New prioritized aggregation operators with priority degrees among priority orders for complex intuitionistic fuzzy information

Abstract

This paper aims to present some novel prioritized aggregation operators for aggregating complex intuitionistic fuzzy values (CIFVs). The aggregation operators are developed by assigning non-negative real numbers called as priority degrees among strict priority levels. The presented work is divided into three folds. The first fold is that uncertainties in the data are represented using CIFVs which have the characteristic of portraying membership and non-membership degrees over the unit disc of the complex plane. The second fold is to present prioritized averaging and geometric operators without priority degrees, prioritized averaging and geometric operators with priority degrees, prioritized ordered weighted averaging and geometric operators with priority degrees based on basic unit interval monotonic (BUM) function for aggregating dependent CIFVs. A number of propositions related to proposed operators are proved. The third fold is that a group decision-making approach based on proposed operators is developed and is applied on a decision-making problem. The results of proposed method are compared with several existing studies. The comparative study results reveal that the proposed approach is valid and gives fair results. The characteristics of the developed method are also compared with several existing approaches which highlight the superiority of the presented work over prevailing techniques. Besides this, the role of the priority degrees on aggregation result is also discussed in detail.

Appendix

Proof of Proposition 1

For $(d_1,d_2, \ldots , d_{n-1})\rightarrow (1,1, \ldots , 1)$ we obtain that $T_v^{(d)}=\prod \nolimits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k}\rightarrow \prod \nolimits _{k=1}^{v-1} S({\mathcal {G}}_k)=T_v$, which gives that $\xi _v^{(d)}\rightarrow \xi _v$. Hence,

$$\begin{aligned}&\lim _{(d_1,d_2, \ldots , d_{n-1})\rightarrow (1,1, \ldots , 1)} \text {CIFPA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big )\\&\quad = \lim _{(d_1,d_2, \ldots , d_{n-1})\rightarrow (1,1, \ldots , 1)}\left( \xi _1^{(d)} {\mathcal {G}}_1 \oplus \xi _2^{(d)} {\mathcal {G}}_2 \right. \\&\qquad \left. \oplus \cdots , \oplus \xi _n^{(d)} {\mathcal {G}}_n \right) \\&\quad = \xi _1 {\mathcal {G}}_1 \oplus \xi _2 {\mathcal {G}}_2 \oplus \cdots , \oplus \xi _n {\mathcal {G}}_n \\&\quad = \text {CIFPA}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) \end{aligned}$$

$\square$

Proof of Proposition 2

For $(d_1,d_2, \ldots , d_{n-1})\rightarrow (0,0, \ldots , 0)$, we have $T_v^{(d)}=\prod \nolimits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k}=1$. It gives that $\xi _v^{(d)}=\frac{T_v^{(d)}}{\sum \nolimits _{v=1}^n T_v^{(d)}}=\frac{1}{n}$. Hence, $\lim _{(d_1,d_2, \ldots , d_{n-1})\rightarrow (0,0, \ldots , 0)} \text {CIFPA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big )=\frac{1}{n} \big ({\mathcal {G}}_1 \oplus {\mathcal {G}}_2 \oplus \cdots \oplus {\mathcal {G}}_n\big ).$ $\square$

Proof of Proposition 3

Since, $d_1 \rightarrow +\infty$. Now, for each $v=2,3, \ldots , n$, we have $T_v^{(d)}=\prod \nolimits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k}=\big (S({\mathcal {G}}_1)\big )^{+\infty }\big (S({\mathcal {G}}_2)\big )^{d_2} \ldots$ $\big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}=0$ because $0< S({\mathcal {G}}_1) < 1$. It gives that $\sum \nolimits _{v=1}^n T_v^{(d)}=T_1^{(d)}=1$ which implies that $\xi _1^{(d)}=\frac{T_1^{(d)}}{\sum \nolimits _{v=1}^n T_1^{(d)}}=1$ and $\xi _v^{(d)}=\frac{T_v^{(d)}}{\sum \nolimits _{v=1}^n T_v^{(d)}}=0$ for each $v=2,3, \ldots , n$. Hence, $\lim _{d_1\rightarrow +\infty } \text {CIFPA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big )={\mathcal {G}}_1$. $\square$

Proof of Proposition 4

Since $(d_1,d_2, \ldots , d_k, d_{k+1}) \rightarrow (0,0 \ldots , 0, +\infty )$ therefore,

$$\begin{aligned} T_v^{(d)}= & {} \prod \limits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k} =\big (S({\mathcal {G}}_1)\big )^{d_1}\big (S({\mathcal {G}}_2)\big )^{d_2} \ldots \big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}\\&\rightarrow \big (S({\mathcal {G}}_1)\big )^{0}\big (S({\mathcal {G}}_2)\big )^{0} \ldots \big (S({\mathcal {G}}_{v-1})\big )^{0}\\= & {} 1 \, \text {for each}\, v=2,3, \ldots , k+1 \\ T_v^{(d)}= & {} \prod \limits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k}=\big (S({\mathcal {G}}_1)\big )^{d_1}\big (S({\mathcal {G}}_2)\big )^{d_2} \ldots \big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}\\&\rightarrow \big (S({\mathcal {G}}_1)\big )^{0}\big (S({\mathcal {G}}_2)\big )^{0} \ldots \big (S({\mathcal {G}}_{k})\big )^{0}\big (S({\mathcal {G}}_{k+1})\big )^{+\infty }\\&\cdots \big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}\\= & {} 0~ \forall ~ v=k+2,k+3, \ldots , n \end{aligned}$$

Thus, $\sum \nolimits _{v=1}^n T_v^{(d)}=k+1$ and $\xi _v^{(d)}=\frac{T_v^{(d)}}{\sum \nolimits _{v=1}^n T_v^{(d)}} \rightarrow \frac{1}{k+1}$ for each $v=1,2, \ldots , k+1$ and $\xi _v^{(d)}=\frac{T_v^{(d)}}{\sum \nolimits _{v=1}^n T_v^{(d)}} \rightarrow \frac{0}{k+1}=0$ $\forall$ $v=k+2,k+3, \ldots , n$. Hence, $\lim _{(d_1,d_2, \ldots , d_k, d_{k+1}) \rightarrow (0,0 \ldots , 0, +\infty )} \text {CIFPA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big )=\frac{1}{k+1}\big ({\mathcal {G}}_1 \oplus {\mathcal {G}}_2 \oplus \cdots \oplus {\mathcal {G}}_{k+1}\big ).$ $\square$

Proof of Proposition 5

Since $(d_1,d_2, \ldots , d_k, d_{k+1}) \rightarrow (1,1 \ldots , 1, +\infty )$ therefore,

$$\begin{aligned} T_v^{(d)}= & {} \prod \limits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k} =\big (S({\mathcal {G}}_1)\big )^{d_1}\big (S({\mathcal {G}}_2)\big )^{d_2} \ldots \big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}\\&\rightarrow S({\mathcal {G}}_1)S({\mathcal {G}}_2) \ldots S({\mathcal {G}}_{v-1})=T_v \, \text {for each}\, v=2,3, \ldots , k+1 \\ T_v^{(d)}= & {} \prod \limits _{k=1}^{v-1} \big (S({\mathcal {G}}_k)\big )^{d_k} =\big (S({\mathcal {G}}_1)\big )^{d_1}\big (S({\mathcal {G}}_2)\big )^{d_2} \ldots \big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}\\&\rightarrow S({\mathcal {G}}_1)S({\mathcal {G}}_2) \ldots S({\mathcal {G}}_{k})\big (S({\mathcal {G}}_{k+1})\big )^{+\infty } \ldots \big (S({\mathcal {G}}_{v-1})\big )^{d_{v-1}}\\= & {} 0 ~ \text {for each}~ v=k+2, k+3, \ldots , n \end{aligned}$$

Thus, $\sum \nolimits _{v=1}^n T_v^{(d)}\rightarrow \sum \nolimits _{v=1}^{k+1} T_v$ and $\xi _v^{(d)}=\frac{T_v^{(d)}}{\sum \nolimits _{v=1}^n T_v^{(d)}} \rightarrow \frac{T_v}{\sum \nolimits _{v=1}^{k+1} T_v}$ for each $v=1,2, \ldots , k+1$ and $\xi _v^{(d)}=\frac{T_v^{(d)}}{\sum \nolimits _{v=1}^n T_v^{(d)}} \rightarrow \frac{0}{\sum \nolimits _{v=1}^{k+1} T_v}=0$ for each $v=k+2,k+3, \ldots , n$. Hence, $\lim _{(d_1,d_2, \ldots , d_k, d_{k+1}) \rightarrow (0,0 \ldots , 0, +\infty )} \text {CIFPA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big )=\text {CIFPA}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_{k+1}\big ).$ $\square$

Proof of the Proposition 6

Since $d_k < d^{\prime }_k$

• (a) When $v \le k$. For convenience, let $S({\mathcal {G}}_v)=\rho _v$. Now,

  

  Similarly, we can obtain that

  

  For simplicity, let $\mathcal {A}=1+\rho ^{d_1}_1+\rho ^{d_1}_1\rho ^{d_2}_2 + \cdots +\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}$ and $\mathcal {B}=\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}+\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d_{k+1}}_{k+1}+ \cdots + \rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{n-1}}_{n-1}$. Then, we obtain that

  $$\begin{aligned} \xi ^{(d)}_v = \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}} ~~~~ \text {and} ~~~~ \xi ^{(d^\prime )}_v = \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d^\prime _{k}}_{k}\mathcal {B}} \end{aligned}$$

  Now, since $d_k < d^{\prime }_k$. It implies that $\rho ^{d_{k}}_{k} \ge \rho ^{d^\prime _{k}}_{k}$ which leads to $\mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}\ge \mathcal {A}+\rho ^{d^\prime _{k}}_{k}\mathcal {B}$. It gives that $\frac{1}{\mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}}\le \frac{1}{ \mathcal {A}+\rho ^{d^\prime _{k}}_{k}\mathcal {B}}$ which implies that, $\frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}} \le \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d^\prime _{k}}_{k}\mathcal {B}}$. Hence, $\xi ^{(d)}_v \le \xi ^{\prime (d^\prime )}_v$ for $v \le k$.

• (b) When $v > k$, we have

  $$\begin{aligned} \xi ^{(d)}_v= & {} \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d_{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}} ~~~~ \text {and} \\ \xi ^{(d^\prime )}_v= & {} \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d^\prime _{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d^\prime _{k}}_{k}\mathcal {B}} \end{aligned}$$

  Since $d_k < d^{\prime }_k$. It implies that $\rho ^{d_{k}}_{k} \ge \rho ^{d^\prime _{k}}_{k}$ which leads to $\frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\le 1$. Now,

  $$\begin{aligned} \xi ^{(d)}_v= & {} \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d_{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}} \\= & {} \frac{\left( \rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d_{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}\right) \left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) }{\left( \mathcal {A}+\rho ^{d_{k}}_{k}\mathcal {B}\right) \left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) } \\= & {} \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d^\prime _{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A}\left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) +\rho ^{d^\prime _{k}}_{k}\mathcal {B} } \\ \end{aligned}$$

  Further, since $\frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\le 1$. It gives that $\mathcal {A} \left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) \le \mathcal {A}$ which leads to $\mathcal {A} \left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) + \rho ^{d^\prime _{k}}_{k}\mathcal {B}\le \mathcal {A} +\rho ^{d^\prime _{k}}_{k}\mathcal {B}$. It implies that

  $$\begin{aligned} \frac{1}{\mathcal {A} \left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) + \rho ^{d^\prime _{k}}_{k}\mathcal {B}} \ge \frac{1}{\mathcal {A} +\rho ^{d^\prime _{k}}_{k}\mathcal {B}} \end{aligned}$$

  and therefore,

  $$\begin{aligned}&\frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d^\prime _{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A} \left( \frac{\rho ^{d^\prime _{k}}_{k}}{\rho ^{d_{k}}_{k}}\right) + \rho ^{d^\prime _{k}}_{k}\mathcal {B}} \\&\quad \ge \frac{\rho ^{d_1}_1 \rho ^{d_2}_2 \ldots \rho ^{d_{k-1}}_{k-1}\rho ^{d^\prime _{k}}_{k}\rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1}}{\mathcal {A} +\rho ^{d^\prime _{k}}_{k}\mathcal {B}} \end{aligned}$$

  Hence, $\xi ^{(d)}_v \ge \xi ^{\prime (d^\prime )}_v$ for $v > k$.

$\square$

Proof of the Proposition 8

Since, $(d_1,d_2, \ldots , d_{n-1})\rightarrow (0,0, \ldots , 0)$. Therefore, for $v=2,3, \ldots , n$, $T_v^{(d)}=\prod \nolimits _{k=1}^{v-1} \rho _k^{d_k}=\rho _1^{d_1} \rho _2^{d_2} \ldots \rho _{v-1}^{d_{v-1}}$ $\rightarrow \rho _1^0 \rho _2^0 \ldots \rho _{v-1}^0=1$ and $T^{(d)}_1=1$. So, $T^{(d)}= \sum \nolimits _{v=1}^n T^{(d)}_v \rightarrow n$. It gives that

$$\begin{aligned}&\frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_v\big \}} T^{(d)}_t}{T^{(d)}} \rightarrow \frac{v}{n}~~ \text {and}~~ \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_{v-1}\big \}} T^{(d)}_t}{T^{(d)}} \rightarrow \frac{v-1}{n} \\&\quad \Rightarrow ~ \xi ^{(d)}_v \rightarrow g\left( \frac{v}{n}\right) -g\left( \frac{v-1}{n}\right) \\ \end{aligned}$$

Hence,

$$\begin{aligned}&\lim _{(d_1,d_2, \ldots , d_{n-1})\rightarrow (0,0, \ldots , 0)} \text {CIFPOWA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) \\&\quad = \lim _{(d_1,d_2, \ldots , d_{n-1})\rightarrow (0,0, \ldots , 0)} \bigoplus \limits _{v=1}^n\xi ^{(d)}_v {\mathcal {G}}_{\sigma (v)}\\&\quad \rightarrow \bigoplus \limits _{v=1}^n \left( g\left( \frac{v}{n}\right) -g\left( \frac{v-1}{n}\right) \right) {\mathcal {G}}_{\sigma (v)} \\&\quad = \text {CIFOWA}_{g}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) \end{aligned}$$

$\square$

Proof of the Proposition 9

Since, $d_1\rightarrow +\infty$. Therefore, for each $v=2,3, \ldots , n$ we have, $T^{(d)}_v=\prod \nolimits _{k=1}^{v-1} \rho ^{d_k}_k=\rho ^{d_1}_1\rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1} \rightarrow \rho ^{\infty }_1 \rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1}=0$ and $T^{(d)}_1=1$. It implies that $T^{(d)}=\sum \nolimits _{v=1}^n T^{(d)}_v=1$. It leads to that

$$\begin{aligned}&\text {if}~~ {\mathcal {G}}_1 \in \mathcal {H}_v ~~ \text {then} ~~ g\left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_v\big \}} T^{(d)}_t}{T^{(d)}}\right) =g(1)=1 ~~ \text {and}~ \\&\quad \text {if}~~{\mathcal {G}}_1 \notin \mathcal {H}_v ~~ \text {then} ~~ g\left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_v\big \}} T^{(d)}_t}{T^{(d)}}\right) =g(0)=0. \end{aligned}$$

Let ${\mathcal {G}}_1={\mathcal {G}}_{\sigma (k)}$ for some $k=1,2, \ldots , n$. Then, using the Eq. (18), we obtain that $\xi ^{(d)}_1 , \xi ^{(d)}_2 , \ldots , \xi ^{(d)}_{k-1} \rightarrow 0$, $\xi ^{(d)}_{k} \rightarrow 1$, $\xi ^{(d)}_{k+1} , \xi ^{(d)}_{k+2}, \ldots , \xi ^{(d)}_{n} \rightarrow 0$. Hence,

$$\begin{aligned}&\lim _{d_1\rightarrow +\infty } \text {CIFPOWA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) \\&\quad = \lim _{d_1\rightarrow +\infty } \bigoplus \limits _{v=1}^n \left( \xi ^{(d)}_v {\mathcal {G}}_{\sigma (v)} \right) \\&\quad \rightarrow \xi ^{(d)}_{k} {\mathcal {G}}_{\sigma (k)} \\&\quad = {\mathcal {G}}_1 \end{aligned}$$

$\square$

Proof of the Proposition 10

Since, $(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (0,0, \ldots , 0, +\infty )$. Therefore, for each $v=2,3, \ldots , k+1$, $T^{(d)}_v=\prod \nolimits _{k=1}^{v-1} \rho ^{d_k}_k$ $=\rho ^{d_1}_1\rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1} \rightarrow \rho ^{0}_1 \rho ^{0}_2 \ldots \rho ^{0}_{v-1}=1$ and when $v=k+2, k+3, \ldots , n$, $T^{(d)}_v=\prod \nolimits _{k=1}^{v-1} \rho ^{d_k}_k=\rho ^{d_1}_1\rho ^{d_2}_2 \ldots \rho ^{d_k}_k \rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1} \rightarrow \rho ^{0}_1 \rho ^{0}_2 \ldots \rho ^{0}_{k} \rho ^{+\infty }_{k+1}\rho ^{d_{k+2}}_{k+2} \ldots \rho ^{d_{v-1}}_{v-1}=0$. It gives that $T^{(d)}=\sum \limits _{v=1}^n T_v^{(d)} \rightarrow k+1$.

Since, $\big (\tau (1),\tau (2), \ldots , \tau (k+1)\big )$ is an arrangement of $(1,2, \ldots , k+1)$ such that $S({\mathcal {G}}_{\tau (v-1)}) \ge S({\mathcal {G}}_{\tau (v)})$ for each $v=2,3, \ldots , k+1$. Also, we have assumed that $\big (\sigma (1), \sigma (2), \ldots , \sigma (n) \big )$ is an arrangement of $(1,2, \ldots , n)$ such that $S({\mathcal {G}}_{\sigma (v-1)}) \ge S({\mathcal {G}}_{\sigma (v)})$ for each $v=2,3, \ldots , n$. As, $\big \{{\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_{k+1}\big \} \subseteq \big \{{\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big \}$ therefore, there must exist natural numbers $1 \le m_1, m_2, \ldots , m_{k+1} \le n$ satisfying ${\mathcal {G}}_{\tau (1)}={\mathcal {G}}_{\sigma (m_1)}$, ${\mathcal {G}}_{\tau (2)}={\mathcal {G}}_{\sigma (m_2)}$, $\ldots$, ${\mathcal {G}}_{\tau (k+1)}={\mathcal {G}}_{\sigma (m_{k+1})}$. Using the Eq. (18), we have

$$\begin{aligned}&\xi ^{(d)}_{m_1} \rightarrow g \left( \frac{1}{k+1}\right) -g \left( \frac{0}{k+1}\right) = \psi _1 \\&\xi ^{(d)}_{m_2} \rightarrow g \left( \frac{2}{k+1}\right) -g \left( \frac{1}{k+1}\right) = \psi _2 \\&\ldots \\&\xi ^{(d)}_{m_{k+1}} \rightarrow g \left( \frac{k+1}{k+1}\right) -g \left( \frac{k}{k+1}\right) = \psi _{k+1} \end{aligned}$$

and all other weights tend to 0. Hence,

$$\begin{aligned}&\lim _{(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (0,0, \ldots , 0, +\infty )} \text {CIFPOWA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) \\&\quad = \lim _{(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (0,0, \ldots , 0, +\infty )} \bigoplus \limits _{v=1}^n \left( \xi ^{(d)}_v {\mathcal {G}}_{\sigma (v)}\right) \\&\quad \rightarrow \psi _1 {\mathcal {G}}_{\sigma (m_1)} \oplus \psi _2 {\mathcal {G}}_{\sigma (m_2)} \oplus \cdots \psi _{k+1} {\mathcal {G}}_{\sigma (m_{k+1})} \\&\quad = \psi _1 {\mathcal {G}}_{\tau (1)} \oplus \psi _2 {\mathcal {G}}_{\tau (2)} \oplus \cdots \psi _{k+1} {\mathcal {G}}_{\tau (k+1)} \\&\quad = \text {CIFOWA}_{g}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_{k+1}\big ) \end{aligned}$$

$\square$

Proof of the Proposition 11

Since, $(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (1,1, \ldots , 1, +\infty )$. Let $\rho _1 \rho _2 \ldots \rho _{v-1}={\mathbb {T}}_v$ for $v=2,3, \ldots , k+1$; ${\mathbb {T}}_1=1$ and ${\mathbb {T}}=\sum \nolimits _{v=1}^{k+1} {\mathbb {T}}_v$. Therefore, for each $v=2,3, \ldots , k+1$, $T^{(d)}_v=\prod \nolimits _{k=1}^{v-1} \rho ^{d_k}_k=\rho ^{d_1}_1\rho ^{d_2}_2 \ldots \rho ^{d_{v-1}}_{v-1} \rightarrow \rho ^{1}_1 \rho ^{1}_2 \ldots \rho ^{1}_{v-1}={\mathbb {T}}_v$. Further, for $v=k+2, k+3, \ldots , n$, $T^{(d)}_v=\prod \nolimits _{k=1}^{v-1} \rho ^{d_k}_k=\rho ^{d_1}_1\rho ^{d_2}_2 \ldots \rho ^{d_k}_k \rho ^{d_{k+1}}_{k+1} \ldots \rho ^{d_{v-1}}_{v-1} \rightarrow \rho ^{1}_1 \rho ^{1}_2 \ldots \rho ^{1}_{k} \rho ^{+\infty }_{k+1}\rho ^{d_{k+2}}_{k+2} \ldots \rho ^{d_{v-1}}_{v-1}=0$. It implies that $T^{(d)}=\sum \nolimits _{v=1}^n T^{(d)}_v=T^{(d)}_1+T^{(d)}_2+ \cdots + T^{(d)}_{k+1}={\mathbb {T}}_1+{\mathbb {T}}_2+{\mathbb {T}}_3+ \cdots +{\mathbb {T}}_{k+1}=\sum \nolimits _{v=1}^{k+1} {\mathbb {T}}_v={\mathbb {T}}$.

Now, assume that $\big (\tau (1),\tau (2), \ldots , \tau (k+1)\big )$ is an arrangement of $(1,2, \ldots , k+1)$ such that $S({\mathcal {G}}_{\tau (v-1)}) \ge S({\mathcal {G}}_{\tau (v)})$ for each $v=2,3, \ldots , k+1$. Further, let $\mathbb {H}_0=\phi$ and $\mathbb {H}_v=\big \{{\mathcal {G}}_{\tau (z)}|z=1,2, \ldots , v\big \}$ for $v=1,2, \ldots , k+1$. Also, we have assumed that $\big (\sigma (1), \sigma (2), \ldots , \sigma (n) \big )$ is an arrangement of $(1,2, \ldots , n)$ such that $S({\mathcal {G}}_{\sigma (v-1)}) \ge S({\mathcal {G}}_{\sigma (v)})$ for each $v=2,3, \ldots , n$. As, $\big \{{\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_{k+1}\big \} \subseteq \big \{{\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big \}$ therefore, there must exist natural numbers $1 \le m_1, m_2, \ldots , m_{k+1} \le n$ satisfying ${\mathcal {G}}_{\tau (1)}={\mathcal {G}}_{\sigma (m_1)}$, ${\mathcal {G}}_{\tau (2)}={\mathcal {G}}_{\sigma (m_2)}$, $\ldots$, ${\mathcal {G}}_{\tau (k+1)}={\mathcal {G}}_{\sigma (m_{k+1})}$. Further, we have

$$\begin{aligned}&\sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_m\big \}} T^{(d)}_t \rightarrow \sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_0\big \}} {\mathbb {T}}_t=0 ~~ \text {for} ~~ m\\&\quad \in \{0,1,2, \ldots m_1-1\}; \\&\sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_m\big \}} T^{(d)}_t \rightarrow \sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_1\big \}} {\mathbb {T}}_t ~~~ \text {for} ~~ m \\&\quad \in \{m_1,m_1+1,m_1+2, \ldots m_2-1\}; \\&\sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_m\big \}} T^{(d)}_t \rightarrow \sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_2\big \}} {\mathbb {T}}_t ~~~ \text {for} ~~ m \\&\quad \in \{m_2,m_2+1,m_2+2, \ldots m_3-1\}; \\&\ldots \\&\sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathcal {H}_m\big \}} T^{(d)}_t \rightarrow \sum \limits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_{k+1}\big \}} {\mathbb {T}}_t ~~~ \text {for} ~~ m \\&\quad \in \{m_{k+1},m_{k+1}+1,m_{k+1}+2, \ldots n\}. \\ \end{aligned}$$

Now, using the Eq. (18), we have

$$\begin{aligned}&\xi ^{(d)}_{m_1} \rightarrow g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_1\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) -g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_0\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \\&\xi ^{(d)}_{m_2} \rightarrow g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_2\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) -g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_1\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \\&\ldots \\&\xi ^{(d)}_{m_{k+1}} \rightarrow g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_{k+1}\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) -g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_k\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \end{aligned}$$

and all other weights tend to 0. Finally, using the Definition 13, we obtain that

$$\begin{aligned}&\lim _{(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (1,1, \ldots , 1, +\infty )}\text {CIFPOWA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) \\&\quad =\lim _{(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (1,1, \ldots , 1, +\infty )}\left( \xi _1^{(d)} {\mathcal {G}}_{\sigma (1)} \oplus \xi _2^{(d)} {\mathcal {G}}_{\sigma (2)} \right. \\&\qquad \left. \oplus \cdots , \oplus \xi _n^{(d)} {\mathcal {G}}_{\sigma (n)}\right) \\&\quad = \xi ^{(d)}_{m_1} {\mathcal {G}}_{\sigma (m_1)} \oplus \xi ^{(d)}_{m_2} {\mathcal {G}}_{\sigma (m_2)} \oplus \cdots \oplus \xi ^{(d)}_{m_{k+1}} {\mathcal {G}}_{\sigma (m_{k+1})} \\&\quad \rightarrow \left( g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_1\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) -g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_0\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \right) {\mathcal {G}}_{\tau (1)} \\&\quad \oplus \left( g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_2\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) -g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_1\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \right) {\mathcal {G}}_{\tau (2)} \\&\qquad \oplus \cdots \oplus \left( g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_{k+1}\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \right. \\&\qquad -\left. g \left( \frac{\sum \nolimits _{t \in \big \{t| {\mathcal {G}}_t \in \mathbb {H}_k\big \}} {\mathbb {T}}_t}{{\mathbb {T}}}\right) \right) {\mathcal {G}}_{\tau (k+1)} \\&\quad = \text {CIFPOWA}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_{k+1}\big ) \end{aligned}$$

Hence, $\lim _{(d_1,d_2, \ldots , d_k, d_{k+1})\rightarrow (1,1, \ldots , 1, +\infty )} \text {CIFPOWA}_{d}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_n\big ) =\text {CIFPOWA}\big ({\mathcal {G}}_1,{\mathcal {G}}_2, \ldots , {\mathcal {G}}_{k+1}\big )$. $\square$