An interval-valued Pythagorean prioritized operator-based game theoretical framework with its applications in multicriteria group decision making

Abstract

Multicriteria decision-making process explicitly evaluates multiple conflicting criteria in decision making. The conventional decision-making approaches assumed that each agent is independent, but the reality is that each agent aims to maximize personal benefit which causes a negative influence on other agents’ behaviors in a real-world competitive environment. In our study, we proposed an interval-valued Pythagorean prioritized operator-based game theoretical framework to mitigate the cross-influence problem. The proposed framework considers both prioritized levels among various criteria and decision makers within five stages. Notably, the interval-valued Pythagorean fuzzy sets are supposed to express the uncertainty of experts, and the game theories are applied to optimize the combination of strategies in interactive situations. Additionally, we also provided illustrative examples to address the application of our proposed framework. In summary, we provided a human-inspired framework to represent the behavior of group decision making in the interactive environment, which is potential to simulate the process of realistic humans thinking.

Appendix

Interval-valued Pythagorean fuzzy prioritized average operators are defined as follows:

Definition 4

Let $a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]$, $[v_{\tilde{P}}^L(a_i), v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, then their aggregated, where $[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]\subset [0,1], [v_{\tilde{P}}^L(a_i)$, $v_{\tilde{P}}^U(a_i)] \subset [0,1]$ and let IVPFPWA $V^n\rightarrow V$. if:

$$\begin{aligned}&{{\mathrm{IVPFPWA}}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^nT_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a_n \end{aligned}$$

The interval-valued Pythagorean fuzzy prioritized weighted average operator is abbreviated as IVPFPWA with $T_i=\prod _{j=1}^{i-1}S(a_j)$$(i= 2, \ldots , n)$, $T_1=1$ and $S(a_j)$ is the score of IVPFS a.

We could obtain the Theorem 1 based on the operations of IVPFSs described in Preliminary.

Theorem 1

Let$a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]$, $\quad [v_{\tilde{P}}^L(a_i)$, $\quad v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, then their aggregated, IVPFPWA is

$$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\left( \left\langle \left[ \sqrt{1-\prod \limits _{i=1}^n\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}}, \right. \right. \right. \\&\qquad \left. \sqrt{1-\prod \limits _{i=1}^n\left( 1-\mu _p^U(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}} \right] ,\\&\qquad \left. \left. \left[ \prod \limits _{i=1}^n\left( v_p^L(a_i)\right) ^\frac{T_i}{\sum _{i=1}^nT_i}, \prod \limits _{i=1}^n\left( v_p^U(a_i)\right) ^\frac{T_i}{\sum _{i=1}^nT_i} \right] \right\rangle \right) \end{aligned}$$

where$T_i=\prod _{j=1}^{i-1}S(a_j)$$(i= 2, \ldots , n)$, $T_1=1$ andS(a) is the score of IVPFSa.

Proof

In the following, we prove the first result follows quickly from Definition 2 and Theorem 1:

$$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^nT_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a_n\\&\quad =\left( \left\langle \left[ \sqrt{1-\prod \limits _{i=1}^n\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}},\right. \right. \right. \\&\qquad \left. \sqrt{1-\prod \limits _{i=1}^n\left( 1-\mu _p^U(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}} \right] , \\&\qquad \left. \left. \left[ \prod \limits _{i=1}^n\left( v_p^L(a_i)\right) ^\frac{T_i}{\sum _{i=1}^nT_i}, \prod \limits _{i=1}^n\left( v_p^U(a_i)\right) ^\frac{T_i}{\sum _{i=1}^nT_i} \right] \right\rangle \right) \end{aligned}$$

by using mathematical induction on n:

• (1) For n = 2, then

  $$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2\right) =\frac{T_1}{\sum _{i=1}^2T_i}a_1+\frac{T_2}{\sum _{i=1}^2T_i}a_2 \\&\quad \frac{T_1}{\sum _{i=1}^2T_i}a_1=\left( \left\langle \left[ \sqrt{1-\left( 1-\mu _p^L(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}},\right. \right. \sqrt{1-\left( 1-\mu _p^U(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}} \right] , \\&\qquad \left. \left. \left[ \left( v_p^L(a_1)\right) ^\frac{T_1}{\sum _{i=1}^2T_i}, \left( v_p^U(a_1)\right) ^\frac{T_1}{\sum _{i=1}^2T_i} \right] \right\rangle \right) \frac{T_2}{\sum _{i=1}^2T_i}a_2=\left( \left\langle \left[ \sqrt{1-\left( 1-\mu _p^L(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}},\right. \right. \right. \\&\qquad \left. \sqrt{1-\left( 1-\mu _p^U(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}} \right] , \left. \left. \left[ \left( v_p^L(a_1)\right) ^\frac{T_2}{\sum _{i=1}^2T_i}, \left( v_p^U(a_2)\right) ^\frac{T_2}{\sum _{i=1}^2T_i} \right] \right\rangle \right) \\&{\mathrm{IVPFPWA}}\left( a_1, a_2\right) = \left( \left\langle \left[ \sqrt{1-\left( 1-\mu _p^L(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}+1-\left( 1-\mu _p^L(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}, \left( 1-\left( 1-\mu _p^L(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\right) \left( 1-\left( 1-\mu _p^L(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}\right) },\right. \right. \right. \\&\quad \quad \left. \sqrt{1-\left( 1-\mu _p^U(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}+1-\left( 1-\mu _p^U(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}, \left( 1-\left( 1-\mu _p^U(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\right) \left( 1-\left( 1-\mu _p^U(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}\right) }\right] , \left[ \left( v_p^L(a_1)\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\left( v_p^L(a_2)\right) ^\frac{T_2}{\sum _{i=1}^2T_i},\right. \\&\quad \quad \left. \left. \left. \left( v_p^U(a_1)\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\left( v_p^U(a_2)\right) ^\frac{T_2}{\sum _{i=1}^2T_i}\right] \right\rangle \right) \\&\quad =\left( \left\langle \left[ \sqrt{1-\left( 1-\mu _p^L(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\left( 1-\mu _p^L(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}},\right. \right. \right. \\&\qquad \left. \sqrt{1-\left( 1-\mu _p^U(a_1)^2\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\left( 1-\mu _p^U(a_2)^2\right) ^\frac{T_2}{\sum _{i=1}^2T_i}}\right] , \\&\quad \quad \left[ \left( v_p^L(a_1)\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\left( v_p^L(a_2)\right) ^\frac{T_2}{\sum _{i=1}^2T_i},\right. \\&\quad \quad \left. \left. \left. \left( v_p^U(a_1)\right) ^\frac{T_1}{\sum _{i=1}^2T_i}\left( v_p^U(a_2)\right) ^\frac{T_2}{\sum _{i=1}^2T_i}\right] \right\rangle \right) \\ \end{aligned}$$

• (2) We suppose it holds for $n=k$, that is

  $$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_k\right) \\&\quad =\frac{T_1}{\sum _{i=1}^kT_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^kT_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^kT_i}a_n\\&\quad =\left( \left\langle \left[ \sqrt{1-\prod \limits _{i=1}^k\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^kT_i}},\right. \right. \right. \\&\qquad \left. \sqrt{1-\prod \limits _{i=1}^k\left( 1-\mu _p^U(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^kT_i}} \right] , \\&\qquad \left. \left. \left[ \prod \limits _{i=1}^k\left( v_p^L(a_i)\right) ^\frac{T_i}{\sum _{i=1}^kT_i}, \prod \limits _{i=1}^k\left( v_p^U(a_i)\right) ^\frac{T_i}{\sum _{i=1}^kT_i} \right] \right\rangle \right) \end{aligned}$$

• (3) We suppose it holds for $n=k+1$, we have

  $$\begin{aligned}&\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}a_{k+1}=\left( \left\langle \left[ \sqrt{1-\left( 1-\mu _p^L\left( a_{k+1}\right) ^2\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}},\right. \right. \right. \\&\quad \quad \left. \sqrt{1-\left( 1-\mu _p^U\left( a_{k+1}\right) ^2\right) ^\frac{T_2}{\sum _{i=1}^{k+1}T_{k+1}}}\right] , \\&\quad \quad \left. \left. \left[ \left( v_p^L(a_{k+1})\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_{k+1}}, \left( v_p^U(a_{k+1})\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_{k+1}} \right] \right\rangle \right) \\&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_{k+1}\right) \\&\quad = \left( \left\langle \left[ \sqrt{1-\prod \limits _{i=1}^k\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^kT_i}+1-\left( 1-\mu _p^U(a_{k+1})^2\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}-\left( 1-\prod \limits _{i=1}^k\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^kT_i}\right) \left( 1-\left( 1-\mu _p^L(a_{k+1})^2\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}\right) },\right. \right. \right. \\&\quad\quad \left. \sqrt{1-\prod \limits _{i=1}^k(1-\mu _p^U(a_i)^2)^\frac{T_i}{\sum _{i=1}^kT_i}+1-(1-\mu _p^U(a_{k+1})^2)^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}-(1-\prod \limits _{i=1}^k(1-\mu _p^U(a_i)^2)^\frac{T_i}{\sum _{i=1}^kT_i})(1-(1-\mu _p^U(a_{k+1})^2)^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i})}\right] \\&\quad \quad \left[ \left( v_P^L(a_{k+1})\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}\prod \limits _{i=1}^k\left( \mu _p^L(a_i)\right) ^\frac{T_i}{\sum _{i=1}^kT_i},\right. \\&\quad \quad \left. \left. \left. \left( v_P^U(a_{k+1})\right) ^\frac{T_{k+1}}{\sum _{i=1}^{k+1}T_i}\prod \limits _{i=1}^k\left( \mu _p^U(a_i)\right) ^\frac{T_i}{\sum _{i=1}^kT_i}\right] \right\rangle \right) \\&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_{k+1}\right) \\&\quad =\frac{T_1}{\sum _{i=1}^{k+1}T_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^{k+1}T_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^{k+1}T_i}a_n\\&\quad =\left( \left\langle \left[ \sqrt{1-\prod \limits _{i=1}^{k+1}\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^{k+1}T_i}},\right. \right. \right. \\&\quad \quad \left. \sqrt{1-\prod \limits _{i=1}^{k+1}\left( 1-\mu _p^U(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^{k+1}T_i}} \right] , \\&\quad \quad \left. \left. \left[ \prod \limits _{i=1}^{k+1}\left( v_p^L(a_i)\right) ^\frac{T_i}{\sum _{i=1}^{k+1}T_i}, \prod \limits _{i=1}^{k+1}\left( v_p^U(a_i)\right) ^\frac{T_i}{\sum _{i=1}^{k+1}T_i} \right] \right\rangle \right) \end{aligned}$$

$\square$

Theorem 2

(Idempotency) Let$a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]$, $\quad [v_{\tilde{P}}^L(a_i)$, $\quad v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, if all$a_i$$(i=1,2,3, \ldots , n))$ are equal ($a_i=a$), then:

$$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) =a\\&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^nT_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a_n \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a \oplus \frac{T_2}{\sum _{i=1}^nT_i}a \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a\\&\quad =\frac{\sum _{i=1}^nT_i}{\sum _{i=1}^nT_i}a=a \end{aligned}$$

Corollary 1

If $a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]$, $[v_{\tilde{P}}^L(a_i)$, $v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, if all $a_i$$(i=1,2,3, \ldots , n))=a^*=([1,1], [0,0])$, then:

$$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad ={\mathrm{IVPFPWA}}\left( a^*, a^*, a^*, \ldots , a^*\right) =([1,1], [0,0]) \end{aligned}$$

After the aggregating, it is also the largest IVPFS.

Proof

In the similar way showed above:

$$\begin{aligned} \begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^nT_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a_n \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a^* \oplus \frac{T_2}{\sum _{i=1}^nT_i}a^* \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a^*\\&\quad =\frac{\sum _{i=1}^nT_i}{\sum _{i=1}^nT_i}a^*=a^* \end{aligned} \end{aligned}$$

Corollary 2

If $a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)], \; [v_{\tilde{P}}^L(a_i)$, $v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, if all$a_i$$(i=1,2,3, \ldots , n))=a_*=([0,0], [1,1])$, then:

$$\begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad ={\mathrm{IVPFPWA}}\left( a_*=, a_*=, a_*=, \ldots , a_*=\right) \\&\quad =([0,0], [1,1]) \end{aligned}$$

After the aggregating, it is also the smallest IVPFS.

Proof

Since $a_1=([0,0], [1,1])$, then we have the score function:

$$\begin{aligned} S(a_1)=0 \end{aligned}$$

Since:

$$\begin{aligned} T_i=\prod \limits _{k=1}^{i-1}S(a_k)\,(i=2, 3, \ldots , n)\hbox { and }T_1=1 \end{aligned}$$

We have

$$\begin{aligned} T_i&= \prod \limits _{k=1}^{i-1}S(a_k)=S(a_1)\times S(a_2)\times S(a_3)\times \cdots S(a_{i-1})\\&\quad =0\times S(a_2)\times S(a_3)\times \cdots S(a_{i-1})\\&\quad =0\,(i=2, 3, \ldots , n) \end{aligned}$$

Thus, $\sum _{i=1}^nT_i=1$

$$\begin{aligned} \begin{aligned}&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\frac{T_1}{\sum _{i=1}^nT_i}a_1 \oplus \frac{T_2}{\sum _{i=1}^nT_i}a_2 \oplus , \ldots , \oplus \frac{T_n}{\sum _{i=1}^nT_i}a_n \\&\quad =\frac{1}{1}a_1 \oplus \frac{0}{1}a_2 \oplus , \ldots , \oplus \frac{0}{1}a_n=\frac{1}{1}a_1=([0,0], [1,1]) \end{aligned} \end{aligned}$$

This reveals that when the criteria owning the highest priority has the smallest IVPFS, then any other criteria could not compensate it even they are all satisfied. $\square$

Theorem 3

(Boundary) If$a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)], \; [v_{\tilde{P}}^L(a_i)$, $\quad v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, and $S(a_i)$ is the scores of IVPFS$a_i$:

$$\begin{aligned} a_*&= \left( \left[ \min _i{\mu _{\tilde{P}}^L(a_i)}, \min _i{\mu _{\tilde{P}}^U(a_i)}\right] ,\left[ \max _i{v_{\tilde{P}}^L(a_i)}, \max _i{v_{\tilde{P}}^U(a_i)}\right] \right) \\ a^*&= \left( \left[ \max _i{\mu _{\tilde{P}}^L(a_i)}, \max _i{\mu _{\tilde{P}}^U(a_i)}\right] ,\left[ \min _i{v_{\tilde{P}}^L(a_i)}, \min _i{v_{\tilde{P}}^U(a_i)}\right] \right) \end{aligned}$$

Then, $a_*\le {{IVPFPWA}}(a_1, a_2, a_3, \ldots , a_n)\le a^*$

Proof

Since $\min _i{\mu _{\tilde{P}}^L(a_i)}\le \mu _{\tilde{P}}^L(a_i)\le \max _i{\mu _{\tilde{P}}^L(a_i)}$, $\min _i{\mu _{\tilde{P}}^U(a_i)}\le \mu _{\tilde{P}}^U(a_i)\le \max _i{\mu _{\tilde{P}}^U(a_i)}$, $\min _i{v_{\tilde{P}}^L(a_i)}\le v_{\tilde{P}}^L(a_i)\le \max _i{v_{\tilde{P}}^L(a_i)}$, $\min _i{v_{\tilde{P}}^U(a_i)}\le v_{\tilde{P}}^U(a_i)\le \max _i{v_{\tilde{P}}^U(a_i)}$

$$\begin{aligned} \prod \limits _{i=1}^n\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}\ge & {} \prod \limits _{i=1}^n\left( 1-\max _i\left( \mu _p^L(a_i)^2\right) \right) ^\frac{T_i}{\sum _{i=1}^nT_i}\\&= 1-\max _i\left( \mu _p^L(a_i)^2\right) \end{aligned}$$

And then

$$\begin{aligned} 1-\prod \limits _{i=1}^n\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i} \le \max _i\left( \mu _p^L(a_i)^2\right) \end{aligned}$$

In the similar way, we have:

$$\begin{aligned} 1-\prod \limits _{i=1}^n\left( 1-\mu _p^L(a_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}\ge & {} \min _i\left( \mu _p^L(a_i)^2\right) , \quad \\ 1-\prod \limits _{i=1}^n\left( 1-\mu _p^L(b_i)^2\right) ^\frac{T_i}{\sum _{i=1}^nT_i}\ge & {} \min _i\left( \mu _p^L(b_i)^2\right) \end{aligned}$$

and

$$\begin{aligned}&\prod \limits _{i=1}^k\left( \min _i\left( v_p^L(a_i)\right) \right) ^\frac{T_i}{\sum _{i=1}^kT_i}\\&\quad \le \prod \limits _{i=1}^k\left( v_p^L(a_i)\right) ^\frac{T_i}{\sum _{i=1}^kT_i}\le \prod \limits _{i=1}^k\left( \max _i\left( v_p^L(a_i)\right) \right) ^\frac{T_i}{\sum _{i=1}^kT_i}\\&\prod \limits _{i=1}^k\left( \min _i\left( v_p^U(a_i)\right) \right) ^\frac{T_i}{\sum _{i=1}^kT_i}\\&\quad \le \prod \limits _{i=1}^k\left( v_p^U(a_i)\right) ^\frac{T_i}{\sum _{i=1}^kT_i}\le \prod \limits _{i=1}^k\left( \max _i\left( v_p^U(a_i)\right) \right) ^\frac{T_i}{\sum _{i=1}^kT_i}\\&{\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =a={\left\{ \left\langle s_{\theta (a_i)},\left[ \mu _{\tilde{P}}^L(a_i), \quad \mu _{\tilde{P}}^U(a_i)\right] , \quad \left[ v_{\tilde{P}}^L(a_i), \quad v_{\tilde{P}}^U(a_i)\right] \right\rangle \right\} } \end{aligned}$$

then

$$\begin{aligned} S(a)&= \frac{\big [2+\left( \mu _P^L\right) ^2+\left( \mu _P^U\right) ^2-\left( v_P^L\right) ^2-\left( v_P^U\right) ^2\big ]}{4}\\\le & {} \frac{\max _i\left( \mu _p^L(a_i)^2\right) +\max _i\left( \mu _p^U(a_i)^2\right) -\min _i\left( v_p^L(a_i)^2\right) - \min _i\left( v_p^U(a_i)^2\right) }{4}\\&= S\left( a^*\right) \\ S(a)&= \frac{\big [2+\left( \mu _P^L\right) ^2+\left( \mu _P^U\right) ^2-\left( v_P^L\right) ^2-\left( v_P^U\right) ^2\big ]}{4}\\\ge & {} \frac{\min _i\left( \mu _p^L(a_i)^2\right) +\min _i\left( \mu _p^U(a_i)^2\right) -\max _i\left( v_p^L(a_i)^2\right) - \max _i\left( v_p^U(a_i)^2\right) }{4}\\&= S\left( a_*\right) \end{aligned}$$

If $S(a_*)< S(a)< S(a^*)$ we could conclude that

$$\begin{aligned} S\left( a_*\right)< {\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) < S\left( a^*\right) \end{aligned}$$

Otherwise, we have $S(a)=S(a^*)$:

$$\begin{aligned} S(a)&= \frac{\big [2+\left( \mu _P^L\right) ^2+\left( \mu _P^U\right) ^2-\left( v_P^L\right) ^2-\left( v_P^U\right) ^2\big ]}{4}\\&= \frac{\max _i\left( \mu _p^L(a_i)^2\right) +\max _i\left( \mu _p^U(a_i)^2\right) -\min _i\left( v_p^L(a_i)^2\right) - \min _i\left( v_p^U(a_i)^2\right) }{4}\\&= S(a^*) \end{aligned}$$

Then, we have

$$\begin{aligned} \mu _{\tilde{P}}^L(a_i)&= \max _i{\mu _{\tilde{P}}^L(a_i)}, \mu _{\tilde{P}}^U(a_i)= \max _i{\mu _{\tilde{P}}^U(a_i)}, v_{\tilde{P}}^L(a_i)\\&= \min _i{v_{\tilde{P}}^L(a_i)}, v_{\tilde{P}}^U(a_i)=\min _i{v_{\tilde{P}}^U(a_i)} \end{aligned}$$

Thus,

$$\begin{aligned} h(a)&= \frac{\big [\left( \mu _P^L\right) ^2+\left( \mu _P^U\right) ^2+\left( v_P^L\right) ^2+\left( v_P^U\right) ^2\big ]}{2}\\&= \frac{\max _i\left( \mu _p^L(a_i)^2\right) +\max _i\left( \mu _p^U(a_i)^2\right) +\min _i\left( v_p^L(a_i)^2\right) + \min _i\left( v_p^U(a_i)^2\right) }{2}\\&= h\left( a^*\right) \end{aligned}$$

Then, we have

$$\begin{aligned} {\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) =a^* \end{aligned}$$

On the other hand, if $S(a)=S(a_*)$:

$$\begin{aligned} S(a)&= \frac{\big [2+\left( \mu _P^L\right) ^2+\left( \mu _P^U\right) ^2-\left( v_P^L\right) ^2-\left( v_P^U\right) ^2\big ]}{4}\\&= \frac{\min _i\left( \mu _p^L(a_i)^2\right) +\min _i\left( \mu _p^U(a_i)^2\right) -\max _i\left( v_p^L(a_i)^2\right) - \max _i\left( v_p^U(a_i)^2\right) }{4}\\&= S\left( a_*\right) \end{aligned}$$

Then we have:

$$\begin{aligned} \mu _{\tilde{P}}^L(a_i)&= \min _i{\mu _{\tilde{P}}^L(a_i)}, \mu _{\tilde{P}}^U(a_i)= \min _i{\mu _{\tilde{P}}^U(a_i)}, v_{\tilde{P}}^L(a_i)=\\&= \max _i{v_{\tilde{P}}^L(a_i)}, v_{\tilde{P}}^U(a_i)=\max _i{v_{\tilde{P}}^U(a_i)} \end{aligned}$$

Therefore

$$\begin{aligned} h(a)&= \frac{\big [\left( \mu _P^L\right) ^2+\left( \mu _P^U\right) ^2+\left( v_P^L\right) ^2+\left( v_P^U\right) ^2\big ]}{2}\\&= \frac{\min _i\left( \mu _p^L(a_i)^2\right) +\min _i\left( \mu _p^U(a_i)^2\right) +\max _i\left( v_p^L(a_i)^2\right) + \max _i\left( v_p^U(a_i)^2\right) }{2}\\&= h\left( a_*\right) \end{aligned}$$

$$\begin{aligned} {\mathrm{IVPFPWA}}\left( a_1, a_2, a_3, \ldots , a_n\right) =a_* \end{aligned}$$

$\square$

Definition 5

Let $a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)], \; [v_{\tilde{P}}^L(a_i)$, $v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, then their aggregated, where $[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]\subset [0,1]$, $[v_{\tilde{P}}^L(a_i), v_{\tilde{P}}^U(a_i)] \subset [0,1]$ and let IVPFPWG $V^n\rightarrow V$. If

$$\begin{aligned} \begin{aligned}&{\mathrm{IVPFPWG}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =a_1^{\frac{T_1}{\sum _{i=1}^nT_i}} \otimes a_2^{\frac{T_2}{\sum _{i=1}^nT_i}} \otimes , \ldots , \otimes a_n^{\frac{T_n}{\sum _{i=1}^nT_i}} \end{aligned} \end{aligned}$$

The interval-valued Pythagorean fuzzy prioritized weighted geometric operator is abbreviated as IVPFPWG with $T_i=\prod _{j=1}^{i-1}S(a_j)$$(i= 2, \ldots , n)$, $T_1=1$ and $S(a_j)$ is the score of IVPFS a.

Theorem 4

Let$a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)], \; [v_{\tilde{P}}^L(a_i)$, $v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, then their aggregated value is still an IVPFS by using IVPFPWG operator and:

$$\begin{aligned}&{\mathrm{IVPFPWG}}\left( a_1, a_2, a_3, \ldots , a_n\right) \\&\quad =\left( \left\langle \left[ \prod \limits _{i=1}^n(\mu _p^L(a_i))^\frac{T_i}{\sum _{i=1}^nT_i}, \prod \limits _{i=1}^n(\mu _p^U(a_i))^\frac{T_i}{\sum _{i=1}^nT_i} \right] ,\right. \right. \\&\qquad \left[ \sqrt{1-\prod \limits _{i=1}^n(1-v_p^L(a_i)^2)^\frac{T_i}{\sum _{i=1}^nT_i}},\right. \\&\quad \quad \left. \left. \left. \sqrt{1-\prod \limits _{i=1}^n(1-v_p^U(a_i)^2)^\frac{T_i}{\sum _{i=1}^nT_i}} \right] \right\rangle \right) \end{aligned}$$

where $T_i=\prod _{j=1}^{i-1}S(a_j)$$(i= 2, \ldots , n)$, $T_1=1$ and S(a) is the score of IVPFSa.

Proof

The proof of Theorem 4 is similar to Theorem 1. $\square$

Theorem 5

(Idempotency) Let$a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]$, $[v_{\tilde{P}}^L(a_i), \; v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, if all$a_i$$(i=1,2,3, \ldots , n))$ are equal ($a_i=a$), then

$$\begin{aligned} {\mathrm{IVPFPWG}}\left( a_1, a_2, a_3, \ldots , a_n\right) =a \end{aligned}$$

Proof

The proof of Theorem 5 is similar to Theorem 2. $\square$

Theorem 6

(Boundary) If$a_i=\{\langle s_{\theta (a_i)},[\mu _{\tilde{P}}^L(a_i), \; \mu _{\tilde{P}}^U(a_i)]$, $[v_{\tilde{P}}^L(a_i), \; v_{\tilde{P}}^U(a_i)]\rangle \}$$(i=1, 2, 3, \ldots , n)$ be a collection of IVPFSs, and$S(a_i)$ is the scores of IVPFS$a_i$:

$$\begin{aligned} a_*&= \left( \left[ \min _i{\mu _{\tilde{P}}^L(a_i)}, \min _i{\mu _{\tilde{P}}^U(a_i)}\right] ,\left[ \max _i{v_{\tilde{P}}^L(a_i)}, \max _i{v_{\tilde{P}}^U(a_i)}\right] \right) \\ a^*&= \left( \left[ \max _i{\mu _{\tilde{P}}^L(a_i)}, \max _i{\mu _{\tilde{P}}^U(a_i)}\right] ,\left[ \min _i{v_{\tilde{P}}^L(a_i)}, \min _i{v_{\tilde{P}}^U(a_i)}\right] \right) \end{aligned}$$

Then $a_*\le {{IVPFPWG}}(a_1, a_2, a_3, \ldots , a_n)\le a^*$.

Proof

The proof of Theorem 6 is similar to Theorem 3. $\square$