Representation Uniqueness of Generalized Triangular Fuzzy Numbers and Generalized Trapezoidal Fuzzy Numbers

1. Introduction

Let $\mathbb{N}$ be the set of all positive integers and let ${\mathbb{R}}^m$ be the m-dimensional Euclidean space. ${\mathbb{R}}^1$ is also written as $\mathbb{R}$.

Usually, the symbols $(a,b,c,d)$ with $a,b,c,d$ in $\mathbb{R}$ represent the elements in ${\mathbb{R}}^4$ and the symbols $(a,b,c)$ with $a,b,c$ in $\mathbb{R}$ represent the elements in ${\mathbb{R}}^3$. In this paper, for each $a,b,c,d$ in $\mathbb{R}$, we use $[a,b,c,d]$ instead of $(a,b,c,d)$ to represent the corresponding element in ${\mathbb{R}}^4$, and use $[a,b,c]$ instead of $(a,b,c)$ to represent the corresponding element in ${\mathbb{R}}^3$.

We use T to denote the set $\{[a,b,c,d]\in {\mathbb{R}}^4:a\le b\le c\le d\}$ and $T_0$ to denote the set $\{[a,b,c,d]\in {\mathbb{R}}^4:a<b\le c<d\}$. Clearly $T_0⫋T$.

We use G to denote the set $\{[a,b,c]\in {\mathbb{R}}^3:a\le b\le c\}$ and $G_0$ to denote the set $\{[a,b,c]\in {\mathbb{R}}^3:a<b<c\}$. Clearly $G_0⫋G$.

In theoretical research and practical applications, triangular fuzzy numbers and trapezoidal fuzzy numbers are often used fuzzy sets [1,2].

Each triangular fuzzy number can be represented as $(a,b,c)$ with $[a,b,c]\in G_0$. Each trapezoidal fuzzy number can be represented as $(a,b,c,d)$ with $[a,b,c,d]\in T_0$.

Naturally, we ask: For each triangular fuzzy number u, whether there is an $[a,b,c]$ which is the unique element of $G_0$ that satisfies $u=(a,b,c)$? For each trapezoidal fuzzy number u, whether there is an $[a,b,c,d]$ which is the unique element of $T_0$ that satisfies $u=(a,b,c,d)$? If these representation uniqueness questions are positively answered, it will bring a lot of convenience to analyze and discuss the triangular fuzzy numbers and the trapezoidal fuzzy numbers.

In this paper, we introduce the concepts of the generalized triangular fuzzy numbers and the generalized trapezoidal fuzzy numbers, which are generalizations of triangular fuzzy numbers and trapezoidal fuzzy numbers, respectively.

The symbols Tag, Tap, Trag and Trap are used to denote the set of triangular fuzzy numbers, the set of trapezoidal fuzzy numbers, the set of generalized triangular fuzzy numbers, and the set of generalized trapezoidal fuzzy numbers, respectively.

We show that for each generalized triangular fuzzy number u, there is an $[a,b,c]$ which is the unique element of G that satisfies $u=(a,b,c)$, and that for each generalized trapezoidal fuzzy number u, there is an $[a,b,c,d]$ which is the unique element of T that satisfies $u=(a,b,c,d)$. As corollaries of these conclusions, we positively answer the above question of representation uniqueness of each triangular fuzzy number and the above question of representation uniqueness of each trapezoidal fuzzy number, and give enhanced versions of these positive answers.

Furthermore, we give several equivalent forms of some conclusions of representation uniqueness. Based on some of these equivalent forms and some basic properties of Tag, Tap, Trag and Trap, we give some relationship among Tag, Tap, Trag and Trap. We also present some properties of Tag, Tap, Trag and Trap without using the conclusions of representation uniqueness. The conclusions of this paper are helpful to the related research of the triangular fuzzy numbers and the trapezoidal fuzzy numbers.

The remainder of this paper is organized as follows. Section 2 reviews some basic concepts related to fuzzy sets, triangular fuzzy numbers and trapezoidal fuzzy numbers, and gives some basic properties of the latter two. In Section 3, we introduce the generalized triangular fuzzy numbers and the generalized trapezoidal fuzzy numbers, and give some properties of the four families of fuzzy sets Tag, Trag, Tap and Trap. Section 4 gives conclusions on representation uniqueness of Tag, Trag, Tap and Trap, respectively. Furthermore, we present several equivalent forms of some of these conclusions of representation uniqueness, and give some relationship among Tag, Trag, Tap and Trap. Section 5 gives some relationships between Tag and Tap, and between Trag and Trap. By last, we draw our conclusions in Section 6.

2. Fuzzy Sets, Triangular Fuzzy Numbers and Trapezoidal Fuzzy Numbers

In this section, we review some basic concepts related to the fuzzy sets, the triangular fuzzy numbers and the trapezoidal fuzzy numbers, and give some basic properties of the latter two. For fuzzy theory and applications, we refer the readers to [1,2,3,4,5,6,7,8,9,10,11,12,13].

Let Y be a nonempty set. The symbol $P\left(Y\right)$ denotes the power set of Y, which is the set of all subsets of Y. The symbol $F\left(Y\right)$ denotes the set of all fuzzy sets in Y, i.e., functions from Y to $[0,1]$. Given $u\in F\left(Y\right)$ and $\alpha \in (0,1]$, the $\alpha$-cut ${\left[u\right]}_{\alpha }$ of u is defined by ${\left[u\right]}_{\alpha }:=\{x\in Y:u\left(x\right)\ge \alpha \}$.

Let Y be a topological space. The symbol $C\left(Y\right)$ denotes the set of all nonempty closed subsets of Y. $K\left(Y\right)$ denotes the set of all nonempty compact subsets of Y. For $u\in F\left(Y\right)$, the 0-cut ${\left[u\right]}_0$ of u is defined by ${\left[u\right]}_0:=\overline{\{x\in Y:u(x)>0\}}$, where $\overline{S}$ denotes the topological closure of S in Y. ${\left[u\right]}_0$ is called the support of u, and is also denoted by supp u.

Some properties of distances on fuzzy sets are discussed in [14,15,16,17].

Definition 2.1. 

We use Tag to denote the set of all triangular fuzzy numbers. $Tag:=\{(a,b,c):[a,b,c]in{G}_0\}$, where, for any $[a,b,c]$ in $G_0$, the triangular fuzzy number $(a,b,c)$ is defined to be the fuzzy set u in $F\left(\mathbb{R}\right)$ given by

$$u\left(x\right)=\left\{\begin{array}{cc}\hfill {\displaystyle \frac{x-a}{b-a}},& ifa\le x\le b,\hfill \\ \hfill {\displaystyle \frac{c-x}{c-b}},& ifb\le x\le c,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [a,c].\hfill \end{array}\right.$$

Definition 2.2. 

We use Tap to denote the set of all trapezoidal fuzzy numbers. $Tap:=\{(a,b,c,d):[a,b,c,d]in{T}_0\}$, where, for any $[a,b,c,d]$ in $T_0$, the trapezoidal fuzzy number $(a,b,c,d)$ is defined to be the fuzzy set u in $F\left(\mathbb{R}\right)$ given by

$$u\left(x\right)=\left\{\begin{array}{cc}\hfill {\displaystyle \frac{x-a}{b-a}},& ifa\le x\le b,\hfill \\ \hfill 1,& ifb\le x\le c,\hfill \\ \hfill {\displaystyle \frac{d-x}{d-c}},& ifc\le x\le d,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [a,d].\hfill \end{array}\right.$$

(ii) (a) Tap=$\{(a,b,c,d):[a,b,c,d]\in T_0\}$. (b) Both the statement “if $u\in$ Tap, then there is an $[a,b,c,d]\in T_0$ satisfying $u=(a,b,c,d)$” and its converse are true. ($u\in \{(a,b,c,d):[a,b,c,d]\in T_0\}$ means that there is an $[a,b,c,d]\in T_0$ satisfying $u=(a,b,c,d)$.) Clearly (a)⇒(b) and (b)⇒(a). (a) is known. So (b) holds.

The statements in this remark are quite obvious. We will use them without citing. If combining a statement (c) and a statement in this remark yields a statement (d), then we will say (c) implies (d).

We say that two fuzzy sets are equal if they have the same membership function.

Remark 2.4. 

(i) For each $a,b,c\in \mathbb{R}$, $(a,b,c)\in$ Tag if and only if $(a,b,b,c)\in$ Tap. (ii) Each triangular fuzzy number $(a,b,c)$ is the trapezoidal fuzzy number $(a,b,b,c)$. (iii) Tag ⊆ Tap. (iv) We can also define Tag based on Tap as follows:

Tag is the set of all triangular fuzzy numbers. Tag := $\{(a,b,c):[a,b,c]\in G_0\}$, where, for any $[a,b,c]$ in $G_0$, the triangular fuzzy number $(a,b,c)$ is defined to be the $(a,b,b,c)$ in Tap.

(v) Each trapezoidal fuzzy number $(a,b,c,d)$ with $b=c$ is the triangular fuzzy number $(a,b,d)$.

Now we give a routine proof of (i). As $Tag=\{(a^{\prime },b^{\prime },c^{\prime }):[a^{\prime },b^{\prime },c^{\prime }]\in G_0\}$, we have (i1) for each $a,b,c\in \mathbb{R}$, $(a,b,c)\in Tag\iff [a,b,c]\in G_0$. As $Tap=\{(a^{\prime },b^{\prime },c^{\prime },d^{\prime }):[a^{\prime },b^{\prime },c^{\prime },d^{\prime }]\in T_0\}$, we have (i2) for each $a,b,c\in \mathbb{R}$, $(a,b,b,c)\in Tap\iff [a,b,b,c]\in T_0$. Clearly for each $a,b,c\in \mathbb{R}$, $[a,b,c]\in G_0$ if and only if $[a,b,b,c]\in T_0$. By (i1) and (i2), this means that (i) holds.

Given $(a,b,c)\in$ Tag. Then, by (i), $(a,b,b,c)\in$ Tap. By Definitions 2.1 and 2.2, $(a,b,c)$ and $(a,b,b,c)$ have the same membership function. Thus $(a,b,c)$ = $(a,b,b,c)$. So (ii) holds. (iii) and (iv) follow immediately from (ii).

We show (v). Given $(a,b,c,d)$ with $b=c$ in Tap. By (i), $(a,b,d)$ is in Tag (see also (I) below). Then, by (ii), $(a,b,d)=(a,b,b,d)$. So (v) holds.

We think that it is okay not to point out the contents such as those mentioned in the following clauses (I) and (II), because they are easy to see.

(I) $(a,b,c)\in$ Tag or $(a,b,b,c)\in$ Tap implies that $a,b,c\in \mathbb{R}$. So (i) can be stated as: $(a,b,c)\in$ Tag if and only if $(a,b,b,c)\in$ Tap.

(II) (ii) implies that if $(a,b,c)\in$ Tag then $(a,b,b,c)\in$ Tap. (v) implies that if $(a,b,b,d)\in$ Tap then $(a,b,d)\in$ Tag. So that (ii) and (v) hold implies that (i) holds. The above proof of (v) indicates that (i) and (ii) hold implies that (v) holds. Below we show that (i) and (v) hold implies that (ii) holds. Given $(a,b,c)$ in Tag. By (i), $(a,b,b,c)$ is in Tap (see also (I)). Then, by (v), $(a,b,b,c)=(a,b,c)$. So (ii) holds.

3. Generalized Triangular Fuzzy Numbers and Generalized Trapezoidal Fuzzy Numbers

In this section, we introduce the generalized triangular fuzzy numbers and the generalized trapezoidal fuzzy numbers. Furthermore, we give some properties of the four families of fuzzy sets Tag, Trag, Tap and Trap including some relationship among them.

Definition 3.1. 

We use Trag to denote the set of all generalized triangular fuzzy numbers. $Trag:=\left\{\right(a,b,c):[a,b,c]inG\}$, where, for any $[a,b,c]$ in G, the generalized triangular fuzzy number $(a,b,c)$ is defined to be the fuzzy set u in $F\left(\mathbb{R}\right)$ in the following way:

$$\begin{array}{c}uisthetriangularfuzzynumber(a,b,c)whena<b<c;\\ u\left(x\right)=\left\{\begin{array}{cc}\hfill {\displaystyle \frac{c-x}{c-b}},& ifb\le x\le c,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [b,c],\hfill \end{array}\right.whena=b<c;\\ u\left(x\right)=\left\{\begin{array}{cc}\hfill {\displaystyle \frac{x-a}{b-a}},& ifa\le x\le b,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [a,b],\hfill \end{array}\right.whena<b=c;\\ u\left(x\right)=\left\{\begin{array}{cc}\hfill 1,& ifx=b,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus \left\{b\right\},\hfill \end{array}\right.whena=b=c.\end{array}$$

We know (a) $Tag=\{(a,b,c):[a,b,c]\in G_0\}$, $Trag=\left\{\right(a,b,c):[a,b,c]\in G\}$, and $G_0⫋G$. In Definition 3.1, for any $[a,b,c]\in G$, we define $(a,b,c)$ in four different cases. The case when $a<b<c$ can be written as the case when $[a,b,c]\in G_0$ because for any $[a,b,c]\in G$, $a<b<c$ if and only if $[a,b,c]\in G_0$. (In fact for any $[a,b,c]\in {\mathbb{R}}^3$, $a<b<c\iff [a,b,c]\in G_0$.) Thus we have (b) for each $[a,b,c]\in G_0$, by Definition 3.1, $(a,b,c)$ in Trag is just the $(a,b,c)$ in Tag. So by (a) and (b), the concept of generalized triangular fuzzy numbers is a kind of generalization of the concept of triangular fuzzy numbers. Hence Tag ⊆ Trag.

Definition 3.2. 

We use Trap to denote the set of all generalized trapezoidal fuzzy numbers. $Trap:=\left\{\right(a,b,c,d):[a,b,c,d]inT\}$, where, for any $[a,b,c,d]$ in T, the generalized trapezoidal fuzzy number $(a,b,c,d)$ is defined to be the fuzzy set u in $F\left(\mathbb{R}\right)$ in the following way:

$$\begin{array}{c}uisthetrapezoidalfuzzynumber(a,b,c,d)whena<b\le c<d;\\ u\left(x\right)=\left\{\begin{array}{cc}\hfill 1,& ifb\le x\le c,\hfill \\ \hfill {\displaystyle \frac{d-x}{d-c}},& ifc\le x\le d,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [b,d],\hfill \end{array}\right.whena=b\le c<d;\\ u\left(x\right)=\left\{\begin{array}{cc}\hfill {\displaystyle \frac{x-a}{b-a}},& ifa\le x\le b,\hfill \\ \hfill 1,& ifb\le x\le c,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [a,c],\hfill \end{array}\right.whena<b\le c=d;\\ u\left(x\right)=\left\{\begin{array}{cc}\hfill 1,& ifb\le x\le c,\hfill \\ \hfill 0,& ifx\in \mathbb{R}\setminus [b,c],\hfill \end{array}\right.whena=b\le c=d.\end{array}$$

We know (a) $Tap=\{(a,b,c,d):[a,b,c,d]\in T_0\}$, $Trap=\left\{\right(a,b,c,d):[a,b,c,d]\in T\}$, and $T_0⫋T$. In Definition 3.2, for any $[a,b,c,d]$ in T, we define $(a,b,c,d)$ in four different cases. The case when $a<b\le c<d$ can be written as the case when $[a,b,c,d]\in T_0$ because for any $[a,b,c,d]\in T$, $a<b\le c<d$ if and only if $[a,b,c,d]\in T_0$. (In fact for any $[a,b,c,d]\in {\mathbb{R}}^4$, $a<b\le c<d\iff [a,b,c,d]\in T_0$.) Thus we have (b) for each $[a,b,c,d]\in T_0$, by Definition 3.2, $(a,b,c,d)$ in Trap is just the $(a,b,c,d)$ in Tap. So the concept of generalized trapezoidal fuzzy numbers is a kind of generalization of the concept of trapezoidal fuzzy numbers. Hence Tap ⊆ Trap.

Remark 3.3. 

(i) (a) Trap=$\left\{\right(a,b,c,d):[a,b,c,d]\in T\}$. (b) Both the statement “if $u\in$ Trap, then there is an $[a,b,c,d]\in T$ satisfying $u=(a,b,c,d)$” and its converse are true. ($u\in \left\{\right(a,b,c,d):[a,b,c,d]\in T\}$ means that there is an $[a,b,c,d]\in T$ satisfying $u=(a,b,c,d)$.) Clearly (a)⇒(b) and (b)⇒(a). (a) is known. So (b) holds.

The statements in this remark are quite obvious. We will use them without citing. If combining a statement (c) and a statement in this remark yields a statement (d), then we will say (c) implies (d).

Remark 3.4. 

(i) For each $a,b,c\in \mathbb{R}$, $(a,b,c)\in$ Trag if and only if $(a,b,b,c)\in$ Trap. (ii) Each generalized triangular fuzzy number $(a,b,c)$ is the generalized trapezoidal fuzzy number $(a,b,b,c)$. (iii) Trag ⊆ Trap. (iv) We can also define Trag based on Trap as follows:

Trag is the set of all generalized triangular fuzzy numbers. Trag := $\left\{\right(a,b,c):[a,b,c]\in G\}$, where, for any $[a,b,c]$ in G, the generalized triangular fuzzy number $(a,b,c)$ is defined to be the $(a,b,b,c)$ in Trap.

(v) Each generalized trapezoidal fuzzy number $(a,b,c,d)$ with $b=c$ is the generalized triangular fuzzy number $(a,b,d)$.

We give a routine proof of (i). As $Trag=\{(a^{\prime },b^{\prime },c^{\prime }):[a^{\prime },b^{\prime },c^{\prime }]\in G\}$, we have (i1) for each $a,b,c\in \mathbb{R}$, $(a,b,c)\in Trag\iff [a,b,c]\in G$. As $Trap=\{(a^{\prime },b^{\prime },c^{\prime },d^{\prime }):[a^{\prime },b^{\prime },c^{\prime },d^{\prime }]\in T\}$, we have (i2) for each $a,b,c\in \mathbb{R}$, $(a,b,b,c)\in Trap\iff [a,b,b,c]\in T$. Clearly for each $a,b,c\in \mathbb{R}$, $[a,b,c]\in G$ if and only if $[a,b,b,c]\in T$. By (i1) and (i2), this means that (i) holds.

Given $(a,b,c)\in$ Trag. Then, by (i), $(a,b,b,c)\in$ Trap. By Definitions 3.1 and 3.2, $(a,b,c)$ and $(a,b,b,c)$ have the same membership function in the four cases $a<b<c$, $a=b<c$, $a<b=c$ and $a=b=c$. Thus $(a,b,c)$ = $(a,b,b,c)$. So (ii) holds. (iii) and (iv) follow immediately from (ii).

We show (v). Given $(a,b,c,d)$ with $b=c$ in Trap. By (i), $(a,b,d)$ is in Trag (see also (I) below). Then, by (ii), $(a,b,d)=(a,b,b,d)$. So (v) holds.

We think that it is okay not to point out the contents such as those mentioned in the following clauses (I) and (II), because they are easy to see.

(I) $(a,b,c)\in$ Trag or $(a,b,b,c)\in$ Trap implies that $a,b,c\in \mathbb{R}$. So (i) can be stated as: $(a,b,c)\in$ Trag if and only if $(a,b,b,c)\in$ Trap.

(II) (ii) implies that if $(a,b,c)\in$ Trag then $(a,b,b,c)\in$ Trap. (v) implies that if $(a,b,b,d)\in$ Trap then $(a,b,d)\in$ Trag. So that (ii) and (v) hold implies that (i) holds. The above proof of (v) indicates that (i) and (ii) hold implies that (v) holds. Below we show that (i) and (v) hold implies that (ii) holds. Given $(a,b,c)$ in Trag. By (i), $(a,b,b,c)$ is in Trap (see also (I)). Then, by (v), $(a,b,b,c)=(a,b,c)$. So (ii) holds.

4. Representation Uniqueness of Generalized Triangular Fuzzy Numbers and Generalized Trapezoidal Fuzzy Numbers

In this section, we show representation uniqueness of Tag, Trag, Tap and Trap, respectively. We give several equivalent forms of some of these conclusions of representation uniqueness. Then we give some relationship among Tag, Trag, Tap and Trap.

For any $[a,b,c,d]$ and $[a_1,b_1,c_1,d_1]$ in ${\mathbb{R}}^4$, $[a,b,c,d]=[a_1,b_1,c_1,d_1]$ means that $a=a_1$, $b=b_1$, $c=c_1$ and $d=d_1$. For any $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ in Trap, $(a,b,c,d)=(a_1,b_1,c_1,d_1)$ means that $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are the same fuzzy set.

For any $[a,b,c]$ and $[a_1,b_1,c_1]$ in ${\mathbb{R}}^3$, $[a,b,c]=[a_1,b_1,c_1]$ means that $a=a_1$, $b=b_1$ and $c=c_1$. For any $(a,b,c)$ and $(a_1,b_1,c_1)$ in Trag, $(a,b,c)=(a_1,b_1,c_1)$ means that $(a,b,c)$ and $(a_1,b_1,c_1)$ are the same fuzzy set.

Theorem 4.1(ii) gives the representation uniqueness of the generalized trapezoidal fuzzy numbers. Theorem 4.1(iv) gives the representation uniqueness of the trapezoidal fuzzy numbers.

Theorem 4.1. 

(i) Let $u=(a,b,c,d)$ be in Trap. Then ${\left[u\right]}_0=[a,d]$ and ${\left[u\right]}_1=[b,c]$. (ii) Let $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ be in Trap. Then $(a,b,c,d)=(a_1,b_1,c_1,d_1)$ if and only if $[a,b,c,d]=[a_1,b_1,c_1,d_1]$. (iii) Let $u=(a,b,c,d)$ be in Tap. Then ${\left[u\right]}_0=[a,d]$ and ${\left[u\right]}_1=[b,c]$. (iv) Let $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ be in Tap. Then $(a,b,c,d)=(a_1,b_1,c_1,d_1)$ if and only if $[a,b,c,d]=[a_1,b_1,c_1,d_1]$.

Proof. 

By Definition 3.2 and easy calculations, we obtain (i). (One way to perform these calculations are to do it based on watching the graphs of the membership functions of $(a,b,c,d)$ in the four cases $a<b\le c<d$, $a=b\le c<d$, $a<b\le c=d$ and $a=b\le c=d$.)

Now we show (ii). If $[a,b,c,d]=[a_1,b_1,c_1,d_1]$, i.e. $a=a_1,b=b_1,c=c_1$ and $d=d_1$, then, by Definition 3.2, $(a,b,c,d)=(a_1,b_1,c_1,d_1)$.

Suppose that $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. Then ${\left[(a,b,c,d)\right]}_1={\left[(a_1,b_1,c_1,d_1)\right]}_1$ and ${\left[(a,b,c,d)\right]}_0={\left[(a_1,b_1,c_1,d_1)\right]}_0$. By (i), this means that $[b,c]=[b_1,c_1]$ and $[a,d]=[a_1,d_1]$. This is equivalent to $a=a_1$, $b=b_1$, $c=c_1$ and $d=d_1$; that is, $[a,b,c,d]=[a_1,b_1,c_1,d_1]$. So (ii) is proved.

As Tap is a subset of Trap, (iii) follows immediately from (i), and (iv) follows immediately from (ii). (iii) is easy and should be known.

□

Proposition 4.2(ii) gives the representation uniqueness of the generalized triangular fuzzy numbers. Proposition 4.2(iv) gives the representation uniqueness of the triangular fuzzy numbers.

Proposition 4.2. 

(i) Let $u=(a,b,c)$ be in Trag. Then ${\left[u\right]}_0=[a,c]$ and ${\left[u\right]}_1=\left\{b\right\}$. (ii) Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Trag. Then $(a,b,c)=(a_1,b_1,c_1)$ if and only if $[a,b,c]=[a_1,b_1,c_1]$. (iii) Let $u=(a,b,c)$ be in Tag. Then ${\left[u\right]}_0=[a,c]$ and ${\left[u\right]}_1=\left\{b\right\}$. (iv) Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Tag. Then $(a,b,c)=(a_1,b_1,c_1)$ if and only if $[a,b,c]=[a_1,b_1,c_1]$.

Proof. 

By Definition 3.1 and easy calculations, we obtain (i). (One way to perform these calculations are to do it based on watching the graphs of the membership functions of $(a,b,c)$ in the four cases $a<b<c$, $a=b<c$, $a<b=c$ and $a=b=c$.)

Now we show (ii). If $[a,b,c]=[a_1,b_1,c_1]$, i.e. $a=a_1$, $b=b_1$ and $c=c_1$, then, by Definition 3.1, $(a,b,c)=(a_1,b_1,c_1)$.

Suppose that $(a,b,c)=(a_1,b_1,c_1)$. Then ${\left[(a,b,c)\right]}_1={\left[(a_1,b_1,c_1)\right]}_1$ and ${\left[(a,b,c)\right]}_0={\left[(a_1,b_1,c_1)\right]}_0$. By (i), this means that $\left\{b\right\}=\left\{b_1\right\}$ and $[a,c]=[a_1,c_1]$. This is equivalent to $a=a_1$, $b=b_1$ and $c=c_1$; that is, $[a,b,c]=[a_1,b_1,c_1]$. So (ii) is proved.

As Tag is a subset of Trag, (iii) follows immediately from (i), and (iv) follows immediately from (ii). (iii) is easy and should be known. □

The above proofs of Theorem 4.1 and Proposition 4.2 are similar.

Remark 4.3. 

Proposition 4.2 is a corollary of Theorem 4.1. This is because for k=i, ii, iii, iv, Proposition 4.2(k) is a corollary of Theorem 4.1(k).

“Theorem 4.1(i)⇒Proposition 4.2(i).” Assume that Theorem 4.1(i) holds. Let $(a,b,c)$ be in Trag. Then ${\left[(a,b,c)\right]}_0={\left[(a,b,b,c)\right]}_0=[a,c]$ (By Remark 3.4(ii), $(a,b,c)$ is the $(a,b,b,c)$ in Trap. So the first = holds. By Theorem 4.1(i), the second = holds.), and ${\left[(a,b,c)\right]}_1={\left[(a,b,b,c)\right]}_1=[b,b]=\left\{b\right\}$ (By Remark 3.4(ii), $(a,b,c)$ is the $(a,b,b,c)$ in Trap. So the first = holds. By Theorem 4.1(i), the second = holds.). So Proposition 4.2(i) holds.

“Theorem 4.1(ii)⇒Proposition 4.2(ii).” Assume that Theorem 4.1(ii) holds. Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Trag. By Remark 3.4(ii), $(a,b,c)$ is the $(a,b,b,c)$ in Trap and $(a_1,b_1,c_1)$ is the $(a_1,b_1,b_1,c_1)$ in Trap. Then $(a,b,c)=(a_1,b_1,c_1)$ means that $(a,b,b,c)=(a_1,b_1,b_1,c_1)$. By Theorem 4.1(ii), $(a,b,b,c)=(a_1,b_1,b_1,c_1)$ means that $[a,b,b,c]=[a_1,b_1,b_1,c_1]$. Clearly $[a,b,b,c]=[a_1,b_1,b_1,c_1]$ means that $[a,b,c]=[a_1,b_1,c_1]$ (see also (I) below). From above, it follows that $(a,b,c)=(a_1,b_1,c_1)$ if and only if $[a,b,c]=[a_1,b_1,c_1]$. So Proposition 4.2(ii) holds.

“Theorem 4.1(iii)⇒Proposition 4.2(iii).” Assume that Theorem 4.1(iii) holds. Let $(a,b,c)$ be in Tag. Then ${\left[(a,b,c)\right]}_0={\left[(a,b,b,c)\right]}_0=[a,c]$ (By Remark 2.4(ii), $(a,b,c)$ is the $(a,b,b,c)$ in Tap. So the first = holds. By Theorem 4.1(iii), the second = holds.), and ${\left[(a,b,c)\right]}_1={\left[(a,b,b,c)\right]}_1=[b,b]=\left\{b\right\}$ (By Remark 2.4(ii), $(a,b,c)$ is the $(a,b,b,c)$ in Tap. So the first = holds. By Theorem 4.1(iii), the second = holds.). So Proposition 4.2(iii) holds.

“Theorem 4.1(iv)⇒Proposition 4.2(iv).” Assume that Theorem 4.1(iv) holds. Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Tag. By Remark 2.4(ii), $(a,b,c)$ is the $(a,b,b,c)$ in Tap and $(a_1,b_1,c_1)$ is the $(a_1,b_1,b_1,c_1)$ in Tap. Then $(a,b,c)=(a_1,b_1,c_1)$ means that $(a,b,b,c)=(a_1,b_1,b_1,c_1)$. By Theorem 4.1(iv), $(a,b,b,c)=(a_1,b_1,b_1,c_1)$ means that $[a,b,b,c]=[a_1,b_1,b_1,c_1]$. Clearly $[a,b,b,c]=[a_1,b_1,b_1,c_1]$ means that $[a,b,c]=[a_1,b_1,c_1]$ (see also (II) below). From above, it follows that $(a,b,c)=(a_1,b_1,c_1)$ if and only if $[a,b,c]=[a_1,b_1,c_1]$. So Proposition 4.2(iv) holds.

The contents in (I) and (II) below are easy to see.

(I) ($\alpha$) For each $l,m,n,l_1,m_1,n_1$ in $\mathbb{R}$, the conditions (I-1) $[l,m,m,n]=[l_1,m_1,m_1,n_1]$, (I-2) $l=l_1$, $m=m_1$ and $n=n_1$, and (I-3) $[l,m,n]=[l_1,m_1,n_1]$, are equivalent.

Clearly (I-1)⇔(I-2) and (I-2)⇔(I-3). This means that (I-1)⇔(I-2)⇔(I-3). So ($\alpha$) holds.

That $a,b,c,a_1,b_1,c_1$ in $\mathbb{R}$ is implicit in the fact that $(a,b,c)$ and $(a_1,b_1,c_1)$ are in Trag. So by ($\alpha$), $[a,b,b,c]=[a_1,b_1,b_1,c_1]$ means that $[a,b,c]=[a_1,b_1,c_1]$.

(II) That $a,b,c,a_1,b_1,c_1$ in $\mathbb{R}$ is implicit in the fact that $(a,b,c)$ and $(a_1,b_1,c_1)$ are in Tag. So by ($\alpha$), $[a,b,b,c]=[a_1,b_1,b_1,c_1]$ means that $[a,b,c]=[a_1,b_1,c_1]$.

Remark 4.4. 

(a) For each u in Trap, $S(u,T)$ is a singleton set, where $S(u,T):=\left\{\right[a,b,c,d]\in T:u=(a,b,c,d\left)\right\}$.

(b) For each u in Trag, $S(u,G)$ is a singleton set, where $S(u,G):=\left\{\right[a,b,c]\in G:u=(a,b,c\left)\right\}$.

Clearly (a) means the following (ā):

(ā) For each u in Trap, there is an $[a,b,c,d]$ which is the unique element of T that satisfies $u=(a,b,c,d)$.

Clearly (b) means the following (b ):

(b ) For each u in Trag, there is an $[a,b,c]$ which is the unique element of G that satisfies $u=(a,b,c)$.

Consider statements (a${\phantom{,}}^{\prime }$) Theorem 4.1(ii), and (b${\phantom{,}}^{\prime }$) Proposition 4.2(ii). It is easy to see that (a)⇔(a${\phantom{,}}^{\prime }$) and (b)⇔(b${\phantom{,}}^{\prime }$) (see the proof below). As (a${\phantom{,}}^{\prime }$) and (b${\phantom{,}}^{\prime }$) hold, it follows that (a) and (b) hold; that is, (ā) and (b ) hold.

Assume that (a${\phantom{,}}^{\prime }$) holds. Let u in Trap. Then there exists an $[a,b,c,d]\in T$ satisfying $u=(a,b,c,d)$. If there is any $[a_1,b_1,c_1,d_1]$ in T satisfying $u=(a_1,b_1,c_1,d_1)$, then $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are in Trap and equal, and hence, by (a${\phantom{,}}^{\prime }$), $[a_1,b_1,c_1,d_1]=[a,b,c,d]$. So (a) holds.

Conversely, assume that (a) holds. The “if” part of (a${\phantom{,}}^{\prime }$) is obvious. A routine proof of this part is given in the proof of Theorem 4.1. The “only if” part of (a${\phantom{,}}^{\prime }$) follows immediately from (a). A routine proof of this part is given as follows. Let $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ be in Trap with $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. Then $[a,b,c,d]$ and $[a_1,b_1,c_1,d_1]$ are in T (In this paper, $(e,f,g,h)$ is well-defined only when $[e,f,g,h]\in T$.). Thus, by (a), $[a,b,c,d]=[a_1,b_1,c_1,d_1]$. Then the “only if” part of (a${\phantom{,}}^{\prime }$) is true. Hence (a${\phantom{,}}^{\prime }$) holds.

So (a)⇔(a${\phantom{,}}^{\prime }$).

Assume that (b${\phantom{,}}^{\prime }$) holds. Let u in Trag. Then there exists an $[a,b,c]\in G$ satisfying $u=(a,b,c)$. If there is any $[a_1,b_1,c_1]$ in G satisfying $u=(a_1,b_1,c_1)$, then $(a,b,c)$ and $(a_1,b_1,c_1)$ are in Trag and equal, and hence, by (b${\phantom{,}}^{\prime }$), $[a_1,b_1,c_1]=[a,b,c]$. So (b) holds.

Conversely, assume that (b) holds. The “if” part of (b${\phantom{,}}^{\prime }$) is obvious. A routine proof of this part is given in the proof of Proposition 4.2. The “only if” part of (b${\phantom{,}}^{\prime }$) follows immediately from (b). A routine proof of this part is given as follows. Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Trag with $(a,b,c)=(a_1,b_1,c_1)$. Then $[a,b,c]$ and $[a_1,b_1,c_1]$ are in G (In this paper, $(e,f,g)$ is well-defined only when $[e,f,g]\in G$.). Thus, by (b), $[a,b,c]=[a_1,b_1,c_1]$. Then the “only if” part of (b${\phantom{,}}^{\prime }$) is true. Hence (b${\phantom{,}}^{\prime }$) holds.

So (b)⇔(b${\phantom{,}}^{\prime }$).

The above proofs of (a)⇔(a${\phantom{,}}^{\prime }$) and (b)⇔(b${\phantom{,}}^{\prime }$) are similar.

Remark 4.5. 

(a) For each u in Tap, $S(u,T)$ is a singleton subset of $T_0$, where $S(u,T):=\left\{\right[a,b,c,d]\in T:u=(a,b,c,d\left)\right\}$.

(b) For each u in Tag, $S(u,G)$ is a singleton subset of $G_0$, where $S(u,G):=\left\{\right[a,b,c]\in G:u=(a,b,c\left)\right\}$.

Clearly (a) means the following (ā):

(ā) For each u in Tap, there is an $[a,b,c,d]$ in $T_0$ which is the unique element of T that satisfies $u=(a,b,c,d)$.

Clearly (b) means the following (b ):

(b ) For each u in Tag, there is an $[a,b,c]$ in $G_0$ which is the unique element of G that satisfies $u=(a,b,c)$.

We show (ā). Let $u\in$ Tap. Then there is an $[a,b,c,d]$ in $T_0$ satisfying $u=(a,b,c,d)$. Note that $u\in$ Trap and $[a,b,c,d]\in T$. By Remark 4.4(ā), this $[a,b,c,d]$ in $T_0$ is the unique element of T that satisfies $u=(a,b,c,d)$. So (ā) holds.

We show (b ). Let $u\in$ Tag. Then there is an $[a,b,c]$ in $G_0$ satisfying $u=(a,b,c)$. Note that $u\in$ Trag and $[a,b,c]\in G$. By Remark 4.4(b ), this $[a,b,c]$ in $G_0$ is the unique element of G that satisfies $u=(a,b,c)$. So (b ) holds.

Obviously (ā) and (b ) hold means that (a) and (b) hold.

From the above proof of (ā), we can see that (ā) is a corollary of Remark 4.4(ā). This means that (a) is a corollary of Remark 4.4(a).

From the above proof of (b ), we can see that (b ) is a corollary of Remark 4.4(b ). This means that (b) is a corollary of Remark 4.4(b).

Below we show that Theorem 4.1(iv)⇔(ā).

Assume that Theorem 4.1(iv) holds. Let $u\in$ Tap. Then there exists an $[a,b,c,d]\in T_0$ satisfying $u=(a,b,c,d)$. If there is any $[a_1,b_1,c_1,d_1]$ in T satisfying $u=(a_1,b_1,c_1,d_1)$, then $(a,b,c,d)=u=(a_1,b_1,c_1,d_1)\in$ Tap, and hence by Theorem 4.1(iv), $[a_1,b_1,c_1,d_1]=[a,b,c,d]$. Thus (ā) holds.

Assume that (ā) holds. Let $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ be in Tap. If $[a,b,c,d]=[a_1,b_1,c_1,d_1]$, i.e. $a=a_1,b=b_1,c=c_1$ and $d=d_1$, then, by Definition 2.2, $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. Suppose that $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. Then $[a,b,c,d]$ and $[a_1,b_1,c_1,d_1]$ are in T (In this paper, $(e,f,g,h)$ is well-defined only when $[e,f,g,h]\in T$.). Thus, by (ā), $[a,b,c,d]=[a_1,b_1,c_1,d_1]$. Hence Theorem 4.1(iv) holds.

So Theorem 4.1(iv)⇔(ā).

Below we show that Proposition 4.2(iv)⇔(b ).

Assume that Proposition 4.2(iv) holds. Let u in Tag. Then there exists an $[a,b,c]\in G_0$ satisfying $u=(a,b,c)$. If there is any $[a_1,b_1,c_1]$ in G satisfying $u=(a_1,b_1,c_1)$, then $(a,b,c)=u=(a_1,b_1,c_1)\in$ Tag, and hence, by Proposition 4.2(iv), $[a_1,b_1,c_1]=[a,b,c]$. Thus (b ) holds.

Conversely, assume that (b ) holds. Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Tag. If $[a,b,c]=[a_1,b_1,c_1]$, i.e. $a=a_1,b=b_1$ and $c=c_1$, then, by Definition 2.1, $(a,b,c)=(a_1,b_1,c_1)$. Suppose that $(a,b,c)=(a_1,b_1,c_1)$. Then $[a,b,c]$ and $[a_1,b_1,c_1]$ are in G (In this paper, $(e,f,g)$ is well-defined only when $[e,f,g]\in G$.). Thus, by (b ), $[a,b,c]=[a_1,b_1,c_1]$. Hence Proposition 4.2(iv) holds.

So Proposition 4.2(iv)⇔(b ).

Remark 4.6. 

(a) For each u in Tap, $S(u,T_0)$ is a singleton set, where $S(u,T_0):=\{[a,b,c,d]\in T_0:u=(a,b,c,d)\}$.

(b) For each u in Tag, $S(u,G_0)$ is a singleton set, where $S(u,G_0):=\{[a,b,c]\in G_0:u=(a,b,c)\}$.

Clearly (a) means the following (ā):

(ā) For each u in Tap, there is an $[a,b,c,d]$ which is the unique element of $T_0$ that satisfies $u=(a,b,c,d)$.

Clearly (b) means the following (b ):

(b ) For each u in Tag, there is an $[a,b,c]$ which is the unique element of $G_0$ that satisfies $u=(a,b,c)$.

Obviously, we can also state (ā) and (b ) as follows:

(ā) For each u in Tap, there is an $[a,b,c,d]$ in $T_0$ which is the unique element of $T_0$ that satisfies $u=(a,b,c,d)$.

(b ) For each u in Tag, there is an $[a,b,c]$ in $G_0$ which is the unique element of $G_0$ that satisfies $u=(a,b,c)$.

Based on these descriptions of (ā) and (b ) or directly, we can see that Remark 4.5(ā) implies (ā) and Remark 4.5(b ) implies (b ). Note that Remark 4.5(ā) and Remark 4.5(b ) are proved. So (ā) and (b ) hold. In other words, (a) and (b) hold.

From the above proof of (ā), we can see that (ā) is a corollary of Remark 4.5(ā). This means that (a) is a corollary of Remark 4.5(a).

From the above proof of (b ), we can see that (b ) is a corollary of Remark 4.5(b ). This means that (b) is a corollary of Remark 4.5(b).

Below we give a routine proof of (ā)⇒(b ). Assume that (ā) holds. Let $u\in$ Tag. Then there is an $[a,b,c]\in G_0$ which satisfies $u=(a,b,c)$. Suppose that there is any $[a_1,b_1,c_1]\in G_0$ satisfying $u=(a_1,b_1,c_1)$. Note that $u\in Tag\subset$ Tap, that both $[a,b,b,c]$ and $[a_1,b_1,b_1,c_1]$ are in $T_0$ (see also (I) below), and that, by Remark 2.4(ii), $(a,b,b,c)=u=(a_1,b_1,b_1,c_1)$. Thus by (ā), $[a,b,b,c]=[a_1,b_1,b_1,c_1]$. This means that $[a,b,c]=[a_1,b_1,c_1]$ (see also ($\alpha$) in Remark 4.3). So (b ) holds.

(I) ($\alpha$) ($\alpha$-1) $[e,f,g]\in G_0$ if and only if $[e,f,f,g]\in T_0$; ($\alpha$-2) $[e,f,g]\in G$ if and only if $[e,f,f,g]\in T$.

Clearly ($\alpha$) holds. As $[a,b,c]$ and $[a_1,b_1,c_1]$ are in $G_0$, by ($\alpha$-1), $[a,b,b,c]$ and $[a_1,b_1,b_1,c_1]$ are in $T_0$.

We call both Remark 4.5(ā) and Remark 4.6(ā) the representation uniqueness of the trapezoidal fuzzy numbers, although Remark 4.5(ā) is an enhanced version of Remark 4.6(ā).

We call both Remark 4.5(b ) and Remark 4.6(b ) the representation uniqueness of the triangular fuzzy numbers, although Remark 4.5(b ) is an enhanced version of Remark 4.6(b ).

Several equivalent forms of Theorem 4.1(ii) are given in the following Remark 4.7.

Remark 4.7. 

We claim the following statements.

(a) Let $(a,b,c,d)\in$ Trap and let A be a subset of T. Define $S:=\{(a_1,b_1,c_1,d_1):[a_1,b_1,c_1,d_1]\in A\}$. Then (a-1) $(a,b,c,d)\in S$ if and only if $[a,b,c,d]\in A$; (a-2) $(a,b,c,d)\notin S$ if and only if $[a,b,c,d]\notin A$.

(b) Let $(a,b,c,d)\in$ Trap and let $A_1$ and $A_2$ be two subsets of T. Define $S_1:=\{(a_1,b_1,c_1,d_1):[a_1,b_1,c_1,d_1]\in A_1\}$ and $S_2:=\{(a_2,b_2,c_2,d_2):[a_2,b_2,c_2,d_2]\in A_2\}$. Then (b-1) $(a,b,c,d)\in S_1\setminus S_2$ if and only if $[a,b,c,d]\in A_1\setminus A_2$; (b-2) $S_1\setminus S_2=\{(f,g,h,k):[f,g,h,k]\in A_1\setminus A_2\}$.

(c) Tap ⫋ Trap, (d) Trag ⫋ Trap, and (e) Tag ⫋ Tap.

First we show (a). To do this, we only need to show (a-1) as (a-1)⇔(a-2). The “if” part of (a-1) is obvious. Suppose that $(a,b,c,d)\in S$. This means that there is an $[a_1,b_1,c_1,d_1]\in A$ such that $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. As $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are in Trap (see also (I) below), by Theorem 4.1(ii), $[a,b,c,d]=[a_1,b_1,c_1,d_1]$. So $[a,b,c,d]\in A$. Thus the “only if” part of (a-1) is proved. So (a-1) holds. Hence (a) is true. (Theorem 4.1(ii)⇒(a-1) is proved in this paragraph.)

Next we show (b). Consider (i) $(a,b,c,d)\in S_1\setminus S_2$, (ii) $(a,b,c,d)\in S_1$ but $(a,b,c,d)\notin S_2$, (iii) $[a,b,c,d]\in A_1$ but $[a,b,c,d]\notin A_2$, and (iv) $[a,b,c,d]\in A_1\setminus A_2$. By (a), (ii)⇔(iii). This means that (i)⇔(iv), as (i)⇔(ii) and (iii)⇔(iv). So (b-1) is proved. ((a)⇒(b-1) is proved in this paragraph.)

(b-2) follows immediately from (b-1). A routine proof of (b-2) is given below.

Let $[f,g,h,k]\in A_1\setminus A_2$. Then $(f,g,h,k)$ in Trap, and by (b-1), $(f,g,h,k)\in S_1\setminus S_2$. Thus $S_1\setminus S_2\supseteq \{(f,g,h,k):[f,g,h,k]\in A_1\setminus A_2\}$. Let $(f,g,h,k)\in S_1\setminus S_2$. Then $(f,g,h,k)\in$ Trap (see also (II) below), and by (b-1), $[f,g,h,k]\in A_1\setminus A_2$. Thus $S_1\setminus S_2\subseteq \{(f,g,h,k):[f,g,h,k]\in A_1\setminus A_2\}$. So (b-2) holds. ((b-1)⇒(b-2) is proved in this paragraph.)

Now we show (c). Trap ∖ Tap = $\{(a,b,c,d):[a,b,c,d]\in T\}\setminus \{(a,b,c,d):[a,b,c,d]\in T_0\}$ = (by (b-2)) $\{(a,b,c,d):[a,b,c,d]\in T\setminus T_0\}\ne \varnothing$ (Clearly $T\setminus T_0\ne \varnothing$. This means the ≠ holds.). Thus Trap ≠ Tap. We have known that Tap ⊆ Trap. So (c) is true.

Now we show (d). Put $T_1:=\{[a,b,c,d]:[a,b,c,d]\in Twithb=c\}$. We can see that Trag = $\{(a,b,c):[a,b,c]\in G\}=\{(a,b,b,c):[a,b,c]\in G\}=\{(a,b,b,c):[a,b,b,c]\in T\}=\{(a,b,c,d):[a,b,c,d]\in T_1\}$, where the second = follows from Remark 3.4(ii), the other =s are easy to see (see also (III) below). Thus Trap ∖ Trag = $\{(a,b,c,d):[a,b,c,d]\in T\}\setminus \{(a,b,c,d):[a,b,c,d]\in T_1\}$ = (by (b-2)) $\{(a,b,c,d):[a,b,c,d]\in T\setminus T_1\}\ne \varnothing$ (Clearly $T\setminus T_1\ne \varnothing$. This means the ≠ holds.). Thus Trap ≠ Trag. We have known that Trag ⊆ Trap. So (d) is true.

Finally we show (e). Put $T_2:=\{[a,b,c,d]:[a,b,c,d]\in T_{0}withb=c\}$. We can see that Tag = $\{(a,b,c):[a,b,c]\in G_0\}=\{(a,b,b,c):[a,b,c]\in G_0\}=\{(a,b,b,c):[a,b,b,c]\in T_0\}=\{(a,b,c,d):[a,b,c,d]\in T_2\}$, where the second = follows from Remark 2.4(ii), the other =s are easy to see (see also (IV) below). Thus Tap ∖ Tag = $\{(a,b,c,d):[a,b,c,d]\in T_0\}\setminus \{(a,b,c,d):[a,b,c,d]\in T_2\}$ = (by (b-2)) $\{(a,b,c,d):[a,b,c,d]\in T_0\setminus T_2\}\ne \varnothing$ (Clearly $T_0\setminus T_2\ne \varnothing$. This means the ≠ holds.). Hence Tap ≠ Tag. We have known that Tag ⊆ Tap. So (e) is true.

(I) Clearly $(a_1,b_1,c_1,d_1)\in$ Trap as $[a_1,b_1,c_1,d_1]\in A\subseteq T$.

Let $l,m,n,t\in \mathbb{R}$. Consider (I-1) $(l,m,n,t)$ is well-defined, (I-2) $[l,m,n,t]\in T$, (I-3) $(l,m,n,t)\in$ Trap. In this paper, $(l,m,n,t)$ is well-defined only when $[l,m,n,t]\in T$; that is, (I-1)⇔(I-2). Clearly (I-2)⇒(I-3) and (I-3)⇒(I-1). So (I-1)⇔(I-2)⇔(I-3). Thus when using (a), (b) or Theorem 4.1(i)(ii), we do not need to verify that a certain $(l,m,n,t)$ is in Trap if it is well-defined. For example, here we do not need to mention that “$(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are in Trap”. This conclusion follows from the fact that $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are well-defined. This fact is implicit in the previously given expression “$(a,b,c,d)=(a_1,b_1,c_1,d_1)$”. That $(a,b,c,d)\in$ Trap is one of the prerequisites of (a).

We think that (I-1)⇔(I-2)⇔(I-3) can be used without citing as it is easy to see. In this paper, we don’t always illustrate that there are multiple ways to prove that a certain element belongs to Trap, as we do here, because it’s easy to see.

Let B be a subset of ${\mathbb{R}}^4$. Consider (I-4) $S_B=\{(l,m,n,t):[l,m,n,t]\in B\}$ is well-defined, (I-5) For each $[l,m,n,t]\in B$, $(l,m,n,t)$ is well-defined, and (I-6) $B\subseteq T$. Clearly (I-4)⇔(I-5) and (I-5)⇔(I-6). This means that (I-4)⇔(I-5)⇔(I-6). Thus, we can conclude that a certain subset D of ${\mathbb{R}}^4$ is included in T if $\left\{\right(l,m,n,t):[l,m,n,t]\in D\}$ is well-defined.

(II) Clearly $S_1\subseteq$ Trap as $A_1\subseteq T$. So $(f,g,h,k)\in S_1\setminus S_2\subseteq S_1\subseteq$ Trap.

In fact $(f,g,h,k)\in$ Trap does not need to be mentioned since it follows from that $(f,g,h,k)$ is well-defined (see (I-1)⇔(I-3) given above), which is implicit in the preceding expression “$(f,g,h,k)\in S_1\setminus S_2$”.

(III) Note that for any $a,b,c\in \mathbb{R}$, $[a,b,b,c]\in T$ is equivalent to $[a,b,c]\in G$. So the third = holds.

Put $A:=\left\{\right(a,b,b,c):[a,b,b,c]\in T\}$ and $B:=\{(a,b,c,d):[a,b,c,d]\in T_1\}$. The fourth = means that (III-1) $A\subseteq B$; that is, for each $[a,b,b,c]\in T$, $(a,b,b,c)\in B$, and (III-2) $B\subseteq A$; that is, for each $[a,b,c,d]\in T_1$, $(a,b,c,d)\in A$. Given $[a,b,b,c]\in T$. Then $[a,b,b,c]\in T_1$ (obviously the converse is true). Thus $(a,b,b,c)\in \{(a,b,c,d):[a,b,c,d]\in T_1\}=B$ (see (III-3) below). Hence (III-1) holds. Given $[a,b,c,d]\in T_1$. This means that $b=c$ and $[a,b,c,d]\in T$. Then $(a,b,c,d)=(a,b,b,d)\in \left\{\right(l,m,m,n):[l,m,m,n]\in T\}=A$ (see (III-4) below). Hence (III-2) holds. So the fourth = holds.

(III-3) Conversely, suppose that $(a,b,b,c)\in B$. Then $(a,b,b,c)\in$ Trap as $(a,b,b,c)$ is well-defined. Also $T_1\subseteq T$. Thus, by (a-1), $[a,b,b,c]\in T_1$.

(III-4) Let $a,b,c,d$ in $\mathbb{R}$. Suppose that $(a,b,c,d)=(a,b,b,d)$. Then $(a,b,c,d)$ and $(a,b,b,d)$ are in Trap as they are well-defined. Thus, by Theorem 4.1(ii), $[a,b,c,d]=[a,b,b,d]$; that is, $b=c$. $(a,b,c,d)$ is well-defined means that $[a,b,c,d]\in T$.

(IV) Note that for any $a,b,c\in \mathbb{R}$, $[a,b,b,c]\in T_0$ is equivalent to $[a,b,c]\in G_0$. So the third = holds.

Put $C:=\{(a,b,b,c):[a,b,b,c]\in T_0\}$ and $D:=\{(a,b,c,d):[a,b,c,d]\in T_2\}$. The fourth = means that (IV-1) $C\subseteq D$; that is, for each $[a,b,b,c]\in T_0$, $(a,b,b,c)\in D$, and (IV-2) $D\subseteq C$; that is, for each $[a,b,c,d]\in T_2$, $(a,b,c,d)\in C$. Given $[a,b,b,c]\in T_0$. Then $[a,b,b,c]\in T_2$ (obviously the converse is true). Thus $(a,b,b,c)\in \{(a,b,c,d):[a,b,c,d]\in T_2\}=D$. Hence (IV-1) holds. Given $[a,b,c,d]\in T_2$. This means that $b=c$ and $[a,b,c,d]\in T_0$. Then $(a,b,c,d)=(a,b,b,d)\in \{(l,m,m,n):[l,m,m,n]\in T_0\}=C$. Hence (IV-2) holds. So the fourth = holds.

(V) (f) The six statements (a), (a-1), (a-2), (b-1), (b-2), and Theorem 4.1(ii) are equivalent.

Clearly (a-1)⇔(a-2). So (a)⇔(a-1)⇔(a-2). (Obviously, the converse is true.) Theorem 4.1(ii)⇒(a-1) and (a)⇒(b-1)⇒(b-2) have been shown in the above contents. To show (f) we only need to show that (b-2)⇒Theorem 4.1(ii), a proof of which is given below.

Assume that (b-2) holds. Let $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ be in Trap. Clearly if $[a,b,c,d]=[a_1,b_1,c_1,d_1]$ then $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. Suppose that $(a,b,c,d)=(a_1,b_1,c_1,d_1)$. Set $S_1:=\left\{(a_1,b_1,c_1,d_1)\right\}$, $S_2:=\left\{(a,b,c,d)\right\}$, $A_1:=\left\{[a_1,b_1,c_1,d_1]\right\}$ and $A_2:=\left\{[a,b,c,d]\right\}$. Clearly $S_1=\{(e,f,g,h):[e,f,g,h]\in A_1\}$ and $S_2=\{(e,f,g,h):[e,f,g,h]\in A_2\}$. We can see that $\{(e,f,g,h):[e,f,g,h]\in A_1\setminus A_2\}=\left(by\right(b-2\left)\right)S_1\setminus S_2=\varnothing$ (see (V-1) below). So $A_1\setminus A_2=\varnothing$; that is, $[a_1,b_1,c_1,d_1]=[a,b,c,d]$ (see (V-2) below). Thus Theorem 4.1(ii) holds. Hence (b-2)⇒Theorem 4.1(ii). So (f) is proved.

(V-1) $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are in Trap means that $[a,b,c,d]$ and $[a_1,b_1,c_1,d_1]$ are in T, which means that $A_1$ and $A_2$ are subsets of T. So (b-2) can be used here. We think that the fact that $A_1$ and $A_2$ are subsets of T does not need to be mentioned as it is implicit in the fact that $(a,b,c,d)$ and $(a_1,b_1,c_1,d_1)$ are in Trap (The contents in the first sentence of this paragraph indicates that these two facts are equivalent.).

(V-2) ($\alpha$) Let B be a subset of ${\mathbb{R}}^4$. Put $S_B:=\{(e,f,g,h):[e,f,g,h]\in B\}$. Then $S_B=\varnothing$ if and only if $B=\varnothing$.

Clearly if $B=\varnothing$ then $S_B=\varnothing$. Suppose that $S_B=\varnothing$. Then $S_B$ is well-defined. This means that $B\subseteq T$ (see (I-4)⇔(I-6)). So if $B\ne \varnothing$, then $S_B\ne \varnothing$, which is a contradiction. Thus $B=\varnothing$. Hence ($\alpha$) holds.

$(a_1,b_1,c_1,d_1)\in$ Trap means that $[a_1,b_1,c_1,d_1]\in T$. So $A_1⫋T$. Thus $A_1\setminus A_2⫋T⫋{\mathbb{R}}^4$. Thus, by ($\alpha$), $\{(e,f,g,h):[e,f,g,h]\in A_1\setminus A_2\}=\varnothing$ if and only if $A_1\setminus A_2=\varnothing$.

It is easy to see that the fact that $A_1\setminus A_2⫋{\mathbb{R}}^4$ is implicit in the fact that $(a_1,b_1,c_1,d_1)\in$ Trap. Also ($\alpha$) can be used directly without citing as it is easy to see. So we think we can directly write $\{(e,f,g,h):[e,f,g,h]\in A_1\setminus A_2\}=\varnothing$ if and only if $A_1\setminus A_2=\varnothing$.

Several equivalent forms of Proposition 4.2(ii) are given in the following Remark 4.8.

Remark 4.8. 

We claim the following statements.

(a) Let $(a,b,c)\in$ Trag and let A be a subset of G. Define $S:=\{(a_1,b_1,c_1):[a_1,b_1,c_1]\in A\}$. Then (a-1) $(a,b,c)\in S$ if and only if $[a,b,c]\in A$; (a-2) $(a,b,c)\notin S$ if and only if $[a,b,c]\notin A$.

(b) Let $(a,b,c)\in$ Trag and let $A_1$ and $A_2$ be two subsets of G. Define $S_1:=\{(a_1,b_1,c_1):[a_1,b_1,c_1]\in A_1\}$ and $S_2:=\{(a_2,b_2,c_2):[a_2,b_2,c_2]\in A_2\}$. Then (b-1) $(a,b,c)\in S_1\setminus S_2$ if and only if $[a,b,c]\in A_1\setminus A_2$; (b-2) $S_1\setminus S_2=\{(f,g,h):[f,g,h]\in A_1\setminus A_2\}$.

(c) Tag ⫋ Trag.

First we show (a). To do this, we only need to show (a-1) as (a-1)⇔(a-2). The “if” part of (a-1) is obvious. Suppose that $(a,b,c)\in S$. This means that there is an $[a_1,b_1,c_1]\in A$ such that $(a,b,c)=(a_1,b_1,c_1)$. As $(a,b,c)$ and $(a_1,b_1,c_1)$ are in Trag (see also (I) below), by Proposition 4.2(ii), $[a,b,c]=[a_1,b_1,c_1]$. So $[a,b,c]\in A$. Thus the “only if” part of (a-1) is proved. So (a-1) holds. Hence (a) is proved. (Proposition 4.2(ii)⇒(a-1) is proved in this paragraph.)

Next we show (b). Consider (i) $(a,b,c)\in S_1\setminus S_2$, (ii) $(a,b,c)\in S_1$ but $(a,b,c)\notin S_2$, (iii) $[a,b,c]\in A_1$ but $[a,b,c]\notin A_2$, and (iv) $[a,b,c]\in A_1\setminus A_2$. By (a), (ii)⇔(iii). This means that (i)⇔(iv), as (i)⇔(ii) and (iii)⇔(iv). So (b-1) is proved. ((a)⇒(b-1) is proved in this paragraph.)

(b-2) follows immediately from (b-1). A routine proof of (b-2) is given below.

Let $[f,g,h]\in A_1\setminus A_2$. Then $(f,g,h)$ in Trag, and by (b-1), $(f,g,h)\in S_1\setminus S_2$. Thus $S_1\setminus S_2\supseteq \{(f,g,h):[f,g,h]\in A_1\setminus A_2\}$. Let $(f,g,h)\in S_1\setminus S_2$. Then $(f,g,h)\in$ Trag (see also (II) below), and by (b-1), $[f,g,h]\in A_1\setminus A_2$. Thus $S_1\setminus S_2\subseteq \{(f,g,h):[f,g,h]\in A_1\setminus A_2\}$. So (b-2) holds. ((b-1)⇒(b-2) is proved in this paragraph.)

Now we show (c). Trag ∖ Tag = $\{(a,b,c):[a,b,c]\in G\}\setminus \{(a,b,c):[a,b,c]\in G_0\}$ = (by (b-2)) $\{(a,b,c):[a,b,c]\in G\setminus G_0\}\ne \varnothing$ (Clearly $G\setminus G_0\ne \varnothing$. This means the ≠ holds.). Thus Trag ≠ Tag. We have known that Tag ⊆ Trag. So (c) is true.

(I) Clearly $(a_1,b_1,c_1)\in$ Trag as $[a_1,b_1,c_1]\in A\subseteq G$.

Let $l,m,n\in \mathbb{R}$. Consider (I-1) $(l,m,n)$ is well-defined, (I-2) $[l,m,n]\in G$, (I-3) $(l,m,n)\in$ Trag. In this paper, $(l,m,n)$ is well-defined only when $[l,m,n]\in G$; that is, (I-1)⇔(I-2). Clearly (I-2)⇒(I-3) and (I-3)⇒(I-1). So (I-1)⇔(I-2)⇔(I-3). Thus when using (a), (b) or Proposition 4.2(i)(ii), we do not need to verify that a certain $(l,m,n)$ is in Trag if it is well-defined. For example, here we do not need to mention that “$(a,b,c)$ and $(a_1,b_1,c_1)$ are in Trag”. This conclusion follows from the fact that $(a,b,c)$ and $(a_1,b_1,c_1)$ are well-defined. This fact is implicit in the previously given expression “$(a,b,c)=(a_1,b_1,c_1)$”. That $(a,b,c)\in$ Trag is one of the prerequisites of (a).

We think that the fact (I-1)⇔(I-2)⇔(I-3) can be used without citing as it is easy to see. In this paper, we do not always illustrate that there are multiple ways to prove that a certain element belongs to Trag, as we do here, because it is easy to see.

Let C be a subset of ${\mathbb{R}}^3$. Consider (I-4) $S_C=\{(l,m,n):[l,m,n]\in C\}$ is well-defined, (I-5) For each $[l,m,n]\in C$, $(l,m,n)$ is well-defined, and (I-6) $C\subseteq G$. Clearly (I-4)⇔(I-5) and (I-5)⇔(I-6). This means that (I-4)⇔(I-5)⇔(I-6). Thus, we can conclude that a certain subset D of ${\mathbb{R}}^3$ is included in G if $\left\{\right(l,m,n):[l,m,n]\in D\}$ is well-defined.

(II) Clearly $S_1\subseteq$ Trag as $A_1\subseteq G$. So $(f,g,h)\in S_1\setminus S_2\subseteq S_1\subseteq$ Trag.

In fact $(f,g,h)\in$ Trag does not need to be mentioned since it follows from that $(f,g,h)$ is well-defined (see (I-1)⇔(I-3) given above), which is implicit in the preceding expression “$(f,g,h)\in S_1\setminus S_2$”.

(III) (d) The six statements (a), (a-1), (a-2), (b-1), (b-2), and Proposition 4.2(ii) are equivalent.

Clearly (a-1)⇔(a-2). So (a)⇔(a-1)⇔(a-2). (Obviously the converse is true.) Proposition 4.2(ii)⇒(a-1) and (a)⇒(b-1)⇒(b-2) have been shown in the above contents. To show (d) we only need to show that (b-2)⇒Proposition 4.2(ii), a proof of which is given below.

Assume that (b-2) holds. Let $(a,b,c)$ and $(a_1,b_1,c_1)$ be in Trag. Clearly if $[a,b,c]=[a_1,b_1,c_1]$ then $(a,b,c)=(a_1,b_1,c_1)$. Suppose that $(a,b,c)=(a_1,b_1,c_1)$. Set $S_1:=\left\{(a_1,b_1,c_1)\right\}$, $S_2:=\left\{(a,b,c)\right\}$, $A_1:=\left\{[a_1,b_1,c_1]\right\}$ and $A_2:=\left\{[a,b,c]\right\}$. Clearly $S_1=\{(e,f,g):[e,f,g]\in A_1\}$ and $S_2=\{(e,f,g):[e,f,g]\in A_2\}$. We can see that $\{(e,f,g):[e,f,g]\in A_1\setminus A_2\}=\left(by\right(b-2\left)\right)S_1\setminus S_2=\varnothing$ (see (V-1) below). So $A_1\setminus A_2=\varnothing$; that is, $[a_1,b_1,c_1]=[a,b,c]$ (see (V-2) below). Thus Proposition 4.2(ii) holds. Hence (b-2)⇒Proposition 4.2(ii). So (d) is proved.

(V-1) $(a,b,c)$ and $(a_1,b_1,c_1)$ are in Trag means that $[a,b,c]$ and $[a_1,b_1,c_1]$ are in G, which means that $A_1$ and $A_2$ are subsets of G. So (b-2) can be used here. We think that the fact that $A_1$ and $A_2$ are subsets of G does not need to be mentioned as it is implicit in the fact that $(a,b,c)$ and $(a_1,b_1,c_1)$ are in Trag (The contents in the first sentence of this paragraph indicates that these two facts are equivalent.).

(V-2) ($\alpha$) Let B be a subset of ${\mathbb{R}}^3$. Put $S_B:=\{(e,f,g):[e,f,g]\in B\}$. Then $S_B=\varnothing$ if and only if $B=\varnothing$.

Clearly if $B=\varnothing$ then $S_B=\varnothing$. Suppose that $S_B=\varnothing$. Then $S_B$ is well-defined. This means that $B\subseteq G$ (see (I-4)⇔(I-6)). So if $B\ne \varnothing$, then $S_B\ne \varnothing$, which is a contradiction. Thus $B=\varnothing$. Hence ($\alpha$) holds.

$(a_1,b_1,c_1)\in$ Trag means that $[a_1,b_1,c_1]\in G$. So $A_1⫋G$. Thus $A_1\setminus A_2⫋G⫋{\mathbb{R}}^3$. Thus, by ($\alpha$), $\{(e,f,g):[e,f,g]\in A_1\setminus A_2\}=\varnothing$ if and only if $A_1\setminus A_2=\varnothing$.

It is easy to see that the fact that $A_1\setminus A_2⫋{\mathbb{R}}^3$ is implicit in the fact that $(a_1,b_1,c_1)\in$ Trag. Also ($\alpha$) can be used directly without citing as it is easy to see. So we think we can directly write $\{(e,f,g):[e,f,g]\in A_1\setminus A_2\}=\varnothing$ if and only if $A_1\setminus A_2=\varnothing$.

Remark 4.9. 

($\alpha$) Suppose that (a)⇒(b). Clearly if (a${\phantom{,}}^{\prime }$)⇒(a) and (b)⇒(b${\phantom{,}}^{\prime }$), then (a${\phantom{,}}^{\prime }$)⇒(b${\phantom{,}}^{\prime }$). Of course, (c)=(c${\phantom{,}}^{\prime }$) is a special case of (c)⇔(c${\phantom{,}}^{\prime }$), (c)⇔(c${\phantom{,}}^{\prime }$) is a special case of (c)⇒(c${\phantom{,}}^{\prime }$).

In this paper, we give some conclusions in the form of “(a)⇒(b)”. These conclusions include “Theorem 4.1(ii)⇒Theorem 4.1(iv)” (see the proof of Theorem 4.1), “Proposition 4.2(ii)⇒Proposition 4.2(iv)” (see the proof of Proposition 4.2) and some conclusions in Remark 4.3. We also give several equivalent forms of Theorem 4.1(ii), Theorem 4.1(iv), Proposition 4.2(ii), Proposition 4.2(iv), respectively. By ($\alpha$), it is easy to obtain various conclusions in the form of “(a)⇒(b)” from certain conclusions in this paper. We will not list them one by one as they are easy to see. Below are a few examples.

We know that Remark 4.4(ā)(⇔Remark 4.4(a))⇔Theorem 4.1(ii), Remark 4.4(b )(⇔Remark 4.4(b))⇔Proposition 4.2(ii), and Theorem 4.1(ii)⇒Proposition 4.2(ii). So Remark 4.4(ā)⇒Remark 4.4(b ).

We know that Remark 4.5(ā)(⇔Remark 4.5(a))⇔Theorem 4.1(iv), Remark 4.5(b )(⇔Remark 4.5(b))⇔Proposition 4.2(iv), and Theorem 4.1(iv)⇒Proposition 4.2(iv). So Remark 4.5(ā)⇒Remark 4.5(b ).

We began to consider the contents of this paper after the corresponding author of this paper independently gave all contents of ChinaXiv:202507.00428 (see https://chinaxiv.org/abs/202507.00428). The corresponding author of this paper also independently gave at least the following contents of this paper: all sentences that contain the expression “the unique element of”, Remarks 4.4, 4.5 and 4.6, and clauses (i), (ii), (iii) and (iv) of Section 6.

5. Some Relationships Between Tag and Tap, and Between Trag and Trap

Let A be a set. A mapping $f:A\to A$ is said to be the identity mapping on A if $f\left(x\right)=x$ for each $x\in A$. A mapping g is said to be an identity mapping if there is a set S and g is the identity mapping on S.

Define Trap${\phantom{,}}^1:=\{(a,b,c,d):(a,b,c,d)\in Trapandb=c\}$. Clearly Trap${\phantom{,}}^1⫋$ Trap and Trap${\phantom{,}}^1=\{(a,b,b,c):(a,b,b,c)\in Trap\}$.

Proposition 5.1. 

(i) Trag = Trap${\phantom{,}}^1$. (ii) Define a mapping $K:Trag\to Tra{p}^1$ as follows: for each $u\in$ Trag, find an $[a,b,c]\in G$ satisfying $u=(a,b,c)$, and then define $K\left(u\right)$ to be $(a,b,b,c)$. Then K is the identity mapping on Trag.

Proof. 

Remark 3.4(ii) implies that Trag ⊆ Trap${\phantom{,}}^1$. Remark 3.4(v) implies that Trap${\phantom{,}}^1$⊆ Trag. (For each $a,b,c\in \mathbb{R}$, $(a,b,b,c)\in$ Trap if and only if $(a,b,b,c)\in$ Trap${\phantom{,}}^1$.) So (i) is true.

We claim the following (a) and (b). (a) K is well-defined; that is, by virtue of K, for each $u\in$ Trag, (a-1) $K\left(u\right)$ is one element, and (a-2) $K\left(u\right)\in Tra{p}^1$. (b) For each $u\in$ Trag, $K\left(u\right)=u$.

Let $u\in$ Trag. Then there is an $[a,b,c]\in G$ satisfying $u=(a,b,c)$. Thus $(a,b,b,c)$ is a value of $K\left(u\right)$ (Formally, $K\left(u\right)$ may have multiple values). By Remark 3.4(ii), $(a,b,c)=u\in$ Trag implies that $(a,b,b,c)\in$ Trap, which means that $(a,b,b,c)\in$ Trap${\phantom{,}}^1$ (see also (I) below), and that $(a,b,c)=(a,b,b,c)$. So to show (a) and (b), we only need to show (c) $K\left(u\right)$ is one element. (Suppose that $K\left(u\right)$ is one element. Then $K\left(u\right)=(a,b,b,c)$. Hence $K\left(u\right)\in$ Trap${\phantom{,}}^1$ and $K\left(u\right)=(a,b,b,c)=(a,b,c)=u$. So (a) and (b) hold.)

Let $[a_1,b_1,c_1]$ be an element of G which satisfies $u=(a_1,b_1,c_1)$. Then $(a_1,b_1,b_1,c_1)$ is a value of $K\left(u\right)$. To show (c), we only need to show that $(a,b,b,c)=(a_1,b_1,b_1,c_1)$. (If this is true, then $K\left(u\right)$ can only be the element $(a,b,b,c)$, and so (c) holds.) Notice that $(a,b,c)=u=(a_1,b_1,c_1)\in$ Trag. Thus $(a,b,b,c)=(a_1,b_1,b_1,c_1)$, as by Remark 3.4(ii), $(a,b,c)=(a,b,b,c)$ and $(a_1,b_1,c_1)=(a_1,b_1,b_1,c_1)$. Hence (c) is proved. So (a) and (b) hold.

Combining (i), (a) and (b) yields that K is the identity mapping on Trag. So (ii) is true. (Clearly that K is the identity mapping on Trag also implies (i), (a) and (b).) The proof is completed.

(I) By Remark 3.4(i), $(a,b,c)=u\in$ Trag (obviously, $a,b,c\in \mathbb{R}$ in this case) implies that $(a,b,b,c)\in$ Trap, which means that $(a,b,b,c)\in$ Trap${\phantom{,}}^1$.

□

Remark 5.2. 

In this remark, the symbols are consistent with those in the above proof of Proposition 5.1.

(i) From the above proof of Proposition 5.1, we can see (i-1) Remark 3.4(ii)(v) imply Proposition 5.1; (i-2) (a) Remark 3.4(ii) implies (a) and (b). Obviously, combining (a) and (b) yields Remark 3.4(ii); Proposition 5.1(ii) implies Remark 3.4(ii)(v).

(ii) The above proof of Proposition 5.1 will become a new proof of Proposition 5.1 if the contents from “Let $[a_1,b_1,c_1]$ be an element of G ” to “Hence (c) is proved.” in it are replaced by the contents in the following clause (ii-1).

(ii-1) By Remark 4.4(b ), $[a,b,c]$ is the unique element of G that satisfies $u=(a,b,c)$. Then $K\left(u\right)$ can only be the element $(a,b,b,c)$. So (c) holds.

(c) can be stated as “Let $u\in$ Trag. Then $K\left(u\right)$ is one element.” (c) holds means that (a-1) holds.

(iii) Clearly Trap${\phantom{,}}^1$= Trag $=\left\{\right(a,b,c):[a,b,c]\in G\}=\left\{\right(a,b,b,c):[a,b,c]\in G\}=\left\{\right(a,b,b,c):[a,b,b,c]\in T\}$, where the third = follows from Remark 3.4(ii) or Proposition 5.1(ii), and the fourth = follows from the fact that $[a,b,c]\in G$ if and only if $[a,b,b,c]\in T$.

Define Tap${\phantom{,}}^1:=\{(a,b,c,d):(a,b,c,d)\in Tapandb=c\}$. Clearly Tap${\phantom{,}}^1⫋$ Tap and Tap${\phantom{,}}^1=\{(a,b,b,c):(a,b,b,c)\in Tap\}$.

Proposition 5.3. 

(i) Tag = Tap${\phantom{,}}^1$. (ii) Define a mapping $L:Tag\to Ta{p}^1$ as follows: for each $u\in$ Tag, find an $[a,b,c]\in G_0$ satisfying $u=(a,b,c)$, and then define $L\left(u\right)$ to be $(a,b,b,c)$. Then L is the identity mapping on Tag.

Proof. 

Remark 2.4(ii) implies that Tag ⊆ Tap${\phantom{,}}^1$. Remark 2.4(v) implies that Tap${\phantom{,}}^1$⊆ Tag. (For each $a,b$ and c in $\mathbb{R}$, $(a,b,b,c)\in$ Tap if and only if $(a,b,b,c)\in$ Tap${\phantom{,}}^1$.) So (i) is true.

We claim the following (a) and (b). (a) L is well-defined; that is, by virtue of L, for each element $u\in$ Tag, (a-1) $L\left(u\right)$ is one element, and (a-2) $L\left(u\right)\in Ta{p}^1$. (b) For each $u\in$ Tag, $L\left(u\right)=u$.

Let $u\in$ Tag. We can find an $[a,b,c]\in G_0$ satisfying $u=(a,b,c)$. Then $(a,b,b,c)$ is a value of $L\left(u\right)$ (Formally, $L\left(u\right)$ may have multiple values). By Remark 2.4(ii), $(a,b,c)=u\in$ Tag implies that $(a,b,b,c)\in$ Tap, which means that $(a,b,b,c)\in$ Tap${\phantom{,}}^1$ (see also (I) below), and that $(a,b,c)=(a,b,b,c)$. So to show (a) and (b), we only need to show (c) $L\left(u\right)$ is one element. (Suppose that $L\left(u\right)$ is one element. Then $L\left(u\right)=(a,b,b,c)$. Hence $L\left(u\right)\in$ Tap${\phantom{,}}^1$ and $L\left(u\right)=(a,b,b,c)=(a,b,c)=u$. So (a) and (b) hold.)

Let $[a_1,b_1,c_1]$ be an element of $G_0$ which satisfies $u=(a_1,b_1,c_1)$. Then $(a_1,b_1,b_1,c_1)$ is a value of $L\left(u\right)$. To show (c), we only need to show that $(a,b,b,c)=(a_1,b_1,b_1,c_1)$. (If this is true, then $L\left(u\right)$ can only be the element $(a,b,b,c)$ in Tap${\phantom{,}}^1$, and so (c) holds.) Notice that $(a,b,c)=u=(a_1,b_1,c_1)\in$ Tag. Thus $(a,b,b,c)=(a_1,b_1,b_1,c_1)$, as by Remark 2.4(ii), $(a,b,c)=(a,b,b,c)$ and $(a_1,b_1,c_1)=(a_1,b_1,b_1,c_1)$. Hence (c) is proved. So (a) and (b) hold.

(i), (a) and (b) hold if and only if (ii) holds. So (ii) is proved as (i), (a) and (b) are proved.

(I) By Remark 2.4(i), $(a,b,c)=u\in$ Tag (obviously, $a,b,c\in \mathbb{R}$ in this case) implies that $(a,b,b,c)\in$ Tap, which means that $(a,b,b,c)\in$ Tap${\phantom{,}}^1$.

□

Remark 5.4. 

In this remark, the symbols are consistent with those in the above proof of Proposition 5.3.

(i) From the above proof of Proposition 5.3, we can see (i-1) Remark 2.4(ii)(v) imply Proposition 5.3; (i-2) Remark 2.4(ii) implies (a) and (b). Obviously, combining (a) and (b) yields Remark 2.4(ii); Proposition 5.3(ii) implies Remark 2.4(ii)(v).

(ii) The above proof of Proposition 5.3 remains true if the contents from “Let $[a_1,b_1,c_1]$ be an element of $G_0$ ” to “Hence (c) is proved.” in it are replaced by the contents in the following clause (ii-1) or by the contents in the following clause (ii-2).

(ii-1) By Remark 4.6(b ), $[a,b,c]$ is the unique element of $G_0$ that satisfies $u=(a,b,c)$. Then, by the definition of L, $L\left(u\right)$ can only be the element $(a,b,b,c)$. Hence (c) holds.

(ii-2) Note that $u\in Tag\subset Trag$ and $G_0\subset G$. Thus, by the definitions of L and K, each value of $L\left(u\right)$ is a value of $K\left(u\right)$, where K is defined in Proposition 5.1. Hence $L\left(u\right)$ is one element, as $K\left(u\right)$ is one element and $(a,b,b,c)$ is a value of $L\left(u\right)$.

(c) can be stated as “Let $u\in$ Tag. Then $L\left(u\right)$ is one element.” (c) holds means that (a-1) holds.

(iii) By Remark 4.5(b ), we have the fact that for each $u\in$ Tag, whether you perform the operation “find an $[a,b,c]\in G_0$ satisfying $u=(a,b,c)$” or the operation “find an $[a,b,c]\in G$ satisfying $u=(a,b,c)$”, the same one element $[a,b,c]$ will be found. (Conversely, this fact also implies Remark 4.5(b ).) So L is invariant if replace “find an $[a,b,c]\in G_0$” by “find an $[a,b,c]\in G$” in the definition of L. In other words, we can also define L as follows:

Define a mapping $L:Tag\to Ta{p}^1$ as follows: for each $u\in$ Tag, find an $[a,b,c]\in G$ satisfying $u=(a,b,c)$, and then define $L\left(u\right)$ to be $(a,b,b,c)$.

And if we define L in this way, then for each $u\in$ Tag, $L\left(u\right)=K\left(u\right)$, and hence $L\left(u\right)$ is one element as $K\left(u\right)$ is one element.

(iv) Clearly Tap${\phantom{,}}^1$= Tag $=\{(a,b,c):[a,b,c]\in G_0\}=\{(a,b,b,c):[a,b,c]\in G_0\}=\{(a,b,b,c):[a,b,b,c]\in T_0\}$, where the third = follows from Remark 2.4(ii) or Proposition 5.3(ii), and the fourth = follows from the fact that $[a,b,c]\in G_0$ if and only if $[a,b,b,c]\in T_0$.

6. Conclusions

In this paper, we give the representation uniqueness of the generalized triangular fuzzy numbers and the representation uniqueness of the generalized trapezoidal fuzzy numbers, which are (i) and (ii) listed below, respectively.

(i) (Remark 4.6(b )) For each u in Trag, there is an $[a,b,c]$ which is the unique element of G that satisfies $u=(a,b,c)$.

(ii) (Remark 4.4(ā)) For each u in Trap, there is an $[a,b,c,d]$ which is the unique element of T that satisfies $u=(a,b,c,d)$.

We show the representation uniqueness of the triangular fuzzy numbers and the representation uniqueness of the trapezoidal fuzzy numbers, which are (iii) and (iv) listed below, respectively.

(iii) (Remark 4.5(b )) For each u in Tag, there is an $[a,b,c]$ in $G_0$ which is the unique element of G that satisfies $u=(a,b,c)$.

(iv) (Remark 4.5(ā)) For each u in Tap, there is an $[a,b,c,d]$ in $T_0$ which is the unique element of T that satisfies $u=(a,b,c,d)$.

We point out that (ii)⇒(i)⇒(iii) and (ii)⇒(iv)⇒(iii) (see Section 4).

Furthermore, we obtain the following relationship among Tag, Trag, Tap and Trap: Tap ⫋ Trap, Trag ⫋ Trap, Tag ⫋ Tap, and Tag ⫋ Trag.

The results of this paper have potential effects on the analysis and applications of the generalized triangular fuzzy numbers and the generalized trapezoidal fuzzy numbers.