Appendix: Proof of the Theorems
Proof of Theorem 1
Here, we prove part 1 and 5. Proofs of 2, 3, 4, and 6 are easy to reader can get by replacing first vector in (p, q)-RLDFN as in 1 and 5, whereas 7 can get by rule 8 in the Definition 11. The rest follows on the similar lines.
1. Let \({\mathcal {A}} = \left( \left( \mu _{A}, \eta _{A} \right) , \left( \alpha _{A}, \beta _{A}\right) \right) \) and \(\mathcal {B} = \left( \left( \mu _{B}, \eta _{B} \right) ,\left( \alpha _{B}, \beta _{B}\right) \right) \) be any (p, q)RLDFNs on U. Where from it can be easily proved that
$$\begin{aligned} \begin{array}{lcl} \left( \mu _{A}, \eta _{A} \right) \oplus \left( \mu _{B}, \eta _{B} \right) &{}=&{} \left( \root p \of { \left( \mu _{A}\right) ^{p} + \left( \mu _{B}\right) ^{p} - \left( \mu _{A}\right) ^{p}\left( \mu _{B}\right) ^{p}}, \eta _{A}\eta _{B} \right) \\ &{}=&{} \left( \root p \of { \left( \mu _{B}\right) ^{p} + \left( \mu _{A}\right) ^{p} - \left( \mu _{B}\right) ^{p}\left( \mu _{A}\right) ^{p}}, \eta _{B}\eta _{A} \right) \\ &{}=&{}\left( \mu _{B}, \eta _{B} \right) \oplus \left( \mu _{A}, \eta _{A} \right) . \end{array} \end{aligned}$$
Similarly, we can prove that \( \left( \alpha _{A}, \beta _{A}\right) \oplus \left( \alpha _{B}, \beta _{B}\right) = \left( \mu _{B}, \eta _{B} \right) \oplus \left( \alpha _{A}, \beta _{A}\right) \). Therefore, we obtain that \({\mathcal {A}}\oplus \mathcal {B} = \mathcal {B}\oplus {\mathcal {A}}\).
5. Let \({\mathcal {A}} = \left( \left( \mu _{A}, \eta _{A} \right) , \left( \alpha _{A}, \beta _{A}\right) \right) \) and \(\mathcal {B} = \left( \left( \mu _{B}, \eta _{B} \right) ,\left( \alpha _{B}, \beta _{B}\right) \right) \) be any (p, q)RLDFNs on U. Where from it can be easily proved that
$$\begin{aligned} \lambda \left( \mu _{A}, \eta _{A} \right) \oplus \lambda \left( \mu _{B}, \eta _{B} \right)= & {} \begin{array}{ccc} \left( \root p \of { 1 - \left( 1 - \left( \mu _{A}\right) ^{p} \right) ^{\lambda }}, \left( \eta _{A}\right) ^{\lambda }\right) \oplus &{}\\ \left( \root p \of { 1 - \left( 1 - \left( \mu _{B}\right) ^{p} \right) ^{\lambda }}, \left( \eta _{B}\right) ^{\lambda } \right) &{} \end{array} \\= & {} \left( \begin{array}{ccc} \root p \of {\begin{array}{ccc} \left( \root p \of { 1- \left( 1- \left( \mu _{A}\right) ^{p}\right) ^{\lambda } }\right) ^{p} + &{} \\ \left( \root p \of { 1- \left( 1- \left( \mu _{B}\right) ^{p} \right) ^{\lambda } }\right) ^{p} + &{} \\ - \left( \begin{array}{ccc} \left( \root p \of { 1- \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda } }\right) ^{p} &{} \\ \left( \root p \of { 1- \left( 1- \left( \mu _{B}\right) ^{p} \right) ^{\lambda } }\right) ^{p} &{} \end{array}\right) &{} \end{array}}, &{} \\ \left( \eta _{A}\right) ^{\lambda }\left( \eta _{B}\right) ^{\lambda }&{} \end{array} \right) \\ \end{aligned}$$
and
$$\begin{aligned} \lambda \left( \left( \mu _{A}, \eta _{A} \right) \oplus \left( \mu _{B}, \eta _{B} \right) \right)= & {} \lambda \left( \root p \of { \left( \mu _{A}\right) ^{p} + \left( \mu _{B}\right) ^{p} - \left( \mu _{A}\right) ^{p}\left( \mu _{B}\right) ^{p}}, \eta _{A}\eta _{B}\right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1 - \left( 1 - \left( \root p \of { \left( \mu _{A}\right) ^{p} + \left( \mu _{B}\right) ^{p} - \left( \mu _{A}\right) ^{p}\left( \mu _{B}\right) ^{p}} \right) ^{p} \right) ^{\lambda }}, &{} \\ \left( \eta _{A}\eta _{B} \right) ^{\lambda } &{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1 - \left( 1 - \left( \left( \mu _{A}\right) ^{p} + \left( \mu _{B}\right) ^{p} - \left( \mu _{A}\right) ^{p}\left( \mu _{B}\right) ^{p} \right) \right) ^{\lambda }}, &{} \\ \left( \eta _{A}\eta _{B} \right) ^{\lambda } &{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1 - \left( \left( 1 - \left( \mu _{A}\right) ^{p} \right) \left( 1 - \left( \mu _{B}\right) ^{p} \right) \right) ^{\lambda }} ,&{} \\ \left( \eta _{A}\eta _{B}\right) ^{\lambda } &{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1- \left( \left( 1- \left( \mu _{A}\right) ^{p} \right) \left( 1- \left( \mu _{B}\right) ^{p} \right) \right) ^{\lambda }},&{} \\ \left( \eta _{A}\eta _{B} \right) ^{\lambda } &{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda }\left( 1- \left( \mu _{B}\right) ^{p} \right) ^{\lambda }}, &{} \\ \left( \eta _{A}\eta _{B} \right) ^{\lambda } &{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { \begin{array}{ccc} 1- \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda } + 1 - \left( 1- \left( \mu _{B}\right) ^{p} \right) ^{\lambda } +&{} \\ - \left( 1- \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda }\right) \left( 1 - \left( 1- \left( \mu _{B}\right) ^{p} \right) ^{\lambda } \right) &{} \end{array}}, &{} \\ \left( \eta _{A}\eta _{B} \right) ^{\lambda } &{} \end{array} \right) , \end{aligned}$$
and so, \( \lambda \left( \mu _{A}, \eta _{A} \right) \oplus \lambda \left( \mu _{B}, \eta _{B} \right) = \lambda \left( \left( \mu _{A}, \eta _{A} \right) \oplus \left( \mu _{B}, \eta _{B} \right) \right) \). Similarly, it follows that \( \lambda \left( \alpha _{A}, \beta _{A} \right) \oplus \lambda \left( \alpha _{B}, \beta _{B} \right) = \lambda \left( \left( \alpha _{A}, \beta _{A} \right) \oplus \left( \alpha _{B}, \beta _{B} \right) \right) \). Therefore, we obtain that \(\lambda \left( {\mathcal {A}}\oplus \mathcal {B}\right) = \lambda {\mathcal {A}} \oplus \lambda \mathcal {B}\).
7. Let \({\mathcal {A}} = \left( \left( \mu _{A}, \eta _{A} \right) , \left( \alpha _{A}, \beta _{A}\right) \right) \) be a (p, q)RLDFN on U. Where from it can be easily proved that
$$\begin{aligned} \begin{array}{lcl} \left( \lambda + \xi \right) \left( \mu _{A}, \eta _{A} \right) &{}=&{} \left( \root p \of { 1 - \left( 1 - \left( \mu _{A}\right) ^{p} \right) ^{\lambda +\xi }}, \left( \eta _{A}\right) ^{\lambda +\xi } \right) \\ &{}=&{} \left( \begin{array}{ccc} \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda }\left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\xi }},&{} \\ \left( \eta _{A} \right) ^{\lambda +\xi } &{} \end{array} \right) \end{array} \end{aligned}$$
and
$$\begin{aligned} \begin{array}{lcl} \lambda \left( \mu _{A}, \eta _{A} \right) \oplus \xi \left( \mu _{A}, \eta _{A} \right) &{}=&{} \begin{array}{ccc} \left( \root p \of { 1 - \left( 1 - \left( \mu _{A}\right) ^{p} \right) ^{\lambda }}, \left( \eta _{A}\right) ^{\lambda } \right) \oplus &{}\\ \left( \root p \of { 1 - \left( 1 - \left( \mu _{A}\right) ^{p} \right) ^{\xi }}, \left( \eta _{A}\right) ^{\xi } \right) &{} \end{array} \\ &{}=&{} \left( \begin{array}{ccc} \root p \of {\begin{array}{ccc} \left( \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda } }\right) ^{p} + &{} \\ \left( \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\xi } }\right) ^{p} + &{} \\ - \left( \begin{array}{ccc} \left( \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda } }\right) ^{p} &{} \\ \left( \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\xi } }\right) ^{p} &{} \end{array}\right) &{} \end{array}}, &{} \\ \left( \eta _{A}\right) ^{\lambda }\left( \eta _{A}\right) ^{\xi }&{} \end{array} \right) \\ &{}=&{} \left( \begin{array}{ccc} \root p \of { \begin{array}{ccc} 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda } + 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\xi } +&{} \\ - \left( 1- \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\lambda }\right) \left( 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\xi } \right) &{} \end{array}}, &{} \\ \left( \eta _{A} \right) ^{\lambda +\xi } &{} \end{array} \right) , \end{array} \end{aligned}$$
and so, \( \left( \lambda + \xi \right) \left( \mu _{A}, \eta _{A} \right) = \lambda \left( \mu _{A}, \eta _{A} \right) \oplus \xi \left( \mu _{A}, \eta _{A} \right) \). It can be similarly proved that \( \left( \lambda + \xi \right) \left( \alpha _{A}, \beta _{A} \right) = \lambda \left( \alpha _{A}, \beta _{A} \right) \oplus \xi \left( \alpha _{A}, \beta _{A} \right) \). Therefore, we obtain that \(\left( \lambda + \xi \right) {\mathcal {A}} = \lambda {\mathcal {A}} \oplus \xi {\mathcal {A}}\). \(\square \)
Proof of Theorem 3
To prove the theorem, we use mathematical induction on n. Therefore, we have the following.
Step 1. Now, for \(n = 2\), we get
$$\begin{aligned} \displaystyle \bigoplus ^{2}_{i = 1}\lambda _{i}\left( \mu _{A_{i}}, \eta _{A_{i}} \right)= & {} \lambda _{1}\left( \mu _{A_{1}}, \eta _{A_{1}} \right) \oplus \lambda _{2}\left( \mu _{A_{2}}, \eta _{A_{2}} \right) \\= & {} \begin{array}{ccc} \left( \root p \of { 1 - \left( 1 - \left( \mu _{A_{1}}\right) ^{p} \right) ^{\lambda _{1}}}, \left( \eta _{A_{1}}\right) ^{\lambda _{1}} \right) \oplus &{} \\ \left( \root p \of { 1 - \left( 1 - \left( \mu _{A_{2}}\right) ^{p} \right) ^{\lambda _{2}}}, \left( \eta _{A_{2}}\right) ^{\lambda _{2}} \right) &{} \end{array}\\= & {} \left( \begin{array}{ccc} \root p \of {\begin{array}{ccc} \left( \root p \of { 1 - \left( 1- \left( \mu _{A_{1}}\right) ^{p} \right) ^{\lambda _{1}}}\right) ^{p} + &{} \\ \left( \root p \of { 1 - \left( 1- \left( \mu _{A_{2}}\right) ^{p} \right) ^{\lambda _{2}} }\right) ^{p} + &{} \\ - \left( \begin{array}{ccc} \left( \root p \of { 1 - \left( 1- \left( \mu _{A_{1}}\right) ^{p} \right) ^{\lambda _{1}}}\right) ^{p} &{} \\ \left( \root p \of { 1 - \left( 1- \left( \mu _{A_{2}}\right) ^{p} \right) ^{\lambda _{2}}}\right) ^{p} &{} \end{array}\right) &{} \end{array}}, &{} \\ \left( \beta _{A_{1}}\right) ^{\lambda _{1}}\left( \eta _{A_{2}}\right) ^{\lambda _{2}}&{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { \begin{array}{ccc} 1- \left( 1- \left( \mu _{A_{1}}\right) ^{p} \right) ^{\lambda _{1}} + 1- \left( 1- \left( \mu _{A_{2}}\right) ^{p} \right) ^{\lambda _{2}} +&{} \\ - \left( 1- \left( 1- \left( \mu _{A_{1}}\right) ^{p} \right) ^{\lambda _{1}}\right) \left( 1- \left( 1- \left( \mu _{A_{2}}\right) ^{p} \right) ^{\lambda _{2}} \right) &{} \end{array}}, &{} \\ \left( \eta _{A_{1}}\right) ^{\lambda _{1}}\left( \eta _{A_{2}}\right) ^{\lambda _{2}}&{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1 - \left( 1- \left( \mu _{A_{1}}\right) ^{p} \right) ^{\lambda _{1}}\left( 1- \left( \mu _{A_{2}}\right) ^{p} \right) ^{\lambda _{2}}}, &{} \\ \displaystyle \prod ^{2}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array} \right) \\= & {} \left( \root p \of { 1 - \displaystyle \prod ^{2}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}}, \displaystyle \prod ^{2}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \right) . \end{aligned}$$
Similarly, \( \displaystyle \bigoplus \nolimits ^{2}_{i = 1}\lambda _{i}\left( \alpha _{A_{i}}, \beta _{A_{i}} \right) = \left( \root p \of { 1 - \displaystyle \prod \nolimits ^{2}_{i = 1}\left( 1- \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}}, \displaystyle \prod \nolimits ^{2}_{i = 1}\left( \beta _{A_{i}} \right) ^{\lambda _{i}} \right) \). Thus, Eq. 11 holds.
Step 2. Suppose that Eq. 11 holds for \(n = k\), that is
$$\begin{aligned} \mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{k} \right) = \left( \begin{array}{ccc} \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{k}_{i = 1}\left( \eta _{A_{i}}\right) ^{\lambda _{i}}&{} \end{array}\right) , &{} \\ \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \alpha _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{k}_{i = 1}\left( \beta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right)&\end{array}\right) . \end{aligned}$$
Step 3. Now, we have to prove that Eq. 11 holds for \(n = k + 1\), based on the operational laws of the (p, q)RLDFNs, we can get
$$\begin{aligned} \displaystyle \bigoplus ^{k+1}_{i = 1}\lambda _{i}\left( \mu _{A_{i}}, \eta _{A_{i}} \right)= & {} \displaystyle \bigoplus ^{k}_{i = 1}\lambda _{i}\left( \mu _{A_{i}}, \eta _{A_{i}} \right) \oplus \lambda _{k+1}\left( \mu _{A_{k+1}}, \eta _{A_{k+1}} \right) \\= & {} \begin{array}{ccc} \left( \root p \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}}, \displaystyle \prod ^{k}_{i = 1}\left( \eta _{A_{i}}\right) ^{\lambda _{i}}\right) &{} \\ \oplus \left( \root p \of { 1 - \left( 1 - \left( \mu _{A_{k+1}}\right) ^{p} \right) ^{\lambda _{k+1}}}, \left( \eta _{A_{k+1}}\right) ^{\lambda _{k+1}} \right) &{} \end{array} \\= & {} \left( \begin{array}{ccc} \root p \of {\begin{array}{ccc} \left( \root p \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{i}\right) ^{p} \right) ^{\lambda _{i}}}\right) ^{p} + &{} \\ \left( \root p \of { 1 - \left( 1- \left( \mu _{k+1}\right) ^{q} \right) ^{\lambda _{k+1}} }\right) ^{p} + &{} \\ - \left( \begin{array}{ccc} \left( \root p \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{i}\right) ^{p} \right) ^{\lambda _{i}}}\right) ^{p} &{} \\ \left( \root p \of { 1 - \left( 1- \left( \mu _{k+1}\right) ^{p} \right) ^{\lambda _{k+1}}}\right) ^{p} &{} \end{array}\right) &{} \end{array}}, &{} \\ \displaystyle \prod ^{k}_{i = 1}\left( \eta _{i}\right) ^{\lambda _{1}}\left( \eta _{k+1}\right) ^{\lambda _{k+1}}&{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { \begin{array}{ccc} 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{i}\right) ^{p} \right) ^{\lambda _{i}} + 1 - \left( 1- \left( \mu _{k+1}\right) ^{p} \right) ^{\lambda _{k+1}} +&{} \\ - \left( 1- \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{i}\right) ^{p} \right) ^{\lambda _{i}}\right) \left( 1 - \left( 1- \left( \mu _{k+1}\right) ^{p} \right) ^{\lambda _{k+1}} \right) &{} \end{array}}, &{} \\ \displaystyle \prod ^{k}_{i = 1}\left( \eta _{A_{i}}\right) ^{\lambda _{i}}\left( \eta _{A_{k+1}}\right) ^{\lambda _{k+1}}&{} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \root p \of { 1- \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}\left( 1- \left( \mu _{A_{k+1}}\right) ^{p} \right) ^{\lambda _{k+1}}}, &{} \\ \displaystyle \prod ^{k+1}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array} \right) \\= & {} \left( \root p \of { 1 - \displaystyle \prod ^{k+1}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}}, \displaystyle \prod ^{k+1}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \right) . \end{aligned}$$
Similarly, \( \displaystyle \bigoplus \nolimits ^{k+1}_{i = 1}\lambda _{i}\left( \alpha _{A_{i}}, \beta _{A_{i}} \right) = \left( \root p \of { 1 - \displaystyle \prod \nolimits ^{k+1}_{i = 1}\left( 1- \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}}, \displaystyle \prod \nolimits ^{k+1}_{i = 1}\left( \beta _{A_{i}} \right) ^{\lambda _{i}} \right) \). Therefore, Eq. 11 holds for \(n = k + 1\), and hence, Eq. 11 holds for any i.
In the following, we will prove that \(\mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n} \right) \) is also a (p, q)RLDFN. Then, since \(0\le \left( \alpha _{A_{i}}\right) ^{p} + \left( \beta _{A_{i}}\right) ^{q} \le 1\), we have
$$\begin{aligned}{} & {} \left( \beta _{A_{i}}\right) ^{q} \le 1-\left( \alpha _{A_{i}}\right) ^{p}\\{} & {} \left( \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \le \left( 1-\left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} \\{} & {} \displaystyle \prod ^{n}_{i = 1}\left( \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \le \displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} \\{} & {} -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} +\displaystyle \prod ^{n}_{i = 1}\left( \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \le 0\\{} & {} 0\le 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} +\displaystyle \prod ^{n}_{i = 1}\left( \left( \beta _{A_{i}}\right) ^{\lambda _{i}} \right) ^{q} \le 1. \end{aligned}$$
Thus, we obtain that
$$\begin{aligned} 0\le \left( 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}\right) \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}} + \displaystyle \prod ^{n}_{i = 1}\left( \left( \beta _{A_{i}}\right) ^{\lambda _{i}} \right) ^{q}\displaystyle \prod ^{n}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \le 1\end{aligned}$$
since
$$\begin{aligned} 0\le \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}}, \displaystyle \prod ^{n}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}}\le 1 \end{aligned}$$
and
$$\begin{aligned} 0\le 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} +\displaystyle \prod ^{n}_{i = 1}\left( \left( \beta _{A_{i}}\right) ^{\lambda _{i}} \right) ^{q} \le 1. \end{aligned}$$
\(\square \)
Proof of Theorem 4
Then, since \({\mathcal {A}}_{1} = {\mathcal {A}}_{2} = \ldots = {\mathcal {A}}_{n} = {\mathcal {A}}\) by Theorem 3, we have
$$\begin{aligned} \begin{array}{lcl} \mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n} \right) &{}=&{} \left( \begin{array}{ccc} \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{n}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right) , &{} \\ \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{n}_{i = 1}\left( \beta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right)&\end{array}\right) \end{array}\\ \begin{array}{lcl} &{}=&{} \left( \begin{array}{ccc} \left( \begin{array}{ccc} \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}},&{} \\ \left( \eta _{A} \right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}&{} \end{array}\right) , &{} \\ \left( \begin{array}{ccc} \root p \of { 1 - \left( 1- \left( \alpha _{A}\right) ^{p} \right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}},&{} \\ \left( \beta _{A} \right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}&{} \end{array}\right) &{} \end{array}\right) \\ &{}=&{} \left( \begin{array}{ccc} \left( \root p \of { 1 - \left( 1- \left( \mu _{A}\right) ^{p} \right) ^{1}}, \left( \eta _{A} \right) ^{1}\right) ,&{} \\ \left( \root p \of { 1 - \left( 1- \left( \alpha _{A}\right) ^{p} \right) ^{1}}, \left( \beta _{A} \right) ^{1}\right) &{} \end{array}\right) \\ &{}=&{} \left( \left( \mu _{A}, \eta _{A} \right) , \left( \alpha _{A}, \beta _{A}\right) \right) \\ &{}=&{}{\mathcal {A}}. \end{array} \end{aligned}$$
Therefore, we obtain that \(\mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots , {\mathcal {A}}_{n} \right) = {\mathcal {A}}\). \(\square \)
Proof of Theorem 5
Then, since \({\mathcal {A}}_{i}\preceq \mathcal {B}_{i}\) for every \(i=1,2,3,\ldots , n\), we have \(\mu _{A_{i}} \le \mu _{B_{i}}\) and \(\eta _{A_{i}} \ge \eta _{B_{i}}\), that is
$$\begin{aligned}{} & {} \left( \mu _{A_{i}}\right) ^{p} \le \left( \mu _{B_{i}}\right) ^{p}\nonumber \\{} & {} 1- \left( \mu _{A_{i}}\right) ^{p} \ge 1- \left( \mu _{B_{i}}\right) ^{p}\nonumber \\{} & {} \left( 1- \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}} \ge \left( 1- \left( \mu _{B_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\{} & {} \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}} \ge \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{B_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\{} & {} 1-\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}} \le 1-\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{B_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\{} & {} \root p \of { 1-\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}} \le \root p \of { 1-\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{B_{i}}\right) ^{p}\right) ^{\lambda _{i}}}. \end{aligned}$$
(19)
From these calculations, we obtain
$$\begin{aligned}{} & {} \left( \eta _{A_{i}}\right) ^{\lambda _{i}} \ge \left( \eta _{B_{i}}\right) ^{\lambda _{i}}\nonumber \\{} & {} \displaystyle \prod ^{n}_{i = 1}\left( \eta _{A_{i}}\right) ^{\lambda _{i}} \ge \displaystyle \prod ^{n}_{i = 1}\left( \eta _{B_{i}}\right) ^{\lambda _{i}}. \end{aligned}$$
(20)
Thus, by Eqs. (19) and (20), we have
$$\begin{aligned} \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{n}_{i = 1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right) \preceq \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \mu _{B_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{n}_{i = 1}\left( \eta _{B_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right) . \end{aligned}$$
(21)
Similarly, we can show that
$$\begin{aligned} \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{n}_{i = 1}\left( \beta _{A_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right) \preceq \left( \begin{array}{ccc} \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \alpha _{B_{i}}\right) ^{p} \right) ^{\lambda _{i}}},&{} \\ \displaystyle \prod ^{n}_{i = 1}\left( \beta _{B_{i}} \right) ^{\lambda _{i}}&{} \end{array}\right) . \end{aligned}$$
(22)
Therefore, by Eqs. (21) and (22), we get \(\mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1}, \ldots ,{\mathcal {A}}_{n} \right) \preceq \mathcal{W}\mathcal{A}\left( \mathcal {B}_{1}, \ldots , \mathcal {B}_{n} \right) \). \(\square \)
Proof of Theorem 6
For the membership grades of \(\mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1},\ldots ,{\mathcal {A}}_{n}\right) \), we have
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}{} & {} \le \left( \mu _{A_{i}}\right) ^{p}\nonumber \\ 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}{} & {} \ge 1 - \left( \mu _{A_{i}}\right) ^{p}\nonumber \\ \left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}{} & {} \ge \left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\ \displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}{} & {} \ge \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\ 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}{} & {} \le 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\ \root p \of { 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}{} & {} \le \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\nonumber \\ \root p \of { 1-\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{ \displaystyle \sum ^{n}_{i = 1}\lambda _{i} }}{} & {} \le \root p \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\nonumber \\ \root p \of { 1 -\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) }{} & {} \le \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\nonumber \\ \displaystyle \bigwedge ^{n}_{i=1}\mu _{A_{i}}{} & {} \le \root p \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}} \end{aligned}$$
(23)
and
$$\begin{aligned} \displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}{} & {} \ge \left( \mu _{A_{i}}\right) ^{p}\nonumber \\ 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}{} & {} \le 1-\left( \mu _{A_{i}}\right) ^{p}\nonumber \\ \left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}{} & {} \le \left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\ \displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}{} & {} \le \displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\ 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}{} & {} \ge 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}\nonumber \\ \root p \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}{} & {} \ge \root p \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\nonumber \\ \root p \of {1-\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}}{} & {} \ge \root p \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\nonumber \\ \root p \of {1-\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \mu _{A_{i}}\right) ^{p}\right) ^{1}}{} & {} \ge \root p \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\nonumber \\ \displaystyle \bigvee ^{n}_{i=1}\mu _{A_{i}}{} & {} \ge \root p \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}. \end{aligned}$$
(24)
Then, by Eqs. (23) and (24), we get
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1}\mu _{A_{i}} \le \root p \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \mu _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\le \displaystyle \bigvee ^{n}_{i=1}\mu _{A_{i}}. \end{aligned}$$
(25)
Similarly we can show that
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1}\alpha _{A_{i}} \le \root p \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \alpha _{A_{i}}\right) ^{p}\right) ^{\lambda _{i}}}\le \displaystyle \bigvee ^{n}_{i=1}\alpha _{A_{i}}. \end{aligned}$$
(26)
For the non-membership grades of \(\mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1},{\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n}\right) \), we have
$$\begin{aligned}{} & {} \displaystyle \bigwedge ^{n}_{i=1} \left( \eta _{A_{i}}\right) ^{\lambda _{i}} \le \left( \eta _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \eta _{A_{i}}\right) ^{\lambda _{i}} \nonumber \\{} & {} \displaystyle \prod ^{n}_{i=1}\displaystyle \bigwedge ^{n}_{i=1} \left( \eta _{A_{i}}\right) ^{\lambda _{i}} \le \displaystyle \prod ^{n}_{i=1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \prod ^{n}_{i=1}\displaystyle \bigvee ^{n}_{i=1} \left( \eta _{A_{i}}\right) ^{\lambda _{i}} \nonumber \\{} & {} \displaystyle \bigwedge ^{n}_{i=1} \left( \eta _{A_{i}}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}} \le \displaystyle \prod ^{n}_{i=1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \eta _{A_{i}}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}} \nonumber \\{} & {} \displaystyle \bigwedge ^{n}_{i=1} \left( \eta _{A_{i}}\right) \le \displaystyle \prod ^{n}_{i=1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \eta _{A_{i}}\right) \nonumber \\{} & {} \displaystyle \bigwedge ^{n}_{i=1} \left( \eta _{A_{i}}\right) \le \displaystyle \prod ^{n}_{i=1}\left( \eta _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \eta _{A_{i}}\right) . \end{aligned}$$
(27)
It can be similarly proved that
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1} \left( \beta _{A_{i}}\right) \le \displaystyle \prod ^{n}_{i=1}\left( \beta _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \beta _{A_{i}}\right) . \end{aligned}$$
(28)
Therefore, by Eqs. (25), (26), (27), and (28), we have \({\mathcal {A}}^{-} \preceq \mathcal{W}\mathcal{A}\left( {\mathcal {A}}_{1},{\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n}\right) \preceq {\mathcal {A}}^{+}\). \(\square \)
Proof of Theorem 7
To prove the Theorem 7, we use mathematical induction on n. For this, we proceed as follows.
Step 1. Now, for \(n = 2\), we have
$$\begin{aligned} \begin{array}{lcl} \displaystyle \bigotimes ^{2}_{i = 1}\left( \mu _{A_{i}}, \eta _{A_{i}} \right) ^{\lambda _{i}} &{}=&{} \left( \mu _{A_{1}}, \eta _{A_{1}} \right) ^{\lambda _{1}}\otimes \left( \mu _{A_{2}}, \eta _{A_{2}} \right) ^{\lambda _{2}}\\ &{}=&{} \begin{array}{ccc} \left( \left( \mu _{1}\right) ^{\lambda _{A_{1}}}, \root q \of { 1 - \left( 1 - \left( \eta _{A_{1}}\right) ^{q} \right) ^{\lambda _{1}}}\right) \otimes &{} \\ \left( \left( \mu _{2}\right) ^{\lambda _{A_{2}}}, \root q \of { 1 - \left( 1 - \left( \eta _{A_{2}}\right) ^{q} \right) ^{\lambda _{2}}} \right) &{} \end{array} \end{array}\qquad \qquad \qquad \qquad \quad \ \\ \begin{array}{lcl} &{}=&{} \left( \begin{array}{ccc} \left( \mu _{A_{1}}\right) ^{\lambda _{1}}\left( \mu _{A_{2}}\right) ^{\lambda _{2}},&{}\\ \root q \of {\begin{array}{ccc} \left( \root q \of { 1- \left( 1- \left( \eta _{A_{1}}\right) ^{q} \right) ^{\lambda _{1}}}\right) ^{q} + &{} \\ \left( \root q \of { 1- \left( 1- \left( \eta _{A_{2}}\right) ^{q} \right) ^{\lambda _{2}} }\right) ^{q} + &{} \\ - \left( \begin{array}{ccc} \left( \root q \of { 1- \left( 1- \left( \eta _{A_{1}}\right) ^{q} \right) ^{\lambda _{1}}}\right) ^{q} &{} \\ \left( \root q \of { 1- \left( 1- \left( \eta _{A_{2}}\right) ^{q} \right) ^{\lambda _{2}}}\right) ^{q} &{} \end{array}\right) &{} \end{array}}&{} \end{array} \right) \\ &{}=&{} \left( \begin{array}{ccc} \left( \mu _{A_{1}}\right) ^{\lambda _{1}}\left( \mu _{A_{2}}\right) ^{\lambda _{2}}, &{}\\ \root q \of { \begin{array}{ccc} 1- \left( 1- \left( \eta _{A_{1}}\right) ^{q} \right) ^{\lambda _{1}} + 1- \left( 1- \left( \eta _{A_{2}}\right) ^{q} \right) ^{\lambda _{2}} +&{}\\ - \left( 1- \left( 1- \left( \eta _{A_{1}}\right) ^{q} \right) ^{\lambda _{1}}\right) \left( 1- \left( 1- \left( \eta _{A_{2}}\right) ^{q} \right) ^{\lambda _{2}} \right) &{} \end{array}} &{} \end{array} \right) \\ &{}=&{} \left( \begin{array}{ccc} \displaystyle \prod ^{2}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, &{}\\ \root q \of { 1- \left( 1- \left( \eta _{A_{1}}\right) ^{q}\right) ^{\lambda _{1}}\left( 1- \left( \eta _{A_{2}}\right) ^{q} \right) ^{\lambda _{2}}} &{} \end{array} \right) \\ &{}=&{}\left( \displaystyle \prod ^{2}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, \root q \of { 1 - \displaystyle \prod ^{2}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}} \right) . \end{array} \end{aligned}$$
Similarly, \( \displaystyle \bigotimes ^{2}_{i = 1}\left( \alpha _{A_{i}}, \beta _{A_{i}} \right) ^{\lambda _{i}} = \left( \displaystyle \prod ^{2}_{i = 1}\left( \alpha _{A_{i}} \right) ^{\lambda _{i}}, \root q \of { 1 - \displaystyle \prod ^{2}_{i = 1}\left( 1- \left( \beta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\right) \). Thus, Eq. (13) holds.
Step 2. Suppose that Eq. (13) holds for \(n = k\), that is
$$\begin{aligned} {\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{k} \right) = \left( \begin{array}{ccc} \left( \begin{array}{ccc} \displaystyle \prod ^{k}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}},&{}\\ \root q \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right) ,&{} \\ \left( \begin{array}{ccc} \displaystyle \prod ^{k}_{i = 1}\left( \alpha _{A_{i}} \right) ^{\lambda _{i}},&{}\\ \root q \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right)&\end{array}\right) . \end{aligned}$$
Step 3. Now, we have to prove that Eq. 13 holds for \(n = k + 1\), based on the operational laws of the (p, q)RLDFNs, we can get
$$\begin{aligned} \displaystyle \bigotimes ^{k+1}_{i = 1}\left( \mu _{A_{i}}, \eta _{A_{i}} \right) ^{\lambda _{i}}_{i}= & {} \displaystyle \bigotimes ^{k}_{i = 1}\left( \mu _{A_{i}}, \eta _{A_{i}} \right) ^{\lambda _{i}}\otimes \left( \mu _{A_{k+1}}, \eta _{A_{k+1}} \right) ^{\lambda _{k+1}}\\= & {} {\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1}, \ldots , {\mathcal {A}}_{k}\right) \otimes \left( \mu _{A_{k+1}}, \eta _{A_{k+1}} \right) ^{\lambda _{k+1}}\\= & {} \begin{array}{ccc} \left( \displaystyle \prod ^{k}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, \root q \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}\right) &{} \\ \otimes \left( \left( \mu _{A_{k+1}}\right) ^{\lambda _{k+1}}, \root q \of { 1 - \left( 1 - \left( \eta _{A_{k+1}}\right) ^{q} \right) ^{\lambda _{k+1}}} \right) &{} \end{array}\\= & {} \left( \begin{array}{ccc} \displaystyle \prod ^{k}_{i = 1}\left( \mu _{A_{i}}\right) ^{\lambda _{1}}\left( \mu _{A_{k+1}}\right) ^{\lambda _{k+1}},&{}\\ \root q \of {\begin{array}{ccc} \left( \root q \of { 1- \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\right) ^{q} + &{} \\ \left( \root q \of { 1- \left( 1- \left( \eta _{A_{k+1}}\right) ^{q} \right) ^{\lambda _{k+1}} }\right) ^{q} + &{} \\ - \left( \begin{array}{ccc} \left( \root q \of { 1- \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}\right) ^{q} &{} \\ \left( \root q \of { 1- \left( 1- \left( \eta _{A_{k+1}}\right) ^{q} \right) ^{\lambda _{k+1}}}\right) ^{q} &{} \end{array}\right)&\end{array}}&\end{array} \right) \\= & {} \left( \begin{array}{ccc} \displaystyle \prod ^{k+1}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, &{}\\ \root q \of { 1 - \displaystyle \prod ^{k}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}\left( 1- \left( \eta _{A_{k+1}}\right) ^{q} \right) ^{\lambda _{k+1}}} &{} \end{array} \right) \\= & {} \left( \displaystyle \prod ^{k+1}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, \root q \of { 1 - \displaystyle \prod ^{k+1}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}} \right) . \end{aligned}$$
Similarly, \( \displaystyle \bigotimes \nolimits ^{k+1}_{i = 1}\left( \alpha _{A_{i}}, \beta _{A_{i}} \right) ^{\lambda _{i}}_{i} = \left( \displaystyle \prod ^{k+1}_{i = 1}\left( \alpha _{A_{i}} \right) ^{\lambda _{i}}, \root q \of { 1 - \displaystyle \prod ^{k+1}_{i = 1}\left( 1- \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}} \right) \). Therefore, Eq. (13) holds for \(n = k+1\), and hence, Eq. (13) holds for any i.
In the following, we will prove that \({\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n} \right) \) is also a (p, q)RLDFN. Then, since \(\left( \alpha _{A_{i}}\right) ^{p} + \left( \beta _{A_{i}}\right) ^{q} \le 1\), we have
$$\begin{aligned}{} & {} \left( \alpha _{A_{i}}\right) ^{p} \le 1-\left( \beta _{A_{i}}\right) ^{q}\\{} & {} \left( \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} \le \left( 1-\left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}\\{} & {} \displaystyle \prod ^{n}_{i = 1}\left( \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} \le \displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \\{} & {} \displaystyle \prod ^{n}_{i = 1}\left( \left( \alpha _{A_{i}}\right) ^{p} \right) ^{\lambda _{i}} - \displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \le 0\\{} & {} 0\le \displaystyle \prod ^{n}_{i = 1}\left( \left( \alpha _{A_{i}}\right) ^{\lambda _{i}} \right) ^{p} + 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \le 1. \end{aligned}$$
Thus, we obtain that
$$\begin{aligned} 0\le \left( 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}\right) \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}} +\displaystyle \prod ^{n}_{i = 1}\left( \left( \alpha _{A_{i}}\right) ^{\lambda _{i}} \right) ^{p}\displaystyle \prod ^{n}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}} \le 1 \end{aligned}$$
since
$$\begin{aligned} 0\le \displaystyle \prod ^{n}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}\le 1 \end{aligned}$$
and
$$\begin{aligned} 0\le \displaystyle \prod ^{n}_{i = 1}\left( \left( \alpha _{A_{i}}\right) ^{\lambda _{i}} \right) ^{p} + 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}} \le 1. \end{aligned}$$
\(\square \)
Proof of Theorem 8
Then, since \({\mathcal {A}}_{1} = {\mathcal {A}}_{2} = \ldots = {\mathcal {A}}_{n} = {\mathcal {A}}\) by Theorem 7, we have
$$\begin{aligned} \begin{array}{lcl} {\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1}, {\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n} \right) &{}=&{} \left( \begin{array}{ccc} \left( \begin{array}{ccc} \displaystyle \prod ^{n}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}},&{}\\ \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right) , &{}\\ \left( \begin{array}{ccc} \displaystyle \prod ^{n}_{i = 1}\left( \alpha _{A_{i}} \right) ^{\lambda _{i}},&{}\\ \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right) &{} \end{array}\right) \\ &{}=&{} \left( \begin{array}{ccc} \left( \begin{array}{ccc} \left( \mu _{A}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}},&{}\\ \root q \of { 1 - \left( 1- \left( \eta _{A}\right) ^{q} \right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}}&{} \end{array}\right) ,&{}\\ \left( \begin{array}{ccc} \left( \alpha _{A}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}},&{}\\ \root q \of { 1 - \left( 1- \left( \beta _{A}\right) ^{q} \right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}}&{} \end{array}\right) &{} \end{array}\right) \\ \end{array}\\ \begin{array}{lcl} &{}=&{} \left( \begin{array}{ccc} \left( \left( \mu _{A} \right) ^{1}, \root q \of { 1 - \left( 1- \left( \eta _{A}\right) ^{q} \right) ^{1}}\right) ,&{}\\ \left( \left( \alpha _{A} \right) ^{1}, \root q \of { 1 - \left( 1- \left( \beta _{A}\right) ^{q} \right) ^{1}}\right) &{} \end{array}\right) \\ &{}=&{} \left( \left( \mu _{A}, \eta _{A} \right) , \left( \alpha _{A}, \beta _{A}\right) \right) \\ &{}=&{}{\mathcal {A}}. \end{array} \end{aligned}$$
Therefore, we obtain that \({\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1},{\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n} \right) = {\mathcal {A}}\). \(\square \)
Proof of Theorem 9
Then, since \({\mathcal {A}}_{i}\preceq \mathcal {B}_{i}\) for every \(i=1,2,3,\ldots , n\), we have \(\mu _{A_{i}} \le \mu _{B_{i}}\), \( \eta _{A_{i}} \ge \eta _{B_{i}}\), \(\alpha _{A_{i}} \le \alpha _{B_{i}}\) and \( \beta _{A_{i}} \ge \beta _{B_{i}}\), that is
$$\begin{aligned} \mu _{A_{i}}{} & {} \ge \mu _{B_{i}}\nonumber \\ \left( \mu _{A_{i}}\right) ^{\lambda _{i}}{} & {} \ge \left( \mu _{B_{i}}\right) ^{\lambda _{i}} \nonumber \\ \displaystyle \prod ^{n}_{i = 1}\left( \mu _{A_{i}}\right) ^{\lambda _{i}}{} & {} \ge \displaystyle \prod ^{n}_{i = 1}\left( \mu _{B_{i}}\right) ^{\lambda _{i}}. \end{aligned}$$
(29)
It can be similarly proved that
$$\begin{aligned} \displaystyle \prod ^{n}_{i = 1}\left( \alpha _{A_{i}}\right) ^{\lambda _{i}} \ge \displaystyle \prod ^{n}_{i = 1}\left( \alpha _{B_{i}}\right) ^{\lambda _{i}}. \end{aligned}$$
(30)
From these calculations, we obtain
$$\begin{aligned} \left( \eta _{A_{i}}\right) ^{q}{} & {} \le \left( \eta _{B_{i}}\right) ^{q}\nonumber \\ 1- \left( \eta _{A_{i}}\right) ^{q}{} & {} \ge 1- \left( \eta _{B_{i}}\right) ^{q}\nonumber \\ \left( 1- \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \ge \left( 1- \left( \eta _{B_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \ge \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{B_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ 1-\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \le 1-\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{B_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ \root q \of { 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}{} & {} \le \root q \of { 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{B_{i}}\right) ^{q}\right) ^{\lambda _{i}}}. \end{aligned}$$
(31)
Similarly, we can prove
$$\begin{aligned} \root q \of { 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \beta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}} \le \root q \of { 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \beta _{B_{i}}\right) ^{q}\right) ^{\lambda _{i}}}. \end{aligned}$$
(32)
Therefore, we obtain that, by Eqs. (29), (30), (31), and (32), we have
$$\begin{aligned} \left( \begin{array}{ccc} \left( \begin{array}{ccc} \displaystyle \prod ^{n}_{i = 1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}}, &{}\\ \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right) , &{}\\ \left( \begin{array}{ccc} \displaystyle \prod ^{n}_{i = 1}\left( \alpha _{A_{i}} \right) ^{\lambda _{i}}, &{}\\ \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \beta _{A_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right)&\end{array}\right) \preceq \left( \begin{array}{ccc} \left( \begin{array}{ccc} \displaystyle \prod ^{n}_{i = 1}\left( \mu _{B_{i}} \right) ^{\lambda _{i}}, &{}\\ \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \eta _{B_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right) , &{}\\ \left( \begin{array}{ccc} \displaystyle \prod ^{n}_{i = 1}\left( \alpha _{B_{i}} \right) ^{\lambda _{i}}, &{}\\ \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1- \left( \beta _{B_{i}}\right) ^{q} \right) ^{\lambda _{i}}}&{} \end{array}\right)&\end{array}\right) , \end{aligned}$$
and hence, \({\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1}, \ldots ,{\mathcal {A}}_{n} \right) \preceq {\mathcal{W}\mathcal{G}}\left( \mathcal {B}_{1}, \ldots , \mathcal {B}_{n} \right) \). \(\square \)
Proof of Theorem 10
For the membership grades of \({\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1},\ldots ,{\mathcal {A}}_{n}\right) \), we have
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1} \left( \mu _{i}\right) ^{\lambda _{i}}{} & {} \le \left( \mu _{i} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \mu _{i}\right) ^{\lambda _{i}}\nonumber \\ \displaystyle \prod ^{n}_{i=1}\displaystyle \bigwedge ^{n}_{i=1} \left( \mu _{A_{i}}\right) ^{\lambda _{i}}{} & {} \le \displaystyle \prod ^{n}_{i=1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \prod ^{n}_{i=1}\displaystyle \bigvee ^{n}_{i=1} \left( \mu _{A_{i}}\right) ^{\lambda _{i}} \nonumber \\ \displaystyle \bigwedge ^{n}_{i=1} \left( \mu _{A_{i}}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}{} & {} \le \displaystyle \prod ^{n}_{i=1}\left( \mu _{A_{i}} \right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \mu _{A_{i}}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}\nonumber \\ \displaystyle \bigwedge ^{n}_{i=1} \left( \mu _{A_{i}}\right){} & {} \le \displaystyle \prod ^{n}_{i=1}\left( \mu _{i}\right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \mu _{A_{i}}\right) . \end{aligned}$$
(33)
Similarly we can prove
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1} \left( \alpha _{A_{i}}\right) \le \displaystyle \prod ^{n}_{i=1}\left( \alpha _{i}\right) ^{\lambda _{i}} \le \displaystyle \bigvee ^{n}_{i=1} \left( \alpha _{A_{i}}\right) . \end{aligned}$$
(34)
For the non-membership grades of \({\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1},\ldots ,{\mathcal {A}}_{n}\right) \), we have
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}{} & {} \le \left( \eta _{A_{i}}\right) ^{q}\nonumber \\ 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}{} & {} \ge 1 - \left( \eta _{A_{i}}\right) ^{q}\nonumber \\ \left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \ge \left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ \displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \ge \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \le 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ \root q \of { 1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}{} & {} \le \root q \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\nonumber \\ \root q \of { 1-\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{ \displaystyle \sum ^{n}_{i = 1}\lambda _{i} }}{} & {} \le \root q \of { 1 - \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\nonumber \\ \root q \of { 1-\left( 1-\displaystyle \bigwedge ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) }{} & {} \le \root q \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\nonumber \\ \displaystyle \bigwedge ^{n}_{i=1}\eta _{A_{i}}{} & {} \le \root q \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}} \end{aligned}$$
(35)
and
$$\begin{aligned} \displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}{} & {} \ge \left( \eta _{A_{i}}\right) ^{q}\nonumber \\ 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}{} & {} \le 1-\left( \eta _{A_{i}}\right) ^{q}\nonumber \\ \left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \le \left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ \displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \le \displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}{} & {} \ge 1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}\nonumber \\ \root q \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}{} & {} \ge \root q \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\nonumber \\ \root q \of {1-\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{\displaystyle \sum ^{n}_{i = 1}\lambda _{i}}}{} & {} \ge \root q \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\nonumber \\ \root q \of {1-\left( 1-\displaystyle \bigvee ^{n}_{i=1}\left( \eta _{A_{i}}\right) ^{q}\right) ^{1}}{} & {} \ge \root q \of {1-\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\nonumber \\ \displaystyle \bigvee ^{n}_{i=1}\eta _{A_{i}}{} & {} \ge \root q \of {1 -\displaystyle \prod ^{n}_{i = 1}\left( 1-\left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}. \end{aligned}$$
(36)
Then, by Eqs. (35) and (36), we get
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1}\eta _{A_{i}} \le \root q \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \eta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\le \displaystyle \bigvee ^{n}_{i=1}\eta _{A_{i}}. \end{aligned}$$
(37)
Similarly, we can show that
$$\begin{aligned} \displaystyle \bigwedge ^{n}_{i=1}\beta _{A_{i}} \le \root q \of { 1- \displaystyle \prod ^{n}_{i = 1}\left( 1 - \left( \beta _{A_{i}}\right) ^{q}\right) ^{\lambda _{i}}}\le \displaystyle \bigvee ^{n}_{i=1}\beta _{A_{i}}. \end{aligned}$$
(38)
Therefore, by Eqs. (33), (34), (37), and (38), we have \({\mathcal {A}}^{-} \preceq {\mathcal{W}\mathcal{G}}\left( {\mathcal {A}}_{1},{\mathcal {A}}_{2},\ldots ,{\mathcal {A}}_{n}\right) \preceq {\mathcal {A}}^{+}\). \(\square \)
