Group decision making based on consistency adjustment of distributed preference relations under asymmetric evaluation grades

Abstract

Group decision making (GDM) is common in real world. Distributed preference relation (DPR) is suitable for capturing the original preference information of decision makers (DMs) in uncertain decision-making environment. The consistency of preference relations is very important to ensure the rationality of decision results. Existing research on the consistency of DPRs assumes that evaluation grades are symmetric on the scale. However, some situations such as risk assessment and investment decisions need to be evaluated with asymmetric and non-uniform evaluation grades. DMs with different cognitive abilities and backgrounds may tend to use multi-granular grade sets, differences in their risk attitudes will lead to different meanings of grades. In this paper, a GDM method with multigranular asymmetric evaluation information is developed. First, a general grade score function is proposed to represent both symmetric and asymmetric grade sets selected by DMs. Due to the limited rationality, it is difficult to ensure that DMs’ subjective preferences are perfectly consistent. Therefore, we establish the consistency of DPRs on a set of asymmetric grades. Considering the willingness of DMs and adjustment magnitude, interactive local adjustment and global adjustment algorithms are presented to achieve acceptable consistency for inconsistent DPR matrices. The different grade sets are unified through conversion between grades, which enables DPRs to be aggregated by evidential reasoning (ER) approach. An example of a decision problem involving investment risk is provided, and the results of comparative analysis demonstrate the validity of the proposed method.

Graphical abstract

Appendix

1.1 Proof of Theorem 1

Proof

To show that $s\left({H}_{n}\right)\in [-\mathrm{1,1}]$ for all $n$ in $\{a, \dots ,b\}$, we need to establish the bounds of the function.

• 1. Lower bound:

First, we need to prove that $s\left({H}_{n}\right)\ge -1$ for all $n$ in $\{a, \dots ,b\}$. From Eq. (7), we have $s\left({H}_{n}\right)=\mu n+\tau \mathrm{sin}\omega n$. The partial derivative of $s\left({H}_{n}\right)$ with respect to $n$ is calculated by

$$\frac{\partial s\left({H}_{n}\right)}{\partial n}= \mu +\omega \tau \mathrm{cos}\omega n$$

Since $\mu +\omega \tau \mathrm{cos}\omega n\ge 0$, then $s\left({H}_{n}\right)$ is monotonically increasing with respect to the variable $n$. The minimum value of $s\left({H}_{n}\right)$ occurs when $n=a$, i.e.

$$s\left({H}_{n}\right)\ge s\left({H}_{a}\right).$$

Given $s\left({H}_{a}\right)\in [-\mathrm{1,1}]$, it is clear that

$$\begin{array}{cc}s\left({H}_{n}\right)\ge -1 \text{ for all }n\text{ in }\{a, \dots ,b\}.\end{array}$$

• 2. Upper bound:

Next, we need to prove that $s\left({H}_{n}\right)\le 1$ for all $n$ in $\{a, \dots ,b\}$. Since $\mu +\omega \tau \mathrm{cos}\omega n\ge 0$, the maximum value of $s\left({H}_{n}\right)$ occurs when $n=b$, i.e.

$$s\left({H}_{n}\right)\le s\left({H}_{b}\right).$$

Given $s\left({H}_{b}\right)\in [-\mathrm{1,1}]$, it is clear that

$$\begin{array}{cc}s\left({H}_{n}\right)\le 1 \text{ for all }n\text{ in }\{a, \dots ,b\}.\end{array}$$

From the lower and upper bounds proofs, we have shown that $s\left({H}_{n}\right)$ is bounded between -1 and 1 for all $n\in \{a, \dots ,b\}$. Therefore, we conclude that $s\left({H}_{n}\right)\in [-\mathrm{1,1}]$ based on Definition 7.

1.2 Proof of Proposition 2

Proof

From Eqs. (17) and (18), we can easily deduce the following results:

For property (1) in Property 2.

If ${r}_{ik}(\mathrm{or} {r}_{kj})=(\mathrm{0,0})$, then

$$\begin{array}{l}{\alpha }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)={\alpha }_{ik}+{\alpha }_{kj}={\alpha }_{kj}\left(\mathrm{or} {\alpha }_{ik}\right),\\ {\beta }_{ij}=\overline{f }\left({\beta }_{ik},{\beta }_{kj}\right)={\beta }_{ik}+{\beta }_{kj}={\beta }_{kj}\left(\mathrm{or} {\beta }_{ik}\right).\end{array}$$

For property (2) in Property 2.

If ${\alpha }_{ik}+{\beta }_{ik}>(\ge )0$ and ${\alpha }_{kj}+{\beta }_{kj}\ge (>)0$, then

$$\begin{array}{l}{\alpha }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)\\ =max\left\{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj},s\left({H}_{{N}_{2}}\right)\right\}\\ =max\left\{{\alpha }_{ik}+\left(1-{\alpha }_{ik}\right){\alpha }_{kj},s\left({H}_{{N}_{2}}\right)\right\}\\ =max\left\{{\alpha }_{ik}\left(1-{\alpha }_{kj}\right)+{\alpha }_{kj},s\left({H}_{{N}_{2}}\right)\right\}.\end{array}$$

Since ${\alpha }_{ik}+\left(1-{\alpha }_{ik}\right){\alpha }_{kj}\ge {\alpha }_{ik}$ and ${\alpha }_{ik}\left(1-{\alpha }_{kj}\right)+{\alpha }_{kj}\ge {\alpha }_{kj}$, then $max\left\{{\alpha }_{ik},{\alpha }_{kj}\right\}\le {\alpha }_{ij}\le s\left({H}_{{N}_{2}}\right)$.

$${\beta }_{ij}=\overline{f }\left({\beta }_{ik},{\beta }_{kj}\right)={-\beta }_{ik}{\beta }_{kj}\ge max({\beta }_{ik},{\beta }_{kj})$$

Hence, ${\alpha }_{ij}+{\beta }_{ij}\ge max\left\{{\alpha }_{ik}+{\beta }_{ik},{\alpha }_{kj}+{\beta }_{kj}\right\}$. Property (3) can be similarly verified.

For properties (4) and (5) in Property 2.

If ${\alpha }_{ik}=s\left({H}_{{N}_{2}}\right)\left({\alpha }_{kj}=s\left({H}_{{N}_{2}}\right)\right)$ and ${\alpha }_{kj}+{\beta }_{kj}\ge 0({\alpha }_{ik}+{\beta }_{ik}\ge 0)$, then

$$\begin{array}{l}\alpha_{ij}=min\{\alpha_{ik}+\alpha_{kj}-\alpha_{ik}\alpha_{kj},{s(H}_{N_2})\}\\\;\;\;\;\;=min\left\{s\left(H_{N_2}\right)+\alpha_{kj}-s\left(H_{N_2}\right)\cdot\alpha_{kj},{s(H}_{N_2})\right\}\left(min\left\{\alpha_{ik}+s\left(H_{N_2}\right)-\alpha_{ik}\cdot s\left(H_{N_2}\right),{s(H}_{N_2})\right\}\right)\\\;\;\;\;\;=s\left(H_{N_2}\right).\end{array}$$

If ${\beta }_{ik}=s({H}_{-{N}_{1}})({\beta }_{kj}=s({H}_{-{N}_{1}}))$ and ${\alpha }_{kj}+{\beta }_{kj}\le 0({\alpha }_{ik}+{\beta }_{ik}\le 0)$, then

$$\begin{array}{l}\beta_{ij}=max\{\beta_{ik}+\beta_{kj}+\beta_{ik}\beta_{kj},{s(H}_{{-N}_1})\}\\\;\;\;\;\;=max\left\{{s(H}_{{-N}_1})+\beta_{kj}+{s(H}_{{-N}_1})\cdot\beta_{kj},{s(H}_{{-N}_1})\right\}\left(min\left\{\beta_{ik}+{s(H}_{{-N}_1})+\beta_{ik}\cdot{s(H}_{{-N}_1}),{s(H}_{{-N}_1})\right\}\right)\\\;\;\;\;\;=s\left(H_{-N_1}\right).\end{array}$$

For properties (6) and (7) in Property 2.

To show that the function $\widehat{f}(\overline{f })$ is monotonically increasing with respect to ${\alpha }_{ik}$ and ${\alpha }_{kj}$(${\beta }_{ik}$ and ${\beta }_{kj}$), we need to demonstrate that for any given values of ${\alpha }_{ik}$ and ${\alpha }_{kj}$(${\beta }_{ik}$ and ${\beta }_{kj}$), if we increase these values, the resulting value of $\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)\left(\left.\overline{f }\left({\beta }_{ik},{\beta }_{kj}\right)\right)\right.$ also increases.

To prove this, we will consider different cases based on the given definition of $\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)$:

• Case 1: ${\alpha }_{ik}+{\beta }_{ik}>\left(\ge \right)0$ and ${\alpha }_{kj}+{\beta }_{kj}\ge \left(>\right)0$

When ${\alpha }_{kj}>{\alpha }_{kj}^{\prime}\left({\alpha }_{ik}>{\alpha }_{ik}^{\prime}\right)$ and ${0<\alpha }_{ik}\left({\alpha }_{kj}\right)<s\left({H}_{{N}_{2}}\right)$, then ${\alpha }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)=min\{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj},{s(H}_{{N}_{2}})\}={\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}$.

If ${\alpha }_{kj}>{\alpha }_{kj}^{\prime}$ and ${0<\alpha }_{ik}<s\left({H}_{{N}_{2}}\right)$

$$\begin{array}{l}{\alpha }_{ij}-{\alpha }_{ij}^{\prime}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)-\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}^{\prime}\right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =\left({\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}\right)-\left({\alpha }_{ik}+{\alpha }_{kj}^{\prime}-{\alpha }_{ik}{\alpha }_{kj}^{\prime}\right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;={\alpha }_{kj}-{\alpha }_{kj}^{\prime}-{\alpha }_{ik}\left({\alpha }_{kj}-{\alpha }_{kj}^{\prime}\right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\left(1-{\alpha }_{kj}\right)\left({\alpha }_{kj}-{\alpha }_{kj}^{\prime}\right)>0.\end{array}$$

The same conclusion can be similarly drawn when ${\alpha }_{ik}>{\alpha }_{ik}^{\prime}$ and $0<{\alpha }_{kj}<s\left({H}_{{N}_{2}}\right)$.

• Case 2: ${\alpha }_{ik}+{\beta }_{ik}<\left(\le \right)0$ and ${\alpha }_{kj}+{\beta }_{kj}\le \left(<\right)0$

If ${\alpha }_{kj}>{\alpha }_{kj}^{\prime}$ and ${0<\alpha }_{ik}<s\left({H}_{{N}_{2}}\right)$ 

$${\alpha }_{ij}-{\alpha }_{ij}^{\prime}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)-\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}^{\prime}\right)={\alpha }_{ik}{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}^{\prime}={\alpha }_{ik}\left({\alpha }_{kj}-{\alpha }_{kj}^{\prime}\right)>0$$

Similarly, it can be concluded that ${\alpha }_{ij}>{\alpha }_{ij}^{\prime}$ when ${\alpha }_{ik}>{\alpha }_{ik}^{\prime}$ and $0<{\alpha }_{kj}<s\left({H}_{{N}_{2}}\right)$.

• Case 3: $\left({\alpha }_{ik},{\beta }_{ik}\right)$ or $({\alpha }_{kj},{\beta }_{kj})=\left(\mathrm{0,0}\right)$

If ${\alpha }_{kj}>{\alpha }_{kj}^{\prime}$ and  ${0<\alpha }_{ik}<s\left({H}_{{N}_{2}}\right)$ 

$${\alpha }_{ij}-{\alpha }_{ij}^{\prime}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)-\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}^{\prime}\right)=\left({\alpha }_{ik}+{\alpha }_{kj}\right)-\left({\alpha }_{ik}+{\alpha }_{kj}^{\prime}\right)={\alpha }_{kj}-{\alpha }_{kj}^{\prime}>0$$

If ${\alpha }_{ik}>{\alpha }_{ik}^{\prime}$ and $0<{\alpha }_{kj}<s\left({H}_{{N}_{2}}\right)$, ${\alpha }_{ij}-{\alpha }_{ij}^{\prime}={\alpha }_{ik}-{\alpha }_{ik}^{\prime}>0$.

• Case 4: $({\alpha }_{ik}+{\beta }_{ik})({\alpha }_{kj}+{\beta }_{kj})<0$ or other situations.

If ${\alpha }_{kj}>{\alpha }_{kj}^{\prime}$ and${0<\alpha }_{ik}<s\left({H}_{{N}_{2}}\right)$ ,

$${\alpha }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)=\frac{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}$$

For properties (6) and (7) in PropertyGiven ${\alpha }_{ik}$, the partial derivative of $\widehat{f}$ with respect to ${\alpha }_{kj}$ is calculated by

$$\begin{array}{l}\frac{\partial \widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)}{\partial {\alpha }_{kj}}=\frac{\left(1-{\alpha }_{ik}\right)\left(2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)\right)-(1-2{\alpha }_{ik})\left({\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}\right)}{{\left(2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)\right)}^{2}}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1-{\alpha }_{ik}+{{\alpha }_{ik}}^{2}}{{\left(2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)\right)}^{2}}>0.\end{array}$$

Thus, we can deduce that ${\alpha }_{ij}>{\alpha }_{ij}^{\prime}$. Similarly, ${\alpha }_{ij}>{\alpha }_{ij}^{\prime}$ holds when ${\alpha }_{ik}>{\alpha }_{ik}^{\prime}$ and $0<{\alpha }_{kj}<s\left({H}_{{N}_{2}}\right)$.

• Case 5: ${\alpha }_{ik}=-{\beta }_{kj}$ and ${\beta }_{ik}=-{\alpha }_{kj}$

When ${\alpha }_{kj}>{\alpha }_{kj}^{\prime}$ and ${0<\alpha }_{ik}<s\left({H}_{{N}_{2}}\right)$, as ${\alpha }_{ij}+{\beta }_{ij}=0$, if ${\alpha }_{ik}+ {\beta }_{ik}\le 0$, then ${\alpha }_{kj}^{\prime}+{\beta }_{ik}<0$, ${\alpha }_{ij}-{\alpha }_{ij}^{\prime}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)-\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}^{\prime}\right)={\alpha }_{ik}\left({\alpha }_{kj}-{\alpha }_{kj}^{\prime}\right)>0$.

If ${\alpha }_{ik}+ {\beta }_{ik}>0$ and ${\alpha }_{kj}^{\prime}+{\beta }_{ik}=0$,

$${\alpha }_{ij}-{\alpha }_{ij}^{\prime}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)-\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}^{\prime}\right)=\left({\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}\right)-\left({\alpha }_{ik}+{\alpha }_{kj}^{\prime}-{\alpha }_{ik}{\alpha }_{kj}^{\prime}\right)>0.$$

If ${\alpha }_{ik}+ {\beta }_{ik}>0$ and ${\alpha }_{kj}^{\prime}+{\beta }_{ik}<0$,

$$\begin{array}{l}{\alpha }_{ij}-{\alpha }_{ij}^{\prime}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)-\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}^{\prime}\right)\\ =\frac{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}-\frac{{\alpha }_{ik}+{\alpha }_{kj}^{\prime}-{\alpha }_{ik}{\alpha }_{kj}^{\prime}}{2-{\alpha }_{ik}{\alpha }_{kj}^{\prime}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}^{\prime}\right)}.\end{array}$$

From case 4, it is clear that ${\alpha }_{ij}>{\alpha }_{ij}^{\prime}$. The same conclusion can be similarly drawn when ${\alpha }_{ik}>{\alpha }_{ik}^{\prime}$ and $0<{\alpha }_{kj}<s\left({H}_{{N}_{2}}\right)$.

In all the cases considered, we have shown that $\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)$ is monotonically increasing with respect to ${\alpha }_{ik}$. As swapping ${\alpha }_{ik}$ and ${\alpha }_{kj}$ does not change the value of $\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)$, the function $\widehat{f}$ is symmetric. Therefore, $\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)$ is monotonically increasing with respect to ${\alpha }_{kj}$. The function $\overline{f }$ is monotonically increasing with respect to ${\beta }_{ik}$ and ${\beta }_{kj}$ can be proved similarly.

For property (8) in Property 2.

If ${\alpha }_{ik}+{\beta }_{ik}={\alpha }_{kj}+{\beta }_{kj}=0$, then

$$\begin{array}{l}{\alpha }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)=\frac{{\alpha }_{ik}{+\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{{2-\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}\\ {\beta }_{ij}=\overline{f }\left({\beta }_{ik},{\beta }_{kj}\right)= \frac{{\beta }_{ik}{+\beta }_{kj}+{\beta }_{ik}{\beta }_{kj}}{{2-\beta }_{ik}{\beta }_{kj}-\left(1+{\beta }_{ik}\right)\left(1+{\beta }_{kj}\right)}=-\frac{{\alpha }_{ik}{+\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{{2-\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}\end{array}$$

Because $\widehat{f}$ and $\overline{f }$ are monotonically increasing, then ${\alpha }_{ij}+{\beta }_{ij}=0$, $min\{{\alpha }_{ik},{\alpha }_{kj}\}\le {\alpha }_{ij}\le 0.5$ and $-0.5\le {\beta }_{ij}\le max\{{\beta }_{ik}, {\beta }_{kj}\}$.

For property (9) in Property 2.

If ${\alpha }_{ik}=-{\beta }_{kj}$ and ${\beta }_{ik}=-{\alpha }_{kj}$, then ${\alpha }_{ik}+{\beta }_{ik}=-({\alpha }_{kj}+{\beta }_{kj})$. If ${\alpha }_{ik}+{\beta }_{ik}=0$, ${\alpha }_{kj}+{\beta }_{kj}=0$; if ${\alpha }_{ik}+{\beta }_{ik}>0$, ${\alpha }_{kj}+{\beta }_{kj}<0$; if ${\alpha }_{ik}+{\beta }_{ik}<0$, ${\alpha }_{kj}+{\beta }_{kj}>0$. Then, we have

$$\begin{array}{l}{\alpha }_{ij}+{\beta }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)+\overline{f }\left({\beta }_{ik},{\beta }_{kj}\right)\\ =\frac{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}+\frac{{\beta }_{ik}{+\beta }_{kj}+{\beta }_{ik}{\beta }_{kj}}{{2-\beta }_{ik}{\beta }_{kj}-\left(1+{\beta }_{ik}\right)\left(1+{\beta }_{kj}\right)}\\ =\frac{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}-\frac{{\alpha }_{ik}+{\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{2-{\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}\\ =0\end{array}$$

For property (10) in Property 2.

If $\left({\alpha }_{ik}+{\beta }_{ik}\right)\left({\alpha }_{kj}+{\beta }_{kj}\right)<0$ and $\left({\alpha }_{ik}+{\beta }_{ik}\right)+\left({\alpha }_{kj}+{\beta }_{kj}\right)>0$, we have $\left|{\alpha }_{ik}\right|+\left|{\alpha }_{kj}\right|>\left|{\beta }_{ik}\right|+\left|{\beta }_{kj}\right|$. Since $\widehat{f}$ and $\overline{f }$ are monotonically increasing, then

$$\begin{array}{l}{\alpha }_{ij}+{\beta }_{ij}=\widehat{f}\left({\alpha }_{ik},{\alpha }_{kj}\right)+\overline{f }\left({\beta }_{ik},{\beta }_{kj}\right)\\ =\frac{{\alpha }_{ik}{+\alpha }_{kj}-{\alpha }_{ik}{\alpha }_{kj}}{{2-\alpha }_{ik}{\alpha }_{kj}-\left(1-{\alpha }_{ik}\right)\left(1-{\alpha }_{kj}\right)}+\frac{{\beta }_{ik}{+\beta }_{kj}+{\beta }_{ik}{\beta }_{kj}}{{2-\beta }_{ik}{\beta }_{kj}-\left(1+{\beta }_{ik}\right)\left(1+{\beta }_{kj}\right)}>0\end{array}$$

Similarly, property (11) in Property 2 is proved.