New extensions of fuzzy sets with applications to rough topology and medical diagnosis

Abstract

One of the most useful expansions of fuzzy sets for coping with information uncertainties is the Fermatean fuzzy sets. Under this environment, in this article, we define a novel extensions of fuzzy sets called n-fuzzy sets and introduce their relationship with intuitionistic fuzzy sets, Pythagorean fuzzy sets and Fermatean fuzzy sets. The n-fuzzy sets can deal with more uncertain situations than intuitionistic fuzzy sets, Pythagorean fuzzy sets and Fermatean fuzzy sets because of their larger range of describing the membership grades. Then, we provide the necessary set of operations for the n-fuzzy sets, as well as their various features. Furthermore, we present the notion of rough n-fuzzy topology. Moreover, we study the concepts of the rough n-fuzzy interior, closure and obtain some of their properties, respectively. Ultimately, we study the Sanchez’s approach for medical diagnosis and extend this concept with the notion of n-fuzzy set.

Appendices

A. Proof of Theorem 2.2

Since for any two numbers $m, r\in [0, 1]$, we have

$$\begin{aligned} m^{5}\le m^{4}\le m^{3}\le m^{2}\le m \end{aligned}$$

and

$$\begin{aligned} r^{5}\le r^{4}\le r^{3}\le r^{2}\le r. \end{aligned}$$

Therefore, we get

$$\begin{aligned}{} & {} m + r \le 1\\{} & {} \Rightarrow m^{2} + r^{2} \le 1 \\{} & {} \Rightarrow m^{3} + r^{3} \le 1 \\{} & {} \Rightarrow m^{4} + r^{4} \le 1\\{} & {} \Rightarrow m^{5} + r^{5} \le 1. \end{aligned}$$

Then, the space of n-fuzzy (for $n= 4$ and 5) membership grades is larger than the space of intuitionistic membership grades, Pythagorean membership grades and Fermatean membership grades.

B. Proof of Theorem 2.4

Let $x_i\in X$ and N be 4-FS. Suppose that that $\pi _{N}(x_i) = 0$ for $x_i\in X$, then we have the following:

• $$\begin{aligned}{} & {} \begin{aligned} (1)&\quad (\delta _{N}(x_i))^{4} + (\theta _{N}(x_i))^{4} =1\\&\quad \Rightarrow -(\delta _{N}(x_i))^{4} = (\theta _{N}(x_i))^{4} -1\\&\quad \Rightarrow -(\delta _{N}(x_i))^{4} = ((\theta _{N}(x_i))^{2})^{2} -1\\&\quad \Rightarrow -(\delta _{N}(x_i))^{4} = ((\theta _{N}(x_i))^{2} -1)((\theta _{N}(x_i))^{2} +1)\\&\quad \Rightarrow \left| (\delta _{N}(x_i))^{4}\right| = \left| ((\theta _{N}(x_i))^{2} -1)((\theta _{N}(x_i))^{2} +1)\right| \\&\quad \Rightarrow \left| \delta _{N}(x_i)\right| ^{4} = \left| ((\theta _{N}(x_i))^{2} -1)((\theta _{N}(x_i))^{2} +1)\right| \\&\quad \Rightarrow \left| \delta _{N}(x_i)\right| = \root 4 \of {\left| ((\theta _{N}(x_i))^{2} -1)((\theta _{N}(x_i))^{2} +1)\right| }. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

C. Proof of Theorem 2.9

From Definition 2.6, we have:

• $$\begin{aligned}&\begin{aligned} (1)\quad N_1 \oplus N_2&= \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}}, \theta _{N_1}\theta _{N_2}\right) \\&\quad \left( \root n \of {\delta _{N_2}^{n}+\delta _{N_1}^{n} -\delta _{N_2}^{n}\delta _{N_1}^{n}}, \theta _{N_2}\theta _{N_1}\right) \\&= N_2 \oplus N_1. \end{aligned} \end{aligned}$$
• $$\begin{aligned}&\begin{aligned} (2) \quad N_1 \otimes N_2&= \left( \delta _{N_1}\delta _{N_2}, \root n \of {\theta _{N_1}^{n}+\theta _{N_2}^{n} -\theta _{N_1}^{n}\theta _{N_2}^{n}}\right) \\&= \left( \delta _{N_2}\delta _{N_1}, \root n \of {\theta _{N_2}^{n}+\theta _{N_1}^{n} -\theta _{N_2}^{n}\theta _{N_1}^{n}}\right) \\&= N_2 \otimes N_1. \end{aligned} \end{aligned}$$
• $$\begin{aligned}&\begin{aligned} (3)\quad N_1\cap N_2&= (min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\\&=(min\{\delta _{N_2},\delta _{N_1}\},max\{\theta _{N_2},\theta _{N_1}\})\\&=N_2\cap N_1. \end{aligned} \end{aligned}$$

• (4) We can proof in a similar fashion to (3).

D. Proof of Theorem 2.10

From Definition 2.6, we have:

• $$\begin{aligned}&\begin{aligned} (1)\quad&(N_1\cap N_2)\cup N_2\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\cup (\delta _{N_2},\theta _{N_2})\\&\quad =(max\{min\{\delta _{N_1},\delta _{N_2}\}, \delta _{N_2}\},\\&\qquad \qquad min\{max\{\theta _{N_1},\theta _{N_2}\},\theta _{N_2}\})\\&\quad = (\delta _{N_2},\theta _{N_2})=N_{2}. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

E. Proof of Theorem 2.11

For the three n-FSs $N, N_1$ and $N_2$, and $\zeta , \zeta _1, \zeta _2 > 0$, we have:

$$\begin{aligned}&\begin{aligned} (1)\quad&\zeta (N_1 \oplus N_2)\\&\quad = \zeta \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}}, \theta _{N_1}\theta _{N_2}\right) \\&\quad = \left( \root n \of {1-(1-\delta _{N_1}^{n}-\delta _{N_2}^{n} +\delta _{N_1}^{n}\delta _{N_2}^{n}})^{\zeta }, (\theta _{N_1}\theta _{N_2})^{\zeta }\right) \\&\quad = \left( \root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }(1-\delta _{N_2}^{n})^{\zeta }}, \theta _{N_1}^{\zeta }\theta _{N_2}^{\zeta }\right) . \end{aligned} \end{aligned}$$

And

$$\begin{aligned}&\begin{aligned}&\zeta N_1 \oplus \zeta N_2\\&\quad = \left( \root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }}, \theta _{N_1}^{\zeta }\right) \oplus \left( \root n \of {1-(1-\delta _{N_2}^{n})^{\zeta }}, \theta _{N_2}^{\zeta }\right) \\&\quad =\left( \root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }+ 1-(1-\delta _{N_2}^{n})^{\zeta }-(1-(1-\delta _{N_1}^{n})^{\zeta })(1-(1-\delta _{N_2}^{n})^{\zeta })}, \theta _{N_1}^{\zeta }\theta _{N_2}^{\zeta }\right) \\&\quad = \left( \root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }(1-\delta _{N_2}^{n})^{\zeta }}, \theta _{N_1}^{\zeta }\theta _{N_2}^{\zeta }\right) =\zeta (N_1 \oplus N_2). \end{aligned}\\&\begin{aligned} (2)\quad&(\zeta _1 + \zeta _2)N\\&\quad = (\zeta _1 + \zeta _2)(\delta _{N}, \theta _{N})= \left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _1 + \zeta _2}}, \theta _{N}^{\zeta _1 + \zeta _2}\right) \\&\quad =\left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _1}(1-\delta _{N}^{n})^{\zeta _2}}, \theta _{N}^{\zeta _1 + \zeta _2}\right) \\&\quad =\left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _1}+ 1-(1-\delta _{N}^{n})^{\zeta _2}-(1-(1-\delta _{N}^{n})^{\zeta _1})(1-(1-\delta _{N}^{n})^{\zeta _2})}, \theta _{N}^{\zeta _1}\theta _{N}^{\zeta _2}\right) \\&\quad =\left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _1}}, \theta _{N}^{\zeta _1}\right) \oplus \left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _2}}, \theta _{N}^{\zeta _2}\right) \\&\quad =\zeta _1 N\oplus \zeta _2 N. \end{aligned} \end{aligned}$$

$$\begin{aligned}&\begin{aligned} (3)\quad&(N_1 \otimes N_2)^{\zeta }\\&\quad = \left( \delta _{N_1}\delta _{N_2}, \root n \of {\theta _{N_1}^{n}+\theta _{N_2}^{n} -\theta _{N_1}^{n}\theta _{N_2}^{n}}\right) ^{\zeta }\\&\quad = \left( (\delta _{N_1}\delta _{N_2})^{\zeta }, \root n \of {1-(1-\theta _{N_1}^{n}-\theta _{N_2}^{n} +\theta _{N_1}^{n}\theta _{N_2}^{n})^{\zeta }}\right) \\&\quad = \left( \delta _{N_1}^{\zeta }\delta _{N_2}^{\zeta }, \root n \of {1-(1-\theta _{N_1}^{n})^{\zeta }(1-\theta _{N_2}^{n})^{\zeta }}\right) \\&\quad = \left( \delta _{N_1}^{\zeta }, \root n \of {1-(1-\theta _{N_1}^{n})^{\zeta }}\right) \\&\qquad \otimes \left( \delta _{N_2}^{\zeta }, \root n \of {1-(1-\theta _{N_2}^{n})^{\zeta }}\right) = N_1^{\zeta } \otimes N_2^{\zeta }. \end{aligned} \end{aligned}$$

$$\begin{aligned}&\begin{aligned} (4)\quad&N^{\zeta _1}\otimes N^{\zeta _2}\\&\quad = \left( \delta _{N}^{\zeta _1}, \root n \of {1-(1-\theta _{N}^{n})^{\zeta _1}}\right) \\&\qquad \otimes \left( \delta _{N}^{\zeta _2}, \root n \of {1-(1-\theta _{N}^{n})^{\zeta _2}}\right) \\&\quad = \left( \delta _{N}^{\zeta _1 + \zeta _2}, \root n \of {1-(1-\theta _{N}^{n})^{\zeta _1 + \zeta _2}}\right) \\&\quad = N^{(\zeta _1+\zeta _2)}. \end{aligned} \end{aligned}$$

F. Proof of Theorem 2.12

For the three n-FSs $N_1, N_2$, $N_3$, and $\zeta > 0$, we have:

• $$\begin{aligned}&\begin{aligned} (1)\quad&N_1\cap (N_2\cap N_3)\\&\quad = (\delta _{N_1}, \theta _{N_1})\cap (min\{\delta _{N_2},\delta _{N_3}\},max\{\theta _{N_2},\theta _{N_3}\})\\&\quad = (min\{\delta _{N_1},min\{\delta _{N_2},\delta _{N_3}\}\},\\&\qquad max\{\theta _{N_1},max\{\theta _{N_2},\theta _{N_3}\}\})\\&\quad = (min\{min\{\delta _{N_1},\delta _{N_2}\},\delta _{N_3}\},\\&\qquad max\{max\{\theta _{N_1},\theta _{N_2}\},\theta _{N_3}\})\\&\quad = (min\{\delta _{N_1},\delta _{N_2}\}, max\{\theta _{N_1},\theta _{N_2}\})\cap (\delta _{N_3}, \theta _{N_3})\\&\quad =(N_1\cap N_2) \cap N_3. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

• $$\begin{aligned}&\begin{aligned} (3)\quad&\zeta (N_1\cup N_2)\\&\quad = \zeta (max\{\delta _{N_1},\delta _{N_2}\},min\{\theta _{N_1},\theta _{N_2}\})\\&\quad = \left( \root n \of {1-(1-max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\})^{\zeta }}, min\{\theta _{N_1}^{\zeta },\theta _{N_2}^{\zeta }\}\right) . \end{aligned} \end{aligned}$$

  And

  $$\begin{aligned}&\begin{aligned}&\zeta N_1\cup \zeta N_2\\&\quad = \left( \root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }}, \theta _{N_1}^{\zeta }\right) \cup \left( \root n \of {1-(1-\delta _{N_2}^{n})^{\zeta }}, \theta _{N_2}^{\zeta }\right) \\&\quad =\left( max\{\root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }}, \root n \of {1-(1-\delta _{N_2}^{n})^{\zeta }}\},min\{\theta _{N_1}^{\zeta }, \theta _{N_2}^{\zeta }\}\right) \\&\quad = \left( \root n \of {1-(1-max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\})^{\zeta }},min\{\theta _{N_1}^{\zeta }, \theta _{N_2}^{\zeta }\}\right) =\zeta (N_1\cup N_2). \end{aligned} \end{aligned}$$

• (4) We can proof in a similar fashion to (3).

G. Proof of Theorem 2.13

For the three n-FSs $N, N_1$, $N_2$, and $\zeta > 0$, we have:

• $$\begin{aligned} \begin{aligned} (1)\quad&(N_1 \cap N_2)^{c}\\&\quad = (min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})^{c}\\&\quad = (max\{\theta _{N_1},\theta _{N_2}\}, min\{\delta _{N_1},\delta _{N_2}\})\\&\quad = (\theta _{N_1},\delta _{N_1})\cup (\theta _{N_2},\delta _{N_2})\\&\quad = N_1^{c} \cup N_2^{c}. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

• $$\begin{aligned} \begin{aligned} (1)\quad&(N_1 \oplus N_2)^{c}\\&\quad = \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}}, \theta _{N_1}\theta _{N_2}\right) ^{c}\\&\quad = \left( \theta _{N_1}\theta _{N_2}, \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}}\right) \\&\quad = (\theta _{N_1},\delta _{N_1})\otimes (\theta _{N_2},\delta _{N_2})\\&\quad = N_1^{c} \otimes N_2^{c}. \end{aligned} \end{aligned}$$

• (4) We can proof in a similar fashion to (3).

• $$\begin{aligned} \begin{aligned} (1)\quad (N^{c})^{\zeta }&= (\theta _{N},\delta _{N})^{\zeta }\\&=\left( \theta _{N}^{\zeta }, \root n \of {1-(1-\delta _{N}^{n})^{\zeta }}\right) \\&=\left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta }}, \theta _{N}^{\zeta }\right) ^{c}\\&=(\zeta N)^{c}. \end{aligned} \end{aligned}$$

• (6) We can proof in a similar fashion to (5).

H. Proof of Theorem 2.14

From Definition 2.6, we have:

• $$\begin{aligned}&\begin{aligned} (1)\quad&(N_1\cap N_2)\oplus N_3\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\oplus (\delta _{N_3},\theta _{N_3})\\&\quad =\left( \root n \of {min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}+\delta _{N_3}^{n}-\delta _{N_3}^{n}min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}},\right. \\&\qquad \left. max\{\theta _{N_1},\theta _{N_2}\}\theta _{N_3}\right) \\&\quad =\left( \root n \of {(1-\delta _{N_3}^{n})min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}+\delta _{N_3}^{n}}, \right. \\&\quad \left. max\{\theta _{N_1}\theta _{N_3},\theta _{N_2}\theta _{N_3}\}\right) . \end{aligned} \end{aligned}$$

  And

  $$\begin{aligned} \begin{aligned}&(N_1\oplus N_3)\cap (N_2\oplus N_3)\\&\quad =\left( \root n \of {\delta _{N_1}^{n}+\delta _{N_3}^{n} -\delta _{N_1}^{n}\delta _{N_3}^{n}}, \theta _{N_1}\theta _{N_3}\right) \\&\qquad \cap \left( \root n \of {\delta _{N_2}^{n}+\delta _{N_3}^{n} -\delta _{N_2}^{n}\delta _{N_3}^{n}}, \theta _{N_2}\theta _{N_3}\right) \\&\quad =\left( min\{\root n \of {\delta _{N_1}^{n}+\delta _{N_3}^{n} -\delta _{N_1}^{n}\delta _{N_3}^{n}},\right. \\&\qquad \left. \root n \of {\delta _{N_2}^{n}+\delta _{N_3}^{n} -\delta _{N_2}^{n}\delta _{N_3}^{n}}\}, max\{\theta _{N_1}\theta _{N_3}, \theta _{N_2}\theta _{N_3}\}\right) \\&\quad =\left( min\{\root n \of {(1-\delta _{N_3}^{n})\delta _{N_1}^{n}+\delta _{N_3}^{n}},\right. \\&\qquad \left. \root n \of {(1-\delta _{N_3}^{n})\delta _{N_2}^{n}+\delta _{N_3}^{n}}\}, max\{\theta _{N_1}\theta _{N_3}, \theta _{N_2}\theta _{N_3}\}\right) \\&\quad =\left( \root n \of {(1-\delta _{N_3}^{n})min\{\delta _{N_1}^{n}, \delta _{N_2}^{n}\}+\delta _{N_3}^{n}},\right. \\&\qquad \left. max\{\theta _{N_1}\theta _{N_3}, \theta _{N_2}\theta _{N_3}\}\right) . \end{aligned} \end{aligned}$$

  Thus, $(N_1\cap N_2)\oplus N_3=(N_1\oplus N_3)\cap (N_2\oplus N_3)$.

• (2) We can proof in a similar fashion to (1).

• $$\begin{aligned} \begin{aligned} (3)\quad&(N_1\cap N_2)\otimes N_3\\&\quad = (min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\otimes N_3\\&\quad =\left( min\{\delta _{N_1},\delta _{N_2}\}\delta _{N_3},\right. \\&\qquad \left. \root n \of {max\{\theta _{N_1}^{n},\theta _{N_2}^{n}\}+\theta _{N_3}^{n}-\theta _{N_3}^{n}max\{\theta _{N_1}^{n},\theta _{N_2}^{n}\}}\right) \\&\quad =\left( min\{\delta _{N_1}\delta _{N_3},\delta _{N_2}\delta _{N_3}\},\right. \\&\qquad \left. \root n \of {(1-\theta _{N_3}^{n})max\{\theta _{N_1}^{n},\theta _{N_2}^{n}\} +\theta _{N_3}^{n}}\right) . \end{aligned} \end{aligned}$$

  And

  $$\begin{aligned} \begin{aligned}&(N_1\otimes N_3)\cap (N_2\otimes N_3)\\&\quad =\left( \delta _{N_1}\delta _{N_3}, \root n \of {\theta _{N_1}^{n}+\theta _{N_3}^{n} -\theta _{N_1}^{n}\theta _{N_3}^{n}}\right) \\&\qquad \cap \left( \delta _{N_2}\delta _{N_3}, \root n \of {\theta _{N_2}^{n}+\theta _{N_3}^{n} -\theta _{N_2}^{n}\theta _{N_3}^{n}}\right) \\&\quad =\left( \delta _{N_1}\delta _{N_3}, \root n \of {(1-\theta _{N_3}^{n})\theta _{N_1}^{n} +\theta _{N_3}^{n}}\right) \\&\qquad \cap \left( \delta _{N_2}\delta _{N_3},\right. \\&\qquad \left. \root n \of {(1-\theta _{N_3}^{n})\theta _{N_2}^{n} +\theta _{N_3}^{n}}\right) \\&\quad =\left( min\{\delta _{N_1}\delta _{N_3},\delta _{N_2}\delta _{N_3}\},\right. \\&\qquad \left. max\left\{ \root n \of {(1-\theta _{N_3}^{n})\theta _{N_1}^{n} +\theta _{N_3}^{n}}, \root n \of {(1-\theta _{N_3}^{n})\theta _{N_2}^{n} +\theta _{N_3}^{n}}\right\} \right) \\&\quad =\left( min\{\delta _{N_1}\delta _{N_3},\delta _{N_2}\delta _{N_3}\},\right. \\&\qquad \left. \root n \of {(1-\theta _{N_3}^{n})max\{\theta _{N_1}^{n},\theta _{N_2}^{n}\} +\theta _{N_3}^{n}}\right) . \end{aligned} \end{aligned}$$

  Thus, $(N_1\cap N_2)\otimes N_3=(N_1\otimes N_3)\cap (N_2\otimes N_3)$.

• (5) We can proof in a similar fashion to (3).

I. Proof of Theorem 2.15

• $$\begin{aligned} \begin{aligned} (1)\quad&N_1\oplus N_2\oplus N_3\\&\quad = (\delta _{N_1},\theta _{N_1})\oplus (\delta _{N_2},\theta _{N_2})\oplus (\delta _{N_3},\theta _{N_3})\\&\quad = \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}}, \theta _{N_1}\theta _{N_2}\right) \oplus (\delta _{N_3},\theta _{N_3})\\&\quad = \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}+\delta _{N_3}^{n}-\delta _{N_3}^{n}(\delta _{N_1}^{n}+\delta _{N_2}^{n}-\delta _{N_1}^{n}\delta _{N_2}^{n})}, \theta _{N_1}\theta _{N_2}\theta _{N_3}\right) \\&\quad = \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n}+\delta _{N_3}^{n} -\delta _{N_1}^{n}\delta _{N_2}^{n}-\delta _{N_1}^{n}\delta _{N_3}^{n}-\delta _{N_2}^{n}\delta _{N_3}^{n}+\delta _{N_1}^{n}\delta _{N_2}^{n}\delta _{N_3}^{n}}, \theta _{N_1}\theta _{N_2}\theta _{N_3}\right) \\&\quad = \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_3}^{n} -\delta _{N_1}^{n}\delta _{N_3}^{n}+\delta _{N_2}^{n}-\delta _{N_2}^{n}(\delta _{N_1}^{n}+\delta _{N_3}^{n}-\delta _{N_1}^{n}\delta _{N_3}^{n})}, \theta _{N_1}\theta _{N_2}\theta _{N_3}\right) \\&\quad = \left( \root n \of {\delta _{N_1}^{n}+\delta _{N_3}^{n} -\delta _{N_1}^{n}\delta _{N_3}^{n}}, \theta _{N_1}\theta _{N_3}\right) \oplus (\delta _{N_2},\theta _{N_2})\\&\quad =N_1\oplus N_3\oplus N_2. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

J. Proof of Theorem 2.20

Let $s(N) = 0$. Then, $(\delta _{N}(x))^{n} = (\theta _{N}(x))^{n}$ implies that $\delta _{N}(x) = \theta _{N}(x)$ for all $x\in X$. Conversely, suppose that $\delta _{N}(x) = \theta _{N}(x)$. It follows immediately that, for all $x\in X$, $(\delta _{N}(x))^{n} = (\theta _{N}(x))^{n}$. Therefore, $(\delta _{N}(x))^{n} - (\theta _{N}(x))^{n}$= 0. Thus, $s(N) = 0$.

K. Proof of Theorem 2.21

• (1) Let $a(N) = 1$. Then, $(\delta _{N}(x))^{n} + (\theta _{N}(x))^{n} = 1$. Since $\pi _{N}(x) =\root n \of {1 - [(\delta _{N}(x))^{n} + (\theta _{N}(x))^{n}]}$, then $\pi _{N}(x)= 0$. Conversely, suppose that $\pi _{N}(x)= 0$. Then, it follow that, $(\delta _{N}(x))^{n} + (\theta _{N}(x))^{n}= 1$ implies $a(N) = 1$.

• (2) Let $a(N) = 0$. Then, $(\delta _{N}(x))^{n} = - (\theta _{N}(x))^{n}$ implies that $\left| \delta _{N}(x)\right| ^{n}= \left| \theta _{N}(x)\right| ^{n}$ and hence $\left| \delta _{N}(x)\right| = \left| \theta _{N}(x)\right|$.

L. Proof of Theorem 2.25

From Definition 2.1, we can obtain $\delta _{N_1},\theta _{N_1},\delta _{N_2},\theta _{N_2}\in [0,1]$, $0\le \delta _{N_1}^{n}+\theta _{N_1}^{n}\le 1$ and $0\le \delta _{N_2}^{n}+\theta _{N_2}^{n}\le 1$, then

• $$\begin{aligned} \begin{aligned} (1)\quad d(N_1, N_2)&=\sqrt{\frac{1}{2}[(\delta _{N_1}^{n}-\delta _{N_2}^{n})^{2}+(\theta _{N_1}^{n}-\theta _{N_2}^{n})^{2}+ (\pi _{N_1}^{n}-\pi _{N_2}^{n})^{2}]}\\&=\sqrt{\frac{1}{2}[(\delta _{N_2}^{n}-\delta _{N_1}^{n})^{2}+(\theta _{N_2}^{n}-\theta _{N_1}^{n})^{2}+ (\pi _{N_2}^{n}-\pi _{N_1}^{n})^{2}]}\\&=d(N_2, N_1). \end{aligned} \end{aligned}$$

• (2) Let $d(N_1, N_2)={{\sqrt{\frac{1}{2}[(\delta _{N_1}^{n}-\delta _{N_2}^{n})^{2}+(\theta _{N_1}^{n}-\theta _{N_2}^{n})^{2}+ (\pi _{N_1}^{n}-\pi _{N_2}^{n})^{2}]}=0}}$, then we get $(\delta _{N_1}^{n}-\delta _{N_2}^{n})=0$, $(\theta _{N_1}^{n}-\theta _{N_2}^{n})=0$ and $(\pi _{N_1}^{n}-\pi _{N_2}^{n})^{2}=0$, and thus $\delta _{N_1}=\delta _{N_2}$, $\theta _{N_1}=\theta _{N_2}$, and $\pi _{N_1}=\pi _{N_2}$, that is, $N_1 = N_2$.

• (3) Obviously $0\le d(N_1, N_2)$. Now,

  $$\begin{aligned} \begin{aligned}&d(N_1, N_2)\\&\quad =\sqrt{\frac{1}{2}[(\delta _{N_1}^{n}-\delta _{N_2}^{n})^{2}+(\theta _{N_1}^{n}-\theta _{N_2}^{n})^{2}+ (\pi _{N_1}^{n}-\pi _{N_2}^{n})^{2}]}\\&\quad =\sqrt{\frac{1}{2}[(\delta _{N_1}^{n}\!-\!\delta _{N_2}^{n})^{2}\!+\!(\theta _{N_1}^{n}\!-\!\theta _{N_2}^{n})^{2}\!+\! [(\delta _{N_1}^{n}\!+\!\delta _{N_2}^{n})\!+\!(\theta _{N_1}^{n}\!+\!\theta _{N_2}^{n})]^{2}]}\\&\quad \le \sqrt{\frac{1}{2}[2\delta _{N_1}^{2n}+2\delta _{N_2}^{2n}+2\theta _{N_1}^{2n}+2\theta _{N_2}^{2n}+4\delta _{N_1}^{n}\theta _{N_1}^{n}+4\delta _{N_2}^{n}\theta _{N_2}^{n}]}\\&\quad =\sqrt{(\delta _{N_1}^{n}+\theta _{N_1}^{n})^{2}+(\delta _{N_2}^{n}+\theta _{N_2}^{n})^{2}}\le \sqrt{2}. \end{aligned} \end{aligned}$$

M. Proof of Theorem 2.27

• (1) Since $\delta _{N_1}\ge \delta _{N_2}$ and $\theta _{N_1}\le min\left\{ \theta _{N_2}, \frac{\theta _{N_2} \pi _{1}}{\pi _{2}}\right\}$, then $\theta _{N_1} \pi _{2}\le \theta _{N_2} \pi _{1} \Rightarrow \theta _{N_1}^{n} \pi _{2}^{n}\le \theta _{N_2}^{n} \pi _{1}^{n} \Rightarrow \theta _{N_1}^{n}\theta _{N_2}^{n}+\theta _{N_1}^{n} \pi _{2}^{n}\le \theta _{N_1}^{n}\theta _{N_2}^{n}+\theta _{N_2}^{n} \pi _{1}^{n} \Rightarrow \theta _{N_1}^{n}(\theta _{N_2}^{n}+\pi _{2}^{n})\le \theta _{N_2}^{n}(\theta _{N_1}^{n}+ \pi _{1}^{n}) \Rightarrow \theta _{N_1}^{n}(1-\delta _{N_2}^{n})\le \theta _{N_2}^{n}(1-\delta _{N_1}^{n}) \Rightarrow \left( \frac{\theta _{N_1}^{n}}{\theta _{N_2}^{n}}\right) ^{\zeta }\le \left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) ^{\zeta }$ or $\left( \frac{\theta _{N_1}^{n}}{\theta _{N_2}^{n}}\right) \le \left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) \Rightarrow 1- \left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) ^{\zeta }+ \left( \frac{\theta _{N_1}^{n}}{\theta _{N_2}^{n}}\right) ^{\zeta }\le 1$ or $1- \left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) + \left( \frac{\theta _{N_1}^{n}}{\theta _{N_2}^{n}}\right) \le 1 \Rightarrow \left( \root n \of {1- \left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) ^{\zeta }}\right) ^{n}+ \left( \frac{\theta _{N_1}^{\zeta }}{\theta _{N_2}^{\zeta }}\right) ^{n}\le 1$ or $\left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}}\right) ^{n}+ \left( \frac{\theta _{N_1}}{\theta _{N_2}}\right) ^{n}\le 1$. Thus, $N_1\ominus N_2$ and $\zeta (N_1\ominus N_2)$ are n-FSs.

• (2) We can proof in a similar fashion to (2).

N. Proof of Theorem 555552.28

• (1) Since $\delta _{N_1}\ge \delta _{N_2}$ and $\theta _{N_1}\le min\left\{ \theta _{N_2}, \frac{\theta _{N_2} \pi _{1}}{\pi _{2}}\right\}$, then from Definitions 2.6 and 2.26, we get $\zeta (N_1\ominus N_2) =\zeta \left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}}, \frac{\theta _{N_1}}{\theta _{N_2}}\right)$ $={{ \left( \root n \of {1-\left( 1-\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}\right) ^{\zeta }}, \left( \frac{\theta _{N_1}}{\theta _{N_2}}\right) ^{\zeta }\right) }}$ $=\left( \root n \of {1-\left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) ^{\zeta }}, \left( \frac{\theta _{N_1}^{\zeta }}{\theta _{N_2}^{\zeta }}\right) \right)$, and $\zeta N_1\ominus \zeta N_2$ $= \left( \root n \of {1-(1-\delta _{N_1}^{n})^{\zeta }}, \theta _{N_1}^{\zeta }\right) \ominus \left( \root n \of {1-(1-\delta _{N_2}^{n})^{\zeta }}, \theta _{N_2}^{\zeta }\right)$ $= \left( \root n \of {\frac{1-(1-\delta _{N_1}^{n})^{\zeta }-1+(1-\delta _{N_2}^{n})^{\zeta }}{1-1+(1-\delta _{N_2}^{n})^{\zeta }}}, \frac{\theta _{N_1}^{\zeta }}{\theta _{N_2}^{\zeta }}\right)$ $= \left( \root n \of {\frac{(1-\delta _{N_2}^{n})^{\zeta }-(1-\delta _{N_1}^{n})^{\zeta }}{(1-\delta _{N_2}^{n})^{\zeta }}}, \frac{\theta _{N_1}^{\zeta }}{\theta _{N_2}^{\zeta }}\right)$ $= \left( \root n \of {1-\left( \frac{1-\delta _{N_1}^{n}}{1-\delta _{N_2}^{n}}\right) ^{\zeta }}, \frac{\theta _{N_1}^{\zeta }}{\theta _{N_2}^{\zeta }}\right)$. Hence, $\zeta (N_1\ominus N_2) = \zeta N_1\ominus \zeta N_2$.

• (2) We can proof in a similar fashion to (1).

• (3) $\zeta _{1} N\ominus \zeta _{2} N = \left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _{1}}}, \theta _{N}^{\zeta _{1}}\right) \ominus \left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _{2}}}, \theta _{N}^{\zeta _{2}}\right) = \left( \root n \of {\frac{1-(1-\delta _{N}^{n})^{\zeta _{1}}-1 +(1-\delta _{N}^{n})^{\zeta _{2}}}{1-1+(1-\delta _{N}^{n})^{\zeta _{2}}}}, \frac{\theta _{N}^{\zeta _{1}}}{\theta _{N}^{\zeta _{2}}}\right) =\left( \root n \of {1-(1-\delta _{N}^{n})^{\zeta _{1}-\zeta _{2}}}, \theta _{N}^{\zeta _{1}-\zeta _{2}}\right) =(\zeta _{1} - \zeta _{2}) N$.

• (4) We can proof in a similar fashion to (3).

O. Proof of Theorem 2.29

From Definitions 2.6 and 2.26, we can get

• (1) $(N_1\ominus N_2)^{c} = \left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}}, \frac{\theta _{N_1}}{\theta _{N_2}}\right) ^{c}=\left( \frac{\theta _{N_1}}{\theta _{N_2}}, \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}}\right) =(\theta _{N_1},\delta _{N_1})\ominus (\theta _{N_2},\delta _{N_2})=N_1^{c}\oslash N_2^{c}$.

• (2) We can proof in a similar fashion to (1).

P. Proof of Theorem 2.30

From Definitions 2.6 and 2.26, we can get

• $$\begin{aligned} \begin{aligned} (1)\quad&(N_1\cup N_2)\ominus (N_1\cap N_2)\\&\quad =(max\{\delta _{N_1},\delta _{N_2}\},min\{\theta _{N_1},\theta _{N_2}\})\\&\qquad \ominus (min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\\&\quad =\left( \root n \of {\frac{(max\{\delta _{N_1},\delta _{N_2}\})^{n}-(min\{\delta _{N_1},\delta _{N_2}\})^{n}}{1-min\{\delta _{N_1},\delta _{N_2}\})^{n}}},\right. \\&\qquad \qquad \left. \frac{min\{\theta _{N_1},\theta _{N_2}\}}{max\{\theta _{N_1},\theta _{N_2}\}}\right) \\&\quad =\left( \root n \of {\frac{max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}-min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}}{1-min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}}}, \frac{\theta _{N_1}}{\theta _{N_2}}\right) \\&\quad =\left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}}, \frac{\theta _{N_1}}{\theta _{N_2}}\right) \\&\quad =N_1\ominus N_2. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

Q. Proof of Theorem 2.31

From Definitions 2.6 and 2.26, we can get

• $$\begin{aligned} \begin{aligned} (1)\quad&(N_1\cap N_2)\ominus (N_1\cup N_2)\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\\&\qquad \ominus (max\{\delta _{N_1},\delta _{N_2}\},min\{\theta _{N_1},\theta _{N_2}\})\\&\quad =\left( \root n \of {\frac{(min\{\delta _{N_1},\delta _{N_2}\})^{n}-(max\{\delta _{N_1},\delta _{N_2}\})^{n}}{1-max\{\delta _{N_1},\delta _{N_2}\})^{n}}},\right. \\&\qquad \qquad \left. \frac{max\{\theta _{N_1},\theta _{N_2}\}}{min\{\theta _{N_1},\theta _{N_2}\}}\right) \\&\quad =\left( \root n \of {\frac{min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}-max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}}{1-max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}}}, \frac{\theta _{N_2}}{\theta _{N_1}}\right) \\&\quad =\left( \root n \of {\frac{\delta _{N_2}^{n}-\delta _{N_1}^{n}}{1-\delta _{N_1}^{n}}}, \frac{\theta _{N_2}}{\theta _{N_1}}\right) \\&\quad =N_2\ominus N_1. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

R. Proof of Theorem 2.32

From Definitions 2.6 and 2.26, we can get

• $$\begin{aligned}&\begin{aligned} (1)\quad&(N_1\ominus N_2)\oplus N_2\\&\quad = \left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}}, \frac{\theta _{N_1}}{\theta _{N_2}}\right) \oplus (\delta _{N_2},\theta _{N_2})\\&\quad =\left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}+\delta _{N_2}^{n} -\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n}}{1-\delta _{N_2}^{n}}\delta _{N_2}^{n}}, \frac{\theta _{N_1}}{\theta _{N_2}}\theta _{N_2}\right) \\&\quad =\left( \root n \of {\frac{\delta _{N_1}^{n}-\delta _{N_2}^{n} +\delta _{N_2}^{n}-\delta _{N_2}^{2n}-\delta _{N_1}^{n}\delta _{N_2}^{n}+\delta _{N_2}^{2n}}{1-\delta _{N_2}^{n}}}, \theta _{N_1}\right) \\&\quad =\left( \root n \of {\frac{\delta _{N_1}^{n}(1-\delta _{N_2}^{n})}{1-\delta _{N_2}^{n}}}, \theta _{N_1}\right) \\&\quad =(\delta _{N_1},\theta _{N_1})=N_1. \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

S. Proof of Theorem 2.34

From Definitions 2.6 and 2.33, we can get

• $$\begin{aligned}&\begin{aligned} (1)\quad&(N_1\cap N_2)\textcircled {a} N_3\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\textcircled {a} (\delta _{N_3},\theta _{N_3})\\&\quad = \left( \frac{min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}+\delta _{N_3}^{n}}{2}, \frac{max\{\theta _{N_1}^{n},\theta _{N_2}\}+\theta _{N_3}^{n}}{2}\right) \\&\quad = \left( min\left\{ \frac{\delta _{N_1}^{n}+\delta _{N_3}^{n}}{2}, \frac{\delta _{N_2}^{n}+\delta _{N_3}^{n}}{2}\right\} , \right. \\&\qquad \left. max\left\{ \frac{\theta _{N_1}^{n}+\theta _{N_3}^{n}}{2}, \frac{\theta _{N_2}^{n}+\theta _{N_3}^{n}}{2}\right\} \right) \\&\quad = \left( \frac{\delta _{N_1}^{n}+\delta _{N_3}^{n}}{2}, \frac{\theta _{N_1}^{n}+\theta _{N_3}^{n}}{2}\right) \bigcap \left( \frac{\delta _{N_2}^{n}+\delta _{N_3}^{n}}{2}, \frac{\theta _{N_2}^{n}+\theta _{N_3}^{n}}{2}\right) \\&\quad =(N_1\textcircled {a} N_3)\cap (N_2\textcircled {a} N_3). \end{aligned} \end{aligned}$$

• (2) We can proof in a similar fashion to (1).

• $$\begin{aligned} \begin{aligned} (3)\quad&(N_1\textcircled {a} N_2)^{c}\\&\quad =\left( \frac{\delta _{N_1}^{n}+\delta _{N_2}^{n}}{2}, \frac{\theta _{N_1}^{n}+\theta _{N_2}^{n}}{2}\right) ^{c}\\&\quad =\left( \frac{\theta _{N_1}^{n}+\theta _{N_2}^{n}}{2}, \frac{\delta _{N_1}^{n}+\delta _{N_2}^{n}}{2}\right) \\&\quad =(\theta _{N_1},\delta _{N_1}) \textcircled {a} (\theta _{N_2},\delta _{N_2})= N_1^{c}\textcircled {a} N_2^{c}. \end{aligned} \end{aligned}$$

• (4) Since for any two real numbers c and d, we have $min(c,d)+max(c,d)= c + d$, and $min(c,d)\cdot max(c,d)= c \cdot d$. Now,

  $$\begin{aligned}&(N_1\cap N_2)\oplus (N_1\cup N_2)\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\oplus (max\{\delta _{N_1},\delta _{N_2}\},min\{\theta _{N_1},\theta _{N_2}\})\\&\quad =(\root n \of {min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}+ max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}-min\{\delta _{N_1}^{n},\delta _{N_2}^{n}\} max\{\delta _{N_1}^{n},\delta _{N_2}^{n}\}},\\&\qquad max\{\theta _{N_1},\theta _{N_2}\}min\{\theta _{N_1},\theta _{N_2}\})=(\root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n}-\delta _{N_1}^{n}\delta _{N_2}^{n}}, \theta _{N_1}\theta _{N_2}). \end{aligned}$$

  And

  $$\begin{aligned}&(N_1\cap N_2)\otimes (N_1\cup N_2)\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\},max\{\theta _{N_1},\theta _{N_2}\})\otimes (max\{\delta _{N_1},\delta _{N_2}\},min\{\theta _{N_1},\theta _{N_2}\})\\&\quad =(min\{\delta _{N_1},\delta _{N_2}\}max\{\delta _{N_1},\delta _{N_2}\},\\&\qquad \root n \of {max\{\theta _{N_1}^{n},\theta _{N_2}^{n}\}+ min\{\theta _{N_1}^{n},\theta _{N_2}^{n}\}-max\{\theta _{N_1}^{n},\theta _{N_2}^{n}\} min\{\theta _{N_1}^{n},\theta _{N_2}^{n}\}})\\&\quad =(\delta _{N_1}\delta _{N_2}, \root n \of {\theta _{N_1}^{n}+\theta _{N_2}^{n}-\theta _{N_1}^{n}\theta _{N_2}^{n}}). \end{aligned}$$

  Thus,

  $$\begin{aligned}&\left( (N_1\cap N_2)\oplus (N_1\cup N_2)\right) \textcircled {a} \left( (N_1\cap N_2)\otimes (N_1\cup N_2)\right) \\&\quad =\left( \root n \of {\delta _{N_1}^{n}+\delta _{N_2}^{n}-\delta _{N_1}^{n}\delta _{N_2}^{n}}, \theta _{N_1}\theta _{N_2}\right) \\&\qquad \textcircled {a} \left( \delta _{N_1}\delta _{N_2}, \root n \of {\theta _{N_1}^{n}+\theta _{N_2}^{n}-\theta _{N_1}^{n}\theta _{N_2}^{n}}\right) \\&\quad =\left( \frac{\delta _{N_1}^{n}+\delta _{N_2}^{n}}{2}, \frac{\theta _{N_1}^{n}+\theta _{N_2}^{n}}{2}\right) \\&\quad =N_1\textcircled {a} N_2. \end{aligned}$$