Ranking of single-valued neutrosophic numbers through the index of optimism and its reasonable properties

Abstract

In this paper an innovative method of ranking neutrosophic number based on the notions of value and ambiguity of a single-valued neutrosophic number is being developed. The method is based on the convex combination of value and ambiguity of truth-membership function with the sum of values and ambiguities of indeterminacy-membership and falsity-membership functions. This convex combination is also termed as an index of optimism. The index of optimism, $\lambda =1$, is termed as optimistic decision-maker as it considers the value and the ambiguity of the truth-membership function, ignoring the contributions from indeterminacy-membership and falsity-membership functions. Similarly, the index of optimism, $\lambda =0$, is termed as pessimistic decision-maker as it considers the values and the ambiguities of the indeterminacy-membership and falsity-membership functions, ignoring the contribution from truth-membership function. Further, the index of optimism, $\lambda =0.5$, is termed as moderate decision-maker as it considers the values and the ambiguities of all the membership functions. The approach is a novel as it completely oath to follow the reasonable properties of a ranking method. It is worth to mention that the current approach consistently ranks the single-valued neutrosophic numbers as well as their corresponding images.

Proofs of the theorems

1.1 Proof of the Theorem 3.2

The proof of the above statements are as follows.

• 1. The proof of this statement is followed immediately.

• 2. Consider the cases when $\widetilde{a}\succ \widetilde{b}$ happens, that is,

  $$\begin{aligned} \widetilde{a}\succ \widetilde{b}\,\text {happens for}\, {\left\{ \begin{array}{ll} &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,\pm 1) \end{array}\right. }. \end{aligned}$$

  Consider the cases when $\widetilde{b}\succ \widetilde{c}$ happens, that is,

  $$\begin{aligned} \widetilde{b}\succ \widetilde{c}\,\text {happens for}\, {\left\{ \begin{array}{ll} &{}{\mathcal {R}}_{\lambda }(\widetilde{b},0,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{c},0,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},0,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{c},0,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{c},0,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{c},0,\pm 1,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,\pm 1,\pm 1) \end{array}\right. } \end{aligned}$$

  Now, to proof this property, it need to discuss these $8\times 8$ cases, which will make the proof tedious. However, one can see the proof trivially, if following claims can be established. Claim 1: Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=0$ in ordering $\widetilde{b}$ and $\widetilde{c}$. Then $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{c}$. The proof of this claim is as follows. Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=0$ in ordering $\widetilde{b}$ and $\widetilde{c}$, then ${\mathcal {V}}(\mu _{\widetilde{b}})\ne {\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{b}})\ne {\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{b}})\ne {\mathcal {V}}(\nu _{\widetilde{c}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{c}})$. So, $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{c}$. Claim 2: Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=\pm 1$ in ordering $\widetilde{b}$ and $\widetilde{c}$. Then $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{c}$. The proof of this claim is as follows. Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=\pm 1$ in ordering $\widetilde{b}$ and $\widetilde{c}$, then ${\mathcal {V}}(\mu _{\widetilde{b}})={\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{b}})={\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{b}})={\mathcal {V}}(\nu _{\widetilde{c}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{c}})$. So, $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{c}$. Claim 3: Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=\pm 1$ in ordering $\widetilde{b}$ and $\widetilde{c}$. then $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{c}$. The proof of this claim is as follows. Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=\pm 1$ in ordering $\widetilde{b}$ and $\widetilde{c}$, then ${\mathcal {V}}(\mu _{\widetilde{b}})={\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{b}})={\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{b}})={\mathcal {V}}(\nu _{\widetilde{c}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{c}})$. So, $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{c}$. Claim 4: Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=0$ in ordering $\widetilde{b}$ and $\widetilde{c}$. Then $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{c}$. The proof of this claim is as follows. Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=0$ in ordering $\widetilde{b}$ and $\widetilde{c}$, then ${\mathcal {V}}(\mu _{\widetilde{b}})\ne {\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{b}})\ne {\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{b}})\ne {\mathcal {V}}(\nu _{\widetilde{c}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{c}})$. So, $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{c}$. From these four claims, it is trivial enough to show that if $\widetilde{a}\succ \widetilde{b}$ and $\widetilde{b}\succ \widetilde{c}$, then $\widetilde{a}\succ \widetilde{c}$. Further, from the definition of $\succeq$, it follows that transitivity also holds for the order relation $\succeq$.

• 3. This statement is followed immediately, as the order relations $\succ$ and $\sim$ particularly based on order relation > and $=$ of real numbers.

• 4. If $\widetilde{a}=\widetilde{b}$, then ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)={\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$. Thus, the statement is followed.

1.2 Proof of the Theorem 3.5

The proof of this theorem, follows immediately if the invariance of $\theta _i$ in ordering $\widetilde{a}$, $\widetilde{b}$ and $\widetilde{a}+\widetilde{c}$, $\widetilde{b}+\widetilde{c}$ can be established. Hence, a claim has to be made. The claim is as follows.

Claim : The value of $\theta _i$ in ordering $\widetilde{a}$ and $\widetilde{b}$ are invariant in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$. The proof of the claim follows from the following eight cases:

Case 1::

Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 2::

Let $\theta _1=0$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _1=0$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 3::

Let $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 4::

Let $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})={\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 5::

Let $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 6::

Let $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 7::

Let $\theta _1=\pm 1$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _1=\pm 1$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

Case 8::

Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})= {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{c}})$. So, $\theta _i=\pm 1$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{c}$.

The above eight cases suggest that $\theta _1$ and $\theta _2$ are invariant in ordering $\widetilde{a}$, $\widetilde{b}$ and $\widetilde{a}+\widetilde{c}$, $\widetilde{b}+\widetilde{c}$. Hence, the claim.

Now, by the Theorem 3.1 it follows that

$$\begin{aligned} {\mathcal {R}}_{\lambda }(\widetilde{a}+\widetilde{c},\theta _1,\theta _2,\theta _3)={\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(\widetilde{c},\theta _1,\theta _2,\theta _3), \end{aligned}$$

and

$$\begin{aligned} {\mathcal {R}}_{\lambda }(\widetilde{b}+\widetilde{c},\theta _1,\theta _2,\theta _3)={\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(\widetilde{c},\theta _1,\theta _2,\theta _3), \end{aligned}$$

Hence, if $\widetilde{a}\succeq \widetilde{b}$, then it is obvious that ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$. This leads to the inequality ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(\widetilde{c},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(\widetilde{c},\theta _1,\theta _2,\theta _3)$, which evidently follows the inequality ${\mathcal {R}}_{\lambda }(\widetilde{a}+\widetilde{c},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b}+\widetilde{c},\theta _1,\theta _2,\theta _3)$. Thus, the result follows immediately.

1.3 Proof of the Theorem 3.9

Let $\widetilde{a}\succeq \widetilde{b}$. Then ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$. Let $k>0$. Then using the Proposition 3.4, it follows that

$$\begin{aligned} {\mathcal {R}}_{\lambda }(k\widetilde{a},\theta _1,\theta _2,\theta _3)=&\lambda \{{\mathcal {V}}(\mu _{k\widetilde{a}})+\theta _1{\mathcal {A}}(\mu _{k\widetilde{a}})\}&\\&+(1-\lambda )\{{\mathcal {V}}(\rho _{k\widetilde{a}})+\theta _2{\mathcal {A}}(\rho _{k\widetilde{a}})+{\mathcal {V}}(\nu _{k\widetilde{a}})+\theta _3{\mathcal {A}}(\nu _{k\widetilde{a}})\}&\\ =&k\lambda \{{\mathcal {V}}(\mu _{\widetilde{a}})+\theta _1{\mathcal {A}}(\mu _{\widetilde{a}})\}&\\&+k(1-\lambda )\{{\mathcal {V}}(\rho _{\widetilde{a}})+\theta _2{\mathcal {A}}(\rho _{\widetilde{a}})+{\mathcal {V}}(\nu _{\widetilde{a}})+\theta _3{\mathcal {A}}(\nu _{\widetilde{a}})\}&\\ =&k{\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3).&\end{aligned}$$

Thus, when $\widetilde{a}\succeq \widetilde{b}$, it follows that ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$. Equivalently, it follows that $k{\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)\ge k{\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$, which can be trivially expressed as ${\mathcal {R}}_{\lambda }(k\widetilde{a},\theta _1,\theta _2,\theta _3) \ge {\mathcal {R}}_{\lambda }(k\widetilde{b},\theta _1,\theta _2,\theta _3)$. So, the result, $k\widetilde{a}\succeq k\widetilde{b}$, follows immediately.

Let $k<0$, assume $k=-m<0$, then the following cases arise.

Case 1::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _i=0$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},0,0)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},0,0)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _i=0$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,0,0)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,0,0)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 2::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _1=0$, $\theta _2=0$ and $\theta _3=\pm 1$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},0,0,\pm 1)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},0,0,\pm 1)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\pm {\mathcal {A}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})\pm {\mathcal {A}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _1=0$, $\theta _2=0$ and $\theta _3=\mp 1$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\mp {\mathcal {A}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})\mp {\mathcal {A}}(\nu _{-m\widetilde{a}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,0,\mp 1)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,0,\mp 1)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 3::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=0$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},0,\pm 1,0)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,0)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\pm {\mathcal {A}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})\pm {\mathcal {A}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _1=0$, $\theta _2=\mp 1$ and $\theta _3=0$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\mp {\mathcal {A}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})\mp {\mathcal {A}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,\mp 1,0)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,\mp 1,0)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 4::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=\pm 1$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},0,\pm 1,\pm 1)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,\pm 1)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\pm {\mathcal {A}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})\pm {\mathcal {A}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\pm {\mathcal {A}}(\nu _{\widetilde{a}}) \ge {\mathcal {V}}(\nu _{\widetilde{b}})\pm {\mathcal {A}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _1=0$, $\theta _2=\mp 1$ and $\theta _3=\mp 1$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\mp {\mathcal {A}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})\mp {\mathcal {A}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\mp {\mathcal {A}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})\mp {\mathcal {A}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,\mp 1,\mp 1)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,\mp 1,\mp 1)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 5::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=0$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,0,0)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,0)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\pm {\mathcal {A}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})\pm {\mathcal {A}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _1=\mp 1$, $\theta _2=0$ and $\theta _3=0$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\mp {\mathcal {A}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})\mp {\mathcal {A}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\mp 1,0,0)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},\mp 1,0,0)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 6::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=\pm 1$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,0,\pm 1)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,\pm 1)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\pm {\mathcal {A}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})\pm {\mathcal {A}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\pm {\mathcal {A}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})\pm {\mathcal {A}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _1=\mp 1$, $\theta _2=0$ and $\theta _3=\mp 1$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\mp {\mathcal {A}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})\mp {\mathcal {A}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\mp {\mathcal {A}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})\mp {\mathcal {A}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\mp 1,0,\mp 1)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},\mp 1,0,\mp 1)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 7::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _1=\pm 1$, $\theta _2=\pm 1$ and $\theta _3=0$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,\pm 1,0)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,0)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\pm {\mathcal {A}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})\pm {\mathcal {A}}(\mu _{\widetilde{a}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\pm {\mathcal {A}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})\pm {\mathcal {A}}(\rho _{\widetilde{a}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _1=\mp 1$, $\theta _2=\mp 1$ and $\theta _3=0$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\mp {\mathcal {A}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})\mp {\mathcal {A}}(\mu _{-m\widetilde{a}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\mp {\mathcal {A}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})\mp {\mathcal {A}}(\rho _{-m\widetilde{a}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\mp 1,\mp 1,0)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},\mp 1,\mp 1,0)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

Case 8::

Let $\widetilde{a}\succeq \widetilde{b}$ for $\theta _i=\pm 1$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,\pm 1,\pm 1)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,\pm 1)$. Now, as $\widetilde{a}\succeq \widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\pm {\mathcal {A}}(\mu _{\widetilde{a}})\ge {\mathcal {V}}(\mu _{\widetilde{b}})\pm {\mathcal {A}}(\mu _{\widetilde{a}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\pm {\mathcal {A}}(\rho _{\widetilde{a}})\ge {\mathcal {V}}(\rho _{\widetilde{b}})\pm {\mathcal {A}}(\rho _{\widetilde{a}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\pm {\mathcal {A}}(\nu _{\widetilde{a}})\ge {\mathcal {V}}(\nu _{\widetilde{b}})\pm {\mathcal {A}}(\nu _{\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$. Thus, $\theta _i=\mp 1$ in ordering $-m\widetilde{a}$ and $-m\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\mp {\mathcal {A}}(\mu _{-m\widetilde{a}})\le {\mathcal {V}}(\mu _{-m\widetilde{b}})\mp {\mathcal {A}}(\mu _{-m\widetilde{a}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\mp {\mathcal {A}}(\rho _{-m\widetilde{a}})\le {\mathcal {V}}(\rho _{-m\widetilde{b}})\mp {\mathcal {A}}(\rho _{-m\widetilde{a}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\mp {\mathcal {A}}(\nu _{-m\widetilde{a}})\le {\mathcal {V}}(\nu _{-m\widetilde{b}})\mp {\mathcal {A}}(\nu _{-m\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\mp 1,\mp 1,\mp 1)\le {\mathcal {R}}_{\lambda }(-m\widetilde{b},\mp 1,\mp 1,\mp 1)$. Hence, the result $-m \widetilde{a}\preceq -m \widetilde{b}$ follows immediately.

1.4 Proof of the Theorem 3.10

Let $k>0$ and $k\widetilde{a}\succeq k\widetilde{b}$. Then ${\mathcal {R}}_{\lambda }(k\widetilde{a},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(k\widetilde{b},\theta _1,\theta _2,\theta _3)$. However, by Proposition 3.4, it follows that $k{\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)\ge k{\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$. Thus, the result follows immediately. If $k<0$, let $k=-m<0$, then $-m\widetilde{a}\succeq -m\widetilde{b}$ implies that ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},\theta _1,\theta _2,\theta _3)$. Now, eight cases arise.

Case 1::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _i=0$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,0,0)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,0,0)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})$and ${\mathcal {V}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},0,0,0)\le {\mathcal {R}}_{\lambda }(\widetilde{b},0,0,0)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 2::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _1=0$, $\theta _2=0$ and $\theta _3=\pm 1$ then ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,0,\pm 1)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,0,\pm 1)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\pm {\mathcal {A}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})\pm {\mathcal {A}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _1=0$, $\theta _2=0$ and $\theta _3=\mp 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})$and ${\mathcal {V}}(\nu _{\widetilde{a}})\mp {\mathcal {A}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})\mp {\mathcal {A}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},0,0,\mp 1)\le {\mathcal {R}}_{\lambda }(\widetilde{b},0,0,\mp 1)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 3::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=0$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,\pm 1,0)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,\pm 1,0)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})\pm {\mathcal {A}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _1=0$, $\theta _2=\mp 1$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})\mp {\mathcal {A}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},0,\mp 1,0)\le {\mathcal {R}}_{\lambda }(\widetilde{b},0,\mp 1,0)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 4::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=\pm 1$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ne {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},0,\pm 1,\pm 1)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},0,\pm 1,\pm 1)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\pm {\mathcal {A}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})\pm {\mathcal {A}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _1=0$, $\theta _2=\mp 1$ and $\theta _3=\mp 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\mp {\mathcal {A}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})\mp {\mathcal {A}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},0,\mp 1,\mp 1)\le {\mathcal {R}}_{\lambda }(\widetilde{b},0,\mp 1,\mp 1)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 5::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=0$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\pm 1,0,0)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},\pm 1,0,0)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})\pm {\mathcal {A}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _1=\mp 1$, $\theta _2=0$ and $\theta _3=0$, in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})\mp {\mathcal {A}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})$and ${\mathcal {V}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},\mp 1,0,0)\le {\mathcal {R}}_{\lambda }(\widetilde{b},\mp 1,0,0)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 6::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=\pm 1$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ne {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\pm 1,0,\pm 1)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},\pm 1,0,\pm 1)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\pm {\mathcal {A}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})\pm {\mathcal {A}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _1=\mp 1$, $\theta _2=0$ and $\theta _3=\mp 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\mp {\mathcal {A}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})\mp {\mathcal {A}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},\mp 1,0,\mp 1)\le {\mathcal {R}}_{\lambda }(\widetilde{b},\mp 1,0,\mp 1)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 7::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _1=\pm 1$, $\theta _2=\pm 1$ and $\theta _3=0$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ne {\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\pm 1,\pm 1,0)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},\pm 1,\pm 1,0)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _1=\mp 1$, $\theta _2=\mp 1$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},\mp 1,\mp 1,0)\le {\mathcal {R}}_{\lambda }(\widetilde{b},\mp 1,\mp 1,0)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

Case 8::

Let $-m\widetilde{a}\succeq -m\widetilde{b}$ for $\theta _i=\pm 1$. Then ${\mathcal {V}}(\mu _{-m\widetilde{a}})={\mathcal {V}}(\mu _{-m\widetilde{b}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})={\mathcal {V}}(\rho _{-m\widetilde{b}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})={\mathcal {V}}(\nu _{-m\widetilde{b}})$ and ${\mathcal {R}}_{\lambda }(-m\widetilde{a},\pm 1,\pm 1,\pm 1)\ge {\mathcal {R}}_{\lambda }(-m\widetilde{b},\pm 1,\pm 1,\pm 1)$. Now, as $-m\widetilde{a}\succeq -m\widetilde{b}$ it follows that ${\mathcal {V}}(\mu _{-m\widetilde{a}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})\ge {\mathcal {V}}(\mu _{-m\widetilde{b}})\pm {\mathcal {A}}(\mu _{-m\widetilde{a}})$, ${\mathcal {V}}(\rho _{-m\widetilde{a}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})\ge {\mathcal {V}}(\rho _{-m\widetilde{b}})\pm {\mathcal {A}}(\rho _{-m\widetilde{a}})$ and ${\mathcal {V}}(\nu _{-m\widetilde{a}})\pm {\mathcal {A}}(\nu _{-m\widetilde{a}})\ge {\mathcal {V}}(\nu _{-m\widetilde{b}})\pm {\mathcal {A}}(\nu _{-m\widetilde{b}})$. Clearly, ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, $\theta _i=\mp 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Further, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})\le {\mathcal {V}}(\mu _{\widetilde{b}})\mp {\mathcal {A}}(\mu _{\widetilde{a}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})\le {\mathcal {V}}(\rho _{\widetilde{b}})\mp {\mathcal {A}}(\rho _{\widetilde{a}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\mp {\mathcal {A}}(\nu _{\widetilde{a}})\le {\mathcal {V}}(\nu _{\widetilde{b}})\mp {\mathcal {A}}(\nu _{\widetilde{b}})$. So, ${\mathcal {R}}_{\lambda }(\widetilde{a},\mp 1,\mp 1,\mp 1)\le {\mathcal {R}}_{\lambda }(\widetilde{b},\mp 1,\mp 1,\mp 1)$. Hence, the result $\widetilde{a}\preceq \widetilde{b}$ follows immediately.

1.5 Proof of the Theorem 3.13

The proof of this theorem, follows immediately if the invariance of $\theta _1$ and $\theta _2$ in ordering $\widetilde{a}$, $\widetilde{b}$ and $\widetilde{a}-\widetilde{c}$, $\widetilde{b}-\widetilde{c}$ can be established. Hence, a claim has to be made. The claim is as follows.

Claim : The value of $\theta _1$ and $\theta _2$ in ordering $\widetilde{a}$ and $\widetilde{b}$ are invariant in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$. The proof of the claim follows from the following eight cases:

Case 1::

Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _i=0$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 2::

Let $\theta _1=0$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _1=0$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 3::

Let $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 4::

Let $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})={\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _1=0$, $\theta _2=\pm 1$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 5::

Let $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=0$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 6::

Let $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _1=\pm 1$, $\theta _2=0$ and $\theta _3=\pm 1$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 7::

Let $\theta _1=\pm 1$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _1=\pm 1$, $\theta _2=\pm 1$ and $\theta _3=0$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

Case 8::

Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then, ${\mathcal {V}}(\mu _{\widetilde{a}})= {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})= {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\mu _{\widetilde{b}-\widetilde{c}})$, ${\mathcal {V}}(\rho _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\rho _{\widetilde{b}-\widetilde{c}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}-\widetilde{c}})= {\mathcal {V}}(\nu _{\widetilde{b}-\widetilde{c}})$. So, $\theta _i=\pm 1$ in ordering $\widetilde{a}-\widetilde{c}$ and $\widetilde{b}-\widetilde{c}$.

The above eight cases suggest that $\theta _1$ and $\theta _2$ are invariant in ordering $\widetilde{a}$, $\widetilde{b}$ and $\widetilde{a}-\widetilde{c}$, $\widetilde{b}-\widetilde{c}$. Hence, the claim.

Now, by the Theorem 3.1 it follows that it follows that

$$\begin{aligned} {\mathcal {R}}_{\lambda }(\widetilde{a}-\widetilde{c},\theta _1,\theta _2,\theta _3)={\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(-\widetilde{c},\theta _1,\theta _2,\theta _3), \end{aligned}$$

and

$$\begin{aligned} {\mathcal {R}}_{\lambda }(\widetilde{b}-\widetilde{c},\theta _1,\theta _2,\theta _3)={\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(-\widetilde{c},\theta _1,\theta _2,\theta _3), \end{aligned}$$

Then, if $\widetilde{a}\succeq \widetilde{b}$, then it is obvious that ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)$. Eventually, it leads to the inequality ${\mathcal {R}}_{\lambda }(\widetilde{a},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(-\widetilde{c},\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b},\theta _1,\theta _2,\theta _3)+{\mathcal {R}}_{\lambda }(-\widetilde{c},\theta _1,\theta _2,\theta _3)$. Thus, evidently it follows that ${\mathcal {R}}_{\lambda }(\widetilde{a}+(-\widetilde{c}),\theta _1,\theta _2,\theta _3)\ge {\mathcal {R}}_{\lambda }(\widetilde{b}+(-\widetilde{c}),\theta _1,\theta _2,\theta _3)$. So, the result follows immediately.

1.6 Proof of the Theorem 3.15

Consider the cases when $\widetilde{a}\succ \widetilde{b}$ happens, that is,

$$\begin{aligned} \widetilde{a}\succ \widetilde{b}\,\text {happens for}\, {\left\{ \begin{array}{ll} &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},0,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},0,\pm 1,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{a},\pm 1,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{b},\pm 1,\pm 1,\pm 1) \end{array}\right. }. \end{aligned}$$

Consider the cases when $\widetilde{c}\succ \widetilde{d}$ happens, that is,

$$\begin{aligned} \widetilde{c}\succ \widetilde{d}\,\text {happens for}\, {\left\{ \begin{array}{ll} &{}{\mathcal {R}}_{\lambda }(\widetilde{c},0,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{d},0,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},0,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{d},0,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},0,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{d},0,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},0,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{d},0,\pm 1,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,0,0)>{\mathcal {R}}_{\lambda }(\widetilde{d},\pm 1,0,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,0,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{d},\pm 1,0,\pm 1)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,\pm 1,0)>{\mathcal {R}}_{\lambda }(\widetilde{d},\pm 1,\pm 1,0)\\ &{}{\mathcal {R}}_{\lambda }(\widetilde{c},\pm 1,\pm 1,\pm 1)>{\mathcal {R}}_{\lambda }(\widetilde{d},\pm 1,\pm 1,\pm 1) \end{array}\right. } \end{aligned}$$

Now, to proof this property, it need to discuss these $8\times 8$ cases, which will make the proof tedious. However, one can see the proof trivially, if following claims can be established.

Claim 1: Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=0$ in ordering $\widetilde{c}$ and $\widetilde{d}$. Then $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$. The proof of this claim is as follows. Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=0$ in ordering $\widetilde{c}$ and $\widetilde{d}$. Then ${\mathcal {V}}(\mu _{\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{d}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{d}})$. So, $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$.

Claim 2: Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=\pm 1$ in ordering $\widetilde{c}$ and $\widetilde{d}$. Then $\theta _i=\pm 1$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$. The proof of this claim is as follows. Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=\pm 1$ in ordering $\widetilde{c}$ and $\widetilde{d}$, then ${\mathcal {V}}(\mu _{\widetilde{c}})={\mathcal {V}}(\mu _{\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{c}})={\mathcal {V}}(\rho _{\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{d}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})={\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})={\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{d}})$. So, $\theta _i=\pm 1$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$.

Claim 3: Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=\pm 1$ in ordering $\widetilde{b}$ and $\widetilde{c}$. Then $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$. The proof of this claim is as follows. Let $\theta _i=0$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})\ne {\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})\ne {\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})\ne {\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=\pm 1$ in ordering $\widetilde{c}$ and $\widetilde{d}$, then ${\mathcal {V}}(\mu _{\widetilde{c}})={\mathcal {V}}(\mu _{\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{c}})={\mathcal {V}}(\rho _{\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{c}})={\mathcal {V}}(\nu _{\widetilde{d}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{d}})$. So, $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$.

Claim 4: Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$, and $\theta _i=0$ in ordering $\widetilde{c}$ and $\widetilde{d}$. Then $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$. The proof of this claim is as follows. Let $\theta _i=\pm 1$ in ordering $\widetilde{a}$ and $\widetilde{b}$. Then ${\mathcal {V}}(\mu _{\widetilde{a}})={\mathcal {V}}(\mu _{\widetilde{b}})$, ${\mathcal {V}}(\rho _{\widetilde{a}})={\mathcal {V}}(\rho _{\widetilde{b}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}})={\mathcal {V}}(\nu _{\widetilde{b}})$. Similarly, if $\theta _i=0$ in ordering $\widetilde{c}$ and $\widetilde{d}$, then ${\mathcal {V}}(\mu _{\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{d}})$. Thus, it follows that ${\mathcal {V}}(\mu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\mu _{\widetilde{b}+\widetilde{d}})$, ${\mathcal {V}}(\rho _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\rho _{\widetilde{b}+\widetilde{d}})$ and ${\mathcal {V}}(\nu _{\widetilde{a}+\widetilde{c}})\ne {\mathcal {V}}(\nu _{\widetilde{b}+\widetilde{d}})$. So, $\theta _i=0$ in ordering $\widetilde{a}+\widetilde{c}$ and $\widetilde{b}+\widetilde{d}$.

From these four claims, it is trivial enough to show that if $\widetilde{a}\succ \widetilde{b}$ and $\widetilde{b}\succ \widetilde{c}$, then $\widetilde{a}+\widetilde{c}\succ \widetilde{b}+\widetilde{d}$.