Appendix 1
The proof of Theorem 1. According to definition 5, we can obtain
$$ n{\delta}_j{p}_{\sigma (j)}=\left\{\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}}\right),\varDelta \left(t{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{n{\delta}_j}\right)\right\} $$
(26)
Thus,
$$ {\left(n{\delta}_j{p}_{\sigma (j)}\right)}^{\lambda_j}=\left\{\begin{array}{c}\varDelta \left(t{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}}\right)}^{\lambda_j}\right),\\ {}\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right)\end{array}\right\} $$
(27)
Thereafter,
$$ \overset{n}{\underset{j=1}{\otimes }}{\left(n{\delta}_j{p}_{\sigma (j)}\right)}^{\lambda_j}=\left\{\begin{array}{c}\varDelta \left(t\prod \limits_{j=1}^n{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}}\right)}^{\lambda_j}\right),\\ {}\varDelta \left(t\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right)\end{array}\right\} $$
(28)
Thereafter,
$$ {\displaystyle \begin{array}{c}\underset{\sigma \in {S}_n}{\oplus}\left(\overset{n}{\underset{j=1}{\otimes }}{\left(n{\delta}_j{p}_{\sigma (j)}\right)}^{\lambda_j}\right)\\ {}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)}\right),\\ {}\varDelta \left(t\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right)\end{array}\right\}\end{array}} $$
(29)
Furthermore,
$$ {\displaystyle \begin{array}{c}\frac{1}{n!}\left(\underset{\sigma \in {S}_n}{\oplus}\left(\overset{n}{\underset{j=1}{\otimes }}{\left(n{\delta}_j{p}_{\sigma (j)}\right)}^{\lambda_j}\right)\right)\\ {}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right),\\ {}\varDelta \left(t{\left(\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right)}^{\frac{1}{n!}}\right)\end{array}\right\}\end{array}} $$
(30)
Therefore,
$$ {\displaystyle \begin{array}{c}2{\mathrm{TLPFPMM}}^{\lambda}\left({p}_1,{p}_2,\dots, {p}_n\right)={\left(\frac{1}{n!}\left(\underset{\sigma \in {S}_n}{\oplus}\left(\overset{n}{\underset{j=1}{\otimes }}{\left(n{\delta}_j{p}_{\sigma (j)}\right)}^{\lambda_j}\right)\right)\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}=\left\{\begin{array}{c}\varDelta \left(t{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\right),\\ {}\varDelta \left(t\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\right)\end{array}\right\}\end{array}} $$
(31)
Appendix 2
The proof of Eq. (12) is a 2TLPFN. Let
$$ {\displaystyle \begin{array}{c}\frac{\varDelta^{-1}\left({s}_{\phi_j},{\varphi}_j\right)}{t}={\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}\frac{\varDelta^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)}{t}=\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\end{array}} $$
Proof. ① Since \( 0\le \frac{\Delta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\le 1, \) we get
$$ 0\le 1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\le 1\kern0.5em and\kern0.5em 0\le \prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\le 1 $$
(32)
Then,
$$ \kern1.65em 0\le {\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}\le 1 $$
(33)
$$ 0\le {\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\le 1 $$
(34)
That means \( 0\le {\Delta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\le t \), so ① is maintained, similarly, we can get \( 0\le {\Delta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\le t \).
-
①
Since\( {\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\le 1 \), we get the following inequality:
$$ {\displaystyle \begin{array}{c}\begin{array}{c}{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\\ {}={\left(1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}+1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\end{array}\\ {}\begin{array}{c}\le {\left(1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}+1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}=1\end{array}\mathrm{i},\mathrm{e}.,\\ {}0\le {\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\le 1\end{array}} $$
That means \( 0\le {\left({\Delta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\right)}^2+{\left({\Delta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\right)}^2\le {t}^2 \),so ② is maintained. Then, we discuss the properties of 2TLPFPMM operator.
Appendix 3
The proof of property 2.
Let \( {p}^{+}=\left({\max}_i\left({S}_{\phi_i},{\varphi}_i\right),{\min}_i\left({S}_{\theta_i},{\vartheta}_i\right)\right)=\left\{{\left({s}_{\phi_i},{\varphi}_i\right)}^{+},{\left({s}_{\theta_i},{\vartheta}_i\right)}^{-}\right\}\left(i=1,2,\dots, n\right) \) and\( {p}^{-}=\left({\min}_i\left({S}_{\phi_i},{\varphi}_i\right),{\max}_i\left({S}_{\theta_i},{\vartheta}_i\right)\right)=\left\{{\left({s}_{\phi_i},{\varphi}_i\right)}^{-},{\left({s}_{\theta_i},{\vartheta}_i\right)}^{+}\right\} \), we can obtain
$$ {\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_i},{\varphi}_i\right)}^{+}}{t}\right)}^2\right)}^{n{\delta}_j}\le {\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_i},{\varphi}_i\right)}^{-}}{t}\right)}^2\right)}^{n{\delta}_j} $$
(35)
Then,
$$ \prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}^{+}}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\ge \prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}^{-}}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j} $$
(36)
Thereafter,
$$ {\displaystyle \begin{array}{c}\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}^{+}}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\\ {}\le \prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}^{-}}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\end{array}} $$
(37)
Furthermore,
$$ {\displaystyle \begin{array}{c}t{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}^{+}}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}\ge t{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}{\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}^{-}}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\end{array}} $$
(38)
That means\( {\Delta}^{-1}{\left({s}_{\phi_i},{\varphi}_i\right)}^{+}\ge {\Delta}^{-1}{\left({s}_{\phi_i},{\varphi}_i\right)}^{-} \). Similarly, we can obtain \( {\Delta}^{-1}\left({s}_{\theta_i},{\vartheta}_i\right)\le {\Delta}^{-1}{\left({s}_{\theta_i},{\vartheta}_i\right)}^{-} \), this indicates p+ ≥ p−. So property 2 is right. It is clear that the 2TLPFPMM operator lacks the property of monotonicity.
Appendix 4
The proof of Theorem 2. According to definition 5, we can obtain
$$ {\displaystyle \begin{array}{c}\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}{p}_{\sigma (j)}\\ {}=\left\{\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}}\right),\varDelta \left(t{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)\right\}\end{array}} $$
(39)
Thus,
$$ {\left(\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}{p}_{\sigma (j)}\right)}^{\lambda_j}=\left\{\begin{array}{c}\varDelta \left(t{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}}\right)}^{\lambda_j}\right),\\ {}\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right)\end{array}\right\} $$
(40)
Thereafter,
$$ {\displaystyle \begin{array}{c}\overset{n}{\underset{j=1}{\otimes }}{\left(\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}{p}_{\sigma (j)}\right)}^{\lambda_j}\\ {}=\left\{\begin{array}{c}\varDelta \left(t\prod \limits_{j=1}^n{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}}\right)}^{\lambda_j}\right),\\ {}\varDelta \left(t\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right)\end{array}\right\}\end{array}} $$
(41)
Thereafter,
$$ {\displaystyle \begin{array}{c}\underset{\sigma \in {S}_n}{\oplus}\left(\overset{n}{\underset{j=1}{\otimes }}{\left(\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}{p}_{\sigma (j)}\right)}^{\lambda_j}\right)\\ {}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}\right),\\ {}\varDelta \left(t\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right)\end{array}\right\}\end{array}} $$
(42)
Furthermore,
$$ {\displaystyle \begin{array}{c}\frac{1}{n!}\left(\underset{\sigma \in {S}_n}{\oplus}\left(\overset{n}{\underset{j=1}{\otimes }}{\left(\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}{p}_{\sigma (j)}\right)}^{\lambda_j}\right)\right)\\ {}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right),\\ {}\varDelta \left(t{\left(\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right)}^{\frac{1}{n!}}\right)\end{array}\right\}\end{array}} $$
(43)
Therefore,
$$ {\displaystyle \begin{array}{c}2{\mathrm{TLPFWPMM}}_w^{\lambda}\left({p}_1,{p}_2,\dots, {p}_n\right)={\left(\frac{1}{n!}\left(\underset{\sigma \in {S}_n}{\oplus}\left(\overset{n}{\underset{j=1}{\otimes }}{\left(\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}{p}_{\sigma (j)}\right)}^{\lambda_j}\right)\right)\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}=\left\{\begin{array}{c}\varDelta \left(t{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\right),\\ {}\varDelta \left(t\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\right)\end{array}\right\}\end{array}} $$
(44)
Appendix 5
The proof of Eq. (16) is a 2TLPFN. Let
$$ {\displaystyle \begin{array}{c}\frac{\varDelta^{-1}\left({s}_{\phi_j},{\varphi}_j\right)}{t}={\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}\frac{\varDelta^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)}{t}=\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\end{array}} $$
Proof. ① Since \( 0\le \frac{\Delta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\le 1, \) we get
$$ 0\le 1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\le 1\kern0.5em and\kern0.5em 0\le \prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\le 1 $$
(45)
Then,
$$ \kern1.65em 0\le {\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}\le 1 $$
(46)
$$ 0\le {\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\le 1 $$
(47)
That means \( 0\le {\Delta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\le t \), so ① is maintained, similarly, we can get \( 0\le {\Delta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\le t \).
② Since \( {\left(\frac{\Delta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\Delta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\le 1 \), we get the following inequality:
$$ {\displaystyle \begin{array}{c}\begin{array}{c}{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\\ {}={\left(1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}+1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\end{array}\\ {}\begin{array}{c}\le {\left(1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}+1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}=1\end{array}\mathrm{i}.\mathrm{e}.,\\ {}0\le {\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\le 1.\end{array}} $$
That means\( 0\le {\left({\varDelta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\right)}^2+{\left({\varDelta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\right)}^2\le {t}^2 \), so ② is maintained.
Appendix 6
The proof of Theorem 3. According to definition 5, we can obtain
$$ {\left({p}_{\sigma (j)}\right)}^{n{\delta}_j}=\left\{\varDelta \left(t{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{n{\delta}_j}\right),\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}}\right)\right\} $$
(48)
Then,
$$ {\lambda}_j{\left({p}_{\sigma (j)}\right)}^{n{\delta}_j}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right),\\ {}\varDelta \left(t{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}}\right)}^{\lambda_j}\right)\end{array}\right\} $$
(49)
Thus,
$$ \overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{n{\delta}_j}\right)=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right),\\ {}\varDelta \left(t\prod \limits_{j=1}^n{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}}\right)}^{\lambda_j}\right)\end{array}\right\} $$
(50)
Therefore,
$$ {\displaystyle \begin{array}{c}\underset{\sigma \in {S}_n}{\otimes}\left(\overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{n{\delta}_j}\right)\right)\\ {}=\left\{\begin{array}{c}\varDelta \left(t\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right),\\ {}\varDelta \left(t\sqrt{1-\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)}\right)\end{array}\right\}\end{array}} $$
(51)
Furthermore,
$$ {\displaystyle \begin{array}{c}{\left(\underset{\sigma \in {S}_n}{\otimes}\left(\overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{n{\delta}_j}\right)\right)\right)}^{\frac{1}{n!}}\\ {}=\left\{\begin{array}{c}\varDelta \left(t{\left(\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}}\right)}^{\frac{1}{n!}}\right),\\ {}\varDelta \left(t\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)\end{array}\right\}\end{array}} $$
(52)
Therefore,
$$ {\displaystyle \begin{array}{c}2{\mathrm{TLPFPDMM}}^{\lambda}\left({p}_1,{p}_2,\cdots, {p}_n\right)=\frac{1}{\sum_{j=1}^n{\lambda}_j}{\left(\underset{\sigma \in {S}_n}{\otimes}\left(\overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{n{\delta}_j}\right)\right)\right)}^{\frac{1}{n!}}\\ {}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\right),\\ {}\varDelta \left(t{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\right)\end{array}\right\}\end{array}} $$
(53)
Appendix 7
The proof of Eq. (21) is a 2TLPFN. Let
$$ {\displaystyle \begin{array}{c}\frac{\varDelta^{-1}\left({s}_{\phi_j},{\varphi}_j\right)}{t}=\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\\ {}\frac{\varDelta^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)}{t}={\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\end{array}} $$
① Since \( 0\le \frac{\Delta^{-1}\left({s}_{\phi_j},{\varphi}_j\right)}{t}\le 1, \) we get
$$ 0\le {\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\le 1\kern0.5em and\kern0.5em 0\le 1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\le 1 $$
(54)
Then
$$ \kern2.31em 0\le {\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\le 1 $$
(55)
$$ 0\le \sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\le 1 $$
(56)
That means \( 0\le {\Delta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\le t \), so ① is maintained, similarly, we can get \( 0\le {\Delta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\le t \).
-
②
Since\( {\left(\frac{\Delta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\Delta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\le 1 \), we can get the following inequality:
$$ {\displaystyle \begin{array}{c}{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\\ {}=\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{2n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\\ {}+{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}\le \sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\\ {}+{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{n{\delta}_j}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}=1\end{array}} $$
That means \( 0\le {\left({\Delta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\right)}^2+{\left({\Delta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\right)}^2\le {t}^2 \), so ② is maintained.
Appendix 8
The proof of Theorem 4. According to definition 5, we can obtain
$$ {\displaystyle \begin{array}{c}{\left({p}_{\sigma (j)}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\\ {}=\left\{\varDelta \left(t{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right),\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}}\right)\right\}\end{array}} $$
(57)
Then,
$$ {\lambda}_j{\left({p}_{\sigma (j)}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right),\\ {}\varDelta \left(t{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}}\right)}^{\lambda_j}\right)\end{array}\right\} $$
(58)
Thus,
$$ \overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right),\\ {}\varDelta \left(t\prod \limits_{j=1}^n{\left(\sqrt{1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}}\right)}^{\lambda_j}\right)\end{array}\right\} $$
(59)
Therefore,
$$ {\displaystyle \begin{array}{c}\underset{\sigma \in {S}_n}{\otimes}\left(\overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)\right)\\ {}=\left\{\begin{array}{c}\varDelta \left(t\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right),\\ {}\varDelta \left(t\sqrt{1-\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}\right)\end{array}\right\}\end{array}} $$
(60)
Furthermore,
$$ {\displaystyle \begin{array}{c}{\left(\underset{\sigma \in {S}_n}{\otimes}\left(\overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)\right)\right)}^{\frac{1}{n!}}\\ {}=\left\{\begin{array}{c}\varDelta \left(t{\left(\prod \limits_{\sigma \in {S}_n}\sqrt{1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}}\right)}^{\frac{1}{n!}}\right),\\ {}\varDelta \left(t\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)\end{array}\right\}\end{array}} $$
(61)
Therefore,
$$ {\displaystyle \begin{array}{c}2{\mathrm{TLPFPWDMM}}^{\lambda}\left({p}_1,{p}_2,\cdots, {p}_n\right)=\frac{1}{\sum_{j=1}^n{\lambda}_j}{\left(\underset{\sigma \in {S}_n}{\otimes}\left(\overset{n}{\underset{j=1}{\oplus }}\left({\lambda}_j{\left({p}_{\sigma (j)}\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)\right)\right)}^{\frac{1}{n!}}\\ {}=\left\{\begin{array}{c}\varDelta \left(t\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\right),\\ {}\varDelta \left(t{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\right)\end{array}\right\}\end{array}} $$
(62)
Appendix 9
The proof of Eq. (25) is a 2TLPFN. Let
$$ {\displaystyle \begin{array}{c}\frac{\varDelta^{-1}\left({s}_{\phi_j},{\varphi}_j\right)}{t}=\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\\ {}\frac{\varDelta^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)}{t}={\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\end{array}} $$
① Since\( 0\le \frac{\Delta^{-1}\left({s}_{\phi_j},{\varphi}_j\right)}{t}\le 1, \)we get.
$$ 0\le {\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\le 1\kern0.5em and\kern0.5em 0\le 1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\le 1 $$
(63)
Then
$$ \kern2.31em 0\le {\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\le 1 $$
(64)
$$ 0\le \sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\le 1 $$
(65)
That means \( 0\le {\Delta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\le t \), so ① is maintained, similarly, we can get \( 0\le {\Delta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\le t \).
② Since\( {\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\le 1 \), we can get the following inequality:
$$ {\displaystyle \begin{array}{c}{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^2+{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\\ {}=\sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\phi_{\sigma (j)}},{\varphi}_{\sigma (j)}\right)}{t}\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\\ {}+{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}\\ {}\le \sqrt{1-{\left(1-\prod \limits_{\sigma \in {S}_n}{\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)}^{\frac{1}{n!}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}}\\ {}+{\left(\sqrt{1-{\left(\prod \limits_{\sigma \in {S}_n}\left(1-\prod \limits_{j=1}^n{\left(1-{\left(1-{\left(\frac{\varDelta^{-1}\left({s}_{\theta_{\sigma (j)}},{\vartheta}_{\sigma (j)}\right)}{t}\right)}^2\right)}^{\frac{2n{\delta}_j{w}_{\sigma (j)}}{\sum_{t=1}^n{\delta}_t{w}_t}}\right)}^{\lambda_j}\right)\right)}^{\frac{1}{n!}}}\right)}^{\frac{1}{\sum_{j=1}^n{\lambda}_j}}=1\end{array}} $$
That means\( 0\le {\left({\varDelta}^{-1}\left({s}_{\phi_j},{\varphi}_j\right)\right)}^2+{\left({\varDelta}^{-1}\left({s}_{\theta_j},{\vartheta}_j\right)\right)}^2\le {t}^2 \), so ② is maintained.
