Generalized Maclaurin symmetric mean aggregation operators based on Archimedean t-norm of the intuitionistic fuzzy soft set information

Abstract

Intuitionistic fuzzy soft set (IFSS) accommodates more uncertainties within the information by considering the parameterization feature than the intuitionistic fuzzy sets and hence its applications are more extensive. Archimedean T-conorm and T-norm (ATT), consists of T-norm and T-conorm classes, is as an essential source to make the comprehensive operational laws. Meanwhile, the Maclaurin symmetric mean (MSM) has a prominent characteristic and the advantage that it can take into account the interrelation between multi-input arguments, including different attributes or different experts. Motivated by these chief characteristics, in this article, we extend the MSM operators to the IFSS based on ATT. In this paper, a method is exploited to solve the multi-criteria decision-making (MCDM) problems under the IFSS environment. To it, firstly, some generalized intuitionistic fuzzy soft operational laws are introduced based on ATT. Secondly, we reveal some averaging and geometric aggregation operators based on MSM operator. Further, some desirable features and particular cases of it are tested and build up with a new technique for illustrating MCDM problems. Finally, an illustration is given to exhibit the methodology and approach’s supremacy is shown through a comparative study with prevailing techniques.

Appendix: Proofs

Proof of Theorem 3:

Proof

For the sake of simplicity, we prove the result for the generator $z(t)=-\log (t)$ and $y(t)=z(1-t).$ Now, for $z(t)=-\log (t),$ Eq. (7) becomes

$$\begin{aligned}&{\text {IFSMSMA}}^{(k,l)}({\mathcal {B}}_{11}, {\mathcal {B}}_{12},\dots , {\mathcal {B}}_{nm})\\&\quad = \left( \begin{aligned}&\left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( 1-\prod \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\< i_{k} \le n \end{array}}\left( 1-\prod \limits _{r=1}^{k} \zeta _{i_r j_s}\right) ^{\frac{1}{C_{n}^{k}}}\right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} , \\&1-\left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( 1-\prod \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\ < i_{k} \le n \end{array}}\left( 1-\prod \limits _{r=1}^{k} \left( 1-\vartheta _{i_r j_s}\right) \right) ^{\frac{1}{C_{n}^{k}}}\right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \end{aligned} \right) \end{aligned}$$

Let

$$\begin{aligned} f(k,l)= & {} \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( 1-\prod \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\ < i_{k} \le n \end{array}}\left( 1-\prod \limits _{r=1}^{k} \zeta _{i_r j_s}\right) ^{\frac{1}{C_{n}^{k}}}\right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \end{aligned}$$

(22)

and

$$\begin{aligned} g(k,l)= & {} 1-\left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( 1-\prod \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\ < i_{k} \le n \end{array}}\left( 1-\prod \limits _{r=1}^{k} \left( 1-\vartheta _{i_r j_s}\right) \right) ^{\frac{1}{C_{n}^{k}}}\right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \end{aligned}$$

(23)

Now, we will prove function f(k, l) is monotonically diminishing w.r.t. k, l. Since by using Lemma 1, we can get

$$\begin{aligned} f(k,l)= & {} \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( 1-\prod \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\< i_{k} \le n \end{array}}\left( 1-\prod \limits _{r=1}^{k} \zeta _{i_r j_s}\right) ^{\frac{1}{C_{n}^{k}}}\right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \\\ge & {} \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( 1-\sum \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\< i_{k} \le n \end{array}}{\frac{1}{C_{n}^{k}}}1-\prod \limits _{r=1}^{k} \zeta _{i_r j_s}\right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \\= & {} \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \sum \limits _{\begin{array}{c} 1\le i_{1}< \cdots \\ < i_{k} \le n \end{array}}{\frac{1}{C_{n}^{k}}}\left( \prod \limits _{r=1}^{k} \zeta _{i_r j_s}\right) \right) ^{\frac{1}{k}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \end{aligned}$$

(24)

We proof the result by taking the contraction method. Assume that the f(k, l) is monotonically increasing w.r.t. the parameter k,  then we get

$$\begin{aligned} f(n,l)> f(n-1, l)> \cdots > f(1,l) \end{aligned}$$

(25)

But from Eq. (24), we have

$$\begin{aligned} f(1, l)\ge & {} \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \sum \limits _{1\le i_{1} \le n}{\frac{1}{C_{n}^{1}}}\prod \limits _{r=1}^{1} \zeta _{i_r j_s}\right) ^{\frac{1}{1}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \\= & {} \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\ < j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \sum \limits _{1\le i_{1} \le n}{\frac{1}{n}} \zeta _{i_1 j_s}\right) \right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \end{aligned}$$

(26)

Also from Eq. (22), we have

$$\begin{aligned} f(n,l)= \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\ < j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \prod \limits _{i=1}^{n} \zeta _{i_r j_s}\right) ^{\frac{1}{n}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \end{aligned}$$

(27)

By using Eqns. (25), (26) and (27), we get

$$\begin{aligned}&\left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \prod \limits _{i=1}^{n} \zeta _{i_r j_s}\right) ^{\frac{1}{n}}\right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \ge \left( 1-\prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \sum \limits _{1\le i_{1} \le n}{\frac{1}{n}} \zeta _{i_1 j_s}\right) \right) ^{\frac{1}{C_{m}^{l}}}\right) ^{\frac{1}{l}} \\&\quad \Leftrightarrow \prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\< j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \prod \limits _{i=1}^{n} \zeta _{i_r j_s}\right) ^{\frac{1}{n}}\right) ^{\frac{1}{C_{m}^{l}}} \le \prod \limits _{\begin{array}{c} 1\le j_{1}< \cdots \\ < j_{l} \le m \end{array}}\left( 1-\prod \limits _{s=1}^{l}\left( \sum \limits _{1\le i_{1} \le n}{\frac{1}{n}} \zeta _{i_1 j_s}\right) \right) ^{\frac{1}{C_{m}^{l}}} \\&\quad \Leftrightarrow \prod \limits _{s=1}^{l}\left( \prod \limits _{i=1}^{n} \zeta _{i_r j_s}\right) ^{\frac{1}{n}} \ge \prod \limits _{s=1}^{l}\left( \sum \limits _{1\le i_{1} \le n}{\frac{1}{n}} \zeta _{i_1 j_s}\right) \\&\quad \Leftrightarrow \left( \prod \limits _{i=1}^{n} \zeta _{i_r j_s}\right) ^{\frac{1}{n}} \ge \sum \limits _{1\le i_{1} \le n}{\frac{1}{n}} \zeta _{i_1 j_s} \end{aligned}$$

Clearly it contradict the inequality given in Lemma 1. Hence, the function f(k, l) is monotonically decreasing w.r.t. the parameter k. Similarly, f(k, l) is monotonically decreasing w.r.t. parameter l and the function g(k, l) is monotonically increasing w.r.t. the parameters k, l. Thus, by applying the definition of score function, we have

$$\begin{aligned} {\text {Sc}}(k,l)=f(k,l)-g(k,l) \end{aligned}$$

(28)

For any value of $k \in [1,n]$ and $k\in {\mathbb {Z}},$ we can obtain

$$\begin{aligned} {\text {Sc}}(k+1,l)-{\text {Sc}}(k,l)&= \left( f(k+1,l)-g(k+1,l)\right) -\left( f(k,l)-g(k,l)\right) \\&= \left( f(k+1,l)-f(k,l)\right) -\left( g(k+1,l)-g(k,l)\right) \end{aligned}$$

Since f(k, l) is monotonically decreasing and g(k, l) is monotonically increasing w.r.t. k,  therefore, we have $f(k+1,l)-f(k,l)<0$ and $g(k+1,l)-g(k,l)> 0.$ This implies ${\text {Sc}}(k+1,l)-{\text {Sc}}(k,l)< 0,$ i.e. for all $k, {\text {Sc}}(k+1,l)<{\text {Sc}}(k,l).$ Thus the IFSMSMA operator is monotonically decreasing w.r.t. parameter k. Similarly, we can prove IFSMSMA operator is monotonically decreasing w.r.t. parameter l. Hence the result. $\square$