}}$, otherwise $x^{\mathrm{*}}=x$. Set $C_x^{\mathrm{*}}=C-x{x}^{\mathrm{*}}$ if $C\ne K_{\mathrm{1}},K_{\mathrm{2}}$, otherwise $C_x^{\mathrm{*}}=C$.

Claim 1 Each of the following holds:

(i) If $v_{\mathrm{0}}{v}_i\in E\left(G\right)$, then $v_{i-\mathrm{1}}x\notin E\left(G\right)$ for every vertex $x\in \left(\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)\backslash \left\{v_{\mathrm{0}}\right\}\right)\bigcup V\left({\zeta }_C\right)$.

(ii) If $v_i{v}_p\in E\left(G\right)$, then $v_{i+\mathrm{1}}x\notin E\left(G\right)$ for every vertex $x\in \left(\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)\backslash \left\{v_p\right\}\right)\bigcup V\left({\zeta }_C\right)$.

Proof   (i) Suppose to the contrary that there is a vertex$x\in \left(\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)\backslash \left\{v_{\mathrm{0}}\right\}\right)\bigcup V\left({\zeta }_C\right)$ such that $v_{i-\mathrm{1}}x$$\in E\left(G\right)$. If $x=v_p$, then ${〈V\left(P\right)〉}_G$ is a cycle and so $G$ has a spanning $\left(t-\mathrm{1}\right)$-ended system, which contradicts the minimality of $\zeta$.

If $x\ne v_p$, then let $Q$ be the element of $\zeta$ containing $x$. Replacing $P$ and $Q$ by

$P^{\text{'}}=\{\begin{array}{ll}Q+x{v}_{i-\mathrm{1}}+\overline{P}(v_{i-\mathrm{1}},v_{\mathrm{0}})+v_{\mathrm{0}}{v}_i+P(v_i,v_p),& Q\in {\zeta }_P;\\ Q_x^{\mathrm{*}}+x{v}_{i-\mathrm{1}}+\overline{P}(v_{i-\mathrm{1}},v_{\mathrm{0}})+v_{\mathrm{0}}{v}_i+P(v_i,v_p),& Q\in {\zeta }_C,\end{array}$

we can obtain a spanning $\left(t-\mathrm{2}\right)$-ended system if $Q\in {\zeta }_P$, or a spanning $\left(t-\mathrm{1}\right)$-ended system if $Q\in {\zeta }_C$, which contradicts the minimality of $\zeta$.

By symmetry, (ii) holds as (i).

Claim 2 If $v_i\in N_{\mathrm{2}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)$ for some $\mathrm{1}\le i\le p-\mathrm{1}$, then (i) $v_i{v}_{\mathrm{0}},v_i{v}_p\in E\left(G\right)$, and (ii) $v_{i-\mathrm{1}}{v}_{\mathrm{0}},v_{i+\mathrm{1}}{v}_p\in E\left(G\right)$ if $\mathrm{2}\le i\le p-\mathrm{2}$.

Proof   (i) Suppose to the contrary that $v_i{v}_{\mathrm{0}}\notin E\left(G\right)$ or $v_i{v}_p\notin E\left(G\right)$. By symmetry, suppose $v_i{v}_p\notin E\left(G\right)$.

Case 1 $v_{\mathrm{0}}{v}_i\in E\left(G\right)$.

There is a vertex $x\in \left(\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)\backslash \left\{v_{\mathrm{0}},v_p\right\}\right)\bigcup V\left({\zeta }_C\right)$ such that $v_{i}x\in E\left(G\right)$. The two neighbors of $v_i$ are $v_{\mathrm{0}}$ and $x$. Then $v_i\ne v_p$ by Lemma 1. Since $\left\{v_{i+\mathrm{1}},v_{\mathrm{0}},x\right\}\subset N_G\left(v\right)$ and $G$ is claw-free, we can get that $v_{i+\mathrm{1}}x\in E\left(G\right)$ by Lemma 1 and Claim 1.

If $Q\in {\zeta }_P$, then replacing $P$ and $Q$ by $P^{\text{'}}=\overline{P}\left(v_p,v_{i+\mathrm{1}}\right)+v_{i+\mathrm{1}}x+Q$ and $Q^{\text{'}}=P\left(v_{\mathrm{0}},v_i\right)+v_i{v}_{\mathrm{0}}$, we can obtain a spanning $\left(t-\mathrm{1}\right)$-ended system, which contradicts the minimality of $\zeta$.

If $Q\in {\zeta }_C$ is a cycle, then let the two neighbors of $x$ in $Q$ be $x^{-}$ and $x^{+}$. Since $\left\{v_i,x^{-},x^{+}\right\}\subset N_G\left(x\right)$ and $G$ is claw-free, we may assume that $x^{-}{v}_i\in E\left(G\right)$ or $x^{-}{x}^{+}\in E\left(G\right)$ by symmetry. If $x^{-}{v}_i\in E\left(G\right)$, then replacing $P$ and $Q$ by $P\left(v_{\mathrm{0}},v_i\right)+v_i{x}^{-}+Q-x{x}^{-}+x{v}_{i+\mathrm{1}}+P\left(v_{i+\mathrm{1}},v_p\right)$, we can obtain a spanning $\left(t-\mathrm{1}\right)$-ended system, which contradicts the minimality of $\zeta$. If $x^{-}{x}^{+}\in E\left(G\right)$, then replacing $P$ and $Q$ by $P^{\text{'}}=P\left(v_{\mathrm{0}},v_i\right)+v_{i}x+x{v}_{i+\mathrm{1}}+P\left(v_{i+\mathrm{1}},v_p\right)$ and $Q^{\text{'}}=Q-x+x^{-}{x}^{+}$, we can obtain a spanning $t$-ended system ${\zeta }^{\text{'}}$ such that $\left|V\left({\zeta }_P^{\text{'}}\right)\right|>\left|V\left({\zeta }_P\right)\right|$, which contradicts $\left(\mathrm{C}\mathrm{1}\right)$.

If $Q\in {\zeta }_C$ is a $K_{\mathrm{1}}$ or $K_{\mathrm{2}}$, then replacing $P$ and $Q$ by $P^{\text{'}}=P\left(v_{\mathrm{0}},v_i\right)+v_{i}x+x{v}_{i+\mathrm{1}}+P\left(v_{i+\mathrm{1}},v_p\right)$ and $Q^{\text{'}}=Q-x$, we can obtain a spanning $t$-ended system ${\zeta }^{\text{'}}$ such that $\left|V\left({\zeta }_P^{\mathrm{\text{'}}}\right)\right|>\left|V\left({\zeta }_P\right)\right|$, which contradicts $\left(\mathrm{C}\mathrm{1}\right)$.

Case 2 $v_{\mathrm{0}}{v}_i\notin E\left(G\right)$.

There are two vertices $x,y\in \left(\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }_P\right)\backslash \left\{v_{\mathrm{0}},v_p\right\}\right)\bigcup V\left({\zeta }_C\right)$ such that $v_{i}x,v_{i}y\in E\left(G\right)$. Since $\left\{v_{i-\mathrm{1}},y,x\right\}\subset N_G\left(v_i\right)$ and $G$ is claw-free, by Lemma 1, we have $v_{i-\mathrm{1}}x\in E\left(G\right)$ or $v_{i-\mathrm{1}}y\in E\left(G\right)$. By symmetry, we may assume $v_{i-\mathrm{1}}x\in E\left(G\right)$. Let $Q$ and $R$ be the elements of $\zeta$ containing $x$ and $y$, respectively.

If $Q\ne R$, then replacing $P,Q$ and $R$ by

$Q^{\text{'}}=\{\begin{array}{ll}P(v_{\mathrm{0}},v_{i-\mathrm{1}})+v_{i-\mathrm{1}}x+Q,& Q\in {\zeta }_P;\\ P(v_{\mathrm{0}},v_{i-\mathrm{1}})+v_{i-\mathrm{1}}x+Q_x^{\mathrm{*}},& Q\in {\zeta }_C,\end{array}$

$R^{\text{'}}=\{\begin{array}{ll}P(v_i,v_p)+v_{i}y+R,& R\in {\zeta }_P;\\ P(v_i,v_p)+v_{i}y+R_x^{\mathrm{*}},& R\in {\zeta }_C,\end{array}$

we can obtain a spanning $t$-ended system ${\zeta }^{\text{'}}$ such that $f\left(Q^{\text{'}}\right)+f\left(R^{\text{'}}\right)\le f\left(R\right)+f\left(P\right)+f\left(Q\right)$. If $f\left(Q^{\text{'}}\right)+f\left(R^{\text{'}}\right)<f\left(R\right)+f\left(P\right)+f\left(Q\right)$, then it contradicts the minimality of $\zeta$.

If $f\left(Q^{\text{'}}\right)+f\left(R^{\text{'}}\right)=f\left(R\right)+f\left(P\right)+f\left(Q\right)$, then it is possible only if $\left\{Q^{\text{'}},R^{\text{'}}\right\}\subset {\zeta }_P$ and $\left\{Q,R\right\}\subset {\zeta }_C$, it contradicts $\left(\mathrm{C}\mathrm{1}\right)$.

If $Q=R$, then $Q$ is a path whose end vertices are $x$ and $y$. Replacing $P$ and $Q$ by $P^{\text{'}}=P\left(v_{\mathrm{0}},v_{i-\mathrm{1}}\right)+v_{i}x+Q+y{v}_i+P\left(v_i,v_p\right)$, we can get a spanning $\left(t-\mathrm{2}\right)$-ended system, which contradicts the minimality of $\zeta$.

(ii) If $\mathrm{2}\le i\le p-\mathrm{2}$, then, since $G$ is claw-free, by (i) $v_{i-\mathrm{1}}{v}_{\mathrm{0}},v_{i+\mathrm{1}}{v}_p\in E\left(G\right)$.

Claim 3 For each $P\in {\zeta }_p$, $d_P\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)\le \left|V\left(P\right)\right|-\mathrm{1}.$ The equality holds if and only if ${〈V\left(P\right)〉}_G$ is a double complete graph.

Proof   By Claim 1, $\left\{v_{\mathrm{0}}\right\}$, $N_P\left(v_{\mathrm{0}}\right)$, $N_P{\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\backslash \left\{v_{\mathrm{0}}\right\}\right)}^{+}$ are pairwise disjoint subsets of $V\left(P\right)$.

Since $\left\{v_{\mathrm{0}}\right\}\bigcup N_P\left(v_{\mathrm{0}}\right)\bigcup N_P{\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\backslash \left\{v_{\mathrm{0}}\right\}\right)}^{+}\subset V\left(P\right)$,we have

$\left|V(P)\right|\ge |\{v_{\mathrm{0}}\}|+|N_G(v_{\mathrm{0}})|+|N_G(\mathrm{E}\mathrm{n}\mathrm{d}(\zeta )\backslash \{v_{\mathrm{0}}{\})}^{+}|$

$=\mathrm{1}+|N_G(v_{\mathrm{0}})|+|N_G(\mathrm{E}\mathrm{n}\mathrm{d}(\zeta )\backslash \{v_{\mathrm{0}}\})|$(1)

$=\mathrm{1}+d_P(\mathrm{E}\mathrm{n}\mathrm{d}(\zeta )).$

Therefore, $d_P\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)\le \left|V\left(P\right)\right|-\mathrm{1}.$

Next, we will show that $d_P\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)=\left|V\left(P\right)\right|-\mathrm{1}$ if and only if ${〈V\left(P\right)〉}_G$ is a double complete graph. Obviously, if ${〈V\left(P\right)〉}_G$ is a double complete graph, then $d_P\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)=\left|V\left(P\right)\right|-\mathrm{1}$.

If $d_P\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)=\left|V\left(P\right)\right|-\mathrm{1}$, then by inequality (1), we have

$V(P)=\{v_{\mathrm{0}}\}\bigcup N_P(v_{\mathrm{0}})\bigcup N_P(\mathrm{E}\mathrm{n}\mathrm{d}(\zeta )\backslash \{v_{\mathrm{0}}{\})}^{+}.$(2)

We can claim that $\left|V\left(P\right)\bigcap N_{\mathrm{2}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)\right|\ge \mathrm{1}$. Otherwise, each vertex $x\in V\left(P\right)\backslash \{v_{\mathrm{0}},v_p\}$ is adjacent to at most one vertex of $\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)$, it follows $d_P\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)\le \left|V\left(P\right)\right|-\mathrm{2}$, a contradiction. By Fact 3 and Claim 2, there exists a vertex $v_i\in \mathrm{I}\mathrm{n}\mathrm{t}\left(P\right)$ such that $v_i\in N_{\mathrm{2}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\right)$, and $v_{\mathrm{0}}{v}_i,v_i{v}_p\in E\left(G\right)$ . Since $G$ is claw-free, $v_{\mathrm{0}}{v}_{i-\mathrm{1}},v_{i+\mathrm{1}}{v}_p\in E\left(G\right)$.

If $v_{\mathrm{0}}{v}_{i-\mathrm{2}}\notin E\left(G\right)$, then by (2), there is a vertex $x\in \mathrm{E}\mathrm{n}\mathrm{d}\left(\zeta \right)\backslash \left\{v_{\mathrm{0}}\right\}$ such that $v_{i-\mathrm{3}}x\in E\left(G\right)$.

Case 1 $x\ne v_p$.

Let $Q$ be the component of $\zeta$ containing $x$. If $Q=K_{\mathrm{1}}$, then since $\left\{v_{i-\mathrm{4}},v_{i-\mathrm{2}},x\right\}\subset N_P\left(v_{i-\mathrm{3}}\right)$ and $G$ is claw-free, we may assume that $v_{i-\mathrm{4}}{v}_{i-\mathrm{2}}\in E\left(G\right)$ or $v_{i-\mathrm{4}}x\in E\left(G\right)$ by symmetry. Replacing $P$ and $Q$ by

$P^{\text{'}}=\{\begin{array}{c}P(v_{\mathrm{0}},v_{i-\mathrm{4}})+v_{i-\mathrm{4}}x+x{v}_{i-\mathrm{3}}+P(v_{i-\mathrm{3}},v_p),\\ x{v}_{i-\mathrm{3}}+v_{i-\mathrm{3}}{v}_{i-\mathrm{2}}+v_{i-\mathrm{2}}{v}_{i-\mathrm{4}}+\overline{P}(v_{i-\mathrm{4}},v_{\mathrm{0}})\\ +v_{\mathrm{0}}{v}_{i-\mathrm{1}}+P(v_{i-\mathrm{1}},v_p),\end{array}\begin{array}{c}{v}_{i-\mathrm{4}}x\in E(G);\\ \\ v_{i-\mathrm{4}}{v}_{i-\mathrm{2}}\in E(G),\end{array}$

we can obtain a spanning $t$-ended system ${\zeta }^{\text{'}}$ such that $\left|V\left({\zeta }_P^{\mathrm{\text{'}}}\right)\right|>\left|V\left({\zeta }_P\right)\right|$, which contradicts $\left(\mathrm{C}\mathrm{1}\right)$.

If $Q\ne K_{\mathrm{1}}$, then replacing $P$ and $Q$ by

$P^{\text{'}}=\{\begin{array}{c}Q+x{v}_{i-\mathrm{3}}+\overline{P}(v_{i-\mathrm{3}},v_{\mathrm{0}})+v_{\mathrm{0}}{v}_{i-\mathrm{1}}+P(v_{i-\mathrm{1}},v_p),\\ Q_x^{*}+x{v}_{i-\mathrm{3}}+\overline{P}(v_{i-\mathrm{3}},v_{\mathrm{0}})+v_{\mathrm{0}}{v}_{i-\mathrm{1}}+P(v_{i-\mathrm{1}},v_p),\end{array}\begin{array}{c}Q\in {\zeta }_P;\\ Q\in {\zeta }_C,\end{array}$

and $Q^{\text{'}}=v_{i-\mathrm{2}}$, we can obtain a spanning t-$\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{d}$ system ${\zeta }^{\text{'}}$ such that $\left|V\left({\zeta }_P^{\mathrm{\text{'}}}\right)\right|>\left|V\left({\zeta }_P\right)\right|$, which contradicts $\left(\mathrm{C}\mathrm{1}\right)$.

Case 2 $x=v_p$.

Consider the path $P^{\text{'}}=v_{i-\mathrm{2}}{v}_{i-\mathrm{3}}\dots v_{\mathrm{0}}{v}_{i-\mathrm{1}}{v}_i{v}_p{v}_{i+\mathrm{1}}{v}_{i+\mathrm{2}}\cdots$

$v_{p-\mathrm{2}}{v}_{p-\mathrm{1}}$. Let ${\zeta }^{\text{'}}=\left(\zeta \backslash \left\{P\right\}\right)\bigcup \left\{P^{\text{'}}\right\}$. By the choice of $\zeta$, there is at least one vertex $v_s\in N_{\mathrm{2}}\left(\mathrm{E}\mathrm{n}\mathrm{d}\left({\zeta }^{\text{'}}\right)\right)\bigcap V\left(P^{\text{'}}\right)$. By Claim 2, $v_{i-\mathrm{2}}{v}_s,v_{p-\mathrm{1}}{v}_s\in E\left(G\right)$. Since $v_{\mathrm{0}}{v}_{i-\mathrm{2}}\notin E\left(G\right)$,$s\ne \mathrm{0}$. Since $v_{i-\mathrm{1}}{v}_{i-\mathrm{2}}\dots v_{\mathrm{0}}{v}_i{v}_p{v}_{i+\mathrm{1}}{v}_{i+\mathrm{2}}\dots v_{p-\mathrm{1}}$ is a path, by Lemma 2 and Claim 2 (i), $s\ne i-\mathrm{1}$. Similarly, $s\ne i+\mathrm{1},p$. If $s=i$, then $v_{i-\mathrm{1}}{v}_{i-\mathrm{2}}{v}_i{v}_{i+\mathrm{1}}\dots v_p{v}_{i-\mathrm{3}}{v}_{i-\mathrm{4}}\dots v_{\mathrm{0}}{v}_{i-\mathrm{1}}$ is a cycle, which contradicts the minimality of $\zeta$.

If $\mathrm{1}\le s\le i-\mathrm{3}$, then by Claim 2 (ii), $v_{s+\mathrm{1}}{v}_{i-\mathrm{2}},v_{s-\mathrm{1}}{v}_{p-\mathrm{1}}$

$\in E\left(G\right)$. Then $P\left(v_{i-\mathrm{3}},v_{s+\mathrm{1}}\right)+v_{s+\mathrm{1}}{v}_{i-\mathrm{2}}$$+v_{i-\mathrm{2}}{v}_s+\overline{P}\left(v_s,v_{\mathrm{0}}\right)+v_{\mathrm{0}}{v}_{i-\mathrm{1}}+P\left(v_{i-\mathrm{1}},v_p\right)+v_p{v}_{i-\mathrm{3}}$ is a cycle, which contradicts the minimality of $\zeta$. If $i+\mathrm{2}\le s\le p-\mathrm{2}$, then by Claim 2 (ii), $v_{s-\mathrm{1}}{v}_{i-\mathrm{2}},v_{s+\mathrm{1}}{v}_{p-\mathrm{1}}\in E\left(G\right)$. By similar argument as the case $\mathrm{1}\le s\le i-\mathrm{3}$, we can get a contradiction.