Linguistic Interval-Valued Pythagorean Fuzzy Sets and Their Application to Multiple Attribute Group Decision-making Process

Abstract

The paper’s aims are to present a novel concept of linguistic interval-valued Pythagorean fuzzy set (LIVPFS) or called a linguistic interval-valued intuitionistic type-2 fuzzy set, which is a robust and trustworthy tool, and to accomplish the imprecise information while solving the decision-making problems. The presented LIVPFS is a generalization of the linguistic Pythagorean fuzzy set, by characterizing the membership and non-membership degrees as the interval-valued linguistic terms to represent the uncertain information. To explore the study, we firstly define some basic operational rules, score and accuracy functions, and the ordering relations of LIVPFS with a brief study of the desirable properties. Based on the stated operational laws, we proposed several weighted averages and geometric aggregating operators to aggregate the linguistic interval-valued Pythagorean fuzzy information. The fundamental inequalities between the proposed operators and their properties are discussed in detail. Finally, a multiple attribute group decision-making (MAGDM) algorithm is promoted to solve the group decision-making problems with uncertain information using linguistic features and the proposed operators. The fundamental inequalities between the proposed operators and their properties are discussed in detail. Also, the illustration of the stated algorithm is given through several numerical examples and compared their performance with the results of the existing algorithms. Based on the stated MAGDM algorithm and the suitable operators, the decision-makers’ can be selected their best alternatives with their own attitude character towards optimism or pessimism choice. The presented LIVPFS is an extension of the several existing sets and is more generalized to utilize the uncertain and imprecise information with a wider range of information. Based on the presented aggregation operators, a decision-maker can select the desired one as per their choices to access the finest alternatives.

Appendix

Proof of Theorem 5

Proof

Based on the operational laws for LIVPFNs as defined in Theorem 1, we can directly obtain the first part. However, to show Eq. 13 holds, we apply the steps of the principle of mathematical induction (PMI) on n as below:

By Definition 15 for ${\mathscr{H}}_{i}(i=1,2)$, we have

$$ \begin{array}{@{}rcl@{}} \omega_{i}\mathcal{H}_{i} &=& \left( \left[ \begin{array}{l} s_{h\sqrt{1-\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}}}, s_{h\sqrt{1-\left( 1-\frac{{b}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}}} \end{array} \right],\right.\\ &&\left.\left[ \begin{array}{l} s_{h\left( \frac{c_{i}}{h}\right)^{\omega_{i}}}, s_{h\left( \frac{d_{i}}{h}\right)^{\omega_{i}}} \end{array} \right] \vphantom{s_{h\sqrt{1-\left( 1-\frac{{a_{i}^{2}}}{h^{2}}\right)^{\omega_{i}}}}}\right) \end{array} $$

Thus, by Definition 16, we have

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1},\mathcal{H}_{2}) = {\omega_{1}}\mathcal{H}_{1} \oplus {\omega_{2}}\mathcal{H}_{2} \\ & =&\left( \left[ \begin{array}{l} s_{h\sqrt{1-\left( 1-\tfrac{{a}_{1}^{2}}{h^{2}}\right)^{\omega_{1}}\left( 1-\tfrac{{a}_{2}^{2}}{h^{2}}\right)^{\omega_{2}}}}, \\ s_{h\sqrt{1-\left( 1-\tfrac{{b}_{1}^{2}}{h^{2}}\right)^{\omega_{1}}\left( 1-\tfrac{{b}_{2}^{2}}{h^{2}}\right)^{\omega_{2}}}} \end{array} \right], \left[ \begin{array}{l} s_{h\left( \tfrac{c_{1}}{h}\right)^{\omega_{2}} \left( \tfrac{c_{2}}{h}\right)^{\omega_{2}}}, \\ s_{h\left( \tfrac{d_{1}}{h}\right)^{\omega_{2}} \left( \tfrac{d_{2}}{h}\right)^{\omega_{2}}} \end{array} \right] \right) \end{array} $$

Hence, it holds for n = 2.

Consider Eq. 13 true for n = k. Then, for n = k + 1, we get

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1},\mathcal{H}_{2},{\ldots} \mathcal{H}_{k+1}) \\ & =&\text{LIVPFWA}(\mathcal{H}_{1},\mathcal{H}_{2},{\dots} \mathcal{H}_{k})\oplus \omega_{k+1} \mathcal{H}_{k+1} \\ &=&\left( \left[ \begin{array}{l} s_{h\sqrt{1-\prod\limits_{i=1}^{k}\left( 1-{a_{i}^{2}}/h^{2}\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{k}\left( 1-{b_{i}^{2}}/h^{2}\right)^{\omega_{i}}}} \end{array} \right], \left[ \begin{array}{l} s_{h\left( \prod\limits_{i=1}^{k}\left( c_{i}/h\right)^{\omega_{i}}\right)}, \\ s_{h\left( \prod\limits_{i=1}^{k}\left( d_{i}/h\right)^{\omega_{i}}\right)} \end{array} \right]\right) \\ &&\oplus \left( \left[ \begin{array}{l} s_{h\sqrt{1-\left( 1-a_{k+1}^{2}/h^{2}\right)^{\omega_{k+1}}}}, \\ s_{h\sqrt{1-\left( 1-b_{k+1}^{2}/h^{2}\right)^{\omega_{k+1}}}} \end{array} \right], \left[ \begin{array}{l} s_{h\left( c_{k+1}/h\right)^{\omega_{k+1}}}, \\ s_{h\left( d_{k+1}/h\right)^{\omega_{k+1}}} \end{array} \right] \right) \\ &=&\left( \left[ \begin{array}{l} s_{h\sqrt{1-\prod\limits_{i=1}^{k+1}\left( 1-{a_{i}^{2}}/h^{2}\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{k+1}\left( 1-{b_{i}^{2}}/h^{2}\right)^{\omega_{i}}}} \end{array} \right], \left[ \begin{array}{l} s_{h\left( \prod\limits_{i=1}^{k+1}\left( c_{i}/h\right)^{\omega_{i}}\right)}, \\ s_{h\left( \prod\limits_{i=1}^{k+1}\left( d_{i}/h\right)^{\omega_{i}}\right)} \end{array} \right] \right) \end{array} $$

i.e., Eq. 13 exists for n = k + 1. Hence, by PMI, Eq. 13 is true for all n. □

Proof of the Theorem 10

Proof

We prove only part (1) here. Since ${\mathscr{H}}_{i}= ([s_{a_{i}},s_{b_{i}}],[s_{c_{i}},s_{d_{i}}] )$, (i = 1, 2, … , n) and ${\mathscr{H}}=([s_{a},s_{b}],[s_{c},s_{d}])$ are LIVPFNs, then we have

$$ \begin{array}{@{}rcl@{}} &&\text{LIVPFWA}(\mathcal{H}_{1}\oplus \mathcal{H}, \mathcal{H}_{2}\oplus \mathcal{H}, \ldots, \mathcal{H}_{n}\oplus \mathcal{H}) \\ & =& \left( \left[\begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a_{i}^{2}}}{h^{2}}\right)\left( 1-\frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{b_{i}^{2}}}{h^{2}}\right)\left( 1-\frac{b^{2}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array} \right], \left[ \begin{array}{c} s_{h\left( \prod\limits_{i=1}^{n}\left( \frac{c_{i}}{h}\frac{c}{h}\right)^{\omega_{i}}\right)},\\ s_{h\left( \prod\limits_{i=1}^{n}\left( \frac{d_{i}}{h}\frac{d}{h}\right)^{\omega_{i}}\right)} \end{array} \right] \right) \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} &&\text{LIVPFWA}(\mathcal{H}_{1}\otimes\mathcal{H},\mathcal{H}_{2}\otimes\mathcal{H},\ldots,\mathcal{H}_{n}\otimes\mathcal{H}) \\ &=& \left( \left[ \begin{array}{l} s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a_{i}^{2}}}{h^{2}}\frac{a^{2}}{h^{2}}\right)^{\omega_{i}}}},\\ s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\frac{{b_{i}^{2}}}{h^{2}}\frac{b^{2}}{h^{2}}\right)^{\omega_{i}}}} \end{array} \right], \left[ \begin{array}{l} s_{h\sqrt{\prod\limits_{i=1}^{n}\left( 1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{c^{2}}{h^{2}}\right)\right)^{\omega_{i}}}},\\ s_{h\sqrt{\prod\limits_{i=1}^{n}\left( 1-\left( 1-\frac{{d}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{d^{2}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array} \right] \right) \end{array} $$

As ${\mathscr{H}}_{i}$ be LIVPFNs, so by its definition, we have a_i, b_i, c_i, d_i ∈ [0, h] which implies that ${a}_{i}^{2}/h^{2}$, ${b}_{i}^{2}/h^{2}$, ${c_{i}^{2}}/h^{2}$, ${d_{i}^{2}}/h^{2}$ ∈ [0, 1]. So by using Lemma 3, we get

$$ \begin{array}{@{}rcl@{}} && \frac{{a}_{i}^{2}}{h^{2}} +\frac{a^{2}}{h^{2}}-\frac{{a}_{i}^{2}}{h^{2}}\frac{a^{2}}{h^{2}}\geq \frac{{a}_{i}^{2}}{h^{2}}\frac{a^{2}}{h^{2}} \\ &\Rightarrow& \prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}\leq \prod\limits_{i=1}^{n}\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\frac{a^{2}}{h^{2}}\right)^{\omega_{i}} \\ &\Rightarrow& h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a_{i}^{2}}}{h^{2}}\right) \left( 1-\frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \\ &&\geq h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a_{i}^{2}}}{h^{2}}\frac{a^{2}}{h^{2}}\right)^{\omega_{i}}} \end{array} $$

(25)

Similarly, we get

$$ \begin{array}{@{}rcl@{}} h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{b_{i}^{2}}}{h^{2}}\right)\left( 1-\frac{b^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \geq h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\frac{{b_{i}^{2}}}{h^{2}}\frac{b^{2}}{h^{2}}\right)^{\omega_{i}}}\\ \end{array} $$

(26)

On the other hand, for ${c}_{i}^{2}/h^{2},c^{2}/h^{2},{d}_{i}^{2}/h^{2},d^{2}/h^{2}\in [0,1]$ for all i and by using Lemma 3, we have $1-\left (1-\frac {{c}_{i}^{2}}{h^{2}}\right )\left (1-\frac {c^{2}}{h^{2}}\right ) \geq \frac {{c}_{i}^{2}}{h^{2}}\frac {c^{2}}{h^{2}}$ and $1-\left (1-\frac {{d}_{i}^{2}}{h^{2}}\right )\left (1-\frac {d^{2}}{h^{2}}\right ) \geq \frac {{d_{i}^{2}}}{h^{2}}\frac {d^{2}}{h^{2}}$ which gives

$$ \begin{array}{@{}rcl@{}} h\sqrt{\prod\limits_{i=1}^{n} \left( 1-\left( 1-\frac{{c_{i}^{2}}}{h^{2}}\right)\left( 1-\frac{c^{2}}{h^{2}}\right) \right)^{\omega_{i}} } \geq h\left( \prod\limits_{i=1}^{n} \left( \frac{{c_{i}^{2}}}{h^{2}}\frac{c^{2}}{h^{2}} \right)^{\omega_{i}}\right)\\ \end{array} $$

(27)

and

$$ \begin{array}{@{}rcl@{}} h\sqrt{\prod\limits_{i=1}^{n} \left( 1-\left( 1-\frac{{d}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{d^{2}}{h^{2}}\right) \right)^{\omega_{i}} } \geq h\left( \prod\limits_{i=1}^{n} \left( \frac{{d}_{i}^{2}}{h^{2}}\frac{d^{2}}{h^{2}} \right)^{\omega_{i}}\right)\\ \end{array} $$

(28)

Therefore, by using Eqs. 25–28 and 9, we get the result. □

Proof of the Theorem 11

Proof

Here, we shall prove only part (1). For any ${\mathscr{H}}_{i}= ([s_{a_{i}},s_{b_{i}}],[s_{c_{i}},s_{d_{i}}])$, (i = 1, 2, … , n) and ${\mathscr{H}}=([s_{a},s_{b}],[s_{c},s_{d}])$, we have

$$ \begin{array}{@{}rcl@{}} &&\text{LIVPFWA}(\lambda\mathcal{H}_{1}\oplus\mathcal{H},\lambda\mathcal{H}_{2}\oplus\mathcal{H},\ldots,\lambda\mathcal{H}_{n}\oplus\mathcal{H}) \\ &=& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{b}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{b^{2}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array}\right], \left[ \begin{array}{c} s_{h\left( \prod\limits_{i=1}^{n}\left( \left( \frac{c_{i}}{h}\right)^{\lambda}\left( \frac{c}{h}\right)\right)^{\omega_{i}}\right)}, \\ s_{h\left( \prod\limits_{i=1}^{n}\left( \left( \frac{d_{i}}{h}\right)^{\lambda}\left( \frac{d}{h}\right)\right)^{\omega_{i}}\right)} \end{array} \right]\right) \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} &&\text{LIVPFWA}(\mathcal{H}_{1}^{\lambda}\otimes\mathcal{H},\mathcal{H}_{2}^{\lambda}\otimes\mathcal{H},\ldots, \mathcal{H}_{n}^{\lambda}\otimes\mathcal{H})\\ &=& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\left( \frac{{b}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{b^{2}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array} \right], \left[ \begin{array}{c} s_{h\sqrt{\prod\limits_{i=1}^{n}\left( 1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{c^{2}}{h^{2}}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{\prod\limits_{i=1}^{n}\left( 1-\left( 1-\frac{{d}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{d^{2}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array} \right] \right) \end{array} $$

Since $0 \leq \left (1-\tfrac {{a_{i}^{2}}}{h^{2}}\right )^{\lambda }\leq 1, 0 \leq \left (1-\frac {a^{2}}{h^{2}}\right ) \leq 1$, $0\leq \left (\frac {{a}_{i}^{2}}{h^{2}}\right )^{\lambda } \leq 1$, $0 \leq \left (\frac {a^{2}}{h^{2}}\right )\leq 1$, then based on the Lemma 2, we have

$$ \begin{array}{@{}rcl@{}} && 1-\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{a^{2}}{h^{2}}\right)\geq \left( \frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{a^{2}}{h^{2}}\right) \\ &\Rightarrow& \prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}\leq \prod\limits_{i=1}^{n}\left( 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}} \\ &\Rightarrow& h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \\ &&\geq h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{a^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \end{array} $$

(29)

Similarly, we can obtain

$$ \begin{array}{@{}rcl@{}} &&h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{b}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{b^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \\ &&\geq h\sqrt{1-\prod\limits_{i=1}^{n}\left( 1-\left( \frac{{b}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{b^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \end{array} $$

(30)

and

$$ \begin{array}{@{}rcl@{}} &&h\sqrt{\prod\limits_{i=1}^{n}\left( 1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{c^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \\ &&\geq h\sqrt{\prod\limits_{i=1}^{n}\left( \left( \frac{{c}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{c^{2}}{h^{2}}\right)\right)^{\omega_{i}} } \end{array} $$

(31)

and

$$ \begin{array}{@{}rcl@{}} && h\sqrt{\prod\limits_{i=1}^{n}\left( 1-\left( 1-\frac{{d}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( 1-\frac{d^{2}}{h^{2}}\right)\right)^{\omega_{i}}}\\ &&\geq h\sqrt{\prod\limits_{i=1}^{n}\left( \left( \frac{{d}_{i}^{2}}{h^{2}}\right)^{\lambda}\left( \frac{d^{2}}{h^{2}}\right)\right)^{\omega_{i}} } \end{array} $$

(32)

Therefore, according to Eqs. 29–32 and by Eq. 9, result (1) holds. □

Proof of the Theorem 13

Proof

For any ${\mathscr{H}}_{i}=([s_{a_{i}}, s_{b_{i}}], [s_{c_{i}}, s_{d_{i}}])$, we have

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\lambda\mathcal{H}_{1}\oplus \mathcal{H}, \lambda\mathcal{H}_{2}\oplus \mathcal{H}, \ldots, \lambda\mathcal{H}_{n}\oplus \mathcal{H}) \\ & =& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-a^{2}/h^{2}\right)\left( 1-{a_{i}^{2}}/h^{2}\right)^{\lambda}\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-b^{2}/h^{2}\right)\left( 1-{b_{i}^{2}}/h^{2}\right)^{\lambda}\right)^{\omega_{i}}}} \end{array} \right],\right.\\ &&\left.\left[ \begin{array}{c} s_{h\prod\limits_{i=1}^{n} \left( \left( c/h\right)\left( c_{i}/h\right)^{\lambda}\right)^{\omega_{i}}}, \\ s_{h\prod\limits_{i=1}^{n} \left( \left( d/h\right)\left( d_{i}/h\right)^{\lambda}\right)^{\omega_{i}}} \end{array} \right] \right) \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1}^{\lambda}\oplus \mathcal{H}, \mathcal{H}_{2}^{\lambda} \oplus \mathcal{H}, \ldots, \mathcal{H}_{n}^{\lambda} \oplus \mathcal{H})\\ &=& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-a^{2}/h^{2}\right)\left( 1-\left( {a_{i}^{2}}/h^{2}\right)^{\lambda}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-b^{2}/h^{2}\right)\left( 1-\left( {b_{i}^{2}}/h^{2}\right)^{\lambda}\right)\right)^{\omega_{i}}}} \end{array} \right],\right.\\ && \left.\left[ \begin{array}{c} s_{h\sqrt{\prod\limits_{i=1}^{n} \left( \left( c^{2}/h^{2}\right)\left( 1- \left( 1-{c_{i}^{2}}/h^{2}\right)^{\lambda}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{\prod\limits_{i=1}^{n} \left( \left( d^{2}/h^{2}\right)\left( 1- \left( 1-{d}_{i}^{2}/h^{2}\right)^{\lambda}\right)\right)^{\omega_{i}}}} \end{array} \right] \right) \end{array} $$

Since 0 ≤ a_i, b_i, c_i, d_i ≤ h which implies that ${a_{i}^{2}}/h^{2}\leq 1$. Now, by using Lemma 4, we get

$$ \begin{array}{@{}rcl@{}} && (1-{a}_{i}^{2}/h^{2})^{\lambda} + ({a}_{i}^{2}/h^{2})^{\lambda} \leq 1 \text{ iff } \lambda\geq 1 \\ \text{and} && (1-{a}_{i}^{2}/h^{2})^{\lambda} + ({a}_{i}^{2}/h^{2})^{\lambda} \geq 1 \text{ iff } 0\leq \lambda\leq 1 \end{array} $$

Thus, for λ ≥ 1, we have $(1-{a_{i}^{2}}/h^{2})^{\lambda } \leq 1 - ({a}_{i}^{2}/h^{2})^{\lambda }$ and hence

$$ \begin{array}{@{}rcl@{}} &&\prod\limits_{i=1}^{n} \left( (1-a^{2}/h^{2})(1-{a_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}} \\ &&\leq \prod\limits_{i=1}^{n} \left( \left( 1-a^{2}/h^{2}\right)\left( 1-\left( {a}_{i}^{2}/h^{2}\right)^{\lambda}\right)\right)^{\omega_{i}} \end{array} $$

which further implies that

$$ \begin{array}{@{}rcl@{}} &&\sqrt{1-\prod\limits_{i=1}^{n} \left( (1-a^{2}/h^{2})(1-{a_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}}} \\ &&\geq \sqrt{1-\prod\limits_{i=1}^{n} \left( (1-a^{2}/h^{2})(1-({a_{i}^{2}}/h^{2})^{\lambda})\right)^{\omega_{i}}} \end{array} $$

Similarly, we can get

$$ \begin{array}{@{}rcl@{}} &&h\sqrt{1-\prod\limits_{i=1}^{n} \left( (1-b^{2}/h^{2})(1-{b}_{i}^{2}/h^{2})^{\lambda}\right)^{\omega_{i}}}\\ &&\geq h\sqrt{1-\prod\limits_{i=1}^{n} \left( (1-b^{2}/h^{2})(1-({b}_{i}^{2}/h^{2})^{\lambda})\right)^{\omega_{i}}} \end{array} $$

Further, for ${c}_{i}^{2}/h^{2}\leq 1$ and by using Lemma 4, we get $(1-{c}_{i}^{2}/h^{2})^{\lambda } + ({c_{i}^{2}}/h^{2})^{\lambda } \leq 1$ iff λ ≥ 1 and $(1-{c_{i}^{2}}/h^{2})^{\lambda } + ({c}_{i}^{2}/h^{2})^{\lambda } \geq 1$ iff 0 ≤ λ ≤ 1. Thus, for λ ≥ 1, we have $(1-{c_{i}^{2}}/h^{2})^{\lambda } + ({c}_{i}^{2}/h^{2})^{\lambda } \leq 1$ which implies that

$$ \begin{array}{@{}rcl@{}} && ({c}_{i}^{2}/h^{2})^{\lambda} \leq 1- (1-{c}_{i}^{2}/h^{2})^{\lambda} \\ &\Rightarrow & \prod\limits_{i=1}^{n} \left( (c^{2}/h^{2})({c_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}} \\&\leq& \prod\limits_{i=1}^{n} \left( (c^{2}/h^{2})(1- (1-{c_{i}^{2}}/h^{2})^{\lambda})\right)^{\omega_{i}} \\ &\Rightarrow & \prod\limits_{i=1}^{n} \left( (c/h)(c_{i}/h)^{\lambda}\right)^{\omega_{i}} \\&\leq& \sqrt{\prod\limits_{i=1}^{n} \left( (c^{2}/h^{2})(1- (1-{c_{i}^{2}}/h^{2})^{\lambda})\right)^{\omega_{i}}} \\ &\Rightarrow & h\prod\limits_{i=1}^{n} \left( (c/h)(c_{i}/h)^{\lambda}\right)^{\omega_{i}} \\&\leq& h\sqrt{\prod\limits_{i=1}^{n} \left( (c^{2}/h^{2})(1- (1-{c_{i}^{2}}/h^{2})^{\lambda})\right)^{\omega_{i}}} \end{array} $$

Similarly, we can get

$$ \begin{array}{@{}rcl@{}} &&h\prod\limits_{i=1}^{n} \left( (d/h)(d_{i}/h)^{\lambda}\right)^{\omega_{i}} \\&\leq& h\sqrt{\prod\limits_{i=1}^{n} \left( (d^{2}/h^{2})(1- (1-{d}_{i}^{2}/h^{2})^{\lambda})\right)^{\omega_{i}}} \end{array} $$

Hence, by Eq. 9, we get the desired result. □

Proof of the Theorem 4

Proof

For LIVPFN ${\mathscr{H}}_{i}=([s_{a_{i}}, s_{b_{i}}], [s_{c_{i}}, s_{d_{i}}])$, we have

$$ \begin{array}{@{}rcl@{}} && \lambda\text{LIVPFWA}(\mathcal{H}_{1},\mathcal{H}_{2},\ldots,\mathcal{H}_{n}) \oplus \mathcal{H} \\ &=& \left( \left[ \begin{array}{l} s_{h\sqrt{1-(1-a^{2}/h^{2})\left( \prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}}}, \\ s_{h\sqrt{1-(1-b^{2}/h^{2})\left( \prod\limits_{i=1}^{n} (1-{b}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}}} \end{array} \right],\right.\\&&\left. \left[ \begin{array}{l} s_{h(c/h)\left( \prod\limits_{i=1}^{n} (c_{i}/h)^{\omega_{i}}\right)^{\lambda}}, \\ s_{h(d/h)\left( \prod\limits_{i=1}^{n} (d_{i}/h)^{\omega_{i}}\right)^{\lambda}} \end{array} \right] \right) \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1},\mathcal{H}_{2},\ldots,\mathcal{H}_{n})^{\lambda} \oplus \mathcal{H} \\ &=& \left( \left[ \begin{array}{c} s_{h\sqrt{1- (1-a^{2}/h^{2})\left( 1-\left( 1-\prod\limits_{i=1}^{n}(1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}\right)}}, \\ s_{h\sqrt{1- (1-b^{2}/h^{2})\left( 1-\left( 1-\prod\limits_{i=1}^{n}(1-{b}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}\right)}} \end{array} \right],\right.\\ &&\left.\left[ \begin{array}{c} s_{h(c/h)\sqrt{1-\left( 1-\prod\limits_{i=1}^{n} ({c}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}}}, \\ s_{h(d/h)\sqrt{1-\left( 1-\prod\limits_{i=1}^{n} ({d}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}}} \end{array} \right] \right) \end{array} $$

Since $\prod \limits _{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega _{i}}\leq 1$ and hence by Lemma 4, we get

$$ \begin{array}{@{}rcl@{}} && \left( 1-\prod\limits_{i=1}^{n} (1-{a_{i}^{2}}/h^{2})^{\omega_{i}}\right)^{\lambda}+\left( \prod\limits_{i=1}^{n} (1-{a_{i}^{2}}/h^{2})^{\omega_{i}}\right)^{\lambda} \\ &&\quad\leq 1 \text{ iff } \lambda \geq 1 \\ \text{and} && \left( 1-\prod\limits_{i=1}^{n} (1-{a_{i}^{2}}/h^{2})^{\omega_{i}}\right)^{\lambda}+\left( \prod\limits_{i=1}^{n} (1-{a_{i}^{2}}/h^{2})^{\omega_{i}}\right)^{\lambda} \\ &&\quad\geq 1 \text{ iff } 0 \leq \lambda \leq 1. \end{array} $$

Thus, for λ ≥ 1, we have

$$ \begin{array}{@{}rcl@{}} & &\left( 1-\prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}+\left( \prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \leq 1 \\ &\Rightarrow & \left( \prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \leq 1 - \left( 1-\prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \\ &\Rightarrow & (1-a^{2}/h^{2})\left( \prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \\ &&\qquad\leq (1-a^{2}/h^{2})\left( 1 - \left( 1-\prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \right) \\ &\Rightarrow & 1-(1-a^{2}/h^{2})\left( \prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \\ && \qquad \geq 1- (1-a^{2}/h^{2})\left( 1 - \left( 1-\prod\limits_{i=1}^{n} (1-{a}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \right) \\ &\Rightarrow & h\sqrt{1-(1-a^{2}/h^{2})\left( \prod\limits_{i=1}^{n} (1-{a_{i}^{2}}/h^{2})^{\omega_{i}}\right)^{\lambda}} \\ && \geq h\sqrt{1- (1-a^{2}/h^{2})\left( 1 - \left( 1-\prod\limits_{i=1}^{n} (1-{a_{i}^{2}}/h^{2})^{\omega_{i}}\right)^{\lambda} \right)} \end{array} $$

Similarly, we can obtain

$$ \begin{array}{@{}rcl@{}} && h\sqrt{1-(1-b^{2}/h^{2})\left( \prod\limits_{i=1}^{n} (1-{b}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}} \\ && \geq h\sqrt{1 - (1-b^{2}/h^{2})\left( 1 \!\!- \left( 1 - \prod\limits_{i=1}^{n} (1-{b}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda} \right)} \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} &&h(c/h)\left( \prod\limits_{i=1}^{n} (c_{i}/h)^{\omega_{i}}\right)^{\lambda} \\&&\leq h(c/h)\sqrt{1-\left( 1-\prod\limits_{i=1}^{n} ({c}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}} \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} &&h(d/h)\left( \prod\limits_{i=1}^{n} (d_{i}/h)^{\omega_{i}}\right)^{\lambda} \\&&\leq h(d/h)\sqrt{1-\left( 1-\prod\limits_{i=1}^{n} ({d}_{i}^{2}/h^{2})^{\omega_{i}}\right)^{\lambda}} \end{array} $$

Thus, based on these inequalities and by Eq. 9, we get the desired result. □

Proof of the Theorem 15

Proof

To prove (1) for any LIVPFN ${\mathscr{H}}_{i}=([s_{a_{i}}, s_{b_{i}}],[s_{c_{i}},$ $s_{d_{i}}])$, we have

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\lambda\mathcal{H}_{1}\otimes \mathcal{H}, \lambda\mathcal{H}_{2}\otimes\mathcal{H},\ldots, \lambda\mathcal{H}_{n} \otimes \mathcal{H}) \\ &=& \left( \left[ \begin{array}{l} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-(A^{2}/h^{2})\left( 1-(1-{a_{i}^{2}}/h^{2})^{\lambda}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-(B^{2}/h^{2})\left( 1-(1-{b_{i}^{2}}/h^{2})^{\lambda}\right)\right)^{\omega_{i}}}} \end{array} \right],\right.\\ &&\left. \left[ \begin{array}{l} s_{h\left( \prod\limits_{i=1}^{n} \sqrt{1-\left( 1-\frac{C^{2}}{h^{2}}\right)\left( 1-\left( \frac{{c_{i}^{2}}}{h^{2}}\right)^{\lambda}\right)} \right)^{\omega_{i}}}, \\ s_{h\left( \prod\limits_{i=1}^{n} \sqrt{1-\left( 1-\frac{D^{2}}{h^{2}}\right)\left( 1-\left( \frac{{d}_{i}^{2}}{h^{2}}\right)^{\lambda}\right)} \right)^{\omega_{i}}}, \end{array} \right] \right) \\ & &\text{LIVPFWA}(\lambda\mathcal{H}_{1}, \lambda\mathcal{H}_{2}, \ldots, \lambda\mathcal{H}_{n}) \\ &=& \left( \left[ \begin{array}{l} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-{a}_{i}^{2}/h^{2}\right)^{\lambda}\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-{b_{i}^{2}}/h^{2}\right)^{\lambda}\right)^{\omega_{i}}}}, \end{array} \right], \left[ \begin{array}{l} s_{h\left( \prod\limits_{i=1}^{n}\left( \frac{c_{i}}{h}\right)^{\lambda}\right)^{\omega_{i}}}, \\ s_{h\left( \prod\limits_{i=1}^{n}\left( \frac{d_{i}}{h}\right)^{\lambda}\right)^{\omega_{i}}} \end{array} \right] \right) \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\lambda\mathcal{H}_{1}\oplus\mathcal{H}, \lambda\mathcal{H}_{2}\oplus\mathcal{H},\ldots, \lambda\mathcal{H}_{n} \oplus\mathcal{H}) \\ &=& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( (A^{2}/h^{2})\left( 1-{a_{i}^{2}}/h^{2}\right)^{\lambda}\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( (B^{2}/h^{2})\left( 1-{b_{i}^{2}}/h^{2}\right)^{\lambda}\right)^{\omega_{i}}}} \end{array} \right],\right.\\ &&\left. \left[ \begin{array}{c} s_{h\left( \prod\limits_{i=1}^{n} \left( \frac{c_{i}}{h}\right)^{\lambda}\left( \frac{C}{h}\right)\right)^{\omega_{i}}}, \\ s_{h\left( \prod\limits_{i=1}^{n} \left( \frac{d_{i}}{h}\right)^{\lambda}\left( \frac{D}{h}\right)\right)^{\omega_{i}}} \end{array} \right] \right) \end{array} $$

We first prove $\text {LIVPFWA}(\lambda {\mathscr{H}}_{1}\otimes {\mathscr{H}}, \lambda {\mathscr{H}}_{2}\otimes {\mathscr{H}}, \ldots , \lambda {\mathscr{H}}_{n}\otimes {\mathscr{H}}) \leq \text {LIVPFWA}(\lambda {\mathscr{H}}_{1}, \lambda {\mathscr{H}}_{2},\ldots , \lambda {\mathscr{H}}_{n})$.

As $A^{2}/h^{2}, 1-{a_{i}^{2}}/h^{2} \in [0,1]$ and hence $(A^{2}/h^{2})\left (1-(1-{a}_{i}^{2}/h^{2})^{\lambda } \right ) \leq 1-(1-{a}_{i}^{2}/h^{2})^{\lambda }$ which implies that $1-A^{2}/h^{2}\left (1-(1-{a}_{i}^{2}/h^{2})^{\lambda }\right ) \geq (1-{a_{i}^{2}}/h^{2})^{\lambda }$ and thus

$$ \begin{array}{@{}rcl@{}} && \prod\limits_{i=1}^{n} \left( 1-A^{2}/h^{2}\left( 1-(1-{a}_{i}^{2}/h^{2})^{\lambda}\right) \right)^{\omega_{i}}\\ &&\quad\geq \prod\limits_{i=1}^{n} \left( (1-{a}_{i}^{2}/h^{2})^{\lambda}\right)^{\omega_{i}} \\ &\Rightarrow & h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-A^{2}/h^{2}\left( 1-(1-{a_{i}^{2}}/h^{2})^{\lambda}\right) \right)^{\omega_{i}}} \\ &&\quad\leq h\sqrt{\prod\limits_{i=1}^{n} 1-\left( (1-{a_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}}} \end{array} $$

Similarly, we can obtain

$$ \begin{array}{@{}rcl@{}} && h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-B^{2}/h^{2}\left( 1-(1-{b}_{i}^{2}/h^{2})^{\lambda}\right) \right)^{\omega_{i}}} \\ &&\quad\leq h\sqrt{\prod\limits_{i=1}^{n} 1-\left( (1-{b_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}}} \\ &\text{and ~~} & h\left( \prod\limits_{i=1}^{n} \sqrt{1-\left( 1-\frac{C^{2}}{h^{2}}\right)\left( 1-\left( \frac{{c}_{i}^{2}}{h^{2}}\right)^{\lambda}\right)} \right)^{\omega_{i}} \\ &&\quad\leq h\left( \prod\limits_{i=1}^{n}\left( \frac{c_{i}}{h}\right)^{\lambda}\right)^{\omega_{i}} \\ &\text{and ~~} & h\left( \prod\limits_{i=1}^{n} \sqrt{1-\left( 1-\frac{D^{2}}{h^{2}}\right)\left( 1-\left( \frac{{d}_{i}^{2}}{h^{2}}\right)^{\lambda}\right)} \right)^{\omega_{i}} \\ &&\quad\leq h\left( \prod\limits_{i=1}^{n}\left( \frac{d_{i}}{h}\right)^{\lambda}\right)^{\omega_{i}} \end{array} $$

According to Eq. 9, we have $\text {LIVPFWA}(\lambda {\mathscr{H}}_{1}\otimes {\mathscr{H}}, \lambda {\mathscr{H}}_{2}\otimes {\mathscr{H}}, \ldots , \lambda {\mathscr{H}}_{n}\otimes {\mathscr{H}})$ ≤ $\text {LIVPFWA}(\lambda {\mathscr{H}}_{1}$, $\lambda {\mathscr{H}}_{2}$, …, $\lambda {\mathscr{H}}_{n})$.

Later, we prove $\text {LIVPFWA}(\lambda {\mathscr{H}}_{1}, \lambda {\mathscr{H}}_{2},\ldots , \lambda {\mathscr{H}}_{n}) \leq \text {LIVPFWA}(\lambda {\mathscr{H}}_{1}\oplus {\mathscr{H}}, \lambda {\mathscr{H}}_{2}\oplus {\mathscr{H}}, \ldots , \lambda {\mathscr{H}}_{n}\oplus {\mathscr{H}})$.

For $A^{2}/h^{2}, {a}_{i}^{2}/h^{2}\leq 1$, we have $(A^{2}/h^{2})(1-{a_{i}^{2}}/h^{2})^{\lambda } \leq (1-{a_{i}^{2}}/h^{2})^{\lambda }$ which implies that $1-\prod \limits _{i=1}^{n}\left ((A^{2}/h^{2})(1-{a}_{i}^{2}/h^{2})^{\lambda }\right )^{\omega _{i}} \geq 1-\prod \limits _{i=1}^{n} \left ((1-{a_{i}^{2}}/h^{2})^{\lambda }\right )^{\omega _{i}}$ and hence, we get

$$ \begin{array}{@{}rcl@{}} &&h\sqrt{1-\prod\limits_{i=1}^{n}\left( (A^{2}/h^{2})(1-{a}_{i}^{2}/h^{2})^{\lambda}\right)^{\omega_{i}}} \\ &&\quad\geq h\sqrt{1-\prod\limits_{i=1}^{n} \left( (1-{a}_{i}^{2}/h^{2})^{\lambda}\right)^{\omega_{i}}}. \end{array} $$

Similarly, we have

$$ \begin{array}{@{}rcl@{}} &&h\sqrt{1-\prod\limits_{i=1}^{n} \left( (B^{2}/h^{2})(1-{b_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}}} \\ &&\geq h\sqrt{1-\prod\limits_{i=1}^{n} \left( (1-{b_{i}^{2}}/h^{2})^{\lambda}\right)^{\omega_{i}}}. \end{array} $$

Now, for ${c_{i}^{2}}/h^{2}, C^{2}/h^{2}\leq 1$, we can derive that

$$ \begin{array}{@{}rcl@{}} & &h\left( \prod\limits_{i=1}^{n} \left( \frac{c_{i}}{h}\right)^{\lambda}\left( \frac{C}{h}\right)\right)^{\omega_{i}} \leq h\left( \prod\limits_{i=1}^{n}\left( \frac{c_{i}}{h}\right)^{\lambda}\right)^{\omega_{i}} \\ &\text{and ~~} & h\left( \prod\limits_{i=1}^{n} \left( \frac{d_{i}}{h}\right)^{\lambda}\left( \frac{D}{h}\right)\right)^{\omega_{i}} \leq h\left( \prod\limits_{i=1}^{n}\left( \frac{d_{i}}{h}\right)^{\lambda}\right)^{\omega_{i}} \end{array} $$

Hence, by Eq. 9, we get $\text {LIVPFWA}(\lambda {\mathscr{H}}_{1},\lambda {\mathscr{H}}_{2},\ldots ,$ $\lambda {\mathscr{H}}_{n}) \leq \text {LIVPFWA}(\lambda {\mathscr{H}}_{1}\oplus {\mathscr{H}}, \lambda {\mathscr{H}}_{2}\oplus {\mathscr{H}}, \ldots , \lambda {\mathscr{H}}_{n}\oplus {\mathscr{H}})$.

Based on these two steps, we get the desired result. □

Proof of Theorem 16

Proof

We shall prove the parts (1) and (5), while others can be proved analogously.

• 1) For two LIVPFNs ${\mathscr{H}}_{i}=([s_{a_{i}}, s_{b_{i}}], [s_{c_{i}}, s_{d_{i}}])$ and $\xi _{i}=([s_{A_{i}}, s_{B_{i}}], [s_{C_{i}}, s_{D_{i}}])$, we have

  $$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1}\oplus\xi_{1},\mathcal{H}_{2}\oplus\xi_{2},\ldots,\mathcal{H}_{n}\oplus\xi_{n}) \\ & =& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-\frac{{a_{i}^{2}}}{h^{2}}\right)\left( 1-\frac{{A_{i}^{2}}}{h^{2}}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-\frac{{b_{i}^{2}}}{h^{2}}\right)\left( 1-\frac{{B_{i}^{2}}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array} \right],\right.\\&&\left. \left[ \begin{array}{c} s_{h\left( \prod\limits_{i=1}^{n}\left( \frac{c_{i}}{h}\right)\left( \frac{C_{i}}{h}\right)\right)^{\omega_{i}}}, \\ s_{h\left( \prod\limits_{i=1}^{n}\left( \frac{d_{i}}{h}\right)\left( \frac{D_{i}}{h}\right)\right)^{\omega_{i}}} \end{array} \right] \right) \end{array} $$

  and

  $$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1}\otimes\xi_{1},\mathcal{H}_{2}\otimes\xi_{2},\ldots,\mathcal{H}_{n}\otimes\xi_{n}) \\ & =& \left( \left[ \begin{array}{c} s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)\left( \frac{{A}_{i}^{2}}{h^{2}}\right)\right)^{\omega_{i}}}}, \\ s_{h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-\left( \frac{{b}_{i}^{2}}{h^{2}}\right)\left( \frac{{B}_{i}^{2}}{h^{2}}\right)\right)^{\omega_{i}}}} \end{array} \right],\right.\\ &&\left. \left[ \begin{array}{c} s_{h\left( \prod\limits_{i=1}^{n}\sqrt{1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{C}_{i}^{2}}{h^{2}}\right)}\right)^{\omega_{i}}}, \\ s_{h\left( \prod\limits_{i=1}^{n}\sqrt{1-\left( 1-\frac{{d}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{D}_{i}^{2}}{h^{2}}\right)}\right)^{\omega_{i}}} \end{array} \right] \right) \end{array} $$

  For ${a_{i}^{2}}/h^{2}, A^{2}/h^{2}\leq 1$ and hence by Lemma 2, we get $1-\left (1-\frac {{a_{i}^{2}}}{h^{2}}\right )\left (1-\frac {{A_{i}^{2}}}{h^{2}}\right ) \geq \left (\frac {{a}_{i}^{2}}{h^{2}}\right )\left (\frac {{A}_{i}^{2}}{h^{2}}\right )$ which implies that

  $$ \begin{array}{@{}rcl@{}} && \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right) \leq 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)\left( \frac{{A}_{i}^{2}}{h^{2}}\right) \\ &\Rightarrow & \prod\limits_{i=1}^{n} \left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}\\ &&\quad \leq \prod\limits_{i=1}^{n} \left( 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)\left( \frac{{A}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}} \\ &\Rightarrow & h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}}\\ && \geq h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-\left( \frac{{a}_{i}^{2}}{h^{2}}\right)\left( \frac{{A}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}} \end{array} $$

  Similarly, we have

  $$ \begin{array}{@{}rcl@{}} &&h\sqrt{1-\prod\limits_{i=1}^{n} \left( \left( 1-\frac{{b}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{B}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}}\\ && \quad\geq h\sqrt{1-\prod\limits_{i=1}^{n} \left( 1-\left( \frac{{b}_{i}^{2}}{h^{2}}\right)\left( \frac{{B}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}} \end{array} $$

  Now, for non-membership components, we have ${c_{i}^{2}}/h^{2}, C^{2}/h^{2}\leq 1$ and again by Lemma 2, we get

  $$ \begin{array}{@{}rcl@{}} && 1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{C}_{i}^{2}}{h^{2}}\right) \geq \left( \frac{{c_{i}^{2}}}{h^{2}}\right)\left( \frac{{C_{i}^{2}}}{h^{2}}\right) \\ &\Rightarrow & \prod\limits_{i=1}^{n} \left( 1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{C}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}\\ && \geq \prod\limits_{i=1}^{n} \left( \left( \frac{{c}_{i}^{2}}{h^{2}}\right)\left( \frac{{C}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}} \end{array} $$
  $$ \begin{array}{@{}rcl@{}} &\Rightarrow & h\prod\limits_{i=1}^{n} \sqrt{\left( 1-\left( 1-\frac{{c}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{C}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}} \\ &&\geq h\prod\limits_{i=1}^{n} \left( \left( \frac{c_{i}}{h}\right)\left( \frac{C_{i}}{h}\right) \right)^{\omega_{i}} \end{array} $$

  Similarly, we get

  $$ \begin{array}{@{}rcl@{}} &&h\prod\limits_{i=1}^{n} \sqrt{\left( 1-\left( 1-\frac{{d}_{i}^{2}}{h^{2}}\right)\left( 1-\frac{{D}_{i}^{2}}{h^{2}}\right) \right)^{\omega_{i}}} \\ &&\geq h\prod\limits_{i=1}^{n} \left( \left( \frac{d_{i}}{h}\right)\left( \frac{D_{i}}{h}\right) \right)^{\omega_{i}} \end{array} $$

  Hence, by Eq. 9, we get desired result.

• 5) For any two LIVPFNs ${\mathscr{H}}_{i}=([s_{a_{i}}, s_{b_{i}}], [s_{c_{i}}, s_{d_{i}}])$ and $\xi _{i}=([s_{A_{i}}, s_{B_{i}}], [s_{C_{i}}, s_{D_{i}}])$, we have

  $$ \begin{array}{@{}rcl@{}} && \text{LIVPFWA}(\mathcal{H}_{1},\mathcal{H}_{2},\ldots,\mathcal{H}_{n}) \otimes \text{LIVPFWA}(\xi_{1},\xi_{2},\ldots,\xi_{n}) \\ & =& \left( \left[ \begin{array}{c} s_{h\sqrt{ \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) } }, \\ s_{h\sqrt{ \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{b}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{B}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) } } \end{array} \right],\right.\\ &&\left.\left[ \begin{array}{c} s_{h\sqrt{ 1-\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{c}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{C}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) } }, \\ s_{h\sqrt{ 1-\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{d}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{D}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) } } \end{array} \right] \right) \end{array} $$

  As $\prod \limits _{i=1}^{n}\left (1-\frac {{a}_{i}^{2}}{h^{2}}\right )^{\omega _{i}}, \prod \limits _{i=1}^{n}\left (1-\frac {{A}_{i}^{2}}{h^{2}}\right )^{\omega _{i}} \leq 1$ and thus by Lemma 2, we have

  $$ \begin{array}{@{}rcl@{}} && 1- \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)^{\omega_{i}} \right) \\ && \qquad \geq \prod\limits_{i=1}^{n}\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\omega_{i}} \prod\limits_{i=1}^{n}\left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)^{\omega_{i}} \\ &\Rightarrow & 1- \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( 1-\frac{{A_{i}^{2}}}{h^{2}}\right)^{\omega_{i}} \right) \\ && \qquad \geq \prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\cdot \left( 1-\frac{{A_{i}^{2}}}{h^{2}}\right)\right)^{\omega_{i}} \\&\Rightarrow & \left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a_{i}^{2}}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)^{\omega_{i}} \right) \\ && \qquad \leq 1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\cdot \left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)\right)^{\omega_{i}} \end{array} $$
  $$ \begin{array}{@{}rcl@{}} &\Rightarrow & h\sqrt{\left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)^{\omega_{i}} \right)} \\ && \qquad \leq h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{a}_{i}^{2}}{h^{2}}\right)\cdot \left( 1-\frac{{A}_{i}^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \end{array} $$

  Similarly, we have

  $$ \begin{array}{@{}rcl@{}} && h\sqrt{\left( 1-\prod\limits_{i=1}^{n}\left( 1-\frac{{b}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( 1-\frac{{B_{i}^{2}}}{h^{2}}\right)^{\omega_{i}} \right)} \\ && \qquad \leq h\sqrt{1-\prod\limits_{i=1}^{n}\left( \left( 1-\frac{{b_{i}^{2}}}{h^{2}}\right)\cdot \left( 1-\frac{{B}_{i}^{2}}{h^{2}}\right)\right)^{\omega_{i}}} \end{array} $$
  $$ \begin{array}{@{}rcl@{}} && h\sqrt{1-\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{c}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{C}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) } \\ && \qquad \geq h\left( \prod\limits_{i=1}^{n}\left( \frac{c_{i}}{h}\right)\left( \frac{C_{i}}{h}\right)\right)^{\omega_{i}} \\ &\text{and ~~ } & h\sqrt{1-\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{d}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right)\left( 1-\prod\limits_{i=1}^{n} \left( \frac{{D}_{i}^{2}}{h^{2}}\right)^{\omega_{i}}\right) } \\ && \qquad \qquad \qquad \geq h\left( \prod\limits_{i=1}^{n}\left( \frac{d_{i}}{h}\right)\left( \frac{D_{i}}{h}\right)\right)^{\omega_{i}} \end{array} $$

  Hence, based on these inequalities and by Eq. 9, the part (5) is proved.

□