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Multiple attributes group decision-making based on trigonometric operators, particle swarm optimization and complex intuitionistic fuzzy values

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Abstract

This paper describes a novel multi-attribute group decision-making (MAGDM) algorithm based on complex intuitionistic fuzzy values (CIFVs). The present work is divided into three parts. In the first part, the uncertainties in the data are expressed in CIFVs with two membership degrees over the unit disc of the complex plane. The second part states some new operational laws based on tangential functions and the aggregation operators to aggregate the different CIFVs. The properties related to the proposed operations and operators are studied. A nonlinear multi-objective optimization model has been formulated by considering the maximizing and minimizing satisfaction and dissatisfaction degrees respectively to derive the attribute weights for the MAGDM problems. The model has been solved using the particle swarm optimization algorithm. Finally, a numerical example illustrates a MAGDM algorithm based on proposed operators. The reliability and effectiveness of the proposed method is explored by comparing it with several general studies.

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Acknowledgments

The author (Harish Garg) is grateful to DST-FIST grant no. SR/FST/MS-1/2017/13 for providing technical support.

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Appendix

Appendix

Proof of the Theorem 1:

Proof

For two CIFNs \({\mathbb {G}}_v=\left( \left( \zeta _v,w_{\zeta _v}\right) , \left( \varphi _v,w_{\varphi _v}\right) \right)\), we have \(0 \le \zeta _v, \varphi _v, w_{\zeta _v}, w_{\varphi _v}\), \(\zeta _v+\varphi _v, w_{\zeta _v}+w_{\varphi _v}\le 1\) for each \(v=1,2\). Further, assume that \({\mathbb {G}}_3={\mathbb {G}}_1 \oplus {\mathbb {G}}_2=\left( \left( \zeta _3,w_{\zeta _3}\right) , \left( \varphi _3,w_{\varphi _3}\right) \right)\).

Now, by using Definition 9, we have \(\zeta _3=\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right)\), \(w_{\zeta _3}=\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _v}\right) \right)\), \(\varphi _3=1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right)\) and \(w_{\varphi _3}=1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _v}\right) \right) \right)\). To prove \({\mathbb {G}}_3\) is CIFN, it is sufficient to show that \(0 \le \zeta _3,w_{\zeta _3},\varphi _3,w_{\varphi _3} \le 1\) such that \(0 \le \zeta _3+w_{\zeta _3}, \varphi _3+w_{\varphi _3} \le 1\). Now, since \(0 \le \zeta _v \le 1\) for each \(v=1,2\). It implies that \(0 \le \frac{\pi }{2}\zeta _v \le \frac{\pi }{2}\) which gives that \(\sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right)\) always takes non-negative value. Also, we know that the principal value branch of \(\tan ^{-1}(x)\) is \(\left[ \frac{-\pi }{2},\frac{\pi }{2}\right]\) for every number x belonging to the extended real system. But since, \(\sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right)\) is always non-negative therefore, \(0 \le \tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\zeta _v\right) \right) \le \frac{\pi }{2}\) which gives that \(0 \le \frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) \le 1\). Hence, \(0 \le \zeta _3 \le 1\). Similarly, it can be obtained that \(0 \le w_{\zeta _3}, \varphi _3,w_{\varphi _3} \le 1\). Further, using the facts that \(\tan (x)\), \(\tan ^{-1}(x)\) are increasing functions and \(0 \le \zeta _v+\varphi _v \le 1\) for each \(v=1,2\), we obtain that

$$\begin{aligned} \zeta _3+\varphi _3= & {} \frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) +1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) \\= & {} 1+\frac{2}{\pi } \left( \tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\zeta _v\right) \right) -\tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) \right) \\\le & {} 1+\frac{2}{\pi } \left( \tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) -\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2} \left( 1-\varphi _v\right) \right) \right) \right) \\= & {} 1 \end{aligned}$$

Also, since \(\zeta _3, \varphi _3 \ge 0\) therefore, \(\zeta _3+\varphi _3 \ge 0\). Hence, we obtain that \(0 \le \zeta _3+\varphi _3 \le 1\). Similarly, it can be proved that \(0 \le w_{\zeta _3}+w_{\varphi _3} \le 1\). Hence, \({\mathbb {G}}_3 ={\mathbb {G}}_1 \oplus {\mathbb {G}}_2\) is a CIFN. In the same way, \({\mathbb {G}}_1 \otimes {\mathbb {G}}_2\) is also CIFN. \(\square\)

Proof of the Theorem 2:

Proof

By Definitions 5 and 9,

$$\begin{aligned} ({\mathbb {G}}_1 \oplus {\mathbb {G}}_2)^c= & {} \left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _v}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _v}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) ^c\\= & {} \left( \begin{aligned}\left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _v}\right) \right) \right) \end{aligned}\right) , \left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _v}\right) \right) \end{aligned}\right) \end{aligned}\right) \\= & {} \left( \left( \varphi _1,w_{\varphi _1}\right) , \left( \zeta _1,w_{\zeta _1}\right) \right) \otimes \left( \left( \varphi _2,w_{\varphi _2}\right) , \left( \zeta _2,w_{\zeta _2}\right) \right) \\= & {} {\mathbb {G}}_1^c \otimes {\mathbb {G}}_2^c \end{aligned}$$

Hence, \(({\mathbb {G}}_1 \oplus {\mathbb {G}}_2)^c={\mathbb {G}}_1^c \otimes {\mathbb {G}}_2^c\). \(\square\)

Proof of the Theorem 5:

Proof

Let \({\mathbb {G}}_v=(\left( \zeta _{G_v},w_{\zeta _{G_v}}\right)\), \(\left( \varphi _{G_v},w_{\varphi _{G_v}}\right) )\) and \({\mathbb {H}}_v=\left( \left( \zeta _{H_v},w_{\zeta _{H_v}}\right) , \left( \varphi _{H_v},w_{\varphi _{H_v}}\right) \right)\). Further, assume that \({\mathbb {G}}_1 \oplus {\mathbb {G}}_2=(\left( \zeta _{G},w_{\zeta _{G}}\right)\), \(\left( \varphi _{G},w_{\varphi _{G}}\right) )\) and \({\mathbb {H}}_1 \oplus {\mathbb {H}}_2=\left( \left( \zeta _{H},w_{\zeta _{H}}\right) , \left( \varphi _{H},w_{\varphi _{H}}\right) \right)\). Since, \({\mathbb {G}}_v \subseteq {\mathbb {H}}_v\). Therefore, by using Definition 5, we obtain that \(\zeta _{G_v} \le \zeta _{H_v}\), \(w_{\zeta _{G_v}} \le w_{\zeta _{H_v}}\), \(\varphi _{G_v} \ge \varphi _{H_v}\) and \(w_{\varphi _{G_v}} \ge w_{\varphi _{H_v}}\) for each \(v=1,2\). Since, \(\zeta _{G_v} \le \zeta _{H_v} \Rightarrow \frac{\pi }{2}\zeta _{G_v} \le \frac{\pi }{2}\zeta _{H_v} \Rightarrow \tan \left( \frac{\pi }{2}\zeta _{G_v}\right) \le \tan \left( \frac{\pi }{2}\zeta _{H_v}\right) \Rightarrow \frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _{G_v}\right) \right) \le \frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _{H_v}\right) \right) \Rightarrow \zeta _G \le \zeta _H\). Also, we have \(\varphi _{G_v} \ge \varphi _{H_v} \Rightarrow 1-\varphi _{G_v} \le 1-\varphi _{H_v} \Rightarrow \frac{\pi }{2}\left( 1-\varphi _{G_v}\right) \le \frac{\pi }{2}\left( 1-\varphi _{H_v}\right) \Rightarrow \tan \left( \frac{\pi }{2}\left( 1-\varphi _{G_v}\right) \right) \le \tan \left( \frac{\pi }{2}\left( 1-\varphi _{H_v}\right) \right) \Rightarrow \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2} \left( 1-\varphi _{G_v}\right) \right) \le \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\left( 1-\varphi _{H_v}\right) \right) \Rightarrow 1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2 \tan \left( \frac{\pi }{2}\left( 1-\varphi _{G_v}\right) \right) \right) \ge 1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _{H_v}\right) \right) \right) \Rightarrow \varphi _{G} \ge \varphi _{H}\). In the similar manner, we can prove that \(w_{\zeta _G} \le w_{\zeta _H}\) and \(w_{\varphi _{G}} \ge w_{\varphi _{H}}\). Hence, by using Definition 5, we have \({\mathbb {G}}_1 \oplus {\mathbb {G}}_2 \subseteq {\mathbb {H}}_1 \oplus {\mathbb {H}}_2\). \(\square\)

Proof of the Theorem 6:

Proof

We apply induction principle on \(\rho\) to prove Eq. (5).

Step 1::

For \(\rho =2\),

$$\begin{aligned}&{\mathbb {G}}_1 \oplus {\mathbb {G}}_1 \\&\quad = \left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _1\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \\&\quad =\left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( 2\tan \left( \frac{\pi }{2}\zeta _1\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( 2\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( 2\tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( 2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \end{aligned}$$

Hence, the Eq. (5) is true for \(\rho =2\).

Step 2::

Assume Eq. (5) true for \(\rho =\rho _0\). Then,

$$\begin{aligned}&(\rho _0+1){\mathbb {G}}_1\\&\quad =\underbrace{{\mathbb {G}}_1\oplus {\mathbb {G}}_1\oplus \ldots \oplus {\mathbb {G}}_1}_{(\rho _0+1) \,\text {times}} \\&\quad = \left( \underbrace{{\mathbb {G}}_1\oplus {\mathbb {G}}_1\oplus \ldots \oplus {\mathbb {G}}_1}_{(\rho _0) \,\text {times}}\right) \oplus {\mathbb {G}}_1 \\&\quad =\left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}\zeta _1\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \\&\qquad \oplus \left( \left( \zeta _{1},w_{\zeta _{1}}\right) ,\left( \varphi _{1},w_{\varphi _{1}}\right) \right) \\&\quad =\left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}\zeta _1\right) +\tan \left( \frac{\pi }{2}\zeta _1\right) \right) ,\\&\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) +\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \\ \end{aligned}\right) , \\ \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) +\tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) ,\\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho _0\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) +\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \\ \end{aligned}\right) \end{aligned}\right) \\&\quad =\left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( (\rho _0+1)\tan \left( \frac{\pi }{2}\zeta _1\right) \right) ,\\&\frac{2}{\pi }\tan ^{-1}\left( (\rho _0+1)\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \\ \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( (\rho _0+1)\tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) ,\\&1-\frac{2}{\pi }\tan ^{-1}\left( (\rho _0+1)\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \\ \end{aligned}\right) \end{aligned}\right) \end{aligned}$$

Thus, the Eq. (5) is true for \(\rho =\rho _0+1\). Hence, by principle of mathematical induction, the Eq. (5) holds for all natural numbers \(\rho\).

\(\square\)

Proof of the Theorem 9:

Proof

By Definitions 5 and 10,

$$\begin{aligned} \left( {\mathbb {G}}_1^c\right) ^{\rho }= & {} \left( \left( \varphi _1, w_{\varphi _1}\right) ,\left( \zeta _1,w_{\zeta _1}\right) \right) ^{\rho } \\= & {} \left( \begin{aligned}\left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \end{aligned}\right) , \left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\zeta _1\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \end{aligned}\right) \end{aligned}\right) \\= & {} \left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\zeta _1\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) ^c\\= & {} \left( \rho {\mathbb {G}}_1\right) ^c \end{aligned}$$

Hence, \(\left( {\mathbb {G}}_1^c\right) ^{\rho }=\left( \rho {\mathbb {G}}_1\right) ^c\). \(\square\)

Proof of the Theorem 10:

Proof

By Definitions 9 and 10,

$$\begin{aligned}&\rho \left( {\mathbb {G}}_1 \oplus {\mathbb {G}}_2\right) \\&\quad = \rho \left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _v}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _v}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \\&\quad = \left( \left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \frac{\pi }{2}\left( \frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) \right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \frac{\pi }{2}\left( \frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _v}\right) \right) \right) \right) , \\ \end{aligned}\right) ,\right. \\&\qquad \qquad \quad \left. \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \frac{\pi }{2}\left( 1-1+\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) \right) \right) ,\\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \frac{\pi }{2}\left( 1-1+\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _v}\right) \right) \right) \right) \right) \\ \end{aligned}\right) \right) \\&\quad =\left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\zeta _v\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}w_{\zeta _v}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-\varphi _v\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \sum \limits _{v=1}^2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _v}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \\&\quad =\left( \begin{aligned} \left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\zeta _1\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-\varphi _1\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \\&\qquad \oplus \left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\zeta _2\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}w_{\zeta _2}\right) \right) \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-\varphi _2\right) \right) \right) , \\&1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _2}\right) \right) \right) \end{aligned}\right) \end{aligned}\right) \\&\quad = \rho {\mathbb {G}}_1 \oplus \rho {\mathbb {G}}_2 \end{aligned}$$

Hence, \(\rho \left( {\mathbb {G}}_1 \oplus {\mathbb {G}}_2\right) =\rho {\mathbb {G}}_1 \oplus \rho {\mathbb {G}}_2\). \(\square\)

Proof of the Theorem 11:

Proof

Take \(c(x,y)=\frac{2}{\pi }\tan ^{-1}\left( x\tan \left( \frac{\pi }{2}y\right) \right)\) and \(d(x,y)=1-\frac{2}{\pi }\tan ^{-1}\left( x\tan \left( \frac{\pi }{2}(1-y)\right) \right)\) for \(x>0\) and \(0 \le y \le 1\). Thus,

$$\begin{aligned}&\frac{\partial c}{\partial x} =\frac{2\tan \left( \frac{\pi }{2}y\right) }{\pi +\pi \left( x\tan \left( \frac{\pi }{2}y\right) \right) ^2} \ge 0 \qquad ; \qquad \frac{\partial c}{\partial y} =\frac{x\sec ^2\left( \frac{\pi }{2}y\right) }{1+\left( x\tan \left( \frac{\pi }{2}y\right) \right) ^2} \ge 0 \\&\frac{\partial d}{\partial x} =\frac{-2\tan \left( \frac{\pi }{2}(1-y)\right) }{\pi +\pi \left( x\tan \left( \frac{\pi }{2}(1-y)\right) \right) ^2} \le 0 \qquad ; \qquad \frac{\partial d}{\partial y} =\frac{x\sec ^2\left( \frac{\pi }{2}(1-y)\right) }{1+\left( x\tan \left( \frac{\pi }{2}(1-y)\right) \right) ^2} \ge 0 \end{aligned}$$

which implies that the function “c” is increasing while “d” is increasing with y and decreasing with x.

Consider two CIFNs \({\mathbb {G}}_v=\left( \left( \zeta _{v},w_{\zeta _{v}}\right) , \left( \varphi _{v},w_{\varphi _{v}}\right) \right)\) (\(v=1,2\)) such that \(\zeta _1 \le \zeta _2\), \(w_{\zeta _1} \le w_{\zeta _2}\), \(\varphi _1 \ge \varphi _2\) and \(w_{\varphi _1} \ge w_{\varphi _2}\), i.e., \({\mathbb {G}}_1 \subseteq {\mathbb {G}}_2\). Now

$$\begin{aligned}&c\left( \rho ,\zeta _1\right) = \frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\zeta _1\right) \right) \le \frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\zeta _2\right) \right) =c\left( \rho ,\zeta _2\right) \\&c\left( \rho ,w_{\zeta _1}\right) = \frac{2}{\pi } \tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \le \frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}w_{\zeta _2}\right) \right) =c\left( \rho ,w_{\zeta _2}\right) \\&d\left( \rho ,\varphi _1\right) = 1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}(1-\varphi _1)\right) \right) \\&\quad \ge 1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}(1-\varphi _2)\right) \right) =d\left( \rho ,\varphi _2\right) \\&d\left( \rho ,w_{\varphi _1}\right) = 1-\frac{2}{\pi } \tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2} \left( 1-w_{\varphi _1}\right) \right) \right) \\&\quad \ge 1-\frac{2}{\pi }\tan ^{-1}\left( \rho \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _2}\right) \right) \right) =d\left( \rho ,w_{\varphi _2}\right) \end{aligned}$$

It implies that \(\rho {\mathbb {G}}_1 \subseteq \rho {\mathbb {G}}_2\). Similarly, it can be obtained that \({\mathbb {G}}_1^\rho \subseteq {\mathbb {G}}_2^\rho\) when \({\mathbb {G}}_1 \subseteq {\mathbb {G}}_2\).

Next, for two reals \(\rho _1\) and \(\rho _2\) such that \(0<\rho _1\le \rho _2\), we have

$$\begin{aligned}&c\left( \rho _1,\zeta _1\right) = \frac{2}{\pi }\tan ^{-1}\left( \rho _1\tan \left( \frac{\pi }{2}\zeta _1\right) \right) \le \frac{2}{\pi }\tan ^{-1}\left( \rho _2\tan \left( \frac{\pi }{2}\zeta _1\right) \right) =c\left( \rho _2,\zeta _1\right) \\&c\left( \rho _1,w_{\zeta _1}\right) = \frac{2}{\pi }\tan ^{-1}\left( \rho _1\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) \le \frac{2}{\pi }\tan ^{-1}\left( \rho _2\tan \left( \frac{\pi }{2}w_{\zeta _1}\right) \right) =c\left( \rho _2,w_{\zeta _1}\right) \\&d\left( \rho _1,\varphi _1\right) = 1-\frac{2}{\pi } \tan ^{-1}\left( \rho _1\tan \left( \frac{\pi }{2}(1-\varphi _1)\right) \right) \\&\quad \ge 1-\frac{2}{\pi }\tan ^{-1}\left( \rho _2\tan \left( \frac{\pi }{2}(1-\varphi _1)\right) \right) =d\left( \rho _2,\varphi _1\right) \\&d\left( \rho _1,w_{\varphi _1}\right) = 1-\frac{2}{\pi }\tan ^{-1}\left( \rho _1\tan \left( \frac{\pi }{2} \left( 1-w_{\varphi _1}\right) \right) \right) \\&\quad \ge 1-\frac{2}{\pi }\tan ^{-1}\left( \rho _2\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _1}\right) \right) \right) =d\left( \rho _2,w_{\varphi _1}\right) \end{aligned}$$

It gives that \(\rho _1{\mathbb {G}}_1 \subseteq \rho _2{\mathbb {G}}_1\). Similarly, \({\mathbb {G}}_1^{\rho _2} \subseteq {\mathbb {G}}_1^{\rho _1}\) when \(\rho _1 \le \rho _2\).

Hence, the result. \(\square\)

Proof of the Property 1:

Proof

Let \({\mathbb {G}}_0=\left( \left( \zeta _0,w_{\zeta _0}\right) , \left( \varphi _0,w_{\varphi _0}\right) \right)\). Since, \({\mathbb {G}}_v={\mathbb {G}}_0 \forall v\). It implies that \(\zeta _v=\zeta _0\), \(w_{\zeta _v}=w_{\zeta _0}\), \(\varphi _v=\varphi _0\) and \(w_{\varphi _v}=w_{\varphi _0} \forall v\). Further, using the fact that \(\sum \limits _{v=1}^n \psi _v=1\) and Eq. (8), we have

$$\begin{aligned}&\text {CIFWA}({\mathbb {G}}_1, {\mathbb {G}}_2, \ldots , {\mathbb {G}}_n) \\&\quad =\left( \left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^n\left( \psi _v\tan \left( \frac{\pi }{2}\zeta _0\right) \right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^n\left( \psi _v\tan \left( \frac{\pi }{2}w_{\zeta _0}\right) \right) \right) \\ \end{aligned}\right) ,\right. \\&\qquad \qquad \left. \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^n\left( \psi _v\tan \left( \frac{\pi }{2}\left( 1-\varphi _0\right) \right) \right) \right) ,\\&1-\frac{2}{\pi }\tan ^{-1}\left( \sum \limits _{v=1}^n\left( \psi _v\tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _0}\right) \right) \right) \right) \\ \end{aligned}\right) \right) \\&\qquad \left( \begin{aligned}\left( \begin{aligned}&\frac{2}{\pi }\tan ^{-1}\left( \tan \left( \frac{\pi }{2}\zeta _0\right) \right) , \\&\frac{2}{\pi }\tan ^{-1}\left( \tan \left( \frac{\pi }{2}w_{\zeta _0}\right) \right) \\ \end{aligned}\right) , \left( \begin{aligned}&1-\frac{2}{\pi }\tan ^{-1}\left( \tan \left( \frac{\pi }{2}\left( 1-\varphi _0\right) \right) \right) ,\\&1-\frac{2}{\pi }\tan ^{-1}\left( \tan \left( \frac{\pi }{2}\left( 1-w_{\varphi _0}\right) \right) \right) \\ \end{aligned}\right) \end{aligned}\right) \\&\quad = \left( \left( \zeta _0,w_{\zeta _0}\right) , \left( \varphi _0,w_{\varphi _0}\right) \right) \end{aligned}$$

Hence, \(\text {CIFWA}({\mathbb {G}}_1, {\mathbb {G}}_2 \ldots {\mathbb {G}}_n)= {\mathbb {G}}_0\). \(\square\)

Proof of the Property 2:

Proof

Since, \({\mathbb {G}}_v \subseteq {\mathbb {H}}_v\) for each v. Then, by Theorem 5, \(\psi _1{\mathbb {G}}_1 \oplus \psi _2{\mathbb {G}}_2 \oplus \ldots \oplus \psi _n{\mathbb {G}}_n \subseteq \psi _1{\mathbb {H}}_1 \oplus \psi _2{\mathbb {H}}_2 \oplus \ldots \oplus \psi _n{\mathbb {H}}_n\). Hence, \(\text {CIFWA}({\mathbb {G}}_1,{\mathbb {G}}_2,\ldots ,{\mathbb {G}}_n) \subseteq \text {CIFWA}({\mathbb {H}}_1,{\mathbb {H}}_2, \ldots , {\mathbb {H}}_n)\). \(\square\)

Proof of the Property 3:

Proof

Since, \({\mathbb {G}}^- \subseteq {\mathbb {G}}_v \subseteq {\mathbb {G}}^+ \forall v\), so by Property 2, \(\text {CIFWA}({\mathbb {G}}^-, {\mathbb {G}}^-\), \(\ldots\), \({\mathbb {G}}^-) \subseteq \text {CIFWA}({\mathbb {G}}_1\), \({\mathbb {G}}_2, \ldots , {\mathbb {G}}_n) \subseteq \text {CIFWA}({\mathbb {G}}^+, {\mathbb {G}}^+, \ldots , {\mathbb {G}}^+)\). Further, by using Property 1, we obtain that \({\mathbb {G}}^- \subseteq \text {CIFWA}({\mathbb {G}}_1, {\mathbb {G}}_2 \ldots {\mathbb {G}}_n) \subseteq {\mathbb {G}}^+\), which completes the proof. \(\square\)

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Rani, D., Garg, H. Multiple attributes group decision-making based on trigonometric operators, particle swarm optimization and complex intuitionistic fuzzy values. Artif Intell Rev 56, 1787–1831 (2023). https://doi.org/10.1007/s10462-022-10208-2

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