**Proof.**

(1) Since *x* ∈ *A<sup>α</sup>*, we have *TjA*(*x*) ≥ *α*. From *A* ⊆ *B*, it follows that *TjA*(*x*) ≤ *TjB*(*x*). Thus, *TjB*(*x*) ≥ *α*. Thus, *x* ∈ *B<sup>α</sup>*. Therefore, *Aα* ⊆ *Bα* for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*). Since *x* ∈ *Aβ*, we have *<sup>I</sup>jA*(*x*) ≤ *β*. From *A* ⊆ *B*, it follows that *<sup>I</sup>jA*(*x*) ≥ *<sup>I</sup>jB*(*x*). Thus, *<sup>I</sup>jB*(*x*) ≤ *β*. Thus, *x* ∈ *Bβ*. Hence, *Aβ* ⊆ *Bβ* for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*).

(2) From *x* ∈ (*A* ∩ *<sup>B</sup>*)*<sup>α</sup>*, we can obtain *TjA*∩*<sup>B</sup>*(*x*) ≥ *α*. Then, min*TjA*(*x*), *TjB*(*x*) ≥ *α*, that is, *TjA*(*x*)<sup>≥</sup> *<sup>α</sup>*,*TjB*(*x*) ≥ *α*. Thus, *x* ∈ *A<sup>α</sup>*, *x* ∈ *B<sup>α</sup>*. Hence, *x* ∈ *Aα* ∩ *B<sup>α</sup>*. On the other hand, since *x* ∈ *Aα* ∩ *B<sup>α</sup>*, we have *x* ∈ *A<sup>α</sup>*, *x* ∈ *B<sup>α</sup>*, that is, *TjA*(*x*) ≥ *α*, *TjB*(*x*) ≥ *α*. Then, min*TjA*(*x*), *TjB*(*x*) ≥ *α*. Thus, *TjA*∩*<sup>B</sup>*(*x*) ≥ *α*. Hence, *x* ∈ (*A* ∩ *<sup>B</sup>*)*<sup>α</sup>*. Based on the above facts, we can check that (*A* ∩ *B*)*<sup>α</sup>* = *Aα* ∩ *Bα* for *j* = 1, 2, ··· ,*l*(*<sup>x</sup>* : *A*, *<sup>B</sup>*). 

Since *x* ∈ (*A* ∩ *<sup>B</sup>*)*β*, we have *<sup>I</sup>jA*∩*<sup>B</sup>*(*x*) ≤ *β*. Then, max*IjA*(*x*), *<sup>I</sup>jB*(*x*)≤ *β*, that is, *<sup>I</sup>jA*(*x*) ≤ *<sup>β</sup>*,*<sup>I</sup>jB*(*x*) ≤ *β*. Then, *x* ∈ *Aβ*, *x* ∈ *Bβ*. Hence, *x* ∈ *Aβ* ∩ *Bβ*. On the other hand, from *x* ∈ *Aβ* ∩ *Bβ*, we have *x* ∈ *Aβ*,*<sup>x</sup>* ∈ *Bβ*. Thus, *<sup>I</sup>jA*(*x*) ≤ *β*, *<sup>I</sup>jB*(*x*) ≤ *β*, that is, max*IjA*(*x*), *<sup>I</sup>jB*(*x*) ≤ *β*. Thus, *<sup>I</sup>jA*∩*<sup>B</sup>*(*x*) ≤ *β*. Thus, *x* ∈(*<sup>A</sup>* ∩ *<sup>B</sup>*)*β*. Therefore, we can check that (*A* ∩ *B*)*β* = *Aβ* ∩ *Bβ* for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*).

(3) From *x* ∈ (*A* ∪ *<sup>B</sup>*)*<sup>α</sup>*, we have *TjA*∪*<sup>B</sup>*(*x*) ≥ *α*. Thus, max*TjA*(*x*), *TjB*(*x*)≥ *α*, that is, *TjA*(*x*) ≥ *α* or *TjB*(*x*) ≥ *α*. Thus, *x* ∈ *Aα* or *x* ∈ *B<sup>α</sup>*. Hence, *x* ∈ *Aα* ∪ *B<sup>α</sup>*. On the other hand, since *x* ∈ *Aα* ∪ *B<sup>α</sup>*, we have *x* ∈ *Aα* or *x* ∈ *B<sup>α</sup>*. Thus, *TjA*(*x*) ≥ *α* or *TjB*(*x*) ≥ *α*, that is, max*TjA*(*x*), *TjB*(*x*) ≥ *α*. Thus,

*TjA*∪*<sup>B</sup>*(*x*) ≥ *α*. Hence, *x* ∈ (*A* ∪ *<sup>B</sup>*)*<sup>α</sup>*. Using the above facts, we can check that (*A* ∪ *B*)*<sup>α</sup>* = *Aα* ∪ *Bα* for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*). Since *x* ∈ (*A* ∪ *<sup>B</sup>*)*γ*, we have *<sup>F</sup>jA*∪*<sup>B</sup>*(*x*) ≤ *γ*, that is, min*FjA*(*x*), *<sup>F</sup>jB*(*x*) ≤ *γ*. Thus, *<sup>F</sup>jA*(*x*) ≤ *γ* or *<sup>F</sup>jB*(*x*) ≤ *γ*. Thus, *x* ∈ *<sup>A</sup>γ* or *x* ∈ *<sup>B</sup>γ*. Hence, *x* ∈ *<sup>A</sup>γ* ∪ *<sup>B</sup>γ*. On the other hand, from *x* ∈ *<sup>A</sup>γ* ∪ *<sup>B</sup>γ*, we have *x* ∈ *<sup>A</sup>γ* or *x* ∈ *<sup>B</sup>γ*. Thus, *<sup>F</sup>jA*(*x*) ≤ *γ* or *<sup>F</sup>jB*(*x*) ≤ *γ*, that is, min*FjA*(*x*), *<sup>F</sup>jB*(*x*) ≤ *γ*.Thus, *<sup>F</sup>jA*∪*<sup>B</sup>*(*x*) ≤ *γ*. Hence, *x* ∈ (*A* ∪ *<sup>B</sup>*)*γ*. Therefore, we can check that (*A* ∪ *<sup>B</sup>*)*γ* = *<sup>A</sup>γ* ∪ *<sup>B</sup>γ* for *j* = 1, 2, ··· ,*l*(*<sup>x</sup>* : *A*, *<sup>B</sup>*). (4) From *x* ∈ 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6*<sup>α</sup>*, we have *Tj*∩*<sup>t</sup>*∈*<sup>T</sup>At*(*x*) ≥ *α*, that is, inf *<sup>t</sup>*∈*<sup>T</sup>TjAt*(*x*) ≥ *α*. Thus, *TjAt*(*x*) ≥ *α* for all *t* ∈ *T*, that is, *x* ∈ (*At*)*<sup>α</sup>* for all *t* ∈ *T*. Hence, *x* ∈ ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)*<sup>α</sup>*. On the other hand, from *x* ∈ ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)*<sup>α</sup>*, it follows that *x* ∈ (*At*)*<sup>α</sup>* for all *t* ∈ *T*. Then, *TjAt*(*x*) ≥ *α* for all *t* ∈ *T*, that is, inf *<sup>t</sup>*∈*<sup>T</sup>TjAt*(*x*) ≥ *α*. Then, *Tj*∩ *t*∈*T At*(*x*) ≥ *α*. Thus, *x* ∈ 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6*<sup>α</sup>*. Based on the above facts, we can check that 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6*<sup>α</sup>* = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)*<sup>α</sup>* for *j* = 1, 2, ··· ,*l*(*l* = max{ *l*(*x* : *At*)|*t* ∈ *<sup>T</sup>*}). Since *x* ∈ 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6*<sup>β</sup>*, we have *Ij*∩*<sup>t</sup>*∈*<sup>T</sup>At*(*x*) ≤ *β*, that is, sup *t*∈*T IjAt*(*x*) ≤ *β*. Thus, *<sup>I</sup>jAt*(*x*) ≤ *β* for all *t* ∈ *T*, that is, *x* ∈ (*At*)*β* for all *t* ∈ *T*. Thus, *x* ∈ ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)*β*. On the other hand, from *x* ∈ ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)*β*, we have *x* ∈ (*At*)*β* for all *t* ∈ *T*. Thus, *<sup>I</sup>jAt*(*x*) ≤ *β* for all *t* ∈ *T*, that is, sup *t*∈*T IjAt*(*x*) ≤ *β*. Thus, *Ij*∩ *t*∈*T At*(*x*) ≤ *β*. Hence, *x* ∈ 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6*<sup>β</sup>*. Therefore, we can check that 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6*<sup>β</sup>* = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)*<sup>β</sup>* for *j* = 1, 2, ··· , *l*(*l* = max {*l* (*x* : *At*)|*t* ∈ *<sup>T</sup>*}).


The (*<sup>α</sup>*, *β*, *<sup>γ</sup>*)-*cut* sets of SVNMS satisfy the following properties. -

**Theorem 2.** *Let A*, *B* ∈ *SVNMS*(*X*)*, α*, *β*, *γ* ∈ [0, 1] *with* 0 ≤ *α* + *β* + *γ* ≤ 3. *Then,*


*(7)* 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)*,* 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*,*<sup>γ</sup>*) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*,*<sup>γ</sup>*)*,* 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*), 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*+) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*+)*,* 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*+,*<sup>γ</sup>*) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*+,*<sup>γ</sup>*)*,* 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*,*<sup>γ</sup>*+) = ∩ *<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*,*<sup>γ</sup>*+), 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*+) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*+)*,* 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*+,*<sup>γ</sup>*+) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*+,*<sup>γ</sup>*+)*; (8)* 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)*,* 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*,*<sup>γ</sup>*) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*,*<sup>γ</sup>*)*,* 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*) ⊇ ∪ *<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*), 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*+) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*+)*,* 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*+,*<sup>γ</sup>*) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*+,*<sup>γ</sup>*)*,* 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*,*<sup>γ</sup>*+) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*,*<sup>γ</sup>*+), 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*+) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*+)*,* 5 ∪*<sup>t</sup>*∈*<sup>T</sup>At*6(*α*+,*β*+,*<sup>γ</sup>*+) ⊇ ∪*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*α*+,*β*+,*<sup>γ</sup>*+)*; (9)* ∩ *t*∈*T A*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*) = *<sup>A</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)*,* ∩ *t*∈*T A*(*<sup>α</sup>t*+,*β<sup>t</sup>*,*γt*) = *<sup>A</sup>*(*α*+,*β*,*<sup>γ</sup>*)*,* ∩ *t*∈*T A*(*<sup>α</sup>t*,*βt*+,*γt*) = *<sup>A</sup>*(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*)*,* ∩ *t*∈*T A*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*+) = *<sup>A</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*+), ∩ *t*∈*T A*(*<sup>α</sup>t*+,*βt*+,*γt*) = *<sup>A</sup>*(*α*+,*β*+,*<sup>γ</sup>*)*,* ∩ *t*∈*T A*(*<sup>α</sup>t*+,*β<sup>t</sup>*,*γt*+) = *<sup>A</sup>*(*α*+,*β*,*<sup>γ</sup>*+)*,* ∩ *t*∈*T A*(*<sup>α</sup>t*,*βt*+,*γt*+) = *<sup>A</sup>*(*<sup>α</sup>*,*β*+,*<sup>γ</sup>*+)*,* ∩ *t*∈*T A*(*<sup>α</sup>t*+,*βt*+,*γt*+) = *<sup>A</sup>*(*α*+,*β*+,*<sup>γ</sup>*+)*,*

*where e α* = ∨*t*∈*Tαt*, *β* = <sup>∧</sup>*t*∈*<sup>T</sup>β<sup>t</sup>*, *γ* = ∧*t*∈*Tγt*.

**Proof.** The proofs of (1)~(4) are obtained directly from Definition 9. We denote,

$$A \cup B = \left\{ \left< \mathbf{x}, \max \left\{ T\_A^j(\mathbf{x}), T\_B^j(\mathbf{x}) \right\}, \min \left\{ I\_A^j(\mathbf{x}), I\_B^j(\mathbf{x}) \right\}, \min \left\{ F\_A^j(\mathbf{x}), F\_B^j(\mathbf{x}) \right\} \right\} \right\},$$

$$A \cap B = \left\{ \left< \mathbf{x}, \min \left\{ T\_A^j(\mathbf{x}), T\_B^j(\mathbf{x}) \right\}, \max \left\{ I\_A^j(\mathbf{x}), I\_B^j(\mathbf{x}) \right\}, \max \left\{ F\_A^j(\mathbf{x}), F\_B^j(\mathbf{x}) \right\} \right\} \right\},$$
 $A \cap B = \left\{ (\mathbf{x} \in A \cup B) \right\}$ 

where *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*).

$$\begin{aligned} \cup\_{t \in T} A\_t &= \left\{ \left\langle \left\langle \mathbf{x}, \sup\_{t \in T} \left\{ T\_{A\_t}^j(\mathbf{x}) \right\} \Big| \,\_{t \in T} \inf\_{t \in T} \left\{ I\_{A\_t}^j(\mathbf{x}) \right\} \Big\rangle\_{t \in T} \inf\_{t \in T} \left\{ F\_{A\_t}^j(\mathbf{x}) \right\} \right\} \right\}, \\\\ \cap\_{t \in T} A\_t &= \left\{ \left\langle \mathbf{x}, \inf\_{t \in T} \left\{ T\_{A\_t}^j(\mathbf{x}) \right\} \mathbf{s} \sup\_{t \in T} \left\{ I\_{A\_t}^j(\mathbf{x}) \right\} \mathbf{s} \sup\_{t \in T} \left\{ F\_{A\_t}^j(\mathbf{x}) \right\} \right\} \right\}. \end{aligned}$$

*j* = 1, 2, ··· , *l* where *l* = max{ *l*(*x* : *At*)|*t* ∈ *<sup>T</sup>*}.

(5) From *x* ∈ (*A* ∩ *<sup>B</sup>*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we have min*TjA*(*x*), *TjB*(*x*)≥ *α*, max*IjA*(*x*), *<sup>I</sup>jB*(*x*)≤ *β*, max*FjA*(*x*), *FjB* (*x*)} ≤ *γ*, that is, *TjA*(*x*) ≥ *α* and *TjB*(*x*) ≥ *<sup>α</sup>*,*<sup>I</sup>jA*(*x*) ≤ *β* and *<sup>I</sup>jB*(*x*) ≤ *<sup>β</sup>*,*<sup>F</sup>jA*(*x*) ≤ *γ* and *<sup>F</sup>jB*(*x*) ≤ *γ*. Thus, *TjA*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*) ≤ *β*, *<sup>F</sup>jA*(*x*) ≤ *γ* and *TjB*(*x*) ≥ *α*, *<sup>I</sup>jB*(*x*) ≤ *β*, *<sup>F</sup>jB*(*x*) ≤ *γ*, that is, *x* <sup>∈</sup>*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), *x* ∈ *B*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). Hence, *x* ∈ *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ∩ *B*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). On the other hand, since *x* ∈ *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ∩ *<sup>B</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we have *x* ∈ *<sup>A</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), *x* ∈ *<sup>B</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), that is, *TjA*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*)<sup>≤</sup> *β*, *<sup>F</sup>jA*(*x*) ≤ *γ* and *TjB*(*x*) ≥ *α*, *<sup>I</sup>jB*(*x*) ≤ *β*, *<sup>F</sup>jB*(*x*) ≤ *γ*. Thus, *TjA*(*x*) ≥ *α* and *TjB*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*) ≤ *β* and *<sup>I</sup>jB*(*x*) ≤ *β*, *<sup>F</sup>jA*(*x*) ≤ *γ* and *<sup>F</sup>jB*(*x*) ≤ *γ*. Hence, min*TjA*(*x*), *TjB*(*x*) ≥ *<sup>α</sup>*,max*IjA*(*x*), *<sup>I</sup>jB*(*x*) ≤ *β*, max*FjA*(*x*), *<sup>F</sup>jB*(*x*) ≤ *γ*. So, *x* ∈ (*A* ∩ *<sup>B</sup>*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). Therefore, (*A* ∩ *<sup>B</sup>*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ∩ *B*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*). (6) Since *x* ∈ *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ∪ *<sup>B</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we have *x* ∈ *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) or *x* ∈ *<sup>B</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), that is, *TjA*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*) ≤

*β*, *<sup>F</sup>jA*(*x*) ≤ *γ* or *TjB*(*x*) ≥ *α*, *<sup>I</sup>jB*(*x*) ≤ *β*, *<sup>F</sup>jB*(*x*) ≤ *γ*. Thus, *TjA*(*x*) ≥ *α* or *TjB*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*) ≤ *β* or *<sup>I</sup>jB*(*x*) ≤ *β*, *<sup>F</sup>jA*(*x*) ≤ *γ* or *<sup>F</sup>jB*(*x*) ≤ *γ*, that is, max*TjA*(*x*), *TjB*(*x*) ≥ *α*, min*IjA*(*x*), *<sup>I</sup>jB*(*x*) ≤

*<sup>β</sup>*,min*FjA*(*x*), *<sup>F</sup>jB*(*x*)<sup>≤</sup> *γ*. Thus, *x* ∈ (*A* ∪ *<sup>B</sup>*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). Therefore, *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ∪ *B*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ⊆ (*A* ∪ *<sup>B</sup>*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*). (7) From *x* ∈ 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we have inf *<sup>t</sup>*∈*<sup>T</sup>TjAt*(*x*) ≥ *α*, sup *t*∈*T IjAt*(*x*) ≤ *β*, sup *t*∈*T FjAt*(*x*) ≤ *γ*, that is, *TjAt*(*x*)<sup>≥</sup> *α*, *<sup>I</sup>jAt*(*x*) ≤ *β*, *<sup>F</sup>jAt*(*x*) ≤ *γ* for all *t* ∈ *T*. Thus, *x* ∈ (*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) for all *t* ∈ *T*. Hence,

*x* ∈ ∩ *<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). On the other hand, for any *x* ∈ ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we have *x* ∈ (*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) for all *t* ∈ *T*,that is, *TjAt*(*x*) ≥ *α*, *<sup>I</sup>jAt*(*x*) ≤ *β*, *<sup>F</sup>jAt*(*x*) ≤ *γ* for all *t* ∈ *T*. Thus,

$$\inf\_{t \in T} \left\{ F\_{A\_t}^j(\mathbf{x}) \right\} \ge \alpha, \sup\_{t \in T} \left\{ I\_{A\_t}^j(\mathbf{x}) \right\} \le \beta, \sup\_{t \in T} \left\{ F\_{A\_t}^j(\mathbf{x}) \right\} \le \gamma.$$

Hence, *x* ∈ 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). Therefore, 5 ∩*<sup>t</sup>*∈*<sup>T</sup>At*6(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = ∩*<sup>t</sup>*∈*<sup>T</sup>*(*At*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) for *j* = 1, 2, ··· , *l*(*x* : *A*, *<sup>B</sup>*). (8)Theproofof(8)issimilartothatof(6).

 (9) Since *x* ∈ ∩*<sup>t</sup>*∈*TA*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*), we have *x* ∈ *A*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*) for all *t* ∈ *T*, that is,

$$T\_A^j(\mathbf{x}) \ge \alpha\_{t\prime} I\_A^j(\mathbf{x}) \le \beta\_{t\prime} F\_A^j(\mathbf{x}) \le \gamma\_t \text{ for all } t \in T.$$

Thus, *TjA*(*x*) ≥ ∨*<sup>t</sup>*∈*Tαt*, *<sup>I</sup>jA*(*x*) ≤ ∧*<sup>t</sup>*∈*<sup>T</sup>β<sup>t</sup>*, *<sup>F</sup>jA*(*x*) ≤ ∧*<sup>t</sup>*∈*Tγt*, that is, *TjA*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*) ≤ *β*, *<sup>F</sup>jA*(*x*) ≤ *γ*. Thus, *x* <sup>∈</sup>*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*). On the other hand, from *x* ∈ *<sup>A</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we have *TjA*(*x*) ≥ *α*, *<sup>I</sup>jA*(*x*) ≤ *β*, *<sup>F</sup>jA*(*x*) ≤ *γ*, that is, *TjA*(*x*) ≥ ∨*<sup>t</sup>*∈*Tαt*, *<sup>I</sup>jA*(*x*) ≤ ∧*<sup>t</sup>*∈*<sup>T</sup>β<sup>t</sup>*, *<sup>F</sup>jA*(*x*) ≤ ∧*<sup>t</sup>*∈*Tγ<sup>t</sup>* for all *t* ∈ *T*. Thus, *TjA*(*x*) ≥ *αt*, *<sup>I</sup>jA*(*x*) ≤ *β<sup>t</sup>*, *<sup>F</sup>jA*(*x*) ≤ *γt* for all *t* ∈ *T*. Thus,*x* ∈ *A*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*) for all *t* ∈ *T*. Hence, *x* ∈ ∩ *t*∈*T A*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*). Therefore, ∩ *t*∈*T A*(*<sup>α</sup>t*,*β<sup>t</sup>*,*γt*) = *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) for *j* = 1, 2, ··· ,*j* = 1, 2, ··· , *l*(*l* = max{ *l*(*x* : *At*)|*t* ∈ *<sup>T</sup>*}). -

**Remark 3.** *In property (6)* (*A* ∪ *<sup>B</sup>*)(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ⊇ *A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) ∪ *<sup>B</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)*, "*⊇*" cannot be strengthened as "*=*". For example, let X* = {*<sup>x</sup>*1, *x*2, *<sup>x</sup>*3}*, A*, *B* ∈ *SVNMS*(*X*) *as follows:*

*A* = {*<sup>x</sup>*1,(0.5, 0.3),(0.1, 0.1),(0.7, 0.8),*<sup>x</sup>*2,(0.7, 0.68, 0.62),(0.3, 0.45, 0.5),(0.34, 0.28, 0.49), *<sup>x</sup>*3,(0.67, 0.5, 0.3),(0.2, 0.3, 0.4),(0.4, 0.5, 0.7)}

$$B = \{ \langle \mathbf{x}\_1, 0.75, 0.2, 0.15 \rangle, \langle \mathbf{x}\_2, (0.43, 0.37, 0.28), (0.5, 0.2, 0.3), (0.7, 0.8, 0.9) \rangle, \\ \langle \mathbf{x}\_3, (1.0, 0.86, 0.79), (0.01, 0.1, 0.2), (0.0, 0.3, 0.2) \rangle \}.$$

*If we choose α* = 0.4*, β* = 0.3*, γ* = 0.5*, then,*

$$A^{(a,\delta,\gamma)} = \{ \langle \mathbf{x}\_1, (1,0), (1,1), (0,0) \rangle, \langle \mathbf{x}\_2, (1,1,1), (1,0,0), (1,1,1) \rangle, \langle \mathbf{x}\_3, (1,0), (1,1,0), (1,1,0) \rangle \},$$

$$B^{(a,\delta,\gamma)} = \{ \langle \mathbf{x}\_1, 1, 1, 1 \rangle, \langle \mathbf{x}\_2, (1,0,0), (0,1,1), (0,0,0) \rangle, \langle \mathbf{x}\_3, (1,1,1), (1,1,1), (1,1,1) \rangle \},$$

$$A \cup B^{(a,\delta,\gamma)} = \{ \langle \mathbf{x}\_1, (1,0), (1,1), (1,0) \rangle, \langle \mathbf{x}\_2, (1,1,1), (1,1,1), (1,1,1) \rangle, \langle \mathbf{x}\_3, (1,1,1), (1,1,1), (1,1,1) \rangle \},$$

$$A^{(a,\delta,\gamma)} \cup B^{(a,\delta,\gamma)} = \{ \langle \mathbf{x}\_1, (1,0), (1,1), (0,0) \rangle, \langle \mathbf{x}\_2, (1,1,1), (0,0,0), (0,0,0) \rangle, \langle \mathbf{x}\_3, (1,1,1), (1,1,0), (1,1,0) \rangle \},$$

$$\text{Obviously}, (A \cup B)^{(a,\delta,\gamma)} \neq A^{(a,\delta,\gamma)} \cup B^{(a,\delta,\gamma)}.$$

In order to ge<sup>t</sup> the decomposition theorem of SVNMS, we also need to introduce the following important concepts.

**Definition 10.** *Let L* = {(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)|*<sup>α</sup>*, *β*, *γ* ∈ [0, <sup>1</sup>], 0 ≤ *α* + *β* + *γ* ≤ <sup>3</sup>}*,* (*<sup>α</sup>*1, *β*1, *<sup>γ</sup>*1) ≤ (*<sup>α</sup>*2, *β*2, *<sup>γ</sup>*2) ⇔ *α*1 ≤ *α*2, *β*1 ≥ *β*2 *, γ*1 ≥ *γ*2 *So, L is a complete lattice, and* (1, 0, 0) *is the biggest element,* (0, 1, 1) *is the smallest element.*

**Definition 11.** *Let* (*<sup>α</sup>*, *β*, *γ*) ∈ *L, B* ∈ 2*X, A* = (*<sup>α</sup>*, *β*, *γ*)*B. And for any x* ∈ *X*,

$$T\_A^j(\mathbf{x}) = \left\{ \begin{array}{c} \mathbf{a}, \mathbf{x} \in B \\ \mathbf{0}, \mathbf{x} \notin B \end{array} \; , \; I\_A^j(\mathbf{x}) = \left\{ \begin{array}{c} \boldsymbol{\beta}, \mathbf{x} \in B \\ \mathbf{1}, \mathbf{x} \notin B \end{array} \; , \; F\_A^j(\mathbf{x}) = \left\{ \begin{array}{c} \boldsymbol{\gamma}, \mathbf{x} \in B \\ \mathbf{1}, \mathbf{x} \notin B \end{array} \; . \right. \right. \end{array} \right. \tag{21}$$

Then, *A* = *x*, *TjA*(*x*), *<sup>I</sup>jA*(*x*), *<sup>F</sup>jA*(*x*)HHH*x* ∈ *X*, *j* = 1, 2, ··· , *l*(*x* : *A*) is a SVNMS on the universe *X*, so we have the definition as follows:

**Definition 12.** *Suppose A* ∈ *SVNMS*(*X*)*,* (*<sup>α</sup>*, *β*, *γ*) ∈ *L, the dot product (truncated product) of* (*<sup>α</sup>*, *β*, *γ*) *and A is defined as*

$$((\mathbf{a}, \beta, \gamma)A)(\mathbf{x}) = \left\{ \left< \mathbf{x}, \mathbf{a} \lor T\_A^j(\mathbf{x}), \beta \land I\_A^j(\mathbf{x}), \ \gamma \land F\_A^j(\mathbf{x}) \right> \middle| \mathbf{x} \in \mathbf{X}, j = 1, 2, \cdots, l(\mathbf{x} : A) \right\}.\tag{22}$$

*That is* (*<sup>α</sup>*, *β*, *γ*)*<sup>A</sup>* ∈ *SVNMS*(*X*).

Now, we can discuss the decomposition theorem of SVNMS based on the definitions and operational properties above.

**Theorem 3.** *Let A be a SVNMS. Then for any* (*<sup>α</sup>*, *β*, *γ*) ∈ *L, we have*

$$A = \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a,\emptyset,\gamma)} = \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a+,\emptyset,\gamma)} = \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a,\emptyset+,\gamma)}$$

$$= \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a,\emptyset,\gamma+)} = \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a+,\emptyset+,\gamma)} = \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a+,\emptyset,\gamma+)}$$

$$= \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma)A^{(a,\emptyset+,\gamma+)} = \bigcup\_{(a,\emptyset,\gamma)\in L} (a,\emptyset,\gamma\cdot)A^{(a+,\emptyset+,\gamma+)} \tag{23}$$

**Proof.** With regard to *A* = ∪ (*<sup>α</sup>*,*β*,*<sup>γ</sup>*)∈*<sup>L</sup>*(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*), we just need to prove *<sup>A</sup>*(*x*) = 5 ∪ (*<sup>α</sup>*,*β*,*<sup>γ</sup>*)∈*<sup>L</sup>*(*<sup>α</sup>*, *β*, *γ*) *<sup>A</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)(*x*) for all *x* ∈ *X*. That is, *<sup>A</sup>*(*x*) = <sup>∨</sup>(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)∈*<sup>L</sup>*(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)(*x*)= <sup>∨</sup>*α*<sup>∈</sup>[0,1]*<sup>α</sup>* ∨ (*A<sup>α</sup>*)*j*(*x*), <sup>∧</sup>*β*∈[0,1]*<sup>β</sup>* ∧ (*Aβ*)*j*(*x*), <sup>∧</sup>*γ*<sup>∈</sup>[0,1]*<sup>γ</sup>* ∧ (*<sup>A</sup>γ*)*j*(*x*) for *x* ∈ *X*. Since *TjA*(*x*) ∈ [0, 1], we have <sup>∨</sup>*α*<sup>∈</sup>[0,1](*α*<sup>∨</sup>(*A<sup>α</sup>*)*j*(*x*)) = ?<sup>∨</sup>*α*<sup>∈</sup>[0,*TjA*(*x*)]*<sup>α</sup>* ∨ (*A<sup>α</sup>*)*j*(*x*)@ ∨ ?<sup>∨</sup>*α*<sup>∈</sup>[*TjA*(*x*),<sup>1</sup>]*<sup>α</sup>* ∨ (*A<sup>α</sup>*)*j*(*x*)@. Indeed, taking *α* ≤ *TjA*(*x*), we have (*A<sup>α</sup>*)*j*(*x*) = 1, otherwise, (*A<sup>α</sup>*)*j*(*x*) = 0. Thus, <sup>∨</sup>*α*<sup>∈</sup>[0,1]*<sup>α</sup>* ∨ (*A<sup>α</sup>*)*j*(*x*) = ∨*α*<sup>∈</sup>[0,*TjA*(*x*)]*<sup>α</sup>* ∨ (*A<sup>α</sup>*)*j*(*x*) =∨*α*<sup>∈</sup>[0,*TjA*(*x*)]*<sup>α</sup>* = *TjA*(*x*). Similarly,<sup>∧</sup>*β*∈[0,1]*<sup>β</sup>* ∧ (*Aβ*)*j*(*x*) = <sup>∧</sup>*β*∈[*IjA*(*x*),<sup>1</sup>]*<sup>β</sup>* ∧ (*Aβ*)*j*(*x*) = <sup>∧</sup>*β*∈[*IjA*(*x*),<sup>1</sup>]*<sup>β</sup>* = *<sup>I</sup>jA*(*x*) and <sup>∧</sup>*γ*<sup>∈</sup>[0,1]*<sup>γ</sup>* ∧ (*<sup>A</sup>γ*)*j*(*x*) =∧*<sup>γ</sup>*<sup>∈</sup>[*FjA*(*x*),<sup>1</sup>]*<sup>γ</sup>* ∧ (*<sup>A</sup>γ*)*j*(*x*) = ∧*<sup>γ</sup>*<sup>∈</sup>[*FjA*(*x*),<sup>1</sup>]*<sup>γ</sup>* = *<sup>F</sup>jA*(*x*). Therefore, <sup>∨</sup>(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)∈*<sup>L</sup>*(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)(*x*) = *TjA*(*x*), *<sup>I</sup>jA*(*x*), *<sup>F</sup>jA*(*x*) = *<sup>A</sup>*(*x*) for *j* = 1, 2, ··· *l*(*x* : *<sup>A</sup>*). -

Next, we use an example to illustrate the idea of the decomposition theorem of SVNMS. **Example 1.** *Let X* = {*<sup>x</sup>*1, *x*2, *<sup>x</sup>*3}*, A* ∈ *SVNMS*(*X*) *as follows:*

*A* = {*<sup>x</sup>*1,(0.6, 0.4),(0.5, 0.3),(0.2, 0.3),*<sup>x</sup>*2, 0.2, 0.4, 0.7,*<sup>x</sup>*3,(0.8, 0.6, 0.5),(0.2, 0.2, 0.3),(0.1, 0.3, 0.4)}.

*We show how A can be represented by180 special SVNMSs using* (*<sup>α</sup>*, *β*, *<sup>γ</sup>*)-*cut sets. According to Definition 9, 11 and 12, we have:*

$$A^{(\mathbf{a},\mathbf{\bar{b}},\gamma)} = \{ \langle \mathbf{x}\_1, (1,1), (0,0), (0,0) \rangle, \langle \mathbf{x}\_2, \mathbf{1}, 0, 0 \rangle, \langle \mathbf{x}\_3, (1,1,1), (1,1,0), (1,0,0) \rangle \},$$

(*<sup>α</sup>*, *β*, *γ*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = {*<sup>x</sup>*1,(0.2, 0.2),(1, <sup>1</sup>),(1, <sup>1</sup>),*<sup>x</sup>*2, 0.2, 1, <sup>1</sup>,*<sup>x</sup>*3,(0.2, 0.2, 0.2),(0.2, 0.2, <sup>1</sup>),(0.1, 1, <sup>1</sup>)}, *where* 0 ≤ *α* ≤ 0.2*,* 0 ≤ *β* ≤ 0.2*,* 0 ≤ *γ* ≤ 0.1*;*

$$A^{(\mathbf{a},\mathbf{\bar{b}},\gamma)} = \{ \langle \mathbf{x}\_1, (1,1), (0,1), (1,0) \rangle, \langle \mathbf{x}\_2, 0, 0, 0 \rangle, \langle \mathbf{x}\_3, (1,1,1), (1,1,1), (1,0,0) \rangle \},$$

(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = {*<sup>x</sup>*1,(0.4, 0.4),(1, 0.3),(0.2, <sup>1</sup>),*<sup>x</sup>*2, 1, 1, <sup>1</sup>,*<sup>x</sup>*3,(0.4, 0.4, 0.4),(0.3, 0.3, 0.3),(0.2, 1, <sup>1</sup>)}, *where* 0.2 < *α* ≤ 0.4*,* 0.2 < *β* ≤ 0.3*,* 0.1 < *γ* ≤ 0.2*;*

$$A^{(a,b,\gamma)} = \{ \langle \mathbf{x}\_1, (1,0), (0,1), (1,1) \rangle, \langle \mathbf{x}\_2, 0, 1, 0 \rangle, \langle \mathbf{x}\_{3\prime}(1,1,1), (1,1,1), (1,1,0) \rangle \}\_{\mathbf{x}\_{3\prime}}$$

(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = {*<sup>x</sup>*1,(0.5, <sup>0</sup>),(1, 0.4),(0.3, 0.3),*<sup>x</sup>*2, 0, 0.4, <sup>1</sup>,*<sup>x</sup>*3,(0.5, 0.5, 0.5),(0.4, 0.4, 0.4),(0.3, 0.3, <sup>1</sup>)}, *where* 0.4 < *α* ≤ 0.5*,* 0.3 < *β* ≤ 0.4*,* 0.2 < *γ* ≤ 0.3*;*

$$A^{(a,b,\gamma)} = \{ \langle \mathbf{x}\_1, (1,0), (1,1), (1,1) \rangle, \langle \mathbf{x}\_2, 0, 1, 0 \rangle, \langle \mathbf{x}\_{3\prime}(1,1,0), (1,1,1), (1,1,1) \rangle \},$$

(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = {*<sup>x</sup>*1,(0.6, <sup>0</sup>),(0.5, 0.5),(0.4, 0.4),*<sup>x</sup>*2, 0, 0.5, <sup>1</sup>,*<sup>x</sup>*3,(0.6, 0.6, <sup>0</sup>),(0.5, 0.5, 0.5),(0.4, 0.4, 0.4)}, *where* 0.5 < *α* ≤ 0.6*,* 0.4 < *β* ≤ 0.5*,* 0.3 < *γ* ≤ 0.4*;*

$$A^{(\mathbf{a},\mathbf{\bar{b}},\gamma)} = \{ \langle \mathbf{x}\_1, (0,0), (1,1), (1,1) \rangle, \langle \mathbf{x}\_2, 0, 1, 1 \rangle, \langle \mathbf{x}\_3, (1,0,0), (1,1,1), (1,1,1) \rangle \},$$

(*<sup>α</sup>*, *β*, *γ*)*A*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*) = {*<sup>x</sup>*1,(0, <sup>0</sup>),(1, <sup>1</sup>),(0.7, 0.7),*<sup>x</sup>*2, 0, 1, 0.7,*<sup>x</sup>*3,(0.8, 0, <sup>0</sup>),(1, 1, <sup>1</sup>),(0.7, 0.7, 0.7)}, *where* 0.6 < *α* ≤ 0.8*,* 0.5 < *β* ≤ 1*,* 0.4 < *γ* ≤ 0.7*;*

$$A^{\langle \mathbf{a}, \mathbf{b}, \gamma \rangle} = \{ \langle \mathbf{x}\_{1\prime}(0,0), (1,1), (1,1) \rangle, \langle \mathbf{x}\_{2\prime}0, \mathbf{1}, 1 \rangle, \langle \mathbf{x}\_{3\prime}(0,0,0), (1,1,1), (1,1,1) \rangle \},$$

$$(\mathbf{a}, \boldsymbol{\beta}, \gamma)A^{(\mathbf{a}, \boldsymbol{\beta}, \gamma)} = \{ \langle \mathbf{x}\_1, (0, 0), (1, 1), (1, 1) \rangle, \langle \mathbf{x}\_2, 0, 1, 1 \rangle, \langle \mathbf{x}\_3, (0, 0, 0), (1, 1, 1), (1, 1, 1) \rangle \},$$

*where* 0.8 < *α* ≤ 1*,* 0.5 < *β* ≤ 1*,* 0.7 < *γ* ≤ 1.

> *Similarly, we can get the rest of the results with special SVNMSs. It is obvious to see,*

$$A = \underset{(\mathfrak{a}, \mathfrak{b}, \gamma) \in L}{\cup} (\mathfrak{a}, \mathfrak{b}, \gamma) \, A^{(\mathfrak{a}, \mathfrak{b}, \gamma)} \,.$$

**Definition 13.** *Suppose H* : *L* → 2*<sup>X</sup> ,* (*<sup>λ</sup>*, *μ*, *ω*) )→ *<sup>H</sup>*(*<sup>λ</sup>*, *μ*, *ω*) *is a mapping, a neutrosophic nested set H can be defined in X if it satisfies the following conditions:*

*(1)* (*<sup>λ</sup>*1, *μ*1, *<sup>ω</sup>*1) ≤ (*<sup>λ</sup>*2, *μ*2, *<sup>ω</sup>*2) ⇒ *<sup>H</sup>*(*<sup>λ</sup>*1, *μ*1, *<sup>ω</sup>*1) ⊇ *<sup>H</sup>*(*<sup>λ</sup>*2, *μ*2, *<sup>ω</sup>*2)*;*

*(2)* <sup>∩</sup>*t*∈*T<sup>H</sup>*(*<sup>λ</sup><sup>t</sup>*, *μ<sup>t</sup>*, *<sup>ω</sup>t*) ⊆ {*H*(*<sup>λ</sup>*, *μ*, *ω*)|*<sup>λ</sup>* < <sup>∨</sup>*t*∈*Tλ<sup>t</sup>* , *μ* > ∧*t*∈*<sup>T</sup>μ<sup>t</sup>*, *ω* > <sup>∧</sup>*t*∈*Tωt*}.

**Remark 4.** *Let SVNL be a set which composed of all neutrosophic nested sets, A* ∈ *SVNMS*(*X*)*, then, all* (*<sup>α</sup>*, *β*, *<sup>γ</sup>*)-*cut sets of A are neutrosophic nested sets.*

**Theorem 4.** *Let A* ∈ *SVNMS*(*X*)*, H* : *L* → 2*<sup>X</sup> ,* (*<sup>α</sup>*, *β*, *γ*) )→ *<sup>H</sup>*(*<sup>α</sup>*, *β*, *<sup>γ</sup>*)*, for any* (*<sup>α</sup>*, *β*, *γ*) ∈ *L satisfy A*(*α*+,*β*+,*<sup>γ</sup>*+) ⊆ *<sup>H</sup>*(*<sup>α</sup>*, *β*, *γ*) ⊆ *<sup>A</sup>*(*<sup>α</sup>*,*β*,*<sup>γ</sup>*)*, then*

$$(\mathcal{I})\quad A = \underset{(\mathfrak{a},\mathfrak{b},\gamma)\in L}{\cup} (\mathfrak{a},\mathfrak{b},\gamma)H(\mathfrak{a},\mathfrak{b},\gamma);$$

