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Group Decision Making Based on Power Heronian Aggregation Operators Under Linguistic Neutrosophic Environment

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Abstract

The power average (PA) operator can overcome some effects of awkward data given by predispose decision makers, and Heronian mean (HM) operator can consider the interrelationship of the aggregated arguments. In order to take the full use of these two kinds of operators, in this article, we combined the PA operator with HM operator and extended them to process linguistic neutrosophic information, and presented the linguistic neutrosophic power Heronian aggregation operator, linguistic neutrosophic power weight Heronian aggregation operator. Further, some properties of these new aggregation operators are investigated and some special cases are discussed. Furthermore, we propose new technique based on these operators for multiple attribute group decision making. Finally, an illustrative example was given to illustrate the effectiveness and advantages of the developed method by comparing with the existing method.

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Acknowledgements

This paper is supported by the National Natural Science Foundation of China (Nos. 71771140, 71471172), the Special Funds of Taishan Scholars Project of Shandong Province (No. ts201511045), Shandong Provincial Social Science Planning Project (Nos. 17BGLJ04, 16CGLJ31 and 16CKJJ27), the Teaching Reform Research Project of Undergraduate Colleges and Universities in Shandong Province (No. 2015Z057), and Key research and development program of Shandong Province (No. 2016GNC110016). The authors also would like to express appreciation to the anonymous reviewers and Editors for their very helpful comments that improved the paper.

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Authors and Affiliations

  1. School of Management Science and Engineering, Shandong University of Finance and Economics, Jinan, 250014, Shandong, China

    Peide Liu

  2. Department of Mathematics and Statistic, International Islamic University, Islamabad, Pakistan

    Tahir Mahmood & Qaisar Khan

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  1. Peide Liu
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  2. Tahir Mahmood
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Correspondence to Peide Liu.

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Appendix 1: Proof of the Theorem 2

Appendix 1: Proof of the Theorem 2

Proof

Firstly, we need to prove the following equation.

$$ \begin{aligned} \sum\limits_{i = 1}^{m} {\sum\limits_{j = i}^{m} {\left( {m\omega_{i} n_{i} } \right)^{P} \otimes \left( {m\omega_{j} n_{j} } \right)^{q} } } \hfill \\ = \left( {f^{* - 1} \left( {1 - \left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = i}^{m} {\left( {1 - (1 - (1 - f^{*} (s_{{\eta_{i} }} ))^{{m\omega_{i} }} )^{p} (1 - (1 - f^{*} (s_{{\eta_{j} }} ))^{{m\omega_{j} }} )^{q} } \right)} } } \right)} \right),} \right. \hfill \\ \quad f^{* - 1} \left( {\left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = i}^{m} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)} \right), \hfill \\ \quad \left. {f^{* - 1} \left( {\left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = i}^{m} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)} \right)} \right). \hfill \\ \end{aligned} $$
(21)

By the operational rules of LNNs defined in (6)–(9), we have

$$ \left( {m\omega_{i} n_{i} } \right)^{p} = \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} } \right),f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} } \right),f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} } \right)} \right)} \right\rangle , $$
$$ \left( {m\omega_{j} n_{j} } \right)^{q} = \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right),f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{j} }} )} \right)^{m\omega j} } \right)^{q} } \right),f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} \right)} \right\rangle . $$
$$ \left( {m\omega _{i} n_{i} } \right)^{p} \otimes \left( {m\omega _{j} n_{j} } \right)^{q} = \left\langle {\left( {f^{{* - 1}} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta _{i} }} )} \right)^{{m\omega _{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta _{j} }} )} \right)^{{m\omega _{j} }} } \right)^{q} } \right),} \right.} \right.f^{{* - 1}} \left( {\left( {1 - \left( {\left( {1 - \left( {f^{*} (s_{{\mu _{i} }} )} \right)^{{m\omega _{i} }} } \right)^{p} } \right)\left( {1 - \left( {f^{*} (s_{{\mu _{j} }} )} \right)^{{m\omega _{j} }} } \right)^{q} } \right)} \right),\left. {\left. {f^{{* - 1}} \left( {\left( {1 - \left( {\left( {1 - \left( {f^{*} (s_{{\nu _{i} }} )} \right)^{{m\omega _{i} }} } \right)^{p} } \right)\left( {1 - \left( {f^{*} (s_{{\nu _{j} }} )} \right)^{{m\omega _{j} }} } \right)^{q} } \right)} \right)} \right)} \right\rangle $$
(22)

(1) When \( m = 2, \) by Eqs. (6) and (22), we have

$$ \begin{aligned} \sum\limits_{i = 1}^{2} {\sum\limits_{j = i}^{2} {\left( {m\omega_{i} n_{i} } \right)^{P} \otimes \left( {m\omega_{j} n_{j} } \right)^{q} } } & = \left( {\left( {2\omega_{1} n_{1} } \right)^{P} \otimes \left( {2\omega_{1} n_{1} } \right)^{q} } \right) \oplus \left( {\left( {2\omega_{1} n_{1} } \right)^{P} \otimes \left( {2\omega_{2} n_{2} } \right)^{q} } \right) \oplus \left( {\left( {2\omega_{2} n_{2} } \right)^{P} \otimes \left( {2\omega_{2} n_{2} } \right)^{q} } \right) \\ & = \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{q} } \right),} \right.} \right.f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{q} } \right), \\ & \;\left. {\left. {f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{q} } \right)} \right)} \right\rangle \oplus \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{q} } \right),} \right.} \right. \\ & \;f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{q} } \right),\text{ }\left. {\left. {f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{1} }} )} \right)^{{2\omega_{1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{q} } \right)} \right)} \right\rangle \\ & \; \oplus \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{q} } \right),} \right.} \right.f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{q} } \right), \\ & \text{ }\left. {\left. {\;f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{2} }} )} \right)^{{2\omega_{2} }} } \right)^{q} } \right)} \right)} \right\rangle . \\ \end{aligned} $$

By using Eq. (6), we get

$$ \begin{aligned} = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{2} {\prod\limits_{j = 1}^{2} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{2\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right),} \right.} \right. \hfill \\ \text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{2} {\prod\limits_{j = 1}^{2} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{2\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{j} }} )} \right)^{{2\omega_{j} }} } \right)^{q} } \right)} } } \right), \hfill \\ \left. {\left. {\text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{2} {\prod\limits_{j = 1}^{2} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{2\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{j} }} )} \right)^{{2\omega_{j} }} } \right)^{q} } \right)} } } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$
(23)

That is, Eq. (21) holds for \( m = 2. \)

(2) Let us assume that Eq. (21) holds for \( m = z. \)

$$ \begin{aligned} = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{z} {\prod\limits_{j = 1}^{z} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{z\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{j} }} )} \right)^{{z\omega_{j} }} } \right)^{q} } \right)} } } \right),} \right.} \right. \hfill \\ \text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{z} {\prod\limits_{j = 1}^{z} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{z\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{j} }} )} \right)^{{z\omega_{j} }} } \right)^{q} } \right)} } } \right), \hfill \\ \left. {\left. {\text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{z} {\prod\limits_{j = 1}^{z} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{z\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{j} }} )} \right)^{{z\omega_{j} }} } \right)^{q} } \right)} } } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$
(24)

Furthermore, when \( m = z + 1 \), we have

$$ \begin{aligned} \sum\limits_{i = 1}^{z + 1} {\sum\limits_{j = i}^{z + 1} {\left( {(z + 1)\omega_{i} n_{i} } \right)^{P} \otimes \left( {(z + 1)\omega_{j} n_{j} } \right)^{q} } } & = \sum\limits_{i = 1}^{z} {\sum\limits_{j = i}^{z} {\left( {\left( {z + 1} \right)\omega_{i} n_{i} } \right)^{p} \otimes \left( {\left( {z + 1} \right)\omega_{j} n_{j} } \right)^{q} \oplus \sum\limits_{i = 1}^{z} {\left( {\left( {z + 1} \right)\omega_{i} n_{i} } \right)^{p} } } } \\ & \quad \otimes \left( {\left( {z + 1} \right)\omega_{z + 1} n_{z + 1} } \right)^{q} \oplus \left( {\left( {z + 1} \right)\omega_{z + 1} n_{z + 1} } \right)^{p} \otimes \left( {\left( {z + 1} \right)\omega_{z + 1} n_{z + 1} } \right)^{q} \\ \end{aligned} $$
(25)

Firstly, we prove that

$$ \begin{aligned} \sum\limits_{i = 1}^{z} {\left( {\left( {z + 1} \right)\omega_{i} n_{i} } \right)^{p} \otimes \left( {\left( {z + 1} \right)\omega_{z + 1} n_{z + 1} } \right)^{q} } = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{z} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{(z + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{z + 1} }} )} \right)^{{(z + 1)\omega_{z + 1} }} } \right)^{q} } \right)} } \right),} \right.} \right. \hfill \\ \quad \text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{z} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{(z + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{z + 1} }} )} \right)^{{(z + 1)\omega_{z + 1} }} } \right)^{q} } \right)} } \right),\left. {\left. {f^{* - 1} \left( {\prod\limits_{i = 1}^{z} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{(z + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{z + 1} }} )} \right)^{{(z + 1)\omega_{z + 1} }} } \right)^{q} } \right)} } \right)} \right)} \right\rangle \hfill \\ \end{aligned} $$
(26)

We shall prove Eq. (26) on mathematical induction on \( z. \)

(a) For \( z = 2, \) we have

$$ \begin{aligned} \sum\limits_{i = 1}^{2} {\left( {\left( {z + 1} \right)\omega_{i} n_{i} } \right)^{p} \otimes \left( {\left( {z + 1} \right)\omega_{z + 1} n_{z + 1} } \right)^{q} } = \left( {\left( {3\omega_{1} n_{1} } \right)^{p} \otimes \left( {3\omega_{3} n_{3} } \right)^{q} } \right) \oplus \left( {\left( {3\omega_{2} n_{2} } \right)^{p} \otimes \left( {3\omega_{3} n_{3} } \right)^{q} } \right) \hfill \\ = \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{1} }} )} \right)^{{3\omega_{1} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right),} \right.} \right.f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{1} }} )} \right)^{{3\omega_{1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right), \hfill \\ \text{ }\left. {\left. {f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{1} }} )} \right)^{{3\omega_{1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right)} \right)} \right\rangle \oplus \left\langle {\left( {f^{* - 1} \left( {\left( {1 - \left( {1 - f^{*} (s_{{\eta_{2} }} )} \right)^{{3\omega_{2} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right),} \right.} \right. \hfill \\ f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{2} }} )} \right)^{{3\omega_{2} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right),\text{ }\left. {\left. {f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{2} }} )} \right)^{{3\omega_{2} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$
$$ \begin{aligned} = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{2} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{3\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right)} } \right),} \right.} \right.f^{* - 1} \left( {\prod\limits_{i = 1}^{2} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{3\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right)} } \right), \hfill \\ \left. {\left. {\text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{2} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{3\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{3} }} )} \right)^{{3\omega_{3} }} } \right)^{q} } \right)} } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$
(27)

(b) Let us assume that Eq. (26) holds for \( z = b \), that is;

$$ \begin{aligned} \sum\limits_{i = 1}^{z} {\left( {\left( {\left( {z + 1} \right)\omega_{i} n_{i} } \right)^{p} \otimes \left( {\left( {z + 1} \right)\omega_{b + 1} n_{b + 1} } \right)^{q} } \right)} \hfill \\ = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{b} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{\left( {b + 1} \right)\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{b + 1} }} )} \right)^{{(b + 1)\omega_{b + 1} }} } \right)^{q} } \right)} } \right),} \right.} \right.f^{* - 1} \left( {\prod\limits_{i = 1}^{b} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{(b + 1)\omega_{i} }} } \right)^{p} } \right.} } \right. \hfill \\ \text{ }\left. {\left( {1 - \left( {f^{*} (s_{{\mu_{b + 1} }} )} \right)^{{(b + 1)\omega_{b + 1} }} } \right)^{q} } \right),\left. {\left. {f^{* - 1} \left( {\prod\limits_{i = 1}^{b} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{(b + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{b + 1} }} )} \right)^{{(b + 1)\omega_{b + 1} }} } \right)^{q} } \right)} } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$
(28)

Then, when \( z = b + 1 \), we have

$$ \begin{aligned} \sum\limits_{i = 1}^{b + 1} {\left( {\left( {\left( {b + 2} \right)\omega_{i} n_{i} } \right)^{p} \otimes \left( {\left( {b + 2} \right)\omega_{b + 2} n_{b + 2} } \right)^{q} } \right)} = \sum\limits_{i = 1}^{b} {\left( {\left( {\left( {b + 2} \right)\omega_{i} n_{i} } \right)^{p} \otimes \left( {\left( {b + 2} \right)\omega_{b + 2} n_{b + 2} } \right)^{q} } \right)} \oplus \left( {\left( {\left( {b + 2} \right)\omega_{b + 1} n_{b + 1} } \right)^{p} \otimes \left( {\left( {b + 2} \right)\omega_{b + 2} n_{b + 2} } \right)^{q} } \right) \hfill \\ = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{b} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{\left( {b + 2} \right)\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)} } \right),f^{* - 1} \left( {\prod\limits_{i = 1}^{b} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{(b + 2)\omega_{i} }} } \right)^{p} } \right.} } \right.} \right.} \right. \hfill \\ \quad \text{ }\left. {\left( {1 - \left( {f^{*} (s_{{\mu_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right),\left. {\text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{b} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{(b + 2)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)} } \right)} \right) \hfill \\ \quad \oplus \left( {f^{* - 1} \left( {1 - \left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{b + 1} }} )} \right)^{{\left( {b + 2} \right)\omega_{b + 1} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)} \right),} \right.f^{* - 1} \left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{b + 1} }} )} \right)^{{(b + 2)\omega_{b + 1} }} } \right)^{p} } \right. \hfill \\ \quad \text{ }\left. {\left( {1 - \left( {f^{*} (s_{{\mu_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)\left. {f^{* - 1} \left( {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{b + 1} }} )} \right)^{{(b + 2)\omega_{b + 1} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)} \right)} \right\rangle \hfill \\ = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{b + 1} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{\left( {b + 2} \right)\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)} } \right),} \right.} \right.f^{* - 1} \left( {\prod\limits_{i = 1}^{b + 1} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{(b + 2)\omega_{i} }} } \right)^{p} } \right.} } \right. \hfill \\ \quad \text{ }\left. {\left( {1 - \left( {f^{*} (s_{{\mu_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right),\left. {\left. {f^{* - 1} \left( {\prod\limits_{i = 1}^{b + 1} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{(b + 2)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{b + 2} }} )} \right)^{{(b + 2)\omega_{b + 2} }} } \right)^{q} } \right)} } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$

Therefore, Eq. (26) is true for \( z = b + 1. \) Hence, Eq. (26) is also true for all \( z. \)

Similarly, we can prove the other parts of Eq. (25).

So Eq. (25) becomes

$$ \begin{aligned} \sum\limits_{i = 1}^{z + 1} {\sum\limits_{j = i}^{z + 1} {\left( {(z + 1)\omega_{i} n_{i} } \right)^{P} \otimes \left( {(z + 1)\omega_{j} n_{j} } \right)^{q} } } = \left\langle {\left( {f^{* - 1} \left( {1 - \prod\limits_{i = 1}^{z + 1} {\prod\limits_{j = 1}^{z + 1} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{(z + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{j} }} )} \right)^{{(z + 1)\omega_{j} }} } \right)^{q} } \right)} } } \right),} \right.} \right. \hfill \\ \text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{z + 1} {\prod\limits_{j = 1}^{z + 1} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{(z + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{j} }} )} \right)^{{(z + 1)\omega_{j} }} } \right)^{q} } \right)} } } \right),\left. {\left. {\text{ }f^{* - 1} \left( {\prod\limits_{i = 1}^{z + 1} {\prod\limits_{j = 1}^{z + 1} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{(z + 1)\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{j} }} )} \right)^{{(z + 1)\omega_{j} }} } \right)^{q} } \right)} } } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$

Therefore, Eq. (21) is true for \( m = z + 1 \). Hence, Eq. (21) is true for all \( m. \)

By Eq. (21), we can prove that Eq. (20) is right. From Eq. (21) and the operational laws defined for LNNs, we have

$$ \begin{aligned} \frac{2}{m(m + 1)}\sum\limits_{i = 1}^{m} {\sum\limits_{j = i}^{m} {\left( {m\omega_{i} n_{i} } \right)^{p} \otimes \left( {m\omega_{j} n_{j} } \right)^{q} } } \hfill \\ = \left\langle {\left( {f^{* - 1} \left( {1 - \left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = 1}^{m} {\left( {1 - \left( {1 - \left( {1 - f^{*} (s_{{\eta_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {1 - f^{*} (s_{{\eta_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)^{{\frac{2}{m(m + 1)}}} } \right),} \right.} \right. \hfill \\ \quad f^{* - 1} \left( {\left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = 1}^{m} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\eta_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\eta_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)^{{\frac{2}{m(m + 1)}}} } \right), \hfill \\ \left. {\left. {\quad f^{* - 1} \left( {\left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = 1}^{m} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\eta_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\eta_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)^{{\frac{2}{m(m + 1)}}} } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$

so,

$$ \begin{aligned} \left( {\frac{2}{m(m + 1)}\sum\limits_{i = 1}^{m} {\sum\limits_{j = i}^{m} {\left( {m\omega_{i} n_{i} } \right)^{p} \otimes \left( {m\omega_{j} n_{j} } \right)^{q} } } } \right)^{{\frac{1}{p + q}}} \hfill \\ = \left\langle {\left( {f^{* - 1} \left( {1 - \left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = i}^{m} {\left( {1 - (1 - (1 - f^{*} (s_{{\eta_{i} }} ))^{{m\omega_{i} }} )^{p} (1 - (1 - f^{*} (s_{{\eta_{j} }} ))^{{m\omega_{j} }} )^{q} } \right)} } } \right)^{{\frac{2}{m(m + 1)}}} } \right)^{{\frac{1}{p + q}}} ,} \right.} \right. \hfill \\ \quad f^{* - 1} \left( {1 - \left( {1 - \left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = i}^{m} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\mu_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\mu_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)^{{\frac{2}{m(m + 1)}}} } \right)^{{\frac{1}{p + q}}} } \right), \hfill \\ \left. {\left. {\quad f^{* - 1} \left( {1 - \left( {1 - \left( {\prod\limits_{i = 1}^{m} {\prod\limits_{j = i}^{m} {\left( {1 - \left( {1 - \left( {f^{*} (s_{{\nu_{i} }} )} \right)^{{m\omega_{i} }} } \right)^{p} \left( {1 - \left( {f^{*} (s_{{\nu_{j} }} )} \right)^{{m\omega_{j} }} } \right)^{q} } \right)} } } \right)^{{\frac{2}{m(m + 1)}}} } \right)^{{\frac{1}{p + q}}} } \right)} \right)} \right\rangle . \hfill \\ \end{aligned} $$

This completes the proof of Theorem 2.□

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Liu, P., Mahmood, T. & Khan, Q. Group Decision Making Based on Power Heronian Aggregation Operators Under Linguistic Neutrosophic Environment. Int. J. Fuzzy Syst. 20, 970–985 (2018). https://doi.org/10.1007/s40815-018-0450-2

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