# A Generalization of Trapezoidal Fuzzy Numbers Based on Modal Interval Theory

^{*}

^{†}

## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

#### 2.1. Modal Intervals

#### 2.2. Fuzzy Numbers

## 3. Modal Interval Trapezoidal Fuzzy Numbers

**Definition**

**1.**

**Definition**

**2.**

- A is a proper-improper MITFN (MITFN${}_{P}^{I}$) if $supp\left(A\right)$ is a proper interval and $core\left(A\right)$ is an improper interval. We denote it by $A=\left(\right)open="("\; close=")">set\left(A\right),\exists ,\forall $;
- A is an improper-proper MITFN (MITFN${}_{I}^{P}$) if $supp\left(A\right)$ is an improper interval and $core\left(A\right)$ is a proper interval. We denote it by $A=\left(\right)open="("\; close=")">set\left(A\right),\forall ,\exists $;
- A is a proper-proper MITFN (MITFN${}_{P}^{P}$) if both the support and core of A are proper intervals. We denote it by $A=\left(\right)open="("\; close=")">set\left(A\right),\exists ,\exists $;
- A is an improper-improper MITFN (MITFN${}_{I}^{I}$) if both the support and core of A are improper intervals. We denote it by $A=\left(\right)open="("\; close=")">set\left(A\right),\forall ,\forall $.

**Proposition**

**1.**

**Proof.**

**Proposition**

**2.**

- A is an MITFN${}_{P}^{I}\iff {a}_{1}\le {a}_{3}\le {a}_{2}\le {a}_{4}$;
- A is an MITFN${}_{I}^{P}\iff {a}_{4}\le {a}_{2}\le {a}_{3}\le {a}_{1}$;
- A is an MITFN${}_{P}^{P}\iff {a}_{1}\le {a}_{2}\le {a}_{3}\le {a}_{4}$;
- A is an MITFN${}_{I}^{I}\iff {a}_{4}\le {a}_{3}\le {a}_{2}\le {a}_{1}$.

**Proof.**

- If A is an MITFN${}_{P}^{I}$, then $supp\left(A\right)=\left(\right)open="["\; close="]">{a}_{1},{a}_{4}$ is proper, that is, ${a}_{1}\le {a}_{4}$ and $core\left(A\right)=\left(\right)open="["\; close="]">{a}_{2},{a}_{3}$ is improper, that is, ${a}_{3}\le {a}_{2}.$As Definition 1 establishes $set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{2},{a}_{3}\subseteq set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{1},{a}_{4}$, in this case, $set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{2},{a}_{3}=\left(\right)open="["\; close="]">{a}_{3},{a}_{2}$ and $set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{1},{a}_{4}=\left(\right)open="["\; close="]">{a}_{1},{a}_{4}$ which means $\left(\right)open="["\; close="]">{a}_{3},{a}_{2}$ $\left(\right)$, that is ${a}_{3}\ge {a}_{1}$ and ${a}_{2}\le {a}_{4}.$From ${a}_{1}\le {a}_{4},$ ${a}_{3}\le {a}_{2},{a}_{3}\ge {a}_{1},{a}_{2}\le {a}_{4}$, it follows that ${a}_{1}\le {a}_{3}\le {a}_{2}\le {a}_{4}.$
- If A is an MITFN${}_{I}^{P}$, $supp\left(A\right)=\left(\right)open="["\; close="]">{a}_{1},{a}_{4}$ is an improper interval, that is, ${a}_{4}\le {a}_{1}$ and $core\left(A\right)=\left(\right)open="["\; close="]">{a}_{2},{a}_{3}$ is a proper interval, that is, ${a}_{2}\le {a}_{3}.$As it must be that $set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{2},{a}_{3}\subseteq set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{1},{a}_{4}$ and $set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{2},{a}_{3}=\left(\right)open="["\; close="]">{a}_{2},{a}_{3}$ $set\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{1},{a}_{4}=\left(\right)open="["\; close="]">{a}_{4},{a}_{1}$ so it follows that $\left(\right)open="["\; close="]">{a}_{2},{a}_{3}$ $\left(\right)$, that is, ${a}_{2}\ge {a}_{4}$ and ${a}_{3}\le {a}_{1}.$From ${a}_{4}\le {a}_{1},$ ${a}_{2}\le {a}_{3},{a}_{2}\ge {a}_{4},{a}_{3}\le {a}_{1}$, it follows that ${a}_{4}\le {a}_{2}\le {a}_{3}\le {a}_{1}.$

**Proposition**

**3.**

**Proof.**

- If both $supp\left(A\right)$ and $core\left(A\right)$ are proper intervals, then ${a}_{1}\le {a}_{4}$ and ${a}_{2}\le {a}_{3}$ so ${a}_{1}+{a}_{2}\le {a}_{3}+{a}_{4}$ and $\left(\right)$ is a proper interval.
- If both $supp\left(A\right)$ and $core\left(A\right)$ are improper intervals, then ${a}_{1}\ge {a}_{4}$ and ${a}_{2}\ge {a}_{3}$ so ${a}_{1}+{a}_{2}\ge {a}_{3}+{a}_{4}$ and $\left(\right)$ is an improper interval.
- If $supp\left(A\right)$ is a proper interval and $core\left(A\right)$ is an improper one, then $set\left(\right)open="("\; close=")">core\left(A\right)$ and $set\left(\right)open="("\; close=")">supp\left(A\right)$. As $set\left(\right)open="("\; close=")">core\left(A\right),$ it holds that ${a}_{3}\ge {a}_{1}$ and ${a}_{2}\le {a}_{4}.$ Thus, $\left(\right)$ is a proper interval.
- If $supp\left(A\right)$ is an improper interval and $core\left(A\right)$ is a proper interval, then $set\left(\right)open="("\; close=")">core\left(A\right)$ and $set\left(\right)open="("\; close=")">supp\left(A\right)$. As $set\left(\right)open="("\; close=")">core\left(A\right),$ it holds that ${a}_{2}\ge {a}_{4}$ and ${a}_{3}\le {a}_{1}.$ Thus, $\left(\right)$ is an improper interval.

**Dual operator**. If $A=\left(\right)open="("\; close=")">{A}^{\prime},{Q}_{1},{Q}_{2}$ $T{I}^{\ast}\left(\mathbb{R}\right)$ the dual operator on A, $dual\left(A\right)$ is defined as:

**Proper and improper operators**. If $A=\left(\right)open="("\; close=")">{A}^{\prime},{Q}_{1},{Q}_{2}$ $T{I}^{\ast}\left(\mathbb{R}\right)$ we define the proper operator on $A,$ as $prop\left(A\right)=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\exists $ and the improper operator on A as $impr\left(A\right)=\left(\right)open="("\; close=")">{A}^{\prime},\forall ,\forall $. Using the canonical notation, if $A=\left(\right)open="("\; close=")">\left(\right)open="["\; close="]">{a}_{1},{a}_{4}$, then:

#### 3.1. Graphical Representation of an MITFN in the Interval Plane

#### 3.2. The Lattice of MITFNs

**Lemma**

**1.**

**Proof.**

**Proposition**

**4.**

- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\exists $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\exists ,\exists $, then$A\subseteq B\iff set\left(\right)open="("\; close=")">supp\left(A\right)$ and $set\left(\right)open="("\; close=")">core\left(A\right)$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\exists $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\forall ,\forall $, then$A\subseteq B\iff A=B=\left(\right)open="("\; close=")">p,p,p,p$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\exists $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\exists ,\forall $, then$A\subseteq B\iff set\left(\right)open="("\; close=")">supp\left(A\right)$ and $core\left(A\right)=core\left(B\right)=\left(\right)open="["\; close="]">p,p$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\exists $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\forall ,\exists $, then$A\subseteq B\iff A=B=\left(\right)open="("\; close=")">p,p,p,p$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\forall ,\forall $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\exists ,\exists $, then$A\subseteq B\iff set\left(\right)open="("\; close=")">supp\left(A\right)\ne \varnothing $ and $set\left(\right)open="("\; close=")">core\left(A\right)\ne \varnothing $;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\forall ,\forall $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\exists ,\forall $, then$A\subseteq B\iff $ $set\left(\right)open="("\; close=")">supp\left(A\right)\ne \varnothing $ and $set\left(\right)open="("\; close=")">core\left(A\right)$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\forall ,\forall $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\forall ,\exists $, then$A\subseteq B\iff set\left(\right)open="("\; close=")">supp\left(A\right)$ and $set\left(\right)open="("\; close=")">core\left(A\right)\ne \varnothing $;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\forall $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\exists ,\forall $, then$A\subseteq B\iff set\left(\right)open="("\; close=")">supp\left(A\right)$ and $set\left(\right)open="("\; close=")">core\left(A\right)$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\exists ,\forall $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\forall ,\exists $, then$A\subseteq B\iff A=B=\left(\right)open="("\; close=")">p,p,p,p$;
- If $A=\left(\right)open="("\; close=")">{A}^{\prime},\forall ,\exists $ and $B=\left(\right)open="("\; close=")">{B}^{\prime},\exists ,\forall $, then$A\subseteq B\iff set\left(\right)open="("\; close=")">supp\left(A\right)\ne \varnothing $ and $core\left(A\right)=core\left(B\right)=\left(\right)open="["\; close="]">p,p$

**Proof.**

**Definition**

**3.**

- $inf\left(\right)open="\{"\; close="\}">A,B$ if $X\subseteq A,X\subseteq B$ and if there exists a $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ such that $D\subseteq A\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $D\subseteq B$, then $D\subseteq X$;
- $sup\left(\right)open="\{"\; close="\}">A,B$ if $A\subseteq Y,B\subseteq Y$ and if there exists a $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ such that $A\subseteq D\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $B\subseteq D$, then $Y\subseteq D.$

**Proposition**

**5.**

**Proof.**

- ${L}^{0}$ and ${L}^{1}$ are proper intervals.As ${L}^{0}$ and ${L}^{1}$ are proper intervals, then A and B are MITFNs${}_{P}^{P}$. Thus, ${A}^{1}\subseteq {A}^{0}$ and ${B}^{1}\subseteq {B}^{0}$ so ${L}^{1}\subseteq {L}^{0}$ and as ${L}^{1}$ and ${L}^{0}$ are proper intervals, $set\left(\right)open="("\; close=")">{L}^{1}$. If $X=\left(\right)open="("\; close=")">{L}^{0},{L}^{1}$, then $X\subseteq A$, $X\subseteq B$ and $X\in T{I}^{\ast}\left(\mathbb{R}\right)$. Moreover, if $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ conforms to $D\subseteq A\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $D\subseteq B$, then ${D}^{0}\subseteq {A}^{0}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}},$ ${D}^{0}\subseteq {B}^{0}$, ${D}^{1}\subseteq {A}^{1}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and ${D}^{1}\subseteq {B}^{1}$ so ${D}^{0}\subseteq {A}^{0}\wedge {B}^{0}$ and ${D}^{1}\subseteq {A}^{1}\wedge {B}^{1}$. That is, $D\subseteq X$; thus, $\left(\right)open="("\; close=")">{L}^{0},{L}^{1}$ Inf$\left(\right)open="\{"\; close="\}">A,B$
- ${L}^{0}$ is a proper interval and ${L}^{1}$ an improper interval.As ${L}^{0}={A}^{0}\wedge {B}^{0}$ is a proper interval, ${L}^{1}={A}^{1}\wedge {B}^{1}$ is an improper interval and $set\left(\right)open="("\; close=")">{A}^{1},$ $set\left(\right)open="("\; close=")">{B}^{1},$ and it follows that ${L}^{1}\subseteq {L}^{0}.$ We distinguish the following two cases according to the inclusion set of ${L}^{1}$ and ${L}^{0}$:
- (a)
- If $set\left(\right)open="("\; close=")">{L}^{1}$, we can proceed as in the first case.
- (b)
- If $set\left(\right)open="("\; close=")">{L}^{1}$, then let us prove that $X=\left(\right)open="("\; close=")">{L}^{0}\wedge {L}^{1},{L}^{1}$ corresponds to $inf\left(\right)open="\{"\; close="\}">A,B$. Notice that, if ${L}^{1}\subseteq {L}^{0}$, then ${L}^{0}\wedge {L}^{1}={L}^{1}$; thus, $X=\left(\right)open="("\; close=")">{L}^{1},{L}^{1}$. It is obvious that $X\in T{I}^{\ast}\left(\mathbb{R}\right)$. Moreover, ${X}^{0}\subseteq {A}^{0}$ and ${X}^{0}\subseteq {B}^{0}$ as ${X}^{0}={L}^{1}\subseteq {L}^{0}$. In a similar way, ${X}^{1}\subseteq {A}^{1}$ and ${X}^{1}\subseteq {B}^{1}$. Therefore, $X\subseteq A$ and $X\subseteq B$.If $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ conforms to $D\subseteq A\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $D\subseteq B$, then ${D}^{1}\subseteq {A}^{1}\wedge {B}^{1}={L}^{1}.$ Notice that, as ${L}^{1}$ is an improper interval, ${D}^{1}$ will also be an improper interval and then $set\left(\right)open="("\; close=")">{L}^{1}$.We must prove that ${D}^{0}\subseteq {L}^{1}$ and, consequently, $D\subseteq X$.
- If ${D}^{0}$ is an improper interval, as $set\left(\right)open="("\; close=")">{D}^{1}$, it follows that ${D}^{0}\subseteq {D}^{1}.$ As ${D}^{1}\subseteq {L}^{1}$, then ${D}^{0}\subseteq {L}^{1}$.
- If ${D}^{0}$ is a proper interval, as $set\left(\right)open="("\; close=")">{D}^{1}={D}^{0}$ and ${D}^{0}\subseteq {L}^{0}$, it follows that $set\left(\right)open="("\; close=")">{D}^{1}$ Using the inclusion $set\left(\right)open="("\; close=")">{L}^{1}$, we obtain $set\left(\right)open="("\; close=")">{L}^{1},$ which contradicts the hypothesis $set\left(\right)open="("\; close=")">{L}^{1}$.

- ${L}^{0}$ is an improper interval and ${L}^{1}$ a proper one.As ${L}^{0}={A}^{0}\wedge {B}^{0}$ is an improper interval, ${L}^{1}={A}^{1}\wedge {B}^{1}$ is a proper interval and $set\left(\right)open="("\; close=")">{A}^{1},$ $set\left(\right)open="("\; close=")">{B}^{1}$, and it follows that $set\left(\right)open="("\; close=")">{L}^{1}$; thus, the demonstration follows as in the first case.
- ${L}^{0}$ and ${L}^{1}$ are improper intervals.
- (a)
- If ${L}^{1}\subseteq {L}^{0}$, then $set\left(\right)open="("\; close=")">{L}^{0}$. Let us prove that $X=\left(\right)open="("\; close=")">{L}^{0}\wedge {L}^{1},{L}^{1}$ corresponds to $inf\left(\right)open="\{"\; close="\}">A,B$. Notice that if ${L}^{1}\subseteq {L}^{0}$, then ${L}^{0}\wedge {L}^{1}={L}^{1}$ and thus $X=\left(\right)open="("\; close=")">{L}^{1},{L}^{1}$. It is obvious that $X\in T{I}^{\ast}\left(\mathbb{R}\right)$. Moreover, ${X}^{0}\subseteq {A}^{0}$ and ${X}^{0}\subseteq {B}^{0}$ as ${X}^{0}={L}^{1}\subseteq {L}^{0}$. In a similar way, ${X}^{1}\subseteq {A}^{1}$ and ${X}^{1}\subseteq {B}^{1}$. Therefore, $X\subseteq A$ and $X\subseteq B$.If $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ conforms to $D\subseteq A\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $D\subseteq B$, then ${D}^{1}\subseteq {A}^{1}\wedge {B}^{1}={L}^{1}$ and ${D}^{0}\subseteq {A}^{0}\wedge {B}^{0}={L}^{0}$. This implies that ${D}^{0}$ and ${D}^{1}$ are improper intervals.Moreover, as $set\left(\right)open="("\; close=")">{D}^{1}$, then due to the modality of ${D}^{0}$ and ${D}^{1}$ it will be the case that ${D}^{0}\subseteq {D}^{1}$ and as ${D}^{1}\subseteq {L}^{1}$, it follows that ${D}^{0}\subseteq {L}^{1}$. Thus, $X=\left(\right)open="("\; close=")">{L}^{0}\wedge {L}^{1},{L}^{1}.$
- (b)
- If ${L}^{0}\subseteq {L}^{1}$, then $set\left(\right)open="("\; close=")">{L}^{1}$. Let us prove that $X=\left(\right)open="("\; close=")">{L}^{0},{L}^{1}$ corresponds to $inf\left(\right)open="\{"\; close="\}">A,B$. If $X=\left(\right)open="("\; close=")">{L}^{0},{L}^{1}$, then $X\subseteq A$, $X\subseteq B$ and $X\in T{I}^{\ast}\left(\mathbb{R}\right)$. Moreover, if $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ conforms to $D\subseteq A\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $D\subseteq B$, then ${D}^{0}\subseteq {A}^{0}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}},$ ${D}^{0}\subseteq {B}^{0}$, ${D}^{1}\subseteq {A}^{1}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and ${D}^{1}\subseteq {B}^{1}$ so ${D}^{0}\subseteq {A}^{0}\wedge {B}^{0}$ and ${D}^{1}\subseteq {A}^{1}\wedge {B}^{1}$. That is, $D\subseteq X$; thus, $\left(\right)open="("\; close=")">{L}^{0},{L}^{1}$ Inf$\left(\right)open="\{"\; close="\}">A,B$
- (c)
- If ${L}^{0}\le {L}^{1}$ or ${L}^{1}\le {L}^{0}$, then let us prove that $X=\left(\right)open="("\; close=")">{L}^{0}\wedge {L}^{1},{L}^{1}$ corresponds to $inf\left(\right)open="\{"\; close="\}">A,B$. Notice that ${L}^{0}\wedge {L}^{1}\subseteq {L}^{1}$ as ${L}^{0}\wedge {L}^{1}$ and ${L}^{1}$ are improper intervals, $set\left(\right)open="("\; close=")">{L}^{1},$ thus $X\in T{I}^{\ast}\left(\mathbb{R}\right)$ and obviously $X\subseteq A$ and $X\subseteq B$.If $D\in T{I}^{\ast}\left(\mathbb{R}\right)$ conforms to $D\subseteq A\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}$and $D\subseteq B$, then ${D}^{1}\subseteq {A}^{1}\wedge {B}^{1}={L}^{1}$ and ${D}^{0}\subseteq {A}^{0}\wedge {B}^{0}={L}^{0}$. As ${L}^{1}$ and ${L}^{0}$ are both improper, ${D}^{1}$ and ${D}^{0}$ will also be improper intervals.Moreover, as $set\left(\right)open="("\; close=")">{D}^{1}$, then due to the modality of ${D}^{0}$ and ${D}^{1}$, it will be the case that ${D}^{0}\subseteq {D}^{1}$ and as ${D}^{1}\subseteq {L}^{1}$, it follows that ${D}^{0}\subseteq {L}^{1}$ and consequently ${D}^{0}\subseteq {L}^{0}\wedge {L}^{1}$. That is $D\subseteq X$, thus $X=\left(\right)open="("\; close=")">{L}^{0}\wedge {L}^{1},{L}^{1}$.

**Proposition**

**6.**

**Proof.**

- If $set\left(\right)open="("\; close=")">{L}^{1}$, then $set\left(\right)open="("\; close=")">{M}^{1}$ and so:$$\begin{array}{c}\hfill sup\left(\right)open="\{"\; close="\}">A,B\\ =& dual\left(\right)open="("\; close=")">du\left(\right)open="("\; close=")">{A}^{0}\wedge du\left(\right)open="("\; close=")">{B}^{0}\hfill & ,du\left(\right)open="("\; close=")">{A}^{1}\end{array}$$
- If $set\left(\right)open="("\; close=")">{L}^{1}$, then$$\begin{array}{c}\hfill sup\left(\right)open="\{"\; close="\}">A,B\\ =& dual\left(\right)open="("\; close=")">du\left(\right)open="("\; close=")">{A}^{0}\wedge du\left(\right)open="("\; close=")">{B}^{0}\hfill & \wedge du\left(\right)open="("\; close=")">{A}^{1}\end{array},du\left(\right)open="("\; close=")">{A}^{1}$$

## 4. Interpretability of the Calculations

**Definition**

**4.**

**Proposition**

**7.**

**Proof.**

**Theorem**

**1.**

**Proof.**

**Corollary**

**1.**

**Corollary**

**2.**

**Definition**

**5.**

**Definition**

**6.**

**Example**

**1.**

- $\forall \alpha \in \left(\right)open="["\; close="]">0,\frac{3}{4}$$a+x=b;$
- $\forall \alpha \in \left(\right)open="["\; close="]">\frac{3}{4},1$$a+x=b.$

- $\forall \alpha \in \left(\right)open="["\; close="]">0,0.5$$a+x=b;$
- $\forall \alpha \in \left(\right)open="["\; close="]">0.5,0.75$$a+x=b;$
- $\forall \alpha \in \left(\right)open="["\; close="]">0.75,1$$a+x=b.$

## 5. Conclusions

## Author Contributions

## Conflicts of Interest

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**MDPI and ACS Style**

Jorba, L.; Adillon, R.
A Generalization of Trapezoidal Fuzzy Numbers Based on Modal Interval Theory. *Symmetry* **2017**, *9*, 198.
https://doi.org/10.3390/sym9100198

**AMA Style**

Jorba L, Adillon R.
A Generalization of Trapezoidal Fuzzy Numbers Based on Modal Interval Theory. *Symmetry*. 2017; 9(10):198.
https://doi.org/10.3390/sym9100198

**Chicago/Turabian Style**

Jorba, Lambert, and Romà Adillon.
2017. "A Generalization of Trapezoidal Fuzzy Numbers Based on Modal Interval Theory" *Symmetry* 9, no. 10: 198.
https://doi.org/10.3390/sym9100198