Hybrid similarity measures of single-valued neutrosophic type-2 fuzzy sets and their application to MCDM based on TOPSIS

Abstract

Similarity measure (SM) formulas are very useful tool for Multi-criteria decision-making (MCDM) problems, machine learning, medical diagnosis, psychology, etc. In this paper, some SM methods are introduced under single-valued type-2 neutrosophic (SVT2N) information, based on the Dice and Jaccard vector measures. Also, Cosine and weighted Cosine SMs between two SVT2N sets are defined for SVT2NSs. Then, motivated by the idea of vector measures, Hybrid SM and weighted Hybrid SM are developed by combining Dice and Cosine SMs. After then, a decision-making method is put forward based on technique for order of preference by similarity to ideal solution (TOPSIS) method. Immediately after, a real example is given to indicate the practicality and effectiveness of the proposed measures. In here, Hybrid SMs are discussed for different values of $\lambda$ and compared with the Hybrid SMs and weighted Hybrid SMs in their own. Furthermore, a comparative analysis is made based on the TOPSIS among the vector measures, distance measures and Hybrid SMs and the agreement between the results with other measures shows that the Hybrid SM is strong and essential to minimize error margin for decision makers.

A Appendices

1.1 A.1

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is open that $\mathbb {D}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Then, let us prove that $\mathbb {D}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$. Since $x^2+y^2\ge 2xy$, we have for all $\tau _i\in \mathfrak {X} (i=1,2,\ldots ,n)$

  $$\begin{aligned}&(\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2\ge 2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\ge 2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2\ge 2 \mathfrak {h}_{\mathfrak {A}_p}(\tau _i) \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\ge 2 \mathfrak {h}_{\mathfrak {A}_s}(\tau _i) \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2\ge 2 \mathfrak {f}_{\mathfrak {A}_p}(\tau _i) \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2\ge 2 \mathfrak {f}_{\mathfrak {A}_s}(\tau _i) \mathfrak {f}_{\mathfrak {B}_s}(\tau _i). \end{aligned}$$

  and $\forall \tau _i\in \mathfrak {X}$

  $$\begin{aligned} \mathbb {D}_{SVT2NS}(\mathfrak {A}, \mathfrak {B}) =\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ 2f_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] } \le 1. \end{aligned}$$

  Thus, for $i=1,2,\ldots ,n$

  $$\begin{aligned}&=\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _1) \mathfrak {t}_{\mathfrak {B}_p}(\tau _1)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _1) \mathfrak {t}_{\mathfrak {B}_s}(\tau _1)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _1)\mathfrak {h}_{\mathfrak {B}_p}(\tau _1) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _1)\mathfrak {h}_{\mathfrak {B}_s}(\tau _1)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _1)\mathfrak {f}_{\mathfrak {B}_p}(\tau _1)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _1)\mathfrak {f}_{\mathfrak {B}_s}(\tau _1)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _1))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _1))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _1))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _1))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _1))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _1))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _1))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _1))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _1))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _1))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _1))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _1))^2 \end{array}\right] } +\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _2) \mathfrak {t}_{\mathfrak {B}_p}(\tau _2)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _2) \mathfrak {t}_{\mathfrak {B}_s}(\tau _2)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _2)\mathfrak {h}_{\mathfrak {B}_p}(\tau _2) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _2)\mathfrak {h}_{\mathfrak {B}_s}(\tau _2)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _2)\mathfrak {f}_{\mathfrak {B}_p}(\tau _2)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _2)\mathfrak {f}_{\mathfrak {B}_s}(\tau _2)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _2))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _2))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _2))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _2))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _2))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _2))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _2))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _2))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _2))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _2))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _2))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _2))^2 \end{array}\right] }\\&\quad + \vdots + \frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _n) \mathfrak {t}_{\mathfrak {B}_p}(\tau _n)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _n) \mathfrak {t}_{\mathfrak {B}_s}(\tau _n)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _n)\mathfrak {h}_{\mathfrak {B}_p}(\tau _n) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _n)\mathfrak {h}_{\mathfrak {B}_s}(\tau _n)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _n)\mathfrak {f}_{\mathfrak {B}_p}(\tau _n)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _n)\mathfrak {f}_{\mathfrak {B}_s}(\tau _n)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _n))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _n))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _n))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _n))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _n))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _n))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _n))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _n))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _n))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _n))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _n))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _n))^2 \end{array}\right] } \le 1+1+\cdots +1 \end{aligned}$$

  Then, if collection is made from side to side;

  $$\begin{aligned}&\sum _{i=1}^n\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +2\mathfrak {h}_{A _s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] }\le n \frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] }\le 1 \end{aligned}$$

• (P2) if $\mathfrak {A}=\mathfrak {B}$,

  $$\begin{aligned}&\frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l}2 (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 + 2(\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2 +2(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+ 2(\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] }{\left[ \begin{array}{l}2 (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 + 2(\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2 +2(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+ 2(\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] }=1 \end{aligned}$$

• (P3) if $\mathfrak {A}$ and $\mathfrak {B}$ are swapped;

  $$\begin{aligned}&\mathbb {D}_{SVT2NS}(\mathfrak {A}, \mathfrak {B}) =\frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] }\\&\quad =\frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {B}_p}(\tau _i) \mathfrak {t}_{\mathfrak {A}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {B}_s}(\tau _i) \mathfrak {t}_{\mathfrak {A}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\mathfrak {h}_{\mathfrak {A}_p}(\tau _i) +2\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)+\\ 2\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] } =\mathbb {D}_{SVT2NS}(\mathfrak {B}, \mathfrak {A}). \end{aligned}$$

1.2 A.2

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is open that $\mathbb {J}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Then, we need only show that $\mathbb {J}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$. We know that for all $x,y\in R$, $x^2+y^2-xy\ge xy$, then for all $\tau _i\in \mathfrak {X}$ we get

  $$\begin{aligned}&(\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2-\mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\ge \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2-\mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\ge \mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2-\mathfrak {h}_{\mathfrak {A}_p}(\tau _i) \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\ge \mathfrak {h}_{\mathfrak {A}_p}(\tau _i) \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2-\mathfrak {h}_{\mathfrak {A}_s}(\tau _i) \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\ge \mathfrak {h}_{\mathfrak {A}_s}(\tau _i) \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2-\mathfrak {f}_{\mathfrak {A}_p}(\tau _i) \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\ge \mathfrak {f}_{\mathfrak {A}_p}(\tau _i) \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2-\mathfrak {f}_{\mathfrak {A}_s}(\tau _i) \mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\ge \mathfrak {f}_{\mathfrak {A}_s}(\tau _i) \mathfrak {f}_{\mathfrak {B}_s}(\tau _i). \end{aligned}$$

  Therefore, for all $\tau _i\in \mathfrak {X},$

  $$\begin{aligned}&\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) \\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }\le 1 \end{aligned}$$

  and for $i=1,2,\ldots ,n$;

  $$\begin{aligned} \sum _{i=1}^n\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }\le n. \end{aligned}$$

  It is conclude that if the operation is applied with $\frac{1}{n}$; $\mathbb {J}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})=$

  $$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }\le 1. \end{aligned}$$

• (P2) if $\mathfrak {A}=\mathfrak {B}$,

  $$\begin{aligned}&\frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2 +(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+ (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2 +(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+ (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] }=1 \end{aligned}$$

• (P3) if SVNT2Ss $\mathfrak {A},\mathfrak {B}$ are swapped;

  $$\begin{aligned}&\mathbb {J}_{SVT2NS}(\mathfrak {A}, \mathfrak {B}) =\frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\ + \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }\\&\quad =\frac{1}{n}\sum _{i=1}^n\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {B}_p}(\tau _i) \mathfrak {t}_{\mathfrak {A}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {B}_s}(\tau _i) \mathfrak {t}_{\mathfrak {A}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\mathfrak {h}_{\mathfrak {A}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {B}_p}(\tau _i) \mathfrak {t}_{\mathfrak {A}_p}(\tau _i)\\ +\mathfrak {t}_{\mathfrak {B}_s}(\tau _i) \mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\\ + \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\\ +\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\\ + \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\\ +\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\\ \end{array}\right] }\\&\quad =\mathbb {J}_{SVT2NS}(\mathfrak {B}, \mathfrak {A}). \end{aligned}$$

1.3 A.3

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is open that $\mathbb {C}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Thus, we need only show that $\mathbb {C}_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$. Here

  $$\begin{aligned}&\sum _{i=1}^n\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] =\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _1) \mathfrak {t}_{\mathfrak {B}_p}(\tau _1)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _1) \mathfrak {t}_{\mathfrak {B}_s}(\tau _1)+\mathfrak {h}_{\mathfrak {A}_p}(\tau _1)\mathfrak {h}_{\mathfrak {B}_p}(\tau _1)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _1)\mathfrak {h}_{\mathfrak {B}_s}(\tau _1)+\mathfrak {f}_{\mathfrak {A}_p}(\tau _1) \mathfrak {f}_{\mathfrak {B}_p}(\tau _1)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _1)\mathfrak {f}_{\mathfrak {B}_s}(\tau _1)\\ \end{array}\right] \\&\quad +\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _2) \mathfrak {t}_{\mathfrak {B}_p}(\tau _2)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _2) \mathfrak {t}_{\mathfrak {B}_s}(\tau _2)+\mathfrak {h}_{\mathfrak {A}_p}(\tau _2)\mathfrak {h}_{\mathfrak {B}_p}(\tau _2)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _2)\mathfrak {h}_{\mathfrak {B}_s}(\tau _2)+\mathfrak {f}_{\mathfrak {A}_p}(\tau _2)\mathfrak {f}_{\mathfrak {B}_p}(\tau _2)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _2)\mathfrak {f}_{\mathfrak {B}_s}(\tau _2)\\ \end{array}\right] + \vdots + \left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _n) \mathfrak {t}_{\mathfrak {B}_p}(\tau _n)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _n) \mathfrak {t}_{\mathfrak {B}_s}(\tau _n)+\mathfrak {h}_{\mathfrak {A}_p}(\tau _n)\mathfrak {h}_{\mathfrak {B}_p}(\tau _n)\\ +\mathfrak {h}_{\mathfrak {A}_s}(\tau _n)\mathfrak {h}_{\mathfrak {B}_s}(\tau _n)+\mathfrak {f}_{\mathfrak {A}_p}(\tau _2)\mathfrak {f}_{\mathfrak {B}_p}(\tau _n)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _n)\mathfrak {f}_{\mathfrak {B}_s}(\tau _n)\\ \end{array}\right] . \end{aligned}$$

  By using Cauchy-Schwarz inequality, we can write the following inequality:

  $$\begin{aligned}&(\mathbb {C}_{SVT2NS}(\mathfrak {A}, \mathfrak {B}))^2 \le \Big [(\mathfrak {t}_{\mathfrak {A}_p}(\tau _1))^2+ (\mathfrak {t}_{\mathfrak {A}_s}(\tau _1))^2+ (\mathfrak {h}_{\mathfrak {A}_p}(\tau _1))^2+ (\mathfrak {h}_{\mathfrak {A}_s}(\tau _1))^2+ (\mathfrak {f}_{\mathfrak {A}_p}(\tau _1))^2\\&\quad +(\mathfrak {f}_{\mathfrak {A}_s}(\tau _1))^2 + (\mathfrak {t}_{\mathfrak {A}_p}(\tau _2))^2 + (\mathfrak {t}_{\mathfrak {A}_s}(\tau _2))^2+ (\mathfrak {h}_{\mathfrak {A}_p}(\tau _2))^2+ (\mathfrak {h}_{\mathfrak {A}_s}(\tau _2))^2\\&\quad + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _2))^2+ (\mathfrak {f}_{\mathfrak {A}_s}(\tau _2))^2 + \cdots + (\mathfrak {t}_{\mathfrak {A}_p}(\tau _n))^2+( \mathfrak {t}_{\mathfrak {A}_s}(\tau _n))^2+ (\mathfrak {h}_{\mathfrak {A}_p}(\tau _n))^2\\&\quad + (\mathfrak {h}_{\mathfrak {A}_s}(\tau _n))^2+( \mathfrak {f}_{\mathfrak {A}_p}(\tau _n))^2+ (\mathfrak {f}_{\mathfrak {A}_s}(\tau _n))^2\Big ]\times \Big [(\mathfrak {t}_{\mathfrak {B}_p}(\tau _1))^2+( \mathfrak {t}_{\mathfrak {B}_s}(\tau _1))^2\\&\quad + (\mathfrak {h}_{\mathfrak {B}_p}(\tau _1))^2+ (\mathfrak {h}_{\mathfrak {B}_s}(\tau _1))^2+ (\mathfrak {f}_{\mathfrak {B}_p}(\tau _1))^2+( \mathfrak {f}_{\mathfrak {B}_s}(\tau _1))^2 + (\mathfrak {t}_{\mathfrak {B}_p}(\tau _2))^2\\&\quad +( \mathfrak {t}_{\mathfrak {B}_s}(\tau _2))^2+( \mathfrak {h}_{\mathfrak {B}_p}(\tau _2))^2+ (\mathfrak {h}_{\mathfrak {B}_s}(\tau _2))^2+ (\mathfrak {f}_{\mathfrak {B}_p}(\tau _2))^2+ (\mathfrak {f}_{\mathfrak {B}_s}(\tau _2))^2 \\&\quad + \cdots + (\mathfrak {t}_{\mathfrak {B}_p}(\tau _n))^2+( \mathfrak {t}_{\mathfrak {B}_s}(\tau _n))^2( \mathfrak {h}_{\mathfrak {B}_p}(\tau _n))^2+( \mathfrak {h}_{\mathfrak {B}_s}(\tau _n))^2+( \mathfrak {f}_{\mathfrak {B}_p}(\tau _n))^2 +(\mathfrak {f}_{\mathfrak {B}_s}(\tau _n))^2\Big ]. \end{aligned}$$

  It is concluded that $\mathbb {C}_{SVNT2FS}(\mathfrak {A}, \mathfrak {B})\le 1$.

• (P2) It can be easily indicated from below proofs.

• (P3) It can be made with similar way to below proofs.

1.4 A.4

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is open that $\mathbb {D}_{WSVNT2FS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Then, we must only show that $\mathbb {D}_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$. Since for all $x, y\in R,$ $x^2+y^2\ge 2xy$, for all $\tau _i \in \mathfrak {X} (i=1,2,\ldots ,n)$,

  $$\begin{aligned}&(\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2\ge 2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\ge 2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2\ge 2 \mathfrak {h}_{\mathfrak {A}_p}(\tau _i) \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\ge 2 \mathfrak {h}_{\mathfrak {A}_s}(\tau _i) \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2\ge 2 \mathfrak {f}_{\mathfrak {A}_p}(\tau _i) \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2\ge 2 \mathfrak {f}_{\mathfrak {A}_s}(\tau _i) \mathfrak {f}_{\mathfrak {B}_s}(\tau _i). \end{aligned}$$

  and

  $$\begin{aligned} \frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] } \le 1. \end{aligned}$$

  (-16)

  If the last inequality is adopted as $\mathcal {D}_i, (i=1,2,\ldots ,n)$, for $w_i\in [0,1]$, $\sum _{i=1}^nw_i=1$ and $\mathcal {D}_i\in [0,1]$ we get the following inequalities

  $$\begin{aligned} w_1\mathcal {D}_1+w_2\mathcal {D}_2+\cdots +w_n\mathcal {D}_n\le w_1+w_2+\cdots +w_n=1 \end{aligned}$$

  and

  $$\begin{aligned}&\sum _{i=1}^n(w_i)\mathcal {D}_i=\sum _{i=1}^n(w_i)\frac{\left[ \begin{array}{l}2 \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +2\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ 2\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+2\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] }\le 1 \end{aligned}$$

• (P2) If $\mathfrak {A}=\mathfrak {B}$, for all $\tau _i\in \mathfrak {X}$

  $$\begin{aligned} \frac{\left[ \begin{array}{l}2 (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 + 2(\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2 +2(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+ 2(\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] }{\left[ \begin{array}{l}2 (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 + 2(\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2 +2(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+ 2(\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+2(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2 \end{array}\right] }\\ =1. \end{aligned}$$

  (-15)

  If we use notation $\mathcal {P}_i$ $(i=1,2,\ldots ,n)$ for -15, we have

  $$\begin{aligned} w_1\mathcal {P}_1+w_2\mathcal {P}_2+\cdots +w_n\mathcal {P}_n \le w_1+w_2+\cdots +w_n=1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \mathbb {D}_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})=\sum _{i=1}^n (w_i)\mathcal {P}_i=1. \end{aligned}$$

• (P3) It can be easily indicated from below proofs.

1.5 A.5

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is open that $\mathbb {J}_{WSVNT2FS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Then, we must show that $\mathbb {J}_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$. For all $x,y\in \mathbb {R}$, $x^2+y^2-xy\ge xy$. Then for all $\tau _i\in \mathfrak {X}$,

  $$\begin{aligned}&(\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2 -\mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\ge \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2 +(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2-\mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\ge \mathfrak {t}_{\mathfrak {A}_s}(\tau _i)\mathfrak {t}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2-\mathfrak {h}_{\mathfrak {A}_p}(\tau _i) \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\ge \mathfrak {h}_{\mathfrak {A}_p}(\tau _i) \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2-\mathfrak {h}_{\mathfrak {A}_s}(\tau _i) \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\ge \mathfrak {h}_{\mathfrak {A}_s}(\tau _i) \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2-\mathfrak {f}_{\mathfrak {A}_p}(\tau _i) \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\ge \mathfrak {f}_{\mathfrak {A}_p}(\tau _i) \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)\\&\quad (\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2-\mathfrak {f}_{\mathfrak {A}_s}(\tau _i) \mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\ge \mathfrak {f}_{\mathfrak {A}_s}(\tau _i) \mathfrak {f}_{\mathfrak {B}_s}(\tau _i). \end{aligned}$$

  Thus,

  $$\begin{aligned}&\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\\ \mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\\ \mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] } \le 1. \end{aligned}$$

  (-15)

  If equation -14 is denoted by $\mathcal {J}_i (i=1,2,\ldots ,n)$, we have

  $$\begin{aligned} w_1\mathcal {J}_1+w_2\mathcal {J}_2+\cdots w_n\mathcal {J}_n\le w_1+w_2+\cdots w_n=1. \end{aligned}$$

  Then,

  $$\begin{aligned}&\mathbb {J}_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})= \sum _{i=1}^n(w_i)\frac{\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] }{\left[ \begin{array}{l} (\mathfrak {t}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {t}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {h}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {h}_{\mathfrak {B}_s}(\tau _i))^2\\ + (\mathfrak {f}_{\mathfrak {A}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_p}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {A}_s}(\tau _i))^2+(\mathfrak {f}_{\mathfrak {B}_s}(\tau _i))^2 \end{array}\right] -\left[ \begin{array}{l} \mathfrak {t}_{\mathfrak {A}_p}(\tau _i) \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)+\\ \mathfrak {t}_{\mathfrak {A}_s}(\tau _i) \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {h}_{\mathfrak {A}_p}(\tau _i)\mathfrak {h}_{\mathfrak {B}_p}(\tau _i) +\\ \mathfrak {h}_{\mathfrak {A}_s}(\tau _i)\mathfrak {h}_{\mathfrak {B}_s}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_p}(\tau _i)\mathfrak {f}_{\mathfrak {B}_p}(\tau _i)+\\ \mathfrak {f}_{\mathfrak {A}_s}(\tau _i)\mathfrak {f}_{\mathfrak {B}_s}(\tau _i)\\ \end{array}\right] } \le 1 \end{aligned}$$

• (P2) It can be easily indicated below proofs.

• (P3) It is open from previous proofs.

1.6 A.6

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is clear that $Hyb_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Then, we must show that $Hyb_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$ as follow. $\lambda \mathbb {D}_{SVT2NS}+(1-\lambda )\mathbb {C}_{SVT2NS}\le \lambda + (1-\lambda ) =1$ from here $Hyb_{SVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$ where $0\le \mathbb {D}_{SVT2NS}\le 1$ and $0\le \mathbb {C}_{SVT2NS}\le 1$ for $\lambda \in [0,1]$.

• (P2) We know that each both SMs are open from below $\mathbb {C}_{SVT2NS}=1$ and $\mathbb {D}_{SVT2NS}=1$ for $\mathfrak {t}_{\mathfrak {A}_p}(\tau _i)= \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)$, $\mathfrak {t}_{\mathfrak {A}_s}(\tau _i)= \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)$, $\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)= \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)$, $\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)= \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)$, $\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)= \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)$ and $\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)= \mathfrak {f}_{\mathfrak {B}_s}(\tau _i)$ and $i=1,\ldots ,n$. Therefore, $Hyb_{SVT2NS}(\mathfrak {A},\mathfrak {B})=1$.

• (P3) The both SMs provide symmetric property from below. So, $Hyb_{SVT2NS}(\mathfrak {A},\mathfrak {B})$ carry symmetric property.

1.7 A.7

• (P1) Since $\mathfrak {A}$ and $\mathfrak {B}$ are SVT2NSs, it is open that $Hyb_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})\ge 0$. Then, we must prove that $Hyb_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$ as follow. $\lambda \mathbb {D}_{WSVT2NS}+(1-\lambda )\mathbb {C}_{SVT2NS}\le \lambda + (1-\lambda ) =1$ from here $Hyb_{WSVT2NS}(\mathfrak {A}, \mathfrak {B})\le 1$ where $0\le \mathbb {D}_{WSVT2NS}\le 1$ and $0\le \mathbb {C}_{WSVT2NS}\le 1$ for $\lambda \in [0,1]$.

• (P2) We know that each both SMs are open from below $\mathbb {C}_{WSVT2NS}=1$ and $\mathbb {D}_{WSVT2NS}=1$ for $\mathfrak {t}_{\mathfrak {A}_p}(\tau _i)= \mathfrak {t}_{\mathfrak {B}_p}(\tau _i)$, $\mathfrak {t}_{\mathfrak {A}_s}(\tau _i)= \mathfrak {t}_{\mathfrak {B}_s}(\tau _i)$, $\mathfrak {h}_{\mathfrak {A}_p}(\tau _i)= \mathfrak {h}_{\mathfrak {B}_p}(\tau _i)$, $\mathfrak {h}_{\mathfrak {A}_s}(\tau _i)= \mathfrak {h}_{\mathfrak {B}_s}(\tau _i)$, $\mathfrak {f}_{\mathfrak {A}_p}(\tau _i)= \mathfrak {f}_{\mathfrak {B}_p}(\tau _i)$ and $\mathfrak {f}_{\mathfrak {A}_s}(\tau _i)= \mathfrak {f}_{\mathfrak {B}_s}(\tau _i)$ and $i=1,\ldots ,n$. Therefore, $Hyb_{WSVT2NS}(\mathfrak {A},\mathfrak {B})=1$.

• (P3) The both SMs provide symmetric property from below. So, $Hyb_{WSVT2NS}(\mathfrak {A},\mathfrak {B})$ satisfies symmetric property.