# The Submodular Inequality of Aggregation Operators

^{*}

## Abstract

**:**

## 1. Introduction

- (1)
- Functional equations can be used to choose an appropriate aggregation operator;
- (2)
- Functional equations can be used to study the properties of fusion methods since they can characterize the corresponding aggregation operators.

## 2. Preliminaries

**Definition**

**1.**

**Definition**

**2.**

**Definition**

**3.**

**Definition**

**4.**

- (1)
- Let N be a strong negation. The binary operator ${A}^{*}:{[0,1]}^{2}\to [0,1]$ defined by$${A}^{*}(x,y)=N\left(A(N\left(x\right),N\left(y\right))\right)$$
- (2)
- Let φ be a bijection on $[0,1]$. The binary operator ${A}_{\phi}:{[0,1]}^{2}\to [0,1]$ defined by$${A}_{\phi}(x,y)={\phi}^{-1}\left(A(\phi \left(x\right),\phi \left(y\right))\right)$$

**Remark**

**1.**

**Definition**

**5.**

**Definition**

**6.**

**Definition**

**7.**

**Definition**

**8.**

## 3. Submodular Inequality of Aggregation Operators under Duality and Isomorphism

**Theorem**

**1.**

- (1)
- $A{\u2aaf}_{sm}B$;
- (2)
- Their N-duals satisfy ${B}^{*}{\u2aaf}_{sm}{A}^{*}$.

**Proof.**

#### Submodular Inequality of Aggregation Operators under Isomorphism

**Theorem**

**2.**

- (1)
- If φ is increasing, then $A{\u2aaf}_{sm}B$ if and only if ${A}_{\phi}{\u2aaf}_{sm}{B}_{\phi}$;
- (2)
- If φ is decreasing, then $A{\u2aaf}_{sm}B$ if and only if ${B}_{\phi}{\u2aaf}_{sm}{A}_{\phi}$.

**Proof.**

- (1)
- If $\phi $ is increasing, then $\phi \left(z\right)\le \phi \left(x\right)$ and$$\begin{array}{ccc}\hfill {A}_{\phi}\left(x,{B}_{\phi}\left(y,z\right)\right)& =& {\phi}^{-1}\left(A(\phi \left(x\right),\phi \circ {\phi}^{-1}\left(B(\phi \left(y\right),\phi \left(z\right))\right))\right)\hfill \\ & =& {\phi}^{-1}\left(A(\phi \left(x\right),B(\phi \left(y\right),\phi \left(z\right)))\right)\hfill \\ & \le & {\phi}^{-1}\left(B(A(\phi \left(x\right),\phi \left(y\right)),\phi \left(z\right))\right)\hfill \\ & =& {\phi}^{-1}\left(B(\phi \circ {\phi}^{-1}\left(A(\phi \left(x\right),\phi \left(y\right))\right),\phi \left(z\right))\right)\hfill \\ & =& {B}_{\phi}\left({A}_{\phi}\left(x,y\right),z\right).\hfill \end{array}$$Thus, ${A}_{\phi}{\u2aaf}_{sm}{B}_{\phi}$.
- (2)
- If $\phi $ is decreasing, then the proof is similar to that of item (1).

## 4. Submodular Inequality of the Ordinal Sum of Conjunctors

**Theorem**

**3.**

**Proof.**

- (1)
- $x,y,z\in [{a}_{i},{b}_{i}],z\le x,i\in I$. For the increasing bijection ${\phi}_{i}:[{a}_{i},{b}_{i}]\to [0,1]$, $x\to \frac{x-{a}_{i}}{{b}_{i}-{a}_{i}}$, and according to the ordinal sum of $A,B$ and Theorem 2, we have$$\begin{array}{c}\hfill A(x,B(y,z))={\phi}_{i}^{-1}\circ {A}_{i}({\phi}_{i}\left({x}^{\prime}\right),{B}_{i}({\phi}_{i}\left({y}^{\prime}\right),{\phi}_{i}\left({z}^{\prime}\right)))\\ \hfill \le {\phi}_{i}^{-1}\circ {B}_{i}({A}_{i}({\phi}_{i}\left({x}^{\prime}\right),{\phi}_{i}\left({y}^{\prime}\right)),{\phi}_{i}\left({z}^{\prime}\right))=B(A(x,y),z).\end{array}$$
- (2)
- $y\le z\le x$.
- $y\notin [{a}_{i},{b}_{i}]$ for any $i\in I$. Hence, $B\left(y,z\right)=y,A\left(x,y\right)=y$, and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=A(x,y)=y=B(y,z)=B\left(A\right(x,y),z).\end{array}$$
- $y\in [{a}_{i},{b}_{i}],z\in [{a}_{i},{b}_{i}],x\notin [{a}_{i},{b}_{i}]$ for some $i\in I$. Hence, $x>{b}_{i},A\left(x,y\right)=y$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=min(x,B(y,z\left)\right)=B(y,z)=B\left(A\right(x,y),z).\end{array}$$
- $y\in [{a}_{i},{b}_{i}],z\notin [{a}_{i},{b}_{i}],x\notin [{a}_{i},{b}_{i}]$ for some $i\in I$. Hence, $B\left(y,z\right)=y,A\left(x,y\right)=y$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=A(x,y)=y=B(y,z)=B\left(A\right(x,y),z).\end{array}$$

- (3)
- $z\le y\le x$.
- $z\notin [{a}_{i},{b}_{i}]$ for any $i\in I$. Hence, $B\left(y,z\right)=z,A\left(x,z\right)=z$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=A(x,z)=z=min\left(A\right(x,y),z)=B\left(A\right(x,y),z).\end{array}$$
- $z\in [{a}_{i},{b}_{i}],y\in [{a}_{i},{b}_{i}],x\notin [{a}_{i},{b}_{i}]$ for some $i\in I$. Hence, $x>{b}_{i},A\left(x,y\right)=y$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=min(x,B(y,z\left)\right)=B(y,z)=B\left(A\right(x,y),z).\end{array}$$
- $z\in [{a}_{i},{b}_{i}],y\notin [{a}_{i},{b}_{i}],x\notin [{a}_{i},{b}_{i}]$ for some $i\in I$. Hence, $B\left(y,z\right)=z,A\left(x,z\right)=z,A(x,y)\ge {b}_{i}$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=A(x,z)\le z=B\left(A\right(x,y),z).\end{array}$$

- (4)
- $z\le x\le y$.
- $z\notin [{a}_{i},{b}_{i}]$ for any $i\in I$. Hence, $B\left(y,z\right)=z,A\left(x,z\right)=z$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=A(x,z)=z=min\left(A\right(x,y),z)=B\left(A\right(x,y),z).\end{array}$$
- $z\in [{a}_{i},{b}_{i}],x\in [{a}_{i},{b}_{i}],y\notin [{a}_{i},{b}_{i}]$ for some $i\in I$. Hence, $y>{b}_{i},B(y,z)=B({b}_{i},z),$$A(x,y)=A(x,{b}_{i})$ and$$\begin{array}{c}\hfill A(x,B(y,z))=A(x,B({b}_{i},z))\le B(A(x,{b}_{i}),z)=B(A(x,y),z).\end{array}$$
- $z\in [{a}_{i},{b}_{i}],x\notin [{a}_{i},{b}_{i}],y\notin [{a}_{i},{b}_{i}]$ for some $i\in I$. Hence, $x,y>{b}_{i},A(x,y)\ge {b}_{i}$ and$$\begin{array}{c}\hfill A(x,B(y,z\left)\right)=A(x,z)=z\le B\left(A\right(x,y),z).\end{array}$$

**Remark**

**2.**

**Example**

**1.**

## 5. Submodular Inequality of T-Norm and T-Conorm

#### 5.1. Submodular Inequality of T-Norm over T-Conorm

**Proposition**

**1.**

**Proof.**

- $y\le z$.$$T\left(x,{S}_{M}\left(y,z\right)\right)=T\left(x,z\right)\le z\le {S}_{M}\left(T\left(x,y\right),z\right).$$
- $y>z$.$$T\left(x,{S}_{M}\left(y,z\right)\right)\phantom{\rule{4.pt}{0ex}}=\phantom{\rule{4.pt}{0ex}}T\left(x,y\right)\le {S}_{M}\left(T\left(x,y\right),z\right).$$

**Example**

**2.**

**Lemma**

**1.**

**Proof.**

**Theorem**

**4.**

**Proof.**

**Example**

**3.**

**Theorem**

**5.**

**Proof.**

- $x\ge {N}_{A}\left(y\right)$.The proof is the inversion of the arguments above.
- $x<{N}_{A}\left(B\left(y,z\right)\right)\le {N}_{A}\left(y\right)$.In this case, $A(x,B(y,z\left)\right)=0$, $B\left(A\right(x,y),z)=B(0,z)=z\ge A(x,B(y,z\left)\right)$.
- ${N}_{A}\left(B\left(y,z\right)\right)\le x<{N}_{A}\left(y\right)$.In this case, $t\left(x\right)+t\left(y\right)>1,A\left(x,y\right)=0,t\left(x\right)+t\left(B\left(y,z\right)\right)\le 1$. By setting $a=t\left(x\right),b=t\left(y\right),c=t\left(z\right)$ in Equation (3), we have$${T}_{h}\left(1,c\right)-{T}_{h}\left(b,c\right)\le 1-b\le a.$$Hence, $c-{h}^{-1}\left(h\left(b\right)+h\left(c\right)\right)\le a$ and $t\left(z\right)-t\circ {g}^{-1}\left(g\left(y\right)+g\left(z\right)\right)\le t\left(x\right)$. We then obtain$$t\left(z\right)\le t\left(x\right)+t\circ {g}^{-1}\left(g\left(y\right)+g\left(z\right)\right),$$$$z\ge {t}^{-1}\left(t\left(x\right)+t\circ {g}^{-1}\left(g\left(y\right)+g\left(z\right)\right)\right).$$Thus, we have$$B\left(A\right(x,y),z)=B(0,z)=z\ge A(x,B(y,z\left)\right).$$

**Theorem**

**6.**

**Theorem**

**7.**

**Proof.**

- $x\ge {N}_{A}\left(y\right),z\le {N}_{B}\left(y\right)$. In this case, we have $t\left(x\right)+t\left(y\right)\le 1$ and $g\left(y\right)+g\left(z\right)\le 1$. Moreover, since $B\left(y,z\right)\ge y$ and $A\left(x,y\right)\le y$, by the monotonicity of t and g, we have $t\left(B\left(y,z\right)\right)+t\left(x\right)\le 1$ and $g\left(A\left(x,y\right)\right)+g\left(z\right)\le 1$. Equation (6) can thus be stated as follows:$${t}^{-1}\left(t\left(x\right)+t\circ {g}^{-1}\left(g\left(y\right)+g\left(z\right)\right)\right)\le {g}^{-1}\left(g\circ {t}^{-1}(t\left(x\right)+t\left(y\right))+g\left(z\right)\right),$$$$t\left(x\right)+t\circ {g}^{-1}\left(g\left(y\right)+g\left(z\right)\right)\ge t\circ {g}^{-1}\left(g\circ {t}^{-1}(t\left(x\right)+t\left(y\right))+g\left(z\right)\right),$$$$u+{h}^{-1}\left(h\left(v\right)+h\left(w\right)\right)\ge {h}^{-1}\left(h\left(u+v\right)+h\left(w\right)\right)$$$$\begin{array}{c}\hfill {h}^{-1}\left(h\left(a\right)+h\left(c\right)\right)-{h}^{-1}\left(h\left(b\right)+h\left(c\right)\right)\le a-b\end{array}$$
- $x\ge {N}_{A}\left(y\right),{N}_{B}\left(y\right)<z\le {N}_{B}\left(A\left(x,y\right)\right)$. In this case, we have $t\left(x\right)+t\left(y\right)\le 1$ and $g\left(y\right)+g\left(z\right)>1$ and $g\circ {t}^{-1}\left(t\left(x\right)+t\left(y\right)\right)+g\left(z\right)\le 1$. Hence, $B\left(y,z\right)=1$, $A\left(x,B\left(y,z\right)\right)=x$. Equation (6) can thus be stated as follows:$$x\le {g}^{-1}\left(g\circ {t}^{-1}(t\left(x\right)+t\left(y\right))+g\left(z\right)\right),$$$$t\left(x\right)\ge t\circ {g}^{-1}\left(g\circ {t}^{-1}(t\left(x\right)+t\left(y\right))+g\left(z\right)\right),$$$$\begin{array}{c}\hfill u\ge {h}^{-1}\left(h\left(u+v\right)+h\left(w\right)\right)\end{array}$$$${T}_{h}\left(a,c\right)-{T}_{h}\left(b,c\right)\le {T}_{h}\left(a,c\right)\le a-b$$
- $x\ge {N}_{A}\left(y\right),z>{N}_{B}\left(A\left(x,y\right)\right)$. In this case, we have $B\left(A(x,y),z\right)=1$ and the result.

- $x\ge {N}_{A}\left(y\right),z\le {N}_{B}\left(y\right).$The proof is the inversion of the arguments above.
- $x\ge {N}_{A}\left(y\right),{N}_{B}\left(y\right)<z\le {N}_{B}\left(A\left(x,y\right)\right)$.The proof is the inversion of the arguments above.
- $x\ge {N}_{A}\left(y\right),{N}_{B}\left(y\right)<{N}_{B}\left(A\left(x,y\right)\right)<z$.In this case, we have $B\left(A\left(x,y\right),z\right)=1$, and the result is trivial.
- $x\le {N}_{A}\left(B\left(y,z\right)\right)<{N}_{A}\left(y\right)$.In this case, we have $A\left(x,B\left(y,z\right)\right)=0$, and the result is trivial.
- ${N}_{A}\left(B\left(y,z\right)\right)<x<{N}_{A}\left(y\right),z<{N}_{B}\left(y\right)$.In this case, the proof is dual for the second case.
- ${N}_{A}\left(B\left(y,z\right)\right)<x<{N}_{A}\left(y\right),z\ge {N}_{B}\left(y\right)$.In this case, we have $A\left(x,y\right)=0,B\left(y,z\right)=1$. Hence, $A\left(x,B\left(y,z\right)\right)=x$ and $B\left(A\left(x,y\right),z\right)=z$; therefore, we shall prove that $x\le z$. Let $b\in [0,1]$ such that $t\left(b\right)=y$. Since h is a decreasing convex function with $h\left(0\right)=1$ and $h\left(1\right)=0$, $\frac{h\left(b\right)-h\left(0\right)}{b}\le h\left(1\right)-h\left(0\right)\le \frac{h\left(1\right)-h(1-b)}{b}$ by Lemma 6.1.1 in [25]. Thus, $h(1-b)\le 1-h\left(b\right)$, i.e., $g\circ {t}^{-1}\left(1-t\left(y\right)\right)\le 1-g\left(y\right)$. Then, ${t}^{-1}\left(1-t\left(y\right)\right)\le {g}^{-1}\left(1-g\left(y\right)\right)$. Hence, we have$$x<{N}_{A}\left(y\right)\le {N}_{B}\left(y\right)\le z$$

**Example**

**4.**

#### 5.2. Submodular Inequality of T-Conorm over T-Norm

**Theorem**

**8.**

**Proof.**

#### 5.3. Submodular Inequality of T-Norm over T-Norm

**Example**

**5.**

- $x+y\le 1$$${T}_{L}\left(x,{T}_{M}\left(y,z\right)\right)=max\left(x+min\left(y,z\right)-1,0\right)=0$$and$${T}_{M}\left({T}_{L}\left(x,y\right),z\right)=min\left(max\left(x+y-1,0\right),z\right)=0;$$
- $x+y>1$ and $z\ge y$$${T}_{L}\left(x,{T}_{M}\left(y,z\right)\right)=max\left(x+min\left(y,z\right)-1,0\right)=x+y-1$$and$${T}_{M}\left({T}_{L}\left(x,y\right),z\right)=min\left(max\left(x+y-1,0\right),z\right)=x+y-1;$$
- $x+y>1$ and $z<y$ and $x+z>1$$${T}_{L}\left(x,{T}_{M}\left(y,z\right)\right)=max\left(x+min\left(y,z\right)-1,0\right)=x+z-1$$and$${T}_{M}\left({T}_{L}\left(x,y\right),z\right)=max\left(x+min\left(y,z\right)-1,0\right)=min\left(x+y-1,z\right);$$
- $x+y>1$ and $z<y$ and $x+z\le 1$$${T}_{L}\left(x,{T}_{M}\left(y,z\right)\right)=max\left(x+min\left(y,z\right)-1,0\right)=0$$and$${T}_{M}\left({T}_{L}\left(x,y\right),z\right)=min(max(x+y-1),z)=min(x+y-1,z)\ge 0.$$

**Proposition**

**2.**

**Proof.**

**Theorem**

**9.**

**Proof.**

**Proposition**

**3.**

**Proof.**

**Example**

**6.**

**Remark**

**3.**

#### 5.4. Submodular Inequality of T-Conorm over T-Conorm

**Example**

**7.**

- $y>x$$${S}_{M}\left(x,{S}_{P}\left(y,z\right)\right)=max\left(x,y+z-y\xb7z\right)=y+z-y\xb7z$$and$${S}_{P}\left({S}_{M}\left(x,y\right),z\right)=max\left(x,y\right)+z-max\left(x,y\right)\xb7z=y+z-y\xb7z;$$
- $y\le x$$${S}_{M}\left(x,{S}_{P}\left(y,z\right)\right)=max\left(x,y+z-y\xb7z\right)$$and$${S}_{P}\left({S}_{M}\left(x,y\right),z\right)=max\left(x,y\right)+z-max\left(x,y\right)\xb7z=x+z-x\xb7z.$$Since $x\u2a7dx+z\left(1-x\right)$ and $y\left(1-z\right)+z\u2a7dx\left(1-z\right)+z$, we have$${S}_{M}\left(x,{S}_{P}\left(y,z\right)\right)\le {S}_{P}\left({S}_{M}\left(x,y\right),z\right).$$

**Proposition**

**4.**

**Proof.**

**Theorem**

**10.**

**Proof.**

**Proposition**

**5.**

**Proof.**

## 6. Conclusions

- (1)
- Some general properties of submodular inequalities in the sense of duality and isomorphism were discussed. The submodular inequality was preserved under the isomorphism of the aggregation operators, while it was reversed under the duality of aggregation operators.
- (2)
- The submodular inequality between the ordinal sum of conjunctors with the same sum and carriers was determined by the submodular inequality between all corresponding sums and conjunctors [23].
- (3)
- The characterization of t-norms and t-conorms in submodular inequalities were presented in terms of the composition of their additive generators. More specifically, in the cases where the Archimedean t-norm was submodular over the Archimedean t-conorm, we offered a characterization based on the convexity of the composition of their additive generators.

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

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Bo, Q.; Li, G.
The Submodular Inequality of Aggregation Operators. *Symmetry* **2022**, *14*, 2354.
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Bo Q, Li G.
The Submodular Inequality of Aggregation Operators. *Symmetry*. 2022; 14(11):2354.
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2022. "The Submodular Inequality of Aggregation Operators" *Symmetry* 14, no. 11: 2354.
https://doi.org/10.3390/sym14112354