Decision support modeling for multiple criteria assessments using a likelihood-based consensus ranking method under Pythagorean fuzzy uncertainty

Abstract

This paper intends to exploit point operator-oriented likelihood measures to constitute a likelihood-based consensus ranking model aimed at conducting multiple criteria decision making encompassing complex uncertain evaluations with Pythagorean fuzzy sets. This paper takes advantage of Pythagorean fuzzy point operators and the scalar functions of upper and lower estimations to formulate a point operator-oriented likelihood measure for preference intensity. On this basis, this paper propounds the notion of penalty weights to characterize dominated relations for acquiring the measurement of comprehensive disagreement and constituting a likelihood-based consensus ranking model. The primary contributions of this study are fourfold. Firstly, two useful point operators are initiated for upper and lower estimations towards Pythagorean membership grades. Secondly, an effective likelihood measure is exploited for determining outranking relations of Pythagorean fuzzy information. Thirdly, a pragmatic concept of penalty weights is proposed for characterizing the dominated relations among alternatives and measuring degrees of comprehensive disagreement. Fourthly, a functional likelihood-based consensus ranking model is constructed for implementing a multiple criteria evaluation with Pythagorean fuzzy uncertainty. Furthermore, a real-life application relating to a financing problem is presented to provide a justification for the practicability of the proposed methodology. This paper executes an analysis of parameters sensitivity and comparative studies for showing more theoretical insights.

Appendices

Appendix A: Detailed proofs

1.1 A.1 Proof of Theorem 1

• (T1.1) In line with Definition 7, one obtains $(\mu_{{M_{\alpha } (P)}} (x))^{2} + (\nu_{{M_{\alpha } (P)}} (x))^{2} = (\mu_{P} (x))^{2} + \alpha (\tau_{P} (x))^{2}$$+ (\nu_{P} (x))^{2}$. Because $(\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} + (\tau_{P} (x))^{2} = 1$, it can be recognized that $0 \le (\mu_{{M_{\alpha } (P)}} (x))^{2} +$$(\nu_{{M_{\alpha } (P)}} (x))^{2} \le 1$. The inequality $0 \le (\mu_{{N_{\beta } (P)}} (x))^{2} + (\nu_{{N_{\beta } (P)}} (x))^{2} \le 1$ is satisfied because $(\mu_{{N_{\beta } (P)}} (x))^{2} +$$(\nu_{{N_{\beta } (P)}} (x))^{2} = (\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} + \beta (\tau_{P} (x))^{2}$ through Definition 8. Thus, (T1.1) is correct.

• (T1.2) It is obvious that $\mu_{P} (x) \le \sqrt {(\mu_{P} (x))^{2} + \alpha (\tau_{P} (x))^{2} }$ from $0 \le \alpha ,\tau_{P} (x) \le 1$; thus, $\mu_{{N_{\beta } (P)}} (x)$$= \mu_{P} (x) \le \mu_{{M_{\alpha } (P)}} (x)$. Analogously, $\nu_{{M_{\alpha } (P)}} (x) = \nu_{P} (x) \le \nu_{{N_{\beta } (P)}} (x)$ is satisfied because $\nu_{P} (x) \le$$\sqrt {(\nu_{P} (x))^{2} + \beta (\tau_{P} (x))^{2} }$. Thus, the correctness of (T1.2) is confirmed.

• (T1.3) As stated in Definition 5, the quasi-ordering $M_{\alpha } (p)\underline { \succ }_{Q} p\underline { \succ }_{Q} N_{\beta } (p)$ holds because $\mu_{{M_{\alpha } (P)}} (x) \ge \mu_{P} (x) \ge \mu_{{N_{\beta } (P)}} (x)$ and $\nu_{{M_{\alpha } (P)}} (x) \le \nu_{P} (x) \le \nu_{{N_{\beta } (P)}} (x)$.

• (T1.4) Applying Definitions 3, 7, and 8, the following is true:

  $$\tau_{{M_{\alpha } (P)}} (x) = \sqrt {1 - (\mu_{P} (x))^{2} - \alpha (\tau_{P} (x))^{2} - (\nu_{P} (x))^{2} } \le \sqrt {1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2} } = \tau_{P} (x),$$
  $$\tau_{{N_{\beta } (P)}} (x) = \sqrt {1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2} - \beta (\tau_{P} (x))^{2} } \le \sqrt {1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2} } = \tau_{P} (x).$$

  Accordingly, this equation indicates that $\max \{ \tau_{{M_{\alpha } (P)}} (x),\tau_{{N_{\beta } (P)}} (x)\} \le \tau_{P} (x)$. Based on Definition 3, it is known that $r_{P} (x) = \sqrt {1 - (\tau_{P} (x))^{2} }$, $r_{{M_{\alpha } (P)}} (x) = \sqrt {1 - (\tau_{{M_{\alpha } (P)}} (x))^{2} }$, and $r_{{N_{\beta } (P)}} (x) = \sqrt {1 - (\tau_{{N_{\beta } (P)}} (x))^{2} }$. It indicates that $\min \{ r_{{M_{\alpha } (P)}} (x),r_{{N_{\beta } (P)}} (x)\} \ge r_{P} (x)$; thus, the property of (T1.4) holds.

• (T1.5) As demonstrated in Chen (2018b), the closer the directions of a commitment (i.e., $d_{P} (x)$, $d_{{M_{\alpha } (P)}} (x)$, and $d_{{N_{\beta } (P)}} (x)$) are to 1 (or 0), the closer the radians (i.e., $\theta_{P} (x)$, $\theta_{{M_{\alpha } (P)}} (x)$, and $\theta_{{N_{\beta } (P)}} (x)$) are to 0 (or $\pi /2$) and the greater the strength of commitment (i.e., $r_{P} (x)$, $r_{{M_{\alpha } (P)}} (x)$, and $r_{{N_{\beta } (P)}} (x)$) is to the supporting (or disapproving) belongingness of x in P. Therefore, the results shown in (T1.2) indicate that $d_{{N_{\beta } (P)}} (x) \le d_{P} (x) \le d_{{M_{\alpha } (P)}} (x)$ and $\theta_{{M_{\alpha } (P)}} (x) \le \theta_{P} (x) \le \theta_{{N_{\beta } (P)}} (x)$ hold. The correctness of (T1.5) is confirmed.

• (T1.6) Based on Definition 3, it is recognized that $p^{c} = (\mu_{{P^{c} }} (x),\nu_{{P^{c} }} (x);r_{{P^{c} }} (x),d_{{P^{c} }} (x)) = (\nu_{P} (x),$$\mu_{P} (x);r_{P} (x),1 - d_{P} (x))$, which also indicates that $\tau_{{P^{c} }} (x) = \tau_{P} (x)$ from $r_{{P^{c} }} (x) = r_{P} (x)$. Applying Definition 7, the upper estimation $M_{\alpha } (p^{c} )$ of the complement of p is given by: $M_{\alpha } (p^{c} ) =$$(\mu_{{M_{\alpha } (P^{c} )}} (x),\nu_{{M_{\alpha } (P^{c} )}} (x);r_{{M_{\alpha } (P^{c} )}} (x),d_{{M_{\alpha } (P^{c} )}} (x))$. The membership grade and the nonmembership grade in $M_{\alpha } (p^{c} )$ are separately derived as follows: $\mu_{{M_{\alpha } (P^{c} )}} (x) = \sqrt {(\mu_{{P^{c} }} (x))^{2} + \alpha (\tau_{{P^{c} }} (x))^{2} } =$$\sqrt {(\nu_{P} (x))^{2} + \alpha (\tau_{P} (x))^{2} } = \nu_{{N_{\alpha } (P)}} (x)$ and $\nu_{{M_{\alpha } (P^{c} )}} (x) = \nu_{{P^{c} }} (x) = \mu_{P} (x) = \mu_{{N_{\alpha } (P)}} (x)$. For the complement $(M_{\alpha } (p^{c} ))^{c}$ of the upper estimation $M_{\alpha } (p^{c} )$, $\mu_{{(M_{\alpha } (P^{c} ))^{c} }} (x) = \nu_{{M_{\alpha } (P^{c} )}} (x) = \mu_{{N_{\alpha } (P)}} (x)$ and $\nu_{{(M_{\alpha } (P^{c} ))^{c} }} (x) = \mu_{{M_{\alpha } (P^{c} )}} (x) = \nu_{{N_{\alpha } (P)}} (x)$. One can verify that $(M_{\alpha } (p^{c} ))^{c} = N_{\alpha } (p)$.

• (T1.7) Based on Definition 8, the lower estimation $N_{\beta } (p^{c} )$ of the complement of p is obtained by $N_{\beta } (p^{c} ) = (\mu_{{N_{\beta } (P^{c} )}} (x),\nu_{{N_{\beta } (P^{c} )}} (x);r_{{N_{\beta } (P^{c} )}} (x),d_{{N_{\beta } (P^{c} )}} (x))$. The membership and nonmembership grades in $N_{\beta } (p^{c} )$ are separately calculated as follows: $\mu_{{N_{\beta } (P^{c} )}} (x) = \mu_{{P^{c} }} (x) = \nu_{P} (x) = \nu_{{M_{\beta } (P)}} (x)$ and $\nu_{{N_{\beta } (P^{c} )}} (x) = \sqrt {(\nu_{{P^{c} }} (x))^{2} + \beta (\tau_{{P^{c} }} (x))^{2} } =$$\sqrt {(\mu_{P} (x))^{2} + \beta (\tau_{P} (x))^{2} } = \mu_{{M_{\beta } (P)}} (x)$. For the complement $(N_{\beta } (p^{c} ))^{c}$ of the lower estimation $N_{\beta } (p^{c} )$, it is apparent that $\mu_{{(N_{\beta } (P^{c} ))^{c} }} (x) = \nu_{{N_{\beta } (P^{c} )}} (x) = \mu_{{M_{\beta } (P)}} (x)$ and $\nu_{{(N_{\beta } (P^{c} ))^{c} }} (x) = \mu_{{N_{\beta } (P^{c} )}} (x) = \nu_{{M_{\beta } (P)}} (x)$, which yields $(N_{\beta } (p^{c} ))^{c} = M_{\beta } (p)$, i.e., (T1.7) is valid. Notably, (T1.6) and (T1.7) demonstrate that the two PF point operators M_α and N_β are dualities. This confirms the truth of Theorem 1.

1.2 A.2 Proof of Theorem 2

In the first place, (T2.2) is trivial. (T2.1) and (T2.3) are demonstrated by virtue of mathematical induction on η. Using the agency of the PF point operator M_α on the recurrent upper estimation $M_{\alpha }^{\eta - 1} (p)$, it is generated that:

$$\mu_{{M_{\alpha }^{\eta } (P)}} (x) = \sqrt {(\mu_{{M_{\alpha }^{\eta - 1} (P)}} (x))^{2} + \alpha (\tau_{{M_{\alpha }^{\eta - 1} (P)}} (x))^{2} } ,$$

$$\nu_{{M_{\alpha }^{\eta } (P)}} (x) = \nu_{{M_{\alpha }^{\eta - 1} (P)}} (x) = \nu_{P} (x),$$

$$r_{{M_{\alpha }^{\eta } (P)}} (x) = \sqrt {(\mu_{{M_{\alpha }^{\eta } (P)}} (x))^{2} + (\nu_{{M_{\alpha }^{\eta } (P)}} (x))^{2} } = \sqrt {(\mu_{{M_{\alpha }^{\eta - 1} (P)}} (x))^{2} + \alpha (\tau_{{M_{\alpha }^{\eta - 1} (P)}} (x))^{2} + (\nu_{P} (x))^{2} } .$$

for $\eta = 0,1,2, \ldots$, in which $\mu_{{M_{\alpha }^{0} (P)}} (x) = \mu_{P} (x)$, $\nu_{{M_{\alpha }^{0} (P)}} (x) = \nu_{P} (x)$, and $\tau_{{M_{\alpha }^{0} (P)}} (x) = \tau_{P} (x)$. Let η = 1. Based on Definition 7 and the condition that $(\tau_{P} (x))^{2} = 1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2}$, one can render:

$$\mu_{{M_{\alpha }^{1} (P)}} (x) = \sqrt {(\mu_{P} (x))^{2} + \alpha (\tau_{P} (x))^{2} } = \sqrt {(\mu_{P} (x))^{2} + \alpha \left( {1 - (\mu_{P} (x))^{2} } \right) - \alpha (\nu_{P} (x))^{2} } ,$$

$$r_{{M_{\alpha }^{1} (P)}} (x) = \sqrt {(\mu_{P} (x))^{2} + \alpha \left( {1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2} } \right) + (\nu_{P} (x))^{2} } = \sqrt {(\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} + \alpha \left( {1 - (\mu_{P} (x))^{2} } \right) - \alpha (\nu_{P} (x))^{2} } .$$

The above results are concordant with the outcome of η = 1 in (T2.1) and (T2.3); that is, the two properties hold for η = 1. Next, let η = 2. It is evident to deduce that $\nu_{{M_{\alpha }^{2} (P)}} (x) = \nu_{{M_{\alpha }^{1} (P)}} (x) = \nu_{P} (x)$. By applying Definitions 7 and 9, it is straightly gained that:

$$\mu_{{M_{\alpha }^{2} (P)}} (x) = \left\{ {(\mu_{P} (x))^{2} + \alpha \left( {1 - (\mu_{P} (x))^{2} } \right) - \alpha (\nu_{P} (x))^{2} + \alpha \left[ {1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2} - \alpha \left( {1 - (\mu_{P} (x))^{2} } \right) + \left. {\left. {\alpha (\nu_{P} (x))^{2} } \right]} \right\}^{0.5} = \sqrt {(\mu_{P} (x))^{2} + \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {2\alpha - \alpha^{2} } \right) - \alpha (\nu_{P} (x))^{2} \left( {2 - \alpha } \right)} ,} \right.} \right.$$

$r_{{M_{\alpha }^{2} (P)}} (x) = \left\{ {(\mu_{P} (x))^{2} + \alpha \left( {1 - (\mu_{P} (x))^{2} } \right) - \alpha (\nu_{P} (x))^{2} + \alpha \left[ {1 - (\mu_{P} (x))^{2} - (\nu_{P} (x))^{2} - \alpha \left( {1 - (\mu_{P} (x))^{2} } \right)\left. {\left. { + \alpha (\nu_{P} (x))^{2} } \right] + (\nu_{P} (x))^{2} } \right\}^{0.5} = \sqrt {(\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} + \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {2\alpha - \alpha^{2} } \right) - \alpha (\nu_{P} (x))^{2} \left( {2 - \alpha } \right)} .} \right.} \right. \,$ The results are consistent with the outcome of η = 2 in (T2.1) and (T2.3); thus, the determination equations are valid in the case of η = 2. Next, assume that (T2.1) and (T2.3) hold for $\eta = \vartheta$. Then,

$$\mu_{{M_{\alpha }^{\vartheta } (P)}} (x) = \sqrt {(\mu_{P} (x))^{2} + \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\vartheta } } \right) - \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\vartheta - 1} {(1 - \alpha )^{k} } } \right)} ,$$

$$r_{{M_{\alpha }^{\vartheta } (P)}} (x) = \sqrt {(\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} + \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\vartheta } } \right) - \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\vartheta - 1} {(1 - \alpha )^{k} } } \right)} ,$$

and $\nu_{{M_{\alpha }^{\vartheta } (P)}} (x) = \nu_{P} (x)$. When $\eta = \vartheta + 1$, one has $\nu_{{M_{\alpha }^{\vartheta + 1} (P)}} (x) = \nu_{P} (x)$; moreover, by way of Definitions 7 and 9, it is yielded that:

$$\mu _{{M_{\alpha }^{{\vartheta + 1}} (P)}} (x) = \sqrt {(\mu _{{M_{\alpha }^{\vartheta } (P)}} (x))^{2} + \alpha (\tau _{{M_{\alpha }^{\vartheta } (P)}} (x))^{2} } = \sqrt {(\mu _{{M_{\alpha }^{\vartheta } (P)}} (x))^{2} + \alpha \left( {1 - (r_{{M_{\alpha }^{\vartheta } (P)}} (x))^{2} } \right)} = \left\{ {(\mu _{P} (x))^{2} + \left( {1 - (\mu _{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\vartheta } } \right) - \alpha (\nu _{P} (x))^{2} \left( {\sum\limits_{{k = 0}}^{{\vartheta - 1}} {(1 - \alpha )^{k} } } \right) + \alpha \left( {1 - (\mu _{P} (x))^{2} } \right) - \alpha (\nu _{P} (x))^{2} - \alpha \left( {1 - (\mu _{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\vartheta } } \right) + \alpha ^{2} (\nu _{P} (x))^{2} \left( {\sum\limits_{{k = 0}}^{{\vartheta - 1}} {(1 - \alpha )^{k} } } \right)} \right\}^{{0.5}} = \left\{ {(\mu _{P} (x))^{2} + \left( {1 - (\mu _{P} (x))^{2} } \right)\left[ {1 - (1 - \alpha )^{\vartheta } + \alpha - \alpha \left( {1 - (1 - \alpha )^{\vartheta } } \right)} \right] + \alpha (\nu _{P} (x))^{2} \left( { - \sum\limits_{{k = 0}}^{{\vartheta - 1}} {(1 - \alpha )^{k} } - 1 + \alpha } \right)} \right\}^{{0.5}} = \sqrt {(\mu _{P} (x))^{2} + \left( {1 - (\mu _{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )(1 - \alpha )^{\vartheta } } \right) - \alpha (\nu _{P} (x))^{2} \left( {(1 - \alpha ) + \sum\limits_{{k = 0}}^{{\vartheta - 1}} {(1 - \alpha )^{k} } } \right)} = \sqrt {(\mu _{P} (x))^{2} + \left( {1 - (\mu _{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{{\vartheta + 1}} } \right) - \alpha (\nu _{P} (x))^{2} \left( {\sum\limits_{{k = 0}}^{\vartheta } {(1 - \alpha )^{k} } } \right)} ,$$

$$r_{{M_{\alpha }^{\vartheta + 1} (P)}} (x) = \sqrt {(\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} + \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\vartheta + 1} } \right) - \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\vartheta } {(1 - \alpha )^{k} } } \right)} .$$

On this basis, (T2.1) and (T2.3) hold for $\eta = \vartheta + 1$. Accordingly, the properties of (T2.1) and (T2.3) are fulfilled corresponding to all η values.

(T2.4) Based on (T2.1), the discrepancy between squared membership grades in the recurrent upper estimations $M_{\alpha }^{\eta } (p)$ and $M_{\alpha }^{\eta - 1} (p)$ for all $\alpha \in [0,1]$ is calculated in this way:

$$\begin{aligned} (\mu_{{M_{\alpha }^{\eta } (P)}} (x))^{2} - (\mu_{{M_{\alpha }^{\eta - 1} (P)}} (x))^{2} & = (\mu_{P} (x))^{2} + \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\eta } } \right) - \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\eta - 1} {(1 - \alpha )^{k} } } \right) - (\mu_{P} (x))^{2} - \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {1 - (1 - \alpha )^{\eta - 1} } \right) + \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\eta - 2} {(1 - \alpha )^{k} } } \right) \\ & = \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {(1 - \alpha )^{\eta - 1} - (1 - \alpha )^{\eta } } \right) - \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\eta - 1} {(1 - \alpha )^{k} } - \sum\limits_{k = 0}^{\eta - 2} {(1 - \alpha )^{k} } } \right) \\ & = \left( {1 - (\mu_{P} (x))^{2} } \right)\left( {(1 - \alpha )^{\eta - 1} - (1 - \alpha )(1 - \alpha )^{\eta - 1} } \right) - \alpha (\nu_{P} (x))^{2} \left( {\sum\limits_{k = 0}^{\eta - 2} {(1 - \alpha )^{k} + (1 - \alpha )^{\eta - 1} } - \sum\limits_{k = 0}^{\eta - 2} {(1 - \alpha )^{k} } } \right) \\ & = \alpha (1 - \alpha )^{\eta - 1} \left( {1 - (\mu_{P} (x))^{2} - \alpha (\nu_{P} (x))^{2} } \right). \\ \end{aligned}$$

It becomes clear that $\alpha \ge 0$, $(1 - \alpha )^{\eta - 1} \ge 0$, and $1 - (\mu_{P} (x))^{2} - \alpha (\nu_{P} (x))^{2} \ge 0$ from $\alpha \in [0,1]$ and $0 \le (\mu_{P} (x))^{2} + (\nu_{P} (x))^{2} \le 1$. This evidences that $(\mu_{{M_{\alpha }^{\eta } (P)}} (x))^{2} - (\mu_{{M_{\alpha }^{\eta - 1} (P)}} (x))^{2} \ge 0$, which reveals that $\mu_{{M_{\alpha }^{\eta } (P)}} (x) \ge \mu_{{M_{\alpha }^{\eta - 1} (P)}} (x)$. On the other side, based on (T2.2), the difference between the non-membership grades in the recurrent upper estimations $M_{\alpha }^{\eta } (p)$ and $M_{\alpha }^{\eta - 1} (p)$ is always equal to zero because $\nu_{{M_{\alpha }^{\eta } (P)}} (x) = \nu_{{M_{\alpha }^{\eta - 1} (P)}} (x) = \nu_{P} (x)$ for each $\alpha \in [0,1]$, which straightly infers that the inequality $\nu_{{M_{\alpha }^{\eta } (P)}} (x) \le \nu_{{M_{\alpha }^{\eta - 1} (P)}} (x)$ is satisfied. From Definition 5, the natural quasi-ordering $M_{\alpha }^{\eta } (p)\underline { \succ }_{Q} M_{\alpha }^{\eta - 1} (p)$ holds because $\mu_{{M_{\alpha }^{\eta } (P)}} (x) \ge \mu_{{M_{\alpha }^{\eta - 1} (P)}} (x)$ and $\nu_{{M_{\alpha }^{\eta } (P)}} (x) \le \nu_{{M_{\alpha }^{\eta - 1} (P)}} (x)$. As a result, (T2.4) is correct.

(T2.5) By using (T2.1), it is yielded that $\lim_{\eta \to \infty } \mu_{{M_{\alpha }^{\eta } (P)}} (x) = \lim_{\eta \to \infty } [(\mu_{P} (x))^{2} + (1 - (\mu_{P} (x))^{2} ) \cdot$$(1 - (1 - \alpha )^{\eta } ) - \alpha (\nu_{P} (x))^{2} (\sum\nolimits_{k = 0}^{\eta - 1} {(1 - \alpha )^{k} } )]^{0.5} = \sqrt {1 - (\nu_{P} (x))^{2} }$. From (T2.2), it is revealed that $\lim_{\eta \to \infty } \nu_{{M_{\alpha }^{\eta } (P)}} (x) = \lim_{\eta \to \infty } \nu_{P} (x) = \nu_{P} (x)$. By applying Definition 2, it indicates that $\lim_{\eta \to \infty } M_{\alpha }^{\eta } (p) =$$(\sqrt {1 - (\nu_{P} (x))^{2} } ,\nu_{P} (x);1,(\pi - 2 \cdot sin^{ - 1} (\nu_{P} (x)))/\pi ))$. Thus, (T2.5) is confirmed, which demonstrates the truth of Theorem 2.

1.3 A.3 Proof of Theorem 4

(T4.1) The values of scalar functions range from 0 to 1 on the basis of Definition 4; thus, $0 \le V(p_{ij} ),V(M_{{\alpha_{ij} }} (p_{ij} )),V(N_{{\beta_{ij} }} (p_{ij} )) \le 1$. In accordance with the property in (T1.5), $d_{ij}^{N} \le d_{ij} \le d_{ij}^{M}$ and $\theta_{ij}^{M} \le \theta_{ij} \le \theta_{ij}^{N}$. Moreover, $r_{ij}^{M} \ge r_{ij}$ and $r_{ij}^{N} \ge r_{ij}$ because $\min \{ r_{ij}^{M} ,r_{ij}^{N} \} \ge r_{ij}$ in (T1.4). It can be deduced that the two inequalities $V(p_{ij} ) \le V(M_{{\alpha_{ij} }} (p_{ij} ))$ and $V(N_{{\beta_{ij} }} (p_{ij} )) \le V(p_{ij} )$ are satisfied because $r_{ij} (d_{ij} - 0.5) \le r_{ij}^{M} (d_{ij}^{M} - 0.5)$ and $r_{ij}^{N} (0.5 - {{2 \cdot \theta_{ij}^{N} } \mathord{\left/ {\vphantom {{2 \cdot \theta_{ij}^{N} } \pi }} \right. \kern-\nulldelimiterspace} \pi }) \le r_{ij} (0.5 - {{2 \cdot \theta_{ij} } \mathord{\left/ {\vphantom {{2 \cdot \theta_{ij} } \pi }} \right. \kern-\nulldelimiterspace} \pi })$, respectively. Thus, $0 \le V(N_{{\beta_{ij} }} (p_{ij} )) \le V(p_{ij} ) \le V(M_{{\alpha_{ij} }} (p_{ij} )) \le 1$.

(T4.2) From (29) and (31), it is known that $r_{ij}^{M} = \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} }$ and $\theta_{ij}^{M} =$${\text{arc}}\cos \left( {{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } } \mathord{\left/ {\vphantom {{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } } {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right. \kern-\nulldelimiterspace} {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)$, respectively. From (34), the following consequence can be determined:

$$V(M_{{\alpha_{ij} }} (p_{ij} )) = \frac{1}{2} + r_{ij}^{M} \cdot \left( {\frac{1}{2} - \frac{{2 \cdot \theta_{ij}^{M} }}{\pi }} \right) = \frac{1}{2} + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right){\text{arc}}\cos \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)} \right].$$

The partial derivative of $V(M_{{\alpha_{ij} }} (p_{ij} ))$ in regard to α_ij is computed in this fashion:

$$\begin{aligned} & \frac{{\partial V(M_{{\alpha_{ij} }} (p_{ij} ))}}{{\partial \alpha_{ij} }} = \frac{{\partial \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\partial \alpha_{ij} }} \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right){\text{arc}}\cos \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)} \right] \\ \, & + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \cdot \frac{{\partial \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right){\text{arc}}\cos \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)} \right]}}{{\partial \alpha_{ij} }} \\ & = \frac{{(\tau_{ij} )^{2} }}{{2\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }} \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right){\text{arc}}\cos \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)} \right] \\ \, & + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \cdot \left\{ { - \left( {\frac{2}{\pi }} \right)\left[ { - {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 - \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)^{2} } }}} \right. \kern-\nulldelimiterspace} {\sqrt {1 - \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)^{2} } }}} \right]} \right\} \\ & = \frac{{(\tau_{ij} )^{2} }}{{2\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }} \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right){\text{arc}}\cos \left( {\frac{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right)} \right] \\ \, & + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \cdot \left[ {{{\left( {\frac{2}{\pi }} \right)} \mathord{\left/ {\vphantom {{\left( {\frac{2}{\pi }} \right)} {\sqrt {1 - \frac{{(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} }}{{(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} }}} }}} \right. \kern-\nulldelimiterspace} {\sqrt {1 - \frac{{(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} }}{{(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} }}} }}} \right]. \\ \end{aligned}$$

Thus, ${{\partial V(M_{{\alpha_{ij} }} (p_{ij} ))} \mathord{\left/ {\vphantom {{\partial V(M_{{\alpha_{ij} }} (p_{ij} ))} {\partial \alpha_{ij} }}} \right. \kern-\nulldelimiterspace} {\partial \alpha_{ij} }} \ge 0$ because ${\text{arc}}\cos \left( {{{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } } \mathord{\left/ {\vphantom {{\sqrt {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } } {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right. \kern-\nulldelimiterspace} {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } }}} \right) \le$$\left( {{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} \right)$ and ${{\left( {{2 \mathord{\left/ {\vphantom {2 \pi }} \right. \kern-\nulldelimiterspace} \pi }} \right)} \mathord{\left/ {\vphantom {{\left( {{2 \mathord{\left/ {\vphantom {2 \pi }} \right. \kern-\nulldelimiterspace} \pi }} \right)} {\sqrt {1 - {{\left( {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)} \mathord{\left/ {\vphantom {{\left( {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)} {\left( {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)}}} }}} \right. \kern-\nulldelimiterspace} {\sqrt {1 - {{\left( {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)} \mathord{\left/ {\vphantom {{\left( {(\mu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)} {\left( {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \alpha_{ij} (\tau_{ij} )^{2} } \right)}}} }} \ge 0$. Thus, $V(M_{{\alpha_{ij} }} (p_{ij} ))$ is monotonically nondecreasing with the allocation parameter α_ij, thereby confirming the truth of (T4.2).

(T4.3) In accordance with (30) and (32), $r_{ij}^{N} = \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} }$ and $\theta_{ij}^{N} =$${\text{arc}}sin\left( {{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } } \mathord{\left/ {\vphantom {{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } } {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right. \kern-\nulldelimiterspace} {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)$, respectively. Applying (35), the following is true:

$$V(N_{{\beta_{ij} }} (p_{ij} )) = \frac{1}{2} + r_{ij}^{N} \cdot \left( {\frac{1}{2} - \frac{{2 \cdot \theta_{ij}^{N} }}{\pi }} \right) = \frac{1}{2} + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right){\text{arc}}sin\left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)} \right].$$

The partial derivative of $V(N_{{\beta_{ij} }} (p_{ij} ))$ compared to β_ij is calculated in this manner:

$$\begin{aligned} & \frac{{\partial V(N_{{\beta_{ij} }} (p_{ij} ))}}{{\partial \beta_{ij} }} = \frac{{\partial \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\partial \beta_{ij} }} \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right)\arcsin \left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)} \right] \\ \, & + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \cdot \frac{{\partial \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right)\arcsin \left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)} \right]}}{{\partial \beta_{ij} }} \\ & = \frac{{(\tau_{ij} )^{2} }}{{2\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }} \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right)\arcsin \left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)} \right] \\ \, & + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \cdot \left\{ { - \left( {\frac{2}{\pi }} \right)\left[ {{1 \mathord{\left/ {\vphantom {1 {\sqrt {1 - \left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)^{2} } }}} \right. \kern-\nulldelimiterspace} {\sqrt {1 - \left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)^{2} } }}} \right]} \right\} \\ & = \frac{{(\tau_{ij} )^{2} }}{{2\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }} \cdot \left[ {\frac{1}{2} - \left( {\frac{2}{\pi }} \right)\arcsin \left( {\frac{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}{{\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right)} \right] \\ \, & + \sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \cdot \left[ { - {{\left( {\frac{2}{\pi }} \right)} \mathord{\left/ {\vphantom {{\left( {\frac{2}{\pi }} \right)} {\sqrt {1 - \frac{{(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} }}{{(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} }}} }}} \right. \kern-\nulldelimiterspace} {\sqrt {1 - \frac{{(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} }}{{(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} }}} }}} \right] \le 0. \\ \end{aligned}$$

Because ${\text{arc}}sin\left( {{{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } } \mathord{\left/ {\vphantom {{\sqrt {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } } {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right. \kern-\nulldelimiterspace} {\sqrt {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } }}} \right) \ge \left( {{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} \right)$ and $- {{\left( {{2 \mathord{\left/ {\vphantom {2 \pi }} \right. \kern-\nulldelimiterspace} \pi }} \right)} \mathord{\left/ {\vphantom {{\left( {{2 \mathord{\left/ {\vphantom {2 \pi }} \right. \kern-\nulldelimiterspace} \pi }} \right)} {}}} \right. \kern-\nulldelimiterspace} {}}$$\sqrt {1 - {{\left( {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \right)} \mathord{\left/ {\vphantom {{\left( {(\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \right)} {\left( {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {(\mu_{ij} )^{2} + (\nu_{ij} )^{2} + \beta_{ij} (\tau_{ij} )^{2} } \right)}}} \le 0$, it is deduced that ${{\partial V(N_{{\beta_{ij} }} (p_{ij} ))} \mathord{\left/ {\vphantom {{\partial V(N_{{\beta_{ij} }} (p_{ij} ))} {\partial \beta_{ij} }}} \right. \kern-\nulldelimiterspace} {\partial \beta_{ij} }} \le 0$. Thus, $V(N_{{\beta_{ij} }} (p_{ij} ))$ is monotonically nonincreasing with the allocation parameter β_ij. Thus, the correctness of (T4.3) is confirmed, which gives substance to the truth of Theorem 4.

1.4 A.4 Proof of Theorem 5

(T5.1) For notational convenience, we denote:

$$\Lambda (p_{ij} ,p_{kj} ) = \frac{{V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} ))}}{{\left( {V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))} \right) + \left( {V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))} \right)}},$$

from which $Lik(p_{ij} \underline { \succ } p_{kj} ) = \max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\}$. The fact that $\max \{ \Lambda (p_{ij} ,p_{kj} ),0\} \ge 0$ yields the outcomes $1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} \le 1$ and $0 \le \max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\} \le 1$. Therefore, the boundedness property of $0 \le Lik(p_{ij} \underline { \succ } p_{kj} ) \le 1$ is satisfied, and (T5.1) is correct.

(T5.2) For necessity, $\max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\} = 0$ from the given condition $Lik(p_{ij} \underline { \succ }$$p_{kj} ) = 0$. Thus, $\max \{ \Lambda (p_{ij} ,p_{kj} ),0\} \ge 1$, which leads to $\Lambda (p_{ij} ,p_{kj} ) \ge 1$. It can be demonstrated that $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \ge$$V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) + V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))$, which leads to $V(N_{{\beta_{kj} }} (p_{kj} )) \ge V(M_{{\alpha_{ij} }} (p_{ij} ))$. For sufficiency, based on the property in (T4.1), $V(M_{{\alpha_{ij} }} (p_{ij} )) -$$V(N_{{\beta_{ij} }} (p_{ij} )) \ge 0$ and $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )) \ge 0$. Thus, $V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) +$$V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )) \ge 0$, from which $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) - (V(N_{{\beta_{kj} }} (p_{kj} )) -$$V(M_{{\alpha_{ij} }} (p_{ij} ))) \ge 0$. By combining the condition $V(N_{{\beta_{kj} }} (p_{kj} )) \ge V(M_{{\alpha_{ij} }} (p_{ij} ))$, $V(M_{{\alpha_{kj} }} (p_{kj} )) -$$V(N_{{\beta_{ij} }} (p_{ij} )) \ge (V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))) + (V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )))$, from which $\Lambda (p_{ij} ,p_{kj} )$$\ge 1$. In addition, $\max \{ \Lambda (p_{ij} ,p_{kj} ),0\} = \Lambda (p_{ij} ,p_{kj} ) \ge 1$ and $1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} \le 0$. Thus, $\max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\} = 0$ (i.e., $Lik(p_{ij} \underline { \succ } p_{kj} ) = 0$), and (T5.2) is valid.

(T5.3) For necessity, the given assumption $Lik(p_{ij} \underline { \succ } p_{kj} ) = 1$ leads to the deduction that $\max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\} = 1$. Logically, $\Lambda (p_{ij} ,p_{kj} ) \le 0$ because $\max \{ \Lambda (p_{ij} ,p_{kj} ),0\} = 0$. Thus, $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le 0$, i.e., $V(N_{{\beta_{ij} }} (p_{ij} )) \ge V(M_{{\alpha_{kj} }} (p_{kj} ))$. For sufficiency, the given assumption $V(N_{{\beta_{ij} }} (p_{ij} )) \ge V(M_{{\alpha_{kj} }} (p_{kj} ))$ indicates that $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le 0$. It is verified that $V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \ge 0$ and $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )) \ge 0$ based on (T4.1). Because $V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) = 0$ and $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )) = 0$ do not occur at the same time, the denominator in $\Lambda (p_{ij} ,p_{kj} )$ is greater than zero on all occasions. Thus, $\Lambda (p_{ij} ,p_{kj} ) \le 0$, which yields $\max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\}$ = 1, namely, $Lik(p_{ij} \underline { \succ } p_{kj} ) = 1$). As a result, (T5.3) is satisfied.

(T5.4) To show the complementarity property, this paper makes the following four assumptions about the scalar functions presented in $\Lambda (p_{ij} ,p_{kj} )$: (1) $V(N_{{\beta_{ij} }} (p_{ij} )) \le V(N_{{\beta_{kj} }} (p_{kj} ))$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) \le V(M_{{\alpha_{kj} }} (p_{kj} ))$; (2) $V(N_{{\beta_{ij} }} (p_{ij} )) \ge V(N_{{\beta_{kj} }} (p_{kj} ))$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge V(M_{{\alpha_{kj} }} (p_{kj} ))$; (3) $V(N_{{\beta_{ij} }} (p_{ij} )) \le V(N_{{\beta_{kj} }} (p_{kj} ))$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge V(M_{{\alpha_{kj} }} (p_{kj} ))$; and (4) $V(N_{{\beta_{ij} }} (p_{ij} )) \ge V(N_{{\beta_{kj} }} (p_{kj} ))$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) \le V(M_{{\alpha_{kj} }} (p_{kj} ))$. We denote $\Lambda (p_{kj} ,p_{ij} ) = (V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} )))/$$(V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) + V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )))$. It is apparent that $\Lambda (p_{ij} ,p_{kj} ) + \Lambda (p_{kj} ,p_{ij} )$$= 1$. In Case (i), in the light of the assumption and the property in (T4.1), $V(N_{{\beta_{ij} }} (p_{ij} )) \le V(M_{{\alpha_{ij} }} (p_{ij} ))$$\le V(M_{{\alpha_{kj} }} (p_{kj} ))$, which indicates that $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \ge 0$. In the event that $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge$$V(N_{{\beta_{kj} }} (p_{kj} ))$, it is readily corroborated that $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le (V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )))$$+ (V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )))$. Thus, $0 \le \Lambda (p_{ij} ,p_{kj} ) \le 1$ and $Lik(p_{ij} \underline { \succ } p_{kj} ) = 1 - \Lambda (p_{ij} ,p_{kj} )$. Furthermore, the condition $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge V(N_{{\beta_{kj} }} (p_{kj} ))$ implies that $0 \le V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} ))$$\le 1$. It is apparent that $0 \le V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le 1$ and $0 \le V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} )) \le 1$, which yields $Lik(p_{kj} \underline { \succ } p_{ij} ) = 1 - \Lambda (p_{kj} ,p_{ij} )$. Accordingly, it is found that:

$$Lik(p_{ij} \underline { \succ } p_{kj} ) + Lik(p_{kj} \underline { \succ } p_{ij} ) = 2 - \frac{{V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) + V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} ))}}{{\left( {V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))} \right) + \left( {V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))} \right)}} = 1.$$

Therefore, (T5.4) holds when $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge V(N_{{\beta_{kj} }} (p_{kj} ))$ in Case (1). When $V(M_{{\alpha_{ij} }} (p_{ij} )) \le$$V(N_{{\beta_{kj} }} (p_{kj} ))$, $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \ge (V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))) + (V(M_{{\alpha_{kj} }} (p_{kj} )) -$$V(N_{{\beta_{kj} }} (p_{kj} )))$. It follows that $\Lambda (p_{ij} ,p_{kj} ) \ge 1$ and $Lik(p_{ij} \underline { \succ } p_{kj} ) = \max \{ 1 - \Lambda (p_{ij} ,p_{kj} ),0\} = 0$. Thus, $V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} )) \le 0$ based on the condition $V(M_{{\alpha_{ij} }} (p_{ij} )) \le V(N_{{\beta_{kj} }} (p_{kj} ))$ and $Lik(p_{kj} \underline { \succ } p_{ij} ) = \max \{ 1 - 0,0\} = 1$, which indicates that $Lik(p_{ij} \underline { \succ } p_{kj} ) + Lik(p_{kj} \underline { \succ } p_{ij} ) = 1$. Therefore, (T5.4) holds when $V(M_{{\alpha_{ij} }} (p_{ij} )) \le V(N_{{\beta_{kj} }} (p_{kj} ))$ in Case (1). Case (2) can be corroborated in an analogous way. Next, it is known that $V(N_{{\beta_{ij} }} (p_{ij} )) \le V(N_{{\beta_{kj} }} (p_{kj} ))$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge V(M_{{\alpha_{kj} }} (p_{kj} ))$ in Case (3). Based on (T4.1), $V(M_{{\alpha_{kj} }} (p_{kj} )) \ge V(N_{{\beta_{kj} }} (p_{kj} ))$, which shows that $V(M_{{\alpha_{kj} }} (p_{kj} ))$$\ge V(N_{{\beta_{kj} }} (p_{kj} )) \ge V(N_{{\beta_{ij} }} (p_{ij} ))$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) \ge V(M_{{\alpha_{kj} }} (p_{kj} )) \ge V(N_{{\beta_{kj} }} (p_{kj} ))$. Thus, $V(M_{{\alpha_{kj} }} (p_{kj} )) -$$V(N_{{\beta_{ij} }} (p_{ij} )) \ge 0$ and $V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} )) \ge 0$ are obtained. It is verified that $(V(M_{{\alpha_{ij} }} (p_{ij} )) -$$V(N_{{\beta_{ij} }} (p_{ij} ))) + (V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))) = (V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} ))) + (V(M_{{\alpha_{ij} }} (p_{ij} )) -$$V(N_{{\beta_{kj} }} (p_{kj} ))) \ge V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \ge 0$. Thus, $0 \le \Lambda (p_{ij} ,p_{kj} ) \le 1$, which implies that $Lik(p_{ij} \underline { \succ } p_{kj} ) = \max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\} = 1 - \Lambda (p_{ij} ,p_{kj} )$. Thereafter, the result $Lik(p_{kj} \underline { \succ } p_{ij} )$$= 1 - \Lambda (p_{kj} ,p_{ij} )$ can be acquired in a similar manner. Thus, $Lik(p_{ij} \underline { \succ } p_{kj} ) + Lik(p_{kj} \underline { \succ } p_{ij} )$$= (1 - \Lambda (p_{ij} ,p_{kj} )) + (1 - \Lambda (p_{kj} ,p_{ij} )) = 1$. Accordingly, (T5.4) is fulfilled in Case (iii), and Case (iv) is corroborated in an analogous manner. Based on these results, the complementarity property is possessed by the proposed PF likelihood measure, and (T5.4) is fulfilled.

(T5.5) When $Lik(p_{ij} \underline { \succ } p_{kj} ) = Lik(p_{kj} \underline { \succ } p_{ij} )$, it is clearly understood that $Lik(p_{ij} \underline { \succ } p_{kj} ) =$$Lik(p_{kj} \underline { \succ } p_{ij} ) = 0.5$ through the agency of (T5.4), and thus, (T5.5) is valid. (T5.6) can be easily inferred from (T5.5).

(T5.7) Based on (T5.4) and (T5.6), $Lik(p_{ij} \underline { \succ } p_{kj} ) + Lik(p_{kj} \underline { \succ } p_{ij} ) = 1$ and $Lik(p_{ij} \underline { \succ } p_{ij} ) = 0.5$, respectively. Accordingly, the following can be derived:

$$\sum\limits_{i = 1}^{m} {\sum\limits_{k = 1}^{m} {Lik(p_{ij} \underline { \succ } p_{kj} )} } = \sum\limits_{i = 1}^{m} {Lik(p_{ij} \underline { \succ } p_{ij} )} + \sum\limits_{i = 1,i < k}^{m} {\left( {Lik(p_{ij} \underline { \succ } p_{kj} ) + Lik(p_{kj} \underline { \succ } p_{ij} )} \right)} = \frac{m}{2} + \frac{m(m - 1)}{2} = \frac{{m^{2} }}{2}.$$

This confirms the correctness of Theorem 5.

1.5 A.5 Proof of Theorem 7

Let us denote $\Lambda (p_{ij} ,p_{lj} ) = (V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{ij} }} (p_{ij} )))/(V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} )) +$$V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{lj} }} (p_{lj} )))$ and $\Lambda (p_{lj} ,p_{ij} ) = (V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{lj} }} (p_{lj} )))/(V(M_{{\alpha_{ij} }} (p_{ij} )) -$$V(N_{{\beta_{ij} }} (p_{ij} )) + V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{lj} }} (p_{lj} )))$ for brevity. Thus, $\Lambda (p_{ij} ,p_{lj} ) + \Lambda (p_{lj} ,p_{ij} ) = 1$. Consider the supposition that $Lik(p_{ij} \underline { \succ } p_{lj} ) \ge 0.5$ and $Lik(p_{lj} \underline { \succ } p_{kj} ) \ge 0.5$. The assumption $Lik(p_{ij} \underline { \succ } p_{lj} ) =$$\max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{lj} ),0\} ,0\} \ge 0.5$ indicates that $V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \ge 0$ and $1 -$$\Lambda (p_{ij} ,p_{lj} ) \ge 0.5$. It is known that $\Lambda (p_{lj} ,p_{ij} ) \ge 0.5$ because $\Lambda (p_{ij} ,p_{lj} ) + \Lambda (p_{lj} ,p_{ij} ) = 1$. It is clear that

$$\Lambda (p_{lj} ,p_{ij} ) = \frac{{V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{lj} }} (p_{lj} ))}}{{\left( {V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{lj} }} (p_{lj} ))} \right) + \left( {V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))} \right)}} \ge \frac{1}{2}.$$

Thus, $0 \le V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{lj} }} (p_{lj} ))$. The assumption 0.5$Lik(p_{lj} \underline { \succ } p_{kj} ) \ge$ (i.e., $\max \{ 1 - \max \{ \Lambda (p_{lj} ,p_{kj} ),0\} ,0\} \ge 0.5$) shows that $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{lj} }} (p_{lj} )) \ge 0$ and $1 - \Lambda (p_{lj} ,p_{kj} ) \ge 0.5$. From $\Lambda (p_{lj} ,p_{kj} ) + \Lambda (p_{kj} ,p_{lj} ) = 1$, one obtains $\Lambda (p_{kj} ,p_{lj} ) \ge 0.5$. To be specific,

$$\Lambda (p_{kj} ,p_{lj} ) = \frac{{V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))}}{{\left( {V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))} \right) + \left( {V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{lj} }} (p_{lj} ))} \right)}} \ge \frac{1}{2}.$$

It can be determined that $0 \le V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{lj} }} (p_{lj} )) \le V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))$. By adding the obtained inequalities $0 \le V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{lj} }} (p_{lj} ))$ and $0 \le$$V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{lj} }} (p_{lj} )) \le V(M_{{\alpha_{lj} }} (p_{lj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))$, it is verified that $0 \le V(M_{{\alpha_{kj} }} (p_{kj} ))$$- V(N_{{\beta_{ij} }} (p_{ij} )) \le V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} ))$. By adding $V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} ))$ to this inequality, the following is obtained: $0 \le V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )){ + }V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )) \le$$V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{kj} }} (p_{kj} )){ + }V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} )).$ This indicates that:

$$0 \le 2\left( {V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} ))} \right) \le \left( {V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))} \right) + \left( {V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))} \right).$$

As a consequence, the following is obtained:

$$0 \le \frac{{V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{ij} }} (p_{ij} ))}}{{\left( {V(M_{{\alpha_{ij} }} (p_{ij} )) - V(N_{{\beta_{ij} }} (p_{ij} ))} \right) + \left( {V(M_{{\alpha_{kj} }} (p_{kj} )) - V(N_{{\beta_{kj} }} (p_{kj} ))} \right)}} \le \frac{1}{2}.$$

Namely, $0 \le \Lambda (p_{ij} ,p_{kj} ) \le 0.5$. Thus, it can be demonstrated that $0 \le \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} \le 0.5$ and $0.5 \le 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} \le 1$. Accordingly, it is shown that $0.5 \le \max \{ 1 - \max \{ \Lambda (p_{ij} ,p_{kj} ),0\} ,0\} \le 1$, which indicates that $Lik(p_{ij} \underline { \succ } p_{kj} ) \ge 0.5$. Therefore, the PF likelihood measure possesses the property of weak transitivity, which demonstrates the truth of Theorem 7.

1.6 A.6 Proof of Theorem 8

(T8.1) According to (T5.1) and (T5.6), it is known that $0 \le Lik(p_{kj} \underline { \succ } p_{ij} ) \le 1$ and $Lik(p_{kj} \underline { \succ } p_{kj} ) = 0.5$, respectively. Thus, the subsequent outcomes can be determined:

$$Pen(p_{ij} ) = \frac{1}{m(m - 1)}\left( {Lik(p_{kj} \underline { \succ } p_{kj} ) + \sum\limits_{i = 1,i \ne k}^{m} {Lik(p_{kj} \underline { \succ } p_{ij} )} + \frac{m}{2} - 1} \right) = \frac{1}{m(m - 1)}\left( {\frac{1}{2} + \sum\limits_{i = 1,i \ne k}^{m} {Lik(p_{kj} \underline { \succ } p_{ij} )} + \frac{m}{2} - 1} \right) \ge \frac{1}{m(m - 1)}\left( {\frac{1}{2} + 0 + \frac{m}{2} - 1} \right) = \frac{1}{2m},$$

$$Pen(p_{ij} ) \le \frac{1}{m(m - 1)}\left( {\frac{1}{2} + \sum\limits_{i = 1,i \ne k}^{m} 1 + \frac{m}{2} - 1} \right) = \frac{1}{m(m - 1)}\left( {\frac{1}{2} + (m - 1) + \frac{m}{2} - 1} \right) = \frac{3}{2m}.$$

Therefore, it follows that ${1 \mathord{\left/ {\vphantom {1 {2m}}} \right. \kern-\nulldelimiterspace} {2m}} \le Pen(p_{ij} ) \le {3 \mathord{\left/ {\vphantom {3 {2m}}} \right. \kern-\nulldelimiterspace} {2m}}$, and (T8.1) is correct.

(T8.2) With the help of the property $Lik(p_{kj} \underline { \succ } p_{ij} ) + Lik(p_{ij} \underline { \succ } p_{kj} ) = 1$ based on (T5.4), one can quickly obtain the following:

$$Pen(p_{ij} ) + \Pr i(p_{ij} ) = \frac{1}{m(m - 1)}\left( {\sum\limits_{k = 1}^{m} {Lik(p_{kj} \underline { \succ } p_{ij} )} + \frac{m}{2} - 1} \right) + \frac{1}{m(m - 1)}\left( {\sum\limits_{k = 1}^{m} {Lik(p_{ij} \underline { \succ } p_{kj} )} + \frac{m}{2} - 1} \right) = \frac{1}{m(m - 1)}\left( {\sum\limits_{k = 1}^{m} {\left( {Lik(p_{kj} \underline { \succ } p_{ij} ) + Lik(p_{ij} \underline { \succ } p_{kj} )} \right)} + m - 2} \right) = \frac{1}{m(m - 1)}\left( {m + m - 2} \right) = \frac{2}{m}.$$

(T8.3) Regarding an evaluative criterion c_j, it can be determined the following:

$$\sum\limits_{i = 1}^{m} {Pen(p_{ij} )} = \sum\limits_{i = 1}^{m} {\frac{1}{m(m - 1)}\left( {\sum\limits_{k = 1}^{m} {Lik(p_{kj} \underline { \succ } p_{ij} )} + \frac{m}{2} - 1} \right)} = \frac{1}{m(m - 1)}\left( {\sum\limits_{i = 1}^{m} {\sum\limits_{k = 1}^{m} {Lik(p_{kj} \underline { \succ } p_{ij} ) + \sum\limits_{i = 1}^{m} {\left( {\frac{m}{2} - 1} \right)} } } } \right) = \frac{1}{m(m - 1)}\left( {\frac{{m^{2} }}{2} + \frac{m(m - 2)}{2}} \right) = \frac{1}{m(m - 1)}\left( {\frac{2m(m - 1)}{2}} \right) = 1.$$

(T8.4) can be trivially proven through the medium of (T8.3), which demonstrates the correctness of Theorem 8.

1.7 A.7 Proof of Theorem 9

Theorem 9 will be demonstrated under the aegis of mathematical induction on n. On the occasion that n = 2, it can be received that $\Delta_{i}^{\phi } = \left[ {\left( {Pen(p_{i1} ) \cdot \left| {\Phi_{i1} - \phi } \right|} \right) \odot w_{1} } \right] \oplus$$\left[ {\left( {Pen(p_{i2} ) \cdot \left| {\Phi_{i2} - \phi } \right|} \right) \odot w_{2} } \right]$. Applying the arithmetic operations in Definition 6, the two-dimensional representation $\Delta_{i}^{\prime\phi }$ ($= (\mu_{\Delta i}^{\phi } ,\nu_{\Delta i}^{\phi } )$) is calculated as follows:

$$\Delta _{i}^{{\prime \phi }} = \left[ {\left( {Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|} \right) \odot w_{1}^{\prime } } \right] \oplus \left[ {\left( {Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|} \right) \odot w_{2}^{\prime } } \right] = \left( {\sqrt {1 - \left( {1 - \left( {\omega _{1} } \right)^{2} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} } ,\left( {\varpi _{1} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} } \right) \oplus \left( {\sqrt {1 - \left( {1 - \left( {\omega _{2} } \right)^{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} } ,\left( {\varpi _{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} } \right) = \left( {\left( {1 - \left( {1 - \left( {\omega _{1} } \right)^{2} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} + 1 - \left( {1 - \left( {\omega _{2} } \right)^{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} - \left( {1 - \left( {1 - \left( {\omega _{1} } \right)^{2} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} } \right) \cdot } \right.} \right.\left. {\left. {\left( {1 - \left( {1 - \left( {\omega _{2} } \right)^{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} } \right)} \right)^{{0.5}} ,\left( {\varpi _{1} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} \cdot \left( {\varpi _{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} } \right) = \left( {\sqrt {1 - \left( {1 - \left( {1 - \left( {1 - \left( {\omega _{1} } \right)^{2} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} } \right)} \right) \cdot \left( {1 - \left( {1 - \left( {1 - \left( {\omega _{2} } \right)^{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} } \right)} \right)} ,} \right.\left. {\left( {\varpi _{1} } \right)^{{Pen(p_{{i1}} ) \cdot \left| {\Phi _{{i1}} - \phi } \right|}} \cdot \left( {\varpi _{2} } \right)^{{Pen(p_{{i2}} ) \cdot \left| {\Phi _{{i2}} - \phi } \right|}} } \right) = \left( {\sqrt {1 - \prod\limits_{{j = 1}}^{2} {\left( {1 - \left( {\omega _{j} } \right)^{2} } \right)^{{Pen(p_{{ij}} ) \cdot \left| {\Phi _{{ij}} - \phi } \right|}} } } ,\prod\limits_{{j = 1}}^{2} {\left( {\varpi _{j} } \right)^{{Pen(p_{{ij}} ) \cdot \left| {\Phi _{{ij}} - \phi } \right|}} } } \right).$$

It can be observed that (43) is fulfilled for n = 2. Suppose that (43) holds for $n = \varsigma$, namely,

$$\Delta_{i}^{\prime \phi } = (\mu_{\Delta i}^{\phi } ,\nu_{\Delta i}^{\phi } ) = \left( {\sqrt {1 - \prod\limits_{j = 1}^{\varsigma } {\left( {1 - \left( {\omega_{j} } \right)^{2} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } ,\prod\limits_{j = 1}^{\varsigma } {\left( {\varpi_{j} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } \right).$$

When $n = \varsigma + 1$, based on Definition 14, it can be acquired that:

$$\Delta_{i}^{\phi } = \left( {\mathop \oplus \limits_{j = 1}^{\varsigma } \left[ {\left( {Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|} \right) \odot w_{j} } \right]} \right) \oplus \left[ {\left( {Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|} \right) \odot w_{\varsigma + 1} } \right].$$

By use of the arithmetic operations with respect to the two-dimensional representation for Pythagorean membership grades, one obtains:

$$\begin{aligned} \Delta_{i}^{\prime \phi } & = \left( {\mathop \oplus \limits_{j = 1}^{\varsigma } \left[ {\left( {Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|} \right) \odot w_{j}^{\prime } } \right]} \right) \oplus \left[ {\left( {Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|} \right) \odot w_{\varsigma + 1}^{\prime } } \right] \\ \, & = \left( {\sqrt {1 - \prod\limits_{j = 1}^{\varsigma } {\left( {1 - \left( {\omega_{j} } \right)^{2} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } ,\prod\limits_{j = 1}^{\varsigma } {\left( {\varpi_{j} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } \right) \\ \, & \oplus \left( {\sqrt {1 - \left( {1 - \left( {\omega_{\varsigma + 1} } \right)^{2} } \right)^{{Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|}} } ,\left( {\varpi_{\varsigma + 1} } \right)^{{Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|}} } \right) \\ \, & = \left( {\left( {1 - \prod\limits_{j = 1}^{\varsigma } {\left( {1 - \left( {\omega_{j} } \right)^{2} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } + 1 - \left( {1 - \left( {1 - \left( {1 - \left( {\omega_{\varsigma + 1} } \right)^{2} } \right)^{{Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|}} } \right)} \right)} \right.} \right. \\ \, & \, \left. { - \left( {1 - \prod\limits_{j = 1}^{\varsigma } {\left( {1 - \left( {\omega_{j} } \right)^{2} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } \right) \cdot \left( {1 - \left( {1 - \left( {1 - \left( {1 - \left( {\omega_{\varsigma + 1} } \right)^{2} } \right)^{{Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|}} } \right)} \right)} \right)} \right)^{0.5} , \\ \, \,\left. {\left( {\prod\limits_{j = 1}^{\varsigma } {\left( {\varpi_{j} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } \right) \cdot \left( {\varpi_{\varsigma + 1} } \right)^{{Pen(p_{i,\varsigma + 1} ) \cdot \left| {\Phi_{i,\varsigma + 1} - \phi } \right|}} } \right) \\ \, & = \left( {\sqrt {1 - \prod\limits_{j = 1}^{\varsigma + 1} {\left( {1 - \left( {\omega_{j} } \right)^{2} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } ,\prod\limits_{j = 1}^{\varsigma + 1} {\left( {\varpi_{j} } \right)^{{Pen(p_{ij} ) \cdot \left| {\Phi_{ij} - \phi } \right|}} } } \right). \\ \end{aligned}$$

It reveals that (43) is satisfied for $n = \varsigma + 1$. As a result, (43) is valid in connection with all n, which confirms the truth of Theorem 9.

1.8 A.8 Proof of Theorem 10

(T10.1) Based on Definition 15 and $0 \le r_{\Delta i}^{\phi } ,d_{\Delta i}^{\phi } \le 1$, it is readily corroborated that $V(\Delta_{i}^{\phi } )$ would be in the range of 0 − 1 (i.e., $0 \le V(\Delta_{i}^{\phi } ) \le 1$).

(T10.2) Concerning necessity, the precondition $V(\Delta_{i}^{\phi } ) = 0$ indicates that $0.5 + r_{\Delta i}^{\phi } \cdot (d_{\Delta i}^{\phi }$$- 0.5) = 0$ (or equivalently, $0.5 + r_{\Delta i}^{\phi } \cdot (0.5 - (2/\pi ) \cdot \theta_{\Delta i}^{\phi } ) = 0$). It is directly validated that $r_{\Delta i}^{\phi } = 1$ and $d_{\Delta i}^{\phi } = 0$ hold due to the conditions $r_{\Delta i}^{\phi } \cdot (d_{\Delta i}^{\phi } - 0.5) = - 0.5$ and $0 \le r_{\Delta i}^{\phi } ,d_{\Delta i}^{\phi } \le 1$. Alternately, the conditions $r_{\Delta i}^{\phi } \cdot (0.5 - (2/\pi ) \cdot \theta_{\Delta i}^{\phi } ) = - 0.5$, $0 \le r_{\Delta i}^{\phi } \le 1$, and $0 \le \theta_{\Delta i}^{\phi } \le \pi /2$ demonstrate the truth of $r_{\Delta i}^{\phi } = 1$ and $\theta_{\Delta i}^{\phi } = \pi /2$. In light of Definition 2, $\mu_{\Delta i}^{\phi } = r_{\Delta i}^{\phi } \cdot \cos (\theta_{\Delta i}^{\phi } ) = 0$ and $\nu_{\Delta i}^{\phi } = r_{\Delta i}^{\phi } \cdot sin(\theta_{\Delta i}^{\phi } ) = 1$. Thus, it can be confirmed that $\Delta_{i}^{\phi } = (0,1;1,0)$. For sufficiency, the condition $\Delta_{i}^{\phi } = (0,1;1,0)$ yields $V(\Delta_{i}^{\phi } ) = 0$ by use of Definition 15. Based on these results, (T10.2) is valid.

(T10.3) For necessity, the precondition $V(\Delta_{i}^{\phi } ) = 1$ indicates that $r_{\Delta i}^{\phi } \cdot (d_{\Delta i}^{\phi } - 0.5) = 0.5$ (or equivalently, $r_{\Delta i}^{\phi } \cdot (0.5 - (2/\pi ) \cdot \theta_{\Delta i}^{\phi } ) = 0.5$). On the basis of Definitions 1 and 2, it is known that $r_{\Delta i}^{\phi } = d_{\Delta i}^{\phi } = 1$ (or equivalently, $r_{\Delta i}^{\phi } = 1$ and $\theta_{\Delta i}^{\phi } = 0$). It is shown that $\Delta_{i}^{\phi } = (1,0;1,1)$ from $\mu_{\Delta i}^{\phi } = r_{\Delta i}^{\phi } \cdot \cos (\theta_{\Delta i}^{\phi } ) = 1$ and $\nu_{\Delta i}^{\phi } = r_{\Delta i}^{\phi } \cdot sin(\theta_{\Delta i}^{\phi } ) = 0$. For sufficiency, it is revealed that $V(\Delta_{i}^{\phi } ) = 1$ when $\Delta_{i}^{\phi } = (1,0;1,1)$. Therefore, (T10.3) is valid.

(T10.4) is known based on Definition 15. In (T10.5), the precondition $\tau_{\Delta i}^{\phi } = 0$ indicates that $r_{\Delta i}^{\phi } = 1$. Thus, $V(\Delta_{i}^{\phi } ) = 0.5 + 1 \cdot (d_{\Delta i}^{\phi } - 0.5) = d_{\Delta i}^{\phi }$, which completes the proof of Theorem 10.

Appendix B: Detailed result tables

Table 1 Computation outcomes related to the PF point operator-oriented upper estimation

Table 2 Computation outcomes related to the PF point operator-oriented lower estimation

Table 3 Results of the penalty weights and their criterion-wise predominating ranks

Table 4 Computation results related to the comprehensive disagreement indicator