Let *H* = *G* − *M*, and we have *V*(*H*) = *V*(*G*) and *δ*(*H*) ≥ *δ*(*G*)−1. It is enough to show that *H* has a fractional (*g*, *f* )factor including *e*. In what follows, we use the counterevidence method to prove this result. Assume that *H* does’t have the desired fractional factor. Then in terms of Theorem 4, there exists a subset *S ⊆ V*(*H*) that satisfies

$$f\left(S\right)+{d}_{H-S}\left(T\right)-g\left(T\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}\text{\hspace{0.17em}}\epsilon \left(S,\text{\hspace{0.17em}}\text{\hspace{0.17em}}T\right)-1$$(1)where *T* = *{x* : *x ∈ V*(*H*)−*S*, *d*_{H}_{−S}(*x*) ≤ *g*(*x*)*}*. If *T* = *∅*, then *d*_{H}_{−S}(*T*) − *g*(*T*) = 0 and by (1) we infer *ε*(*S*, *T*) − 1 ≥ *f* (*S*) ≥ (*a* + *Δ*)*|S|* ≥ 2*|S|* ≥ *ε*(*S*, *T*), a contradiction. Thus, we have *T* ≠*∅*.

Set *D* = *V*(*G*) − *S ∪ T* and *E*_{G}(*T*) = *{e* : *e* = *xy ∈ E*(*G*), *x*, *y ∈ T}*. Since *M* is an *m*-matching, we deduce

$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{H-S}\left(T\right)={d}_{G-M-S}\left(T\right)\\ =\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{G-S}\left(T\right)-\left(2\left|M\cap {E}_{G}\left(T\right)\right|+\left|M\cap {E}_{G}\left(T,\text{\hspace{0.17em}}\text{\hspace{0.17em}}D\right)\right|\right)\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{G-S}\left(T\right)-\mathrm{min}\left\{2m,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|T\right|\right\}.\end{array}$$(2)Now, we discuss the following three cases according to the value of *|S|*.

**Case 1**. *|S|* = 0.

In this case, we have *ε*(*S*, *T*) = 0 and $\delta \left(H\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta \left(G\right)-1\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(i-1\right)\left(b-\Delta \right)\left(b+2\right)+1}{a}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}>b-\Delta .$By means of (1), we yield

$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(S\right)+{d}_{H-S}\left(T\right)-g\left(T\right)\\ =\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{H}\left(T\right)-g\left(T\right)={\displaystyle \sum _{x\in T}\left({d}_{H}\left(x\right)-g\left(x\right)\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\displaystyle \sum _{x\in T}\left(\delta \left(H\right)-\left(b-\Delta \right)\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}0,\end{array}$$a contradiction.

**Case 2**. *|S|* = 1.

In this case, we have *ε*(*S*, *T*) ≤ 1 and $\delta \left(G\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(i-1\right)\left(b-\Delta \right)\left(b+2\right)+1}{a}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}>\text{\hspace{0.17em}}\text{\hspace{0.17em}}b-\Delta +1.$In view of the integrity of *δ*(*G*), we have *δ*(*G*) ≥ *b* − *Δ* + 2. In light of (1) and (2), we derive

$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\epsilon \left(S,\text{\hspace{0.17em}}\text{\hspace{0.17em}}T\right)-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge f\left(S\right)+{d}_{H-S}\left(T\right)-g\left(T\right)\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(S\right)+{d}_{G-S}\left(T\right)-\mathrm{min}\left\{\left|T\right|,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2m\right\}-g\left(T\right)\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(S\right)+{d}_{G-S}\left(T\right)-\left|T\right|-g\left(T\right)\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(S\right)+{d}_{G}\left(T\right)-2\left|T\right|-g\left(T\right)\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(S\right)+\delta \left(G\right)\left|T\right|-2\left|T\right|-g\left(T\right)\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(a+\Delta \right)\left|S\right|+\left(b-\Delta +2\right)\left|T\right|-2\left|T\right|-\left(b-\Delta \right)\left|T\right|\\ =\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(a+\Delta \right)\left|S\right|=a+\Delta \text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\end{array}$$a contradiction.

**Case 3**. *|S|* ≥ 2.

Note that *ε*(*S*, *T*) ≤ 2 and *T* ≠*∅*. Let *d*_{1} = min*{d*_{G}_{−S}(*x*) : *x ∈ T}* and choose *x*_{1} *∈ T* such that *d*_{G}_{−S}(*x*_{1}) = *d*_{1}. If *T* − *N*_{T}[*x*_{1}] ≠*∅*, let *d*_{2} = min*{d*_{G}_{−S}(*x*) : *x ∈ T* − *N*_{T}[*x*_{1}]*}* and choose *x*_{2} *∈ T* − *N*_{T}[*x*_{1}] such that *d*_{G}_{−S}(*x*_{2}) = *d*_{2}. Continue this step, if *z* ≥ 2 and $T\backslash \left({\cup}_{j=1}^{z-1}{N}_{T}\left[{x}_{j}\right]\right)\ne \overline{)0},$let

$${d}_{z}=\mathrm{min}\left\{{d}_{G-S}\left(x\right)|x\in T\backslash \left({\cup}_{j=1}^{z-1}{N}_{T}\left[{x}_{j}\right]\right)\right\},$$and select ${x}_{z}\in T\backslash \left({\cup}_{j=1}^{z-1}{N}_{T}\left[{x}_{j}\right]\right)$with *d*_{G}_{−S}(*x*_{z}) = *d*_{z}. Thus, we construct a sequence 0 ≤ *d*_{1} ≤ *d*_{2} ≤ ·· ·≤ *d*_{π} ≤ *b* − *Δ* + 1 and an independent set *{x*_{1}, *x*_{2}, · · · , *x*_{π}}⊆ T.

Now, we the next lemma which is similar to Lemma 5 in Gao et al. [8].

#### Lemma 1

In the above conditions, we have

$$\left|T\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\{\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i-1\right)\left(b+2\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{G-S}\left(x\right)=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{any}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in T,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i-1\right)\left(b+2\right)+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{otherwise}\text{.}\text{\hspace{0.17em}}\end{array}$$**Proof**. Note that *|S|* + *d*_{1} = *|S|* + *d*_{G}_{−S}(*x*_{1})≥ *d*_{G}(*x*_{1})≥ *δ*(*G*). We verify that

$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|S\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta \left(G\right)-{d}_{1}\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(i-1\right)\left(b-\Delta \right)\left(b+2\right)+1}{a+\Delta}+1-{d}_{1}.\end{array}$$(3)Firstly, we justify that $\left|T\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i=1\right)\left(b+2\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{G-S}\left(x\right)=1$for any *x ∈ T*. In this case, we get *d*_{1} = 1.

Assume that *|T|* ≤ (*i* − 1)(*b* + 2) − 1. By means of (2), (3), *d*_{1} = 1 and *i* ≥ 2, we yield

$$\begin{array}{rcl}& \phantom{\rule{1em}{0ex}}& f(S)+{d}_{H-S}(T)-g(T)\\ & \ge & f(S)+{d}_{G-S}(T)-min\{|T|,2m\}-g(T)\\ & \ge & f(S)+{d}_{G-S}(T)-|T|-g(T)\\ & \ge & (a+\mathrm{\Delta})|S|+|T|-|T|-(b-\mathrm{\Delta})|T|\\ & =& (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})|T|\\ & \ge & (a+\mathrm{\Delta})(\frac{(i-1)(b-\mathrm{\Delta})(b+2)+1}{a+\mathrm{\Delta}}+1-{d}_{1})\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta})((i-1)(b+2)-1)\\ & =& (a+\mathrm{\Delta})\frac{(i-1)(b-\mathrm{\Delta})(b+2)+1}{a+\mathrm{\Delta}}\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta})((i-1)(b+2)-1)\\ & =& b-\mathrm{\Delta}+1>a\ge 2\ge \epsilon (S,T),\end{array}$$which contradicts (1). Therefore, the first part of Lemma 1 is established.

Next, we check that *|T|* ≥ (*i* − 1)(*b* + 2) + 1 in the rest cases. Assume that *|T|* ≤ (*i* − 1)(*b* + 2) in this case, and the discussion can be divided into three situations in terms of the value of *d*_{1}.

**Case 1**. *d*_{1} = 0.

In view of (1), (3) and *d*_{1} = 0, we yield

$$\begin{array}{rcl}& \phantom{\rule{1em}{0ex}}& \epsilon (S,T)-1\ge f(S)+{d}_{H-S}(T)-g(T)\\ & \ge & f(S)-g(T)\ge (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})|T|\\ & \ge & (a+\mathrm{\Delta})(\frac{(i-1)(b-\mathrm{\Delta})(b+2)+1}{a+\mathrm{\Delta}}+1-{d}_{1})\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta})(i-1)(b+2)\\ & =& a+\mathrm{\Delta}+1>a\ge 2\ge \epsilon (S,T),\end{array}$$a contradiction.

**Case 2**. *d*_{1} = 1.

Following from (1), (3), *d*_{1} = 1 and *ε*(*S*, *T*) ≤ 2, we get

$$\begin{array}{rcl}& \phantom{\rule{1em}{0ex}}& 1\ge \epsilon (S,T)-1\ge f(S)+{d}_{H-S}(T)-g(T)\\ & \ge & f(S)-g(T)\ge (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})|T|\\ & \ge & (a+\mathrm{\Delta})(\frac{(i-1)(b-\mathrm{\Delta})(b+2)+1}{a+\mathrm{\Delta}}+1-{d}_{1})\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta})(i-1)(b+2)\\ & =& 1.\end{array}$$It implies that all the “≥” should be “=”, and it further reveals that *d*_{H}_{−S}(*T*) = 0, i.e., *d*_{H}_{−S}(*x*) = 0 for each *x ∈ T*.

Since *d*_{1} = 1, we confirm that *d*_{G}_{−S}(*x*) = 1 for each *x ∈ T* which contradicts the assumption that there exist *x ∈ T* satisfying *d*_{G}_{−S}(*x*) ≠1.

**Case 3**. 2 ≤ *d*_{1} ≤ *b* − *Δ* + 1.

Using (1), (2), (3), *i* ≥ 2 and2 ≤ *d*_{1} ≤ *b*−*Δ*+1, we derive

$$\begin{array}{rcl}& \phantom{\rule{1em}{0ex}}& \epsilon (S,T)-1\ge f(S)+{d}_{H-S}(T)-g(T)\\ & \ge & f(S)+{d}_{G-S}(T)-min\{|T|,2m\}-g(T)\\ & \ge & f(S)+{d}_{G-S}(T)-|T|-g(T)\\ & \ge & (a+\mathrm{\Delta})|S|+{d}_{1}|T|-|T|-(b-\mathrm{\Delta})|T|\\ & =& (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta}-{d}_{1}+1)|T|\\ & \ge & (a+\mathrm{\Delta})(\frac{(i-1)(b-\mathrm{\Delta})(b+2)+1}{a+\mathrm{\Delta}}+1-{d}_{1})\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta}-{d}_{1}+1)(i-1)(b+2)\\ & =& ({d}_{1}-1)((i-1)(b+2)-(a+\mathrm{\Delta}))+1\\ & \ge & ({d}_{1}-1)((b+2)-(a+\mathrm{\Delta}))+1\\ & >& ({d}_{1}-1)+1={d}_{1}\\ & \ge & 2\ge \epsilon (S,T),\end{array}$$a contradiction.

Therefore, we infer that *|T|* ≥ (*i* − 1)(*b* + 2) + 1 in other cases.

The next lemma shows that the desired independent subset exists in *T*.

#### Lemma 2

In the previous setting, there exists an independent subset *{x*_{1}, *x*_{2}, · · · , *x*_{i}}⊆ T.

**Proof**. If *d*_{G}_{−S}(*x*) = 1 for any *x ∈ T*, then in terms of Lemma 1, we have *|T|* ≥ (*i* − 1)(*b* + 2). In light of *b* ≥ 2 and *d*_{G}_{−S}(*x*) = 1 for any *x ∈ T*, we determine that *d*_{G}_{−S}(*x*) ≤ *b*−1 for any *x ∈ T*. Combining this with *|T|* ≥ (*i* − 1)(*b* + 2) = *b*(*i*−1)+2(*i*−1) ≥ *b*(*i*−1)+2, we conclude that there exists an independent subset *{x*_{1}, *x*_{2}, · · · , *x*_{π}}⊆ T for *π* = *i*.

In other cases, it holds that *|T|* ≥ (*i* − 1)(*b* + 2) + 1 by means of Lemma 1. It is clear that *d*_{G}_{−S}(*x*) ≤ *d*_{H}_{−S}(*x*) + 1 ≤ *g*(*x*) + 1 ≤ *b* − *Δ* +1 ≤ *b* + 1 for any *x ∈ T*. Thus, we can find an independent subset, *{x*_{1}, *x*_{2}, · · · , *x*_{π}}⊆ T for *π* = *i*.

The proof of Lemma 2 is completed.

By lemma 2 and the condition of Theorem 3, we have

$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(b-\Delta \right)n+2}{a+b}+i\\ \begin{array}{ll}\le \hfill & \left|{N}_{G}\left({x}_{1}\right)\cup {N}_{G}\left({x}_{2}\right)\cup \cdots \cup {N}_{G}\left({x}_{i}\right)\right|\hfill \\ \le \hfill & \left|S\right|+{\displaystyle \sum _{j=1}^{i}{d}_{j}}\hfill \end{array}\end{array}$$and

$$\left|S\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(b-\Delta \right)n+2}{a+b}+i-{\displaystyle \sum _{j=1}^{i}{d}_{j}}.$$(4)The next lemma presents the lower bound of *d*_{i}.

**Proof**. Suppose that *d*_{i} = 0 or *d*_{i} = 1. In terms of (4) and *|S|* + *|T|* ≤ *n*, we ensure that

$$\begin{array}{rcl}& \phantom{\rule{1em}{0ex}}& f(S)+{d}_{H-S}(T)-g(T)\\ & \ge & f(S)-g(T)\\ & \ge & (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})|T|\\ & \ge & (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})(n-|S|)\\ & =& (a+b)|S|-(b-\mathrm{\Delta})n\\ & \ge & (a+b)(\frac{(b-\mathrm{\Delta})n+2}{a+b}+i-\sum _{j=1}^{i}{d}_{j})\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta})n\\ & \ge & (a+b)\frac{(b-\mathrm{\Delta})n+2}{a+b}-(b-\mathrm{\Delta})n\\ & =& 2\ge \epsilon (S,T),\end{array}$$which contradicts (1). Hence we get *d*_{i} ̸*∈ {*0, 1*}*. It is easy to check that

$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|{N}_{T}\left[{x}_{j}\right]\right|-\left|{N}_{T}\left[{x}_{j}\right]\cap \left({\cup}_{z=1}^{j-1}{N}_{T}\left[{x}_{z}\right]\right)\right|\\ \ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}j=2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}3,\cdots ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i-1,\end{array}$$(5)$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|{\cup}_{z=1}^{j}{N}_{T}\left[{x}_{z}\right]\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}\text{\hspace{0.17em}}{\displaystyle \sum _{z=1}^{j}\left|{N}_{T}\left[{x}_{z}\right]\right|}\\ \begin{array}{ll}\le \hfill & {\displaystyle \sum _{z=1}^{j}\left({d}_{G-S}\left({x}_{z}\right)+1\right)}\hfill \\ =\hfill & {\displaystyle \sum _{z=1}^{j}\left({d}_{z}+1\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}j=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\cdots ,i.}\hfill \end{array}\end{array}$$(6)and

$$\left|T\right|-\left|{\cup}_{j=1}^{i}{N}_{T}\left[{x}_{j}\right]\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}1.$$(7)It follows from (1), (2), (5), (6), *n* ≥ *|S|* + *|T|*, Lemma 3 and *d*_{i} ≤ *b* − *Δ* + 1 that

$$\begin{array}{rcl}& \phantom{\rule{1em}{0ex}}& 1\ge \epsilon (S,T)-1\\ & \ge & f(S)+{d}_{H-S}(T)-g(T)\\ & \ge & f(S)+{d}_{G-S}(T)-min\{|T|,2m\}-g(T)\\ & \ge & f(S)+{d}_{G-S}(T)-g(T)-2m\\ & \ge & (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})|T|+{d}_{1}|{N}_{T}[{x}_{1}]|\\ & \phantom{\rule{1em}{0ex}}& +{d}_{2}(|{N}_{T}[{x}_{2}]|-|{N}_{T}[{x}_{2}]\cap {N}_{T}[{x}_{1}]|)+\cdots \\ & \phantom{\rule{1em}{0ex}}& +{d}_{i-1}(|{N}_{T}[{x}_{i-1}]|-|{N}_{T}[{x}_{i-1}]\cap ({\cup}_{j=1}^{i-2}{N}_{T}[{x}_{j}])|)\\ & \phantom{\rule{1em}{0ex}}& +{d}_{i}(|T|-|{\cup}_{j=1}^{i-1}{N}_{T}[{x}_{j}])||)-2m\end{array}$$$$\begin{array}{rcl}& \ge & (a+\mathrm{\Delta})|S|-(b-\mathrm{\Delta})|T|+{d}_{1}|{N}_{T}[{x}_{1}]|\\ & \phantom{\rule{1em}{0ex}}& +{d}_{2}(|{N}_{T}[{x}_{2}]|-|{N}_{T}[{x}_{2}]\cap {N}_{T}[{x}_{1}]|)+\cdots \\ & \phantom{\rule{1em}{0ex}}& +{d}_{i-1}(|{N}_{T}[{x}_{i-1}]|-|{N}_{T}[{x}_{i-1}]\cap ({\cup}_{j=1}^{i-2}{N}_{T}[{x}_{j}])|)\\ & \phantom{\rule{1em}{0ex}}& +({d}_{i}-1)(|T|-|{\cup}_{j=1}^{i-1}{N}_{T}[{x}_{j}])||)-2m+1\\ & \ge & (a+\mathrm{\Delta})|S|+({d}_{1}-{d}_{i}+1)|{N}_{T}[{x}_{1}]|+\sum _{j=2}^{i-1}{d}_{j}\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta}+1-{d}_{i})|T|-({d}_{i}-1)\sum _{j=2}^{i-1}|{N}_{T}[{x}_{j}]|\\ & \phantom{\rule{1em}{0ex}}& -2m+1\\ & \ge & (a+\mathrm{\Delta})|S|+({d}_{1}-{d}_{i})({d}_{1}+1)+1+\sum _{j=2}^{i-1}{d}_{j}\\ & \phantom{\rule{1em}{0ex}}& -(b-\mathrm{\Delta}+1-{d}_{i})|T|-({d}_{i}-1)\sum _{j=2}^{i-1}({d}_{j}+1)\\ & \phantom{\rule{1em}{0ex}}& -2m+1\\ & =& (a+\mathrm{\Delta})|S|+{d}_{1}({d}_{1}-1)+\sum _{j=1}^{i-1}{d}_{j}-2m+1\\ & \phantom{\rule{1em}{0ex}}& -({d}_{i}-1)\sum _{j=1}^{i-1}({d}_{j}+1)-(b-\mathrm{\Delta}+1-{d}_{i})|T|\\ & \ge & (a+\mathrm{\Delta})|S|+{d}_{1}({d}_{1}-1)+\sum _{j=1}^{i-1}{d}_{j}-2m+1\\ & \phantom{\rule{1em}{0ex}}& -({d}_{i}-1)\sum _{j=1}^{i-1}({d}_{j}+1)-(b-\mathrm{\Delta}+1-{d}_{i})(n-|S|)\\ & =& (a+b+1-{d}_{i})|S|+{d}_{1}({d}_{1}-1)+\sum _{j=1}^{i-1}{d}_{j}-2m\\ & \phantom{\rule{1em}{0ex}}& -({d}_{i}-1)\sum _{j=1}^{i-1}({d}_{j}+1)-(b-\mathrm{\Delta}+1-{d}_{i})n+1,\end{array}$$which implies

$$\left(a+b+1-{d}_{i}\right)\left|S\right|+{d}_{1}\left({d}_{1}-1\right)+{\displaystyle \sum _{j=1}^{i-1}{d}_{j}}$$(8)$$-\left({d}_{i}-1\right){\displaystyle \sum _{j=1}^{i-1}\left({d}_{j}+1\right)}-\left(b-\Delta +1-{d}_{i}\right)n-2m\le 0.$$By means of (4), (8), Lemma 3, *i* ≥ 2, *d*_{1} ≤ *d*_{2} ≤ ··· ≤ *d*_{i} ≤ *b* − *Δ* + 1 and $n\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(a+b\right)\left(i\left(a+b+1\right)+2m-2\right)+2}{a+\Delta},$we have

$$\begin{array}{rcl}0& \ge & (a+b+1-{d}_{i})|S|+{d}_{1}({d}_{1}-1)+\sum _{j=1}^{i-1}{d}_{j}\\ & \phantom{\rule{1em}{0ex}}& -({d}_{i}-1)\sum _{j=1}^{i-1}({d}_{j}+1)-(b-\mathrm{\Delta}+1-{d}_{i})n-2m\\ & \ge & (a+b+1-{d}_{i})(\frac{(b-\mathrm{\Delta})n+2}{a+b}+i-\sum _{j=1}^{i}{d}_{j})\\ & \phantom{\rule{1em}{0ex}}& -({d}_{i}-2)\sum _{j=1}^{i-1}{d}_{j}-({d}_{i}-1)(i-1)\\ & \phantom{\rule{1em}{0ex}}& -n(b-\mathrm{\Delta}+1-{d}_{i})-2m\\ & \ge & (a+b+1-{d}_{i})(\frac{(b-\mathrm{\Delta})n+2}{a+b}+i-i{d}_{i})\\ & \phantom{\rule{1em}{0ex}}& -({d}_{i}-2){d}_{i}(i-1)-({d}_{i}-1)(i-1)\\ & \phantom{\rule{1em}{0ex}}& -n(b-\mathrm{\Delta}+1-{d}_{i})-2m\\ & =& ({d}_{i}-1)(\frac{(a+\mathrm{\Delta})n-2}{a+b}-i(a+b+1)+{d}_{i})\\ & \phantom{\rule{1em}{0ex}}& -2m+i+1\\ & \ge & ({d}_{i}-1)(\frac{(a+\mathrm{\Delta})n-2}{a+b}-i(a+b+1)+2)\\ & \phantom{\rule{1em}{0ex}}& -2m+i+1\\ & \ge & ({d}_{i}-1)((i(a+b+1)+2m-2)-i(a+b+1)\\ & \phantom{\rule{1em}{0ex}}& +2)-2m+i+1\\ & =& 2m({d}_{i}-2)+i+1\ge 3,\end{array}$$a contradiction.

Therefore, the desired theorem is proved.

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.