The Enumeration of (⊙,∨)-Multiderivations on a Finite MV-Chain

Abstract

In this paper, $(\odot ,\vee )$-multiderivations on an MV-algebra A are introduced, the relations between $(\odot ,\vee )$-multiderivations and $(\odot ,\vee )$-derivations are discussed. The set $\mathrm{MD}\left(A\right)$ of $(\odot ,\vee )$-multiderivations on A can be equipped with a preorder, and $\left(\mathrm{MD}\right(A)/\sim ,\preccurlyeq )$ can be made into a partially ordered set with respect to some equivalence relation ∼. In particular, for any finite MV-chain $L_n$, $(\mathrm{MD}\left(L_n\right)/\sim ,\preccurlyeq )$ becomes a complete lattice. Finally, a counting principle is built to obtain the enumeration of $\mathrm{MD}\left(L_n\right)$.

1. Introduction

The concept of derivation originating from analysis has been delineated for a variety of algebraic structures which come in analogy with the Leibniz rule

$$\frac{d}{dx}\left(fg\right)=\frac{d}{dx}\left(f\right)g+f\frac{d}{dx}\left(g\right).$$

Posner [1] introduced the derivation on prime rings $(R,+,·)$ as a mapping d from R to R such that for all $x,y\in R$:

$$\left(1\right)d(x·y)=d\left(x\right)·y+x·d\left(y\right),\left(2\right)d(x+y)=d\left(x\right)+d\left(y\right).$$

It implies that

$$\left(3\right)d\left(1\right)=0,\left(4\right)d\left(0\right)=0,$$

which are the 0-ary version of $\left(1\right)$ and $\left(2\right)$, respectively.

The derivations on lattices $(L,\vee ,\wedge )$ were defined in [2] by Szász and were developed in [3] by Ferrari as a map d from L to L such that for all elements $x,y$ in L:

$$\left(i\right)d(x\wedge y)=\left(d\right(x)\wedge y)\vee (x\wedge d(y\left)\right),\left(ii\right)d(x\vee y)=d\left(x\right)\vee d\left(y\right).$$

Xin et al. [4,5] investigated the derivations on a lattice satisfying only condition (i). In fact, a derivation d on L with both the Leibniz rule (i) and the linearity (ii) implies that $d\left(x\right)=x\wedge u$ for some $u\in L$ [6] (Proposition 2.5). If u is the maximum of a lattice, then such a derivation is actually the identity. It seems that this is an important reason for the derivations on, for instance, BCI-algebra [7], residuated lattices [8], basic algebra [9], L-algebra [10], and differential lattices [6], which are defined with the unique requirement of the Leibniz rule (i) (for the discussion in detail, cf. Section 2).

The derivation on an MV-algebra (A, ⊕,*, 0) was firstly introduced by Alshehri [11] as a mapping d from A to A satisfying an ($\odot ,\oplus$)-condition: $\forall x,y\in A$,

$$d(x\odot y)=\left(d\right(x)\odot y)\oplus (x\odot d(y\left)\right),$$

where $x\odot y$ is defined to be ${(x^{\ast }\oplus y^{\ast })}^{\ast }$. Then, several derivations on MV-algebras have been considered in [12,13,14,15]. However, the interplay of the ring operations · and + is more similar to the interplay between the MV-operations ⊙ and ∨ rather than that between the MV-operations ⊙ and ⊕. In fact, the main interplay between · and + in rings is the distributivity of · over +. In MV-algebras, ⊙ distributes over ∨, as in rings, while it is not true that ⊙ distributes over ⊕. It is also true that ⊙ distributes over ∧, but ∨ is preferable because the identity element of ∨ is absorbing for ⊙, that is, $0\odot x=0$ for any element x in an MV-algebra A, as in rings, while the same is not true for ∧. Therefore, the (⊙, ∨)-derivation on MV-algebras [16] is a nature improvement of Alshehri’s celebrated work [11] of the (⊙, ⊕)-derivation (cf. Section 2 for more discussion).

Let E and F be nonempty sets. A multifunction f: $E\to \mathrm{\Delta }\left(F\right)$ is a map (or function) from E into $\mathrm{\Delta }\left(F\right)$, the collection of nonempty subsets of F. The multifunction [17] is also known as set-valued function [18]. Significantly, multifunctions have many diverse and interesting applications in control problems [19,20] and mathematical economics [21,22]. Motivated by the role played by derivations on MV-algebras and the work of multiderivations on lattices [23], it is imperative to undertake a systematic study of the corresponding algebraic structure for derivations on MV-algebras.

This article is a continuation of work on $(\odot ,\vee )$-multiderivations based on the nature (⊙, ∨)-derivation on MV-algebras [16], that is, a set-valued generalization of point-valued $(\odot ,\vee )$-derivations. Section 2 starts with a review of the $(\odot ,\vee )$-derivations on an MV-algebra A. In Section 3, we first define a natural preorder on $\mathrm{\Delta }\left(A\right)$ that $M⪯N$ iff for every $m\in M$ there exists $n\in N$ such that $m\le n$. Then, we introduce $(\odot ,\vee )$-multiderivations on MV-algebras. The relations between $(\odot ,\vee )$-derivations and $(\odot ,\vee )$-multiderivations on an MV-algebra are given (Propositions 5–7). In Section 4, we investigate the set of $(\odot ,\vee )$-multiderivations $\mathrm{MD}\left(A\right)$ on an MV-algebra A. Let $\sigma ,{\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$. Define $\sigma \preccurlyeq {\sigma }^{\prime }$ if $\sigma \left(x\right)⪯{\sigma }^{\prime }\left(x\right)$ for any $x\in A$, and an equivalence relation ∼ on $\mathrm{MD}\left(A\right)$ by $\sigma \sim {\sigma }^{\prime }$ iff $\sigma \preccurlyeq {\sigma }^{\prime }$ and ${\sigma }^{\prime }\preccurlyeq \sigma$. Then, $\left(\mathrm{MD}\right(A)/\sim ,\preccurlyeq )$ is a poset. For an n-element MV-chain $L_n$, we show that $(\mathrm{MD}\left(L_n\right)/\sim ,\preccurlyeq )$ is isomorphic to the complete lattice $\mathrm{Der}\left(L_n\right)$, the underlying set of $(\odot ,\vee )$-derivations on $L_n$ (Theorem 1), so we deduce that $|\mathrm{MD}\left(L_n\right)/\sim |=|\mathrm{Der}\left(L_n\right)|$, then [16] (Theorem 3.11) can be applied. Moreover, we define an equivalence relation ∼ on $\mathrm{\Delta }\left(A\right)$, and present the fact that the poset $\mathrm{\Delta }(L_n\times L_2)/\sim$ is isomorphic to the complete lattice $\mathrm{Der}\left(L_{n+1}\right)$ (Proposition 11). However, the cardinalities of different equivalence classes with respect to the equivalence relation ∼ are different in general (Example 5). In Section 5, by building a counting principle (Theorem 3) for $(\odot ,\vee )$-multiderivations on an n-element MV-chain $L_n$, we finally obtain the enumeration of $\mathrm{MD}\left(L_n\right)$: $(7·3^{n-1}-2^{n+2}+1)/2$.

Notation. Throughout this paper, A denotes an MV-algebra; $\left|X\right|$ denotes the cardinality of a set X; $\mathrm{\Delta }\left(X\right)$ denotes the set of nonempty subsets of a set X; ⊔ means disjoint union; $\mathbb{N}$ denotes the set of natural numbers; “iff” is the abbreviation for “if and only if”.

2. Preliminaries

Definition 1 

([24]). An algebra (A, ⊕, *, 0) is an MV-algebra if the following axioms are satisfied:

(MV1) (associativity) $x\oplus (y\oplus z)=(x\oplus y)\oplus z.$

(MV2) (commutativity) $x\oplus y=y\oplus x.$

(MV3) (existence of the unit 0) $x\oplus 0=x.$

(MV4) (involution) $x^{\ast \ast }=x.$

(MV5) (maximal element $0^{\ast }$) $x\oplus 0^{\ast }=0^{\ast }.$

(MV6) (Łukasiewicz axiom) ${(x^{\ast }\oplus y)}^{\ast }\oplus y={(y^{\ast }\oplus x)}^{\ast }\oplus x.$

Define $1=0^{\ast }$ and the natural order on A as follows: $y\ge x$ iff $x\odot y^{\ast }=0$. Then, the interval $[a,b]=\{r\in A\mid a\le r\le b\}$ for any $a,b\in A$ and $a\le b$. Note that A is a bounded distributive lattice with respect to the natural order [24] (Proposition 1.5.1) with 0, 1, and

$$x\vee y=(x\odot y^{\ast })\oplus y,x\wedge y=x\odot (x^{\ast }\oplus y).$$

(1)

An MV-chain is an MV-algebra which is linearly ordered with respect to the natural order.

Example 1 

([24]). Let $L=[0,1]$ be the real unit interval. Define

$$x\oplus y=\min \{1,x+y\} and x^{\ast } = 1 - for any x,y \in L.$$

Then (L,⊕, *, 0) is an MV-chain. Note that $x\odot y=\max \{0,x+y-1\}$.

Example 2. 

For every $2\le n\in {\mathbb{N}}_{+}$, let

$$L_n=\left\{0,{\displaystyle \frac{1}{n-1}},{\displaystyle \frac{2}{n-1}},\cdots ,{\displaystyle \frac{n-2}{n-1}},1\right\}.$$

Then the n-element subset $L_n$ is an MV-subalgebra of L.

Lemma 1. 

([24,25]). If A is an MV-algebra, then the following statements are true $\forall x,y,z\in A$:

1. 

$x\oplus y\ge x\vee y\ge x\ge x\wedge y\ge x\odot y$.

2. 

$x\oplus y=0$ iff $x=y=0$. $x\odot y=1$ iff $x=y=1$.

3. 

If $y\ge x$, then $y\vee z\ge x\vee z$, $y\wedge z\ge x\wedge z$.

4. 

If $y\ge x$, then $y\oplus z\ge x\oplus z$, $y\odot z\ge x\odot z$.

5. 

$y\ge x$ iff $x^{\ast }\ge y^{\ast }$.

6. 

$x\odot (y\wedge z)=(x\odot y)\wedge (x\odot z)$.

7. 

$x\odot (y\vee z)=(x\odot y)\vee (x\odot z)$.

8. 

$x\odot y\le z$ iff $x\le y^{\ast }\oplus z$.

Let $\mathrm{\Omega }$ be an index set. The direct product ${\prod }_{i\in \mathrm{\Omega }}{A}_i$ [24] of a family of MV-algebras ${\left\{A_i\right\}}_{i\in \mathrm{\Omega }}$ is the MV-algebra with cartesian product of the family and pointwise MV-operations. We denote $A_1\times A_2\times \cdots \times A_n$ when $\mathrm{\Omega }$ is a positive integer n. We call $a\in A$ idempotent if $a\oplus a=a$. Let $\mathbf{B}\left(A\right)$ be the set of idempotent elements of A and $B_{2^n}$ be the $2^n$-element Boolean algebra. Note that $B_4$ is actually $L_2\times L_2$ [24].

Lemma 2 

([24], Proposition 3.5.3). Let A be a subalgebra of $[0,1]$. Let $A^{+}=\{x\in A\mid$$x>0\}$ and $a=\inf A^{+}$ be the infimum of $A^{+}$. If $a=0$, then A is a dense subchain of $[0,1]$. If $a>0$, then $A=L_n$ for some $n\ge 2$.

Defintion 2 

([16]). If A is an MV-algebra, then a map d from A to A is an$(\odot ,\vee )$-derivation on A if $\forall x,y\in A$,

$$d(x\odot y)=\left(d\right(x)\odot y)\vee (x\odot d(y\left)\right).$$

(2)

Let $\mathrm{Der}\left(A\right)$ be the set of $(\odot ,\vee )$-derivations on A. For $X=\{x_1,x_2,\cdots ,x_n\}$ and a map $d:X\to X$, we shall write d as

$$\left(\begin{array}{cccc}{x}_1& x_2& \cdots & x_n\\ d\left(x_1\right)& d\left(x_2\right)& \cdots & d\left(x_n\right)\end{array}\right).$$

The mappings $\mathrm{I}{\mathrm{d}}_A$ and ${\mathbf{0}}_A$, defined by $\mathrm{I}{\mathrm{d}}_A\left(x\right)=x$ and ${\mathbf{0}}_A\left(x\right)=0$$(\forall x\in A)$, respectively, are $(\odot ,\vee )$-derivations on A. For $u\in A$, the operator ${\chi }^{\left(u\right)}\left(x\right):=\left\{\begin{array}{cc}u,\hfill & \mathrm{if}x=1\hfill \\ x.\hfill & \mathrm{otherwise}\hfill \end{array}\right.\in \mathrm{Der}\left(A\right)$. More examples are given in [16].

Proposition 1 

([16]). If A is an MV-algebra and $d\in \mathrm{Der}\left(A\right)$, then the followings hold for all $x,y\in A$:

1. 

$0=d\left(0\right)$.

2. 

$x\ge d\left(x\right)$.

3. 

If $d\left(x\right)=x$, then $d\left(y\right)=y$ for $y\le x$.

Remark 1. 

Now let us give some explanations of the naturality of an $(\mathrm{\odot },\mathrm{\vee })$-derivation in Definition 2. The interplay of the ring operations · and + is more similar to the interplay between the MV-operations ⊙ and ∨ rather than that between the MV-operations ⊙ and ⊕.

Next we discuss why we include only Equation (2). Recall that $d\left(0\right)=0$ is the 0-ary version of $d(x+y)=d\left(x\right)+d\left(y\right)$ in derivations on a ring. For MV-algebras, $d\left(0\right)=0$ is the 0-ary version of (a); see Proposition 1 (1). $d\left(1\right)=0$ is the 0-ary version of $d(x·y)=d\left(x\right)·d\left(y\right)$ in derivations on a ring. Hence, it seems that the most faithful and natural derivation notion on A as a translation of the ring-theoretic notion of derivation (cf. Introduction) would include:

(a) 

$d(x\odot y)=\left(d\right(x)\odot y)\vee (x\odot d(y\left)\right),$

(b) 

$d\left(1\right)=0$,

(c) 

$d(x\vee y)=d\left(x\right)\vee d\left(y\right)$,

(d) 

$d\left(0\right)=0$.

However, (b) and (c) imply that d is trivial (note that (a) is automatically assumed).

Lemma 3. 

If A is an MV-algebra and $d:A\to A$ is a map satisfying (a), (b) and (c) for any $x,y\in A$. Then, $d={\mathbf{0}}_A$.

Proof. 

Assume $x\le y$, it follows from (c) that $d\left(y\right)=d(x\vee y)=d\left(x\right)\vee d\left(y\right)$ and thus $d\left(x\right)\le d\left(y\right)$. Together with (b) $d\left(1\right)=0$, we have $d\left(x\right)=0$ for any $x\in A$ since $x\le 1$. Hence, $d={\mathbf{0}}_A$. □

Next, we consider what will happen if the condition (b′) $d\left(1\right)=1$ replaces (b) $d\left(1\right)=0$.

Lemma 4. 

If $d:A\to A$ is a mapping from an MV-algebra A to A with (a) and (b′) for any $x,y\in A$, then, $d=\mathrm{I}{\mathrm{d}}_A$.

Proof. 

Assume d satisfies (a) and (b′). We obtain that d satisfies Proposition 1 (3) since d satisfies (a). Both with (b′) $d\left(1\right)=1$, we obtain $d\left(x\right)=x$ for any $x\in A$. Therefore, $d=I{d}_A$. □

Recall that for a given $a\in A$, a principal $(\odot ,\vee )$-derivation $d_a$ on A [16] is defined by $d_a\left(x\right):=a\odot x$ for all $x\in A$. An $(\odot ,\vee )$-derivation d is isotone [16] if $\forall x,y\in A,$$y\ge x$ implies that $d\left(y\right)\ge d\left(x\right)$. Note that ${\mathbf{0}}_A$ and $I{d}_A$ are both principal and isotone. More generally, we obtain the following.

Proposition 2 

([16] (Proposition 3.19)). Let A be an MV-algebra and d be a map satisfying (a) and (b”). Then, the followings are equivalent:

1. 

d is isotone;

2. 

$d\left(1\right)\odot x=d\left(x\right)$ for all $x\in A$;

3. 

$d\left(x\right)\vee d\left(y\right)=d(x\vee y)$.

If d satisfies (b), then the principal derivations on MV-algebra A will not be included, expect ${\mathbf{0}}_A$. Even identity derivations $\mathrm{I}{\mathrm{d}}_A$ will not be within our scope of consideration. Hence, the scope of the study will be significantly narrowed.

Remark 2. 

Note that d is isotone if d satisfies (c). In fact, if $x\le y$, then $d\left(y\right)=d(x\vee y)=d\left(x\right)\vee d\left(y\right)$ and thus $d\left(x\right)\le d\left(y\right)$. The isotone case is a special case of d, thus the scope of research will be narrowed. This case has been partially studied in [16], Section 3.3.

Therefore, we use the derivation meaning from Definition 2 in our series papers since [16] on.

3. $(\odot ,\vee )$-Multiderivations on an MV-Algebra

Let X and Y be two nonempty sets. Recall that a set-valued function or multivalued function (for short, multifunction) F between X and Y is a map $F:X\to \mathrm{\Delta }\left(Y\right)$. The set $F\left(x\right)$ is called the image of x under F (cf. [26], Appendix A).

Definition 3. 

Let A be an MV-algebra and $M,N\in \mathrm{\Delta }\left(A\right)$. We define four binary operations $\oplus ,\odot ,\vee ,\wedge$ and an unary operation * on $\mathrm{\Delta }\left(A\right)$ by:

$$M★N=\{m★n\mid m\in M, n\in N\}\mathit{and}{M}^{\ast }=\{m^{\ast }\mid m\in M\}$$

where $★\in \{\oplus ,\odot ,\vee ,\wedge \}$.

Remark 3. 

1. 

Note that $M\vee N$ means the pointwise $m\vee n$ operation from Equation (1) of sets, which is different from the supremum of M and N. $M\wedge N$ has a similar meaning.

2. 

We abbreviate $M★\left\{x\right\}$ and ${\left\{x\right\}}^{\ast }$ by $M★x$ and $x^{\ast }$, respectively. But if $\left\{x\right\}$ appears by itself such as $M⪯\left\{x\right\}$, we still use $\left\{x\right\}$.

We define a binary relation $M⪯N$ iff for every $m\in M$ there exists $n\in N$ such that $m\le n$. Denote $M\prec N$ if $M⪯N$ and $M\ne N$.

Then, ⪯ is a preorder on $\mathrm{\Delta }\left(A\right)$. In fact, the reflexivity and transitivity of ⪯ are clear. However, ⪯ does not satisfy antisymmetry in general. In fact, ⪯ satisfies antisymmetry iff the MV-algebra A is trivial: If A is trivial, we have $\mathrm{\Delta }\left(A\right)=\left\{\right\{0\left\}\right\}$ and $\left\{0\right\}⪯\left\{0\right\}$. Hence, ⪯ satisfies antisymmetry. Conversely, suppose A is nontrivial, we have $A\ne \left\{1\right\}$, but $\left\{1\right\}⪯A$ and $A⪯\left\{1\right\}$, a contradiction.

Lemma 5. 

Let A be an MV-algebra and $x,a,b,c,e,f\in A$. Then, the followings hold:

1. 

If $x\le b\odot c$, then there exists $t\in A$ such that $t\le b$ and $x=t\odot c$.

2. 

If $x\le b\vee c$, then there exist $t,s\in A$ such that $t\le b,s\le c$ and $x=t\vee s$.

3. 

$[a,b]\odot c=[a\odot c,b\odot c]$.

4. 

$[a,b]\vee [e,f]=[a\vee e,b\vee f]$.

Proof. 

$\left(1\right)$ Assume $x\le b\odot c$, then

$$x=(b\odot c)\wedge x=(b\odot c)\odot ({(b\odot c)}^{\ast }\oplus x)=b\odot ({(b\odot c)}^{\ast }\oplus x)\odot c.$$

Thus, we may choose $t=b\odot ({(b\odot c)}^{\ast }\oplus x)$.

$\left(2\right)$ Assume $x\le b\vee c$. Recall that A is a distributive lattice. So

$$x=(b\vee c)\wedge x=(b\wedge x)\vee (c\wedge x).$$

Hence, we can obtain $x=t\vee s$ by taking $t=b\wedge x,s=c\wedge x$.

$\left(3\right)$ For each $x\in [a,b]$, we obtain $a\odot c\le x\odot c\le b\odot c$ by Lemma 1 (4). Thus, $[a,b]\odot c\subseteq [a\odot c,b\odot c]$. It suffices to prove that $[a\odot c,b\odot c]\subseteq [a,b]\odot c$. For any $a\odot c\le x\le b\odot c$, by (1) there is $t=b\odot ({(b\odot c)}^{\ast }\oplus x)\le b$ such that $x=t\odot c$. If we can prove $a\le t$, then the result follows immediately. Note that

$$t=b\odot ({(b\odot c)}^{\ast }\oplus x)=b\odot (b^{\ast }\oplus c^{\ast }\oplus x)=b\wedge (c^{\ast }\oplus x).$$

Since $a\odot c\le x$, we have $a\le c^{\ast }\oplus x$ by Lemma 1 (8). Together with $a\le b$, we obtain $a\le b\wedge (c^{\ast }\oplus x)=t$. Thus, we conclude that $[a,b]\odot c=[a\odot c,b\odot c]$.

$\left(4\right)$ For any $t\in [a,b],s\in [e,f]$, we have $a\vee e\le t\vee s\le b\vee f$ by Lemma 1 (3). Thus, $[a,b]\vee [e,f]\subseteq [a\vee e,b\vee f]$. It is enough to prove that $[a\vee e,b\vee f]\subseteq [a,b]\vee [e,f]$. For any $a\vee e\le x\le b\vee f$, there exist $t,s\in A$ such that

$$t=b\wedge x\le b,s=f\wedge x\le f\mathrm{and}x=t\vee s$$

by (2). If we can prove $a\le t$ and $e\le s$, then the result follows. Note that since $a\le b$ and $a\le a\vee e\le x$, we have $a\le b\wedge x=t$. Similarly, $e\le s$. Therefore, $[a\vee e,b\vee f]=[a,b]\vee [e,f]$. □

The following result holds for any MV-algebra A since it is a distributive lattice under the natural order.

Lemma 6 

([23] (Lemma 2.1)). Let L be a lattice and $M,N,P,Q\in \mathrm{\Delta }\left(L\right)$. Then, the following statements hold:

1. 

$M\wedge N⪯M⪯M\vee N$.

2. 

If $M⪯N$ and $P⪯Q$, then $M\wedge P⪯N\wedge Q$ and $M\vee P⪯N\vee Q$. In particular, $M⪯N$ implies $M\wedge P⪯N\wedge P$.

3. 

$M\subseteq M\wedge M,M\subseteq M\vee M$. If M is a sublattice of L, then $M=M\vee M$.

4. 

$M\vee N=N\vee M$.

5. 

$(M\vee N)\vee P=M\vee (N\vee P)$.

6. 

If $M\vee N\subseteq M$, then $N⪯M$.

7. 

If L is distributive, then $(M\vee N)\wedge P\subseteq (M\wedge P)\vee (N\wedge P)$.

Remark 4. 

1. 

Note that the converse inclusion of Lemma 6 (3), i.e., $M\wedge M\subseteq M$ and $M\vee M\subseteq M$, does not hold in general. For example, consider the Boolean lattice $B_4=\{0,a,b,1\}$ (see Figure 1), $M=\{a,b\}\subseteq B_4$, then $0=a\wedge b\in M\wedge M$ and $1=a\vee b\in M\vee M$, but $0,1\notin M$.

Figure 1. Hasse diagram of $B_4$.

2. 

The converse of Lemma 6 (6), i.e., $N⪯M$ implies $M\vee N\subseteq M$ may not hold. For example, in $L_3$, let $N=\{0,\frac{1}{2}\},M=\{0,1\}$. We have $N⪯M$ but $M\vee N=\{0,\frac{1}{2},1\}⊈M$.

3. 

The converse inclusion of Lemma 6 (7) holds if P is a singleton but need not hold in general. This is slightly different from [23]. For example, let $B_8=\{0,a,b,c,u,v,w,1\}$ be the 8-element Boolean lattice as Figure 2, $M=\left\{u\right\},N=\left\{w\right\}$ and $P=\{a,b,c\}$. We can check that $u=a\vee b=(u\wedge a)\vee (w\wedge b)\in (M\wedge P)\vee (N\wedge P)$ but $u\notin P=(M\vee N)\wedge P$.

Figure 2. Hasse diagram of $B_8$.

According to Lemma 1, one obtains

Lemma 7. 

Assume that A is an MV-algebra, $M,N,P,Q\in \mathrm{\Delta }\left(A\right)$, and $m\in M$. Then, the following statements hold:

1. 

If $M⪯N$ and $P⪯Q$, then $M\oplus P⪯N\oplus Q$ and $M\odot P⪯N\odot Q$. In particular, $M⪯N$ implies $M\oplus P⪯N\oplus P$ and $M\odot P⪯N\odot P$.

2. 

$m\odot (P\vee Q)=(m\odot P)\vee (m\odot Q)$.

3. 

$m\odot (P\cup Q)=(m\odot P)\cup (m\odot Q)$.

4. 

$M\odot N⪯M\wedge N⪯M⪯M\vee N⪯M\oplus N$.

5. 

If $M\oplus N\subseteq M$, then $N⪯M$.

Proof. 

$\left(1\right)$ Suppose $M⪯N$ and $P⪯Q$. For any $x=m\oplus p\in M\oplus P$, there are $n\in N$ and $q\in Q$ such that $m\le n$ and $p\le q$. It follows from Lemma 1 (4) that $m\oplus p\le m\oplus q\le n\oplus q$, where $n\oplus q\in N\oplus Q$. Thus, $M\oplus P⪯N\oplus Q$. Similarly, we have $M\odot P⪯N\odot Q$. In particular, we obtain $M\oplus P⪯N\oplus P$ and $M\odot P⪯N\odot P$.

$\left(2\right)$ For any $p\in P$ and $q\in Q$, we have $m\odot (p\vee q)=(m\odot p)\vee (m\odot q)\in (m\odot P)\vee (m\odot Q)$ by Lemma 1 (7). Thus, $m\odot (P\vee Q)\subseteq (m\odot P)\vee (m\odot Q)$. The reverse inclusion can be verified similarly. Therefore, $m\odot (P\vee Q)=(m\odot P)\vee (m\odot Q)$.

(3) We have $x\in m\odot (P\cup Q)$, iff there is $y\in P\cup Q$ such that $x=m\odot y$, iff there is $y\in P$ or $y\in Q$ such that $x=m\odot y$, iff $x\in m\odot P$ or $x\in m\odot Q$, iff $x\in (m\odot P)\cup (m\odot Q)$. Hence, $m\odot (P\cup Q)=(m\odot P)\cup (m\odot Q)$.

$\left(4\right)$ For any $m\in M$ and $n\in N$, we know $m\odot n\le m\wedge n\le m\le m\vee n\le m\oplus n$ by Lemma 1 (1). The result follows immediately.

$\left(5\right)$ Assume $M\oplus N\subseteq M$, then for any $n\in N$, there exists $m\in M$ such that $m\oplus n\in M$. So by Lemma 1 (1) we obtain $n\le m\oplus n$. Therefore, $N⪯M$. □

To study whether (Δ(A), ⊕, *, {0}) is an MV-algebra, we first give

Lemma 8. 

If A is an MV-algebra, then, for any $M,N,P\in \mathrm{\Delta }\left(A\right)$, the followings hold:

1. 

$(M\oplus N)\oplus P=M\oplus (N\oplus P)$.

2. 

$M\oplus N=N\oplus M$.

3. 

$M\oplus 0=M$.

4. 

$M^{\ast \ast }=M$.

5. 

$M\oplus 0^{\ast }=\left\{0^{\ast }\right\}$.

Proof. 

(1)–(5) follow from (MV1)–(MV5), respectively. □

Remark 5. 

Since (MV1)–(MV5) are satisfied on $\mathrm{\Delta }\left(A\right)$, it is natural to consider whether (MV6) ${(M^{\ast }\oplus N)}^{\ast }\oplus N={(N^{\ast }\oplus M)}^{\ast }\oplus M$ holds on $\mathrm{\Delta }\left(A\right)$. The answer is no. For example, let $M=\left\{\frac{1}{2}\right\}$ and $N=\{0,1\}$ on three-element MV-chain $L_3$. It is easy to see that ${({\frac{1}{2}}^{\ast }\oplus \{0,1\})}^{\ast }\oplus \{0,1\}=\{0,\frac{1}{2}\}\oplus \{0,1\}=\{0,\frac{1}{2},1\}\ne \{\frac{1}{2},1\}={({\{0,1\}}^{\ast }\oplus \frac{1}{2})}^{\ast }\oplus \frac{1}{2}$. That is, ${(M^{\ast }\oplus N)}^{\ast }\oplus N\ne {(N^{\ast }\oplus M)}^{\ast }\oplus M$.

If A is a nontrivial MV-algebra, and $\phi :A\to \mathrm{\Delta }\left(A\right)$ is a multifunction on A. $\phi$ is called additive and negative, if $\phi (x\oplus y)=\phi \left(x\right)\oplus \phi \left(y\right)$ and $\phi \left(x^{\ast }\right)={\left(\phi \left(x\right)\right)}^{\ast }$ for all $x,y\in A$, respectively.

Proposition 3. 

Let A be an MV-algebra and $\phi :A\to \mathrm{\Delta }\left(A\right)$ be a multifunction on A. If φ is additive and negative, then (φ(A), ⊕, *, φ(0)) is an MV-algebra, where $\phi \left(A\right)=\left\{\phi \right(x)\mid x\in A\}$.

Proof. 

It is sufficient to prove (MV3), (MV5) and (MV6), since we know that (φ(A), ⊕, *, φ(0)) satisfies (MV1), (MV2) and (MV4) by Lemma 8. Since $\phi$ is additive and negative, it follows that $\phi \left(x\right)\oplus \phi \left(0\right)=\phi (x\oplus 0)=\phi \left(x\right)$ and $\phi \left(x\right)\oplus \phi {\left(0\right)}^{\ast }=\phi (x\oplus 0^{\ast })=\phi \left(0^{\ast }\right)=\phi {\left(0\right)}^{\ast }$. Furthermore, ${(\phi {\left(x\right)}^{\ast }\oplus \phi \left(y\right))}^{\ast }\oplus \phi \left(y\right)=\phi {(x^{\ast }\oplus y)}^{\ast }\oplus \phi \left(y\right)=\phi ({(x^{\ast }\oplus y)}^{\ast }\oplus y)=\phi ({(y^{\ast }\oplus x)}^{\ast }\oplus x)=\phi {(y^{\ast }\oplus x)}^{\ast }\oplus \phi \left(x\right)={(\phi {\left(y\right)}^{\ast }\oplus \phi \left(x\right))}^{\ast }\oplus \phi \left(x\right)$ for any $x,y\in A$. Thus, (φ(A), ⊕, *, φ(0)) is an MV-algebra. □

Now let us define the $(\odot ,\vee )$-multiderivation.

Definition 4. 

If A is an MV-algebra, a multifunction $\sigma :A\to \mathrm{\Delta }\left(A\right)$ is called an $(\odot ,\vee )$-multiderivation on A if

$$\sigma (x\odot y)=\left(\sigma \right(x)\odot y)\vee (x\odot \sigma (y\left)\right)$$

(3)

for all $x,y\in A$. Denote the set of $(\odot ,\vee )$-multiderivations on A by $\mathrm{MD}\left(A\right)$.

Example 3. (i) Consider the MV-chain $L_4$. We define a multifunction σ on $L_4$ by $\sigma \left(0\right)=\left\{0\right\}$, $\sigma \left(\frac{1}{3}\right)=\{0,\frac{1}{3}\}$, $\sigma \left(\frac{2}{3}\right)=\{0,\frac{2}{3}\},\sigma \left(1\right)=\{0,1\}$. Then, we can check σ is an $(\odot ,\vee )$-multiderivation on $L_4$. In fact, $\sigma ={\beta }_1$ (see Corollary 1).

(ii) Consider the standard MV-algebra $L=[0,1]$. We define a multifunction $\sigma :L\to \mathrm{\Delta }\left(L\right)$ by $\sigma \left(x\right)=[0,x]$ for all $x\in L$. Then, we can verify that σ is an $(\odot ,\vee )$-multiderivation on L (see Proposition 6).

(iii) Let A be an MV-algebra and $S\subseteq A$ be a subalgebra of A. Define a multifunction ${\sigma }_S$ on A by ${\sigma }_S\left(x\right)=x\odot S$, $\forall x\in A$, then ${\sigma }_S\in \mathrm{MD}\left(A\right)$, which is called aprincipal $(\odot ,\vee )$-multiderivation. In fact, for any $x,y\in A$, since the subalgebra S must be a sublattice of A, it follows that $S=S\vee S$ by Lemma 6 (3). According to Lemma 7 (2), we immediately have ${\sigma }_S(x\odot y)=x\odot y\odot S=x\odot y\odot (S\vee S)=(x\odot y\odot S)\vee (x\odot y\odot S)=({\sigma }_S\left(x\right)\odot y)\vee (x\odot {\sigma }_S\left(y\right))$.

Proposition 4. 

If A is an MV-algebra and $\sigma \in \mathrm{MD}\left(A\right)$. Then, the followings hold for all $x,y\in A$,

1. 

$\sigma \left(0\right)=\left\{0\right\}$.

2. 

$\sigma \left(x\right)⪯\left\{x\right\}$.

3. 

$\sigma \left(x\right)\odot \sigma \left(y\right)⪯\sigma (x\odot y)⪯\sigma \left(x\right)\vee \sigma \left(y\right)$.

4. 

$x\odot \sigma \left(1\right)⪯\sigma \left(x\right)$.

5. 

If I is a lower set of A, then $\sigma \left(x\right)\subseteq I$ holds for any $x\in I$.

6. 

Let $1\in \sigma \left(1\right)$. Then, $x\in \sigma \left(x\right)$.

Proof. 

(1) Taking $x=y=0$ in Equation (3), we obtain $\sigma \left(0\right)=\sigma (0\odot 0)=\left(\sigma \right(0)\odot 0)\vee (0\odot \sigma (0\left)\right)=\left\{0\right\}$.

(2) Since $x\odot x^{\ast }=0$, we know that $\left\{0\right\}=\sigma \left(0\right)=\sigma (x\odot x^{\ast })=(\sigma \left(x\right)\odot x^{\ast })\vee (x\odot \sigma \left(x^{\ast }\right))$ by $\left(1\right)$. So $\sigma \left(x\right)\odot x^{\ast }=\left\{0\right\}$ and we obtain $\sigma \left(x\right)⪯\left\{x\right\}$.

(3) By Lemma 6 (3), we have $\sigma \left(x\right)\odot \sigma \left(y\right)\subseteq \left(\sigma \right(x)\odot \sigma (y\left)\right)\vee \left(\sigma \right(x)\odot \sigma (y\left)\right)$. Moreover, $\sigma \left(x\right)\odot \sigma \left(y\right)⪯\sigma \left(x\right)\odot y$ and $\sigma \left(x\right)\odot \sigma \left(y\right)⪯x\odot \sigma \left(y\right)$ by (2) and Lemma 7 (1). Thus,

$$\sigma \left(x\right)\odot \sigma \left(y\right)\subseteq \left(\sigma \right(x)\odot \sigma (y\left)\right)\vee \left(\sigma \right(x)\odot \sigma (y\left)\right)⪯\left(\sigma \right(x)\odot y)\vee (x\odot \sigma (y\left)\right)=\sigma (x\odot y)$$

by Lemma 6 (2). Moreover, by Lemma 7 (1) and Lemma 6 (2) we have

$$\sigma (x\odot y)=\left(\sigma \right(x)\odot y)\vee (x\odot \sigma (y\left)\right)⪯\sigma \left(x\right)\vee \sigma \left(y\right).$$

(4) Since $x=1\odot x$, it follows that $\sigma \left(x\right)=\sigma (1\odot x)=\sigma \left(x\right)\vee (x\odot \sigma (1\left)\right)$ by Equation (3). Then, we can obtain $x\odot \sigma \left(1\right)⪯\sigma \left(x\right)$ by Lemma 6 (6).

(5) For any $x\in I$, we know $\sigma \left(x\right)⪯\left\{x\right\}$ by (2). It induces that $y\le x$ holds for any $y\in \sigma \left(x\right)$. Then, $y\in I$ since I is a lower set. Thus, $\sigma \left(x\right)\subseteq I$.

(6) Since $1\in \sigma \left(1\right)$, there must exist $y\in \sigma \left(x\right)$ such that $x=x\odot 1\le y$ by (4). Moreover, by (2) we know $y\le x$ always holds for y. Hence, we obtain $x=y$ and $x\in \sigma \left(x\right)$. □

Now, let us explore the relations between $(\odot ,\vee )$-derivation d and $(\odot ,\vee )$-multiderivation $\sigma$ on A.

On the one hand, given an $(\odot ,\vee )$-derivation d on A, how can we construct an $(\odot ,\vee )$-multiderivation on A? We get started with a direct construction. Assume $d\in \mathrm{Der}\left(A\right)$. Define a multifunction $\alpha :A\to \mathrm{\Delta }\left(A\right)$ as follows:

$$\alpha \left(x\right)=\left\{d\right(x\left)\right\}\mathrm{for}\mathrm{any}x\in A.$$

Then, $\alpha \in \mathrm{MD}\left(A\right)$.

Proposition 5. 

If A is an MV-algebra and $d\in \mathrm{Der}\left(A\right)$, define a multifunction $\beta :A\to \mathrm{\Delta }\left(A\right)$ on A as follows

$$\beta \left(x\right):=\{0,d(x\left)\right\}.$$

Then, $\beta \in \mathrm{MD}\left(A\right)$ iff $d\left(x\right)\odot y=x\odot d\left(y\right)$ holds for any $x,y\in A$ with $d\left(x\right)\odot y>0$ and $x\odot d\left(y\right)>0$.

Proof. 

Assuming $\beta \in \mathrm{MD}\left(A\right)$, it follows that

$$\begin{array}{cc}\hfill \{0,d(x\odot y\left)\right\}& =\beta (x\odot y)\hfill \\ \hfill & =\left(\beta \right(x)\odot y)\vee (x\odot \beta (y\left)\right)\hfill \\ \hfill & =\left(\right\{0,d\left(x\right)\}\odot y)\vee (x\odot \{0,d\left(y\right)\left\}\right)\hfill \\ \hfill & =\{0,d(x)\odot y\}\vee \{0,x\odot d(y\left)\right\}\hfill \\ \hfill & =\{0,d(x)\odot y,x\odot d(y),d(x\odot y\left)\right\}\hfill \end{array}$$

for any $x,y\in A$. From the chain of equalities, we know that $d\left(x\right)\odot y,x\odot d\left(y\right)\in \{0,d(x\odot y\left)\right\}$. If both $d\left(x\right)\odot y>0$ and $x\odot d\left(y\right)>0$, then $d\left(x\right)\odot y=d(x\odot y)=x\odot d\left(y\right)$.

Conversely, let $x,y\in A$.

Then,

$$\beta (x\odot y)=\{0,d(x\odot y\left)\right\}$$

and

$$\left(\beta \right(x)\odot y)\vee (x\odot \beta (y\left)\right)=\{0,d(x)\odot y,x\odot d(y),d(x\odot y\left)\right\}.$$

There are only two cases:

If $d\left(x\right)\odot y=0$ or $x\odot d\left(y\right)=0$, without loss of generality, assume that $d\left(x\right)\odot y=0$. Then,

$$d(x\odot y)=0\vee (x\odot d(y\left)\right)=x\odot d\left(y\right).$$

Thus, $\left(\beta \right(x)\odot y)\vee (x\odot \beta (y\left)\right)=\{0,d(x\odot y\left)\right\}=\beta (x\odot y)$.

If $d\left(x\right)\odot y=x\odot d\left(y\right)$, then

$$d(x\odot y)=d\left(x\right)\odot y=x\odot d\left(y\right).$$

Thus, $\left(\beta \right(x)\odot y)\vee (x\odot \beta (y\left)\right)=\{0,d(x\odot y\left)\right\}=\beta (x\odot y)$.

Consequently, we infer $\beta \in \mathrm{MD}\left(A\right)$. □

Corollary 1. 

If A is an MV-algebra, and $a\in A$, a multifunction ${\beta }_a:A\to \mathrm{\Delta }\left(A\right)$ on A is defined as follows

$${\beta }_a\left(x\right):=\{0,d_a\left(x\right)\}.$$

Then ${\beta }_a\in \mathrm{MD}\left(A\right)$.

Proof. 

If $d=d_a$ in Proposition 5, then for any $x,y\in A$, we know $d\left(x\right)\odot y=a\odot x\odot y=x\odot d\left(y\right)$. Hence, we infer that ${\beta }_a\in \mathrm{MD}\left(A\right)$ by Proposition 5. □

Remark 6. 

The conclusion is not necessarily true for general $(\odot ,\vee )$-derivations. For example, $d=\left(\begin{array}{cccc}0& \frac{1}{3}& \frac{2}{3}& 1\\ 0& \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right)$ is an $(\odot ,\vee )$-derivation on $L_4$. But $\beta (\frac{2}{3}\odot 1)=\{0,\frac{2}{3}\}\ne \{0,\frac{1}{3},\frac{2}{3}\}=\{0,\frac{2}{3}\}\vee \{0,\frac{1}{3}\}=(\{0,\frac{2}{3}\}\odot 1)\vee (\frac{2}{3}\odot \{0,\frac{2}{3}\})=(\beta \left(\frac{2}{3}\right)\odot 1)\vee (\frac{2}{3}\odot \beta \left(1\right))$.

Proposition 6. 

Let A be an MV-algebra and $d\in \mathrm{Der}\left(A\right)$. Define a multifunction $\gamma :A\to \mathrm{\Delta }\left(A\right)$ on A as follows

$$\gamma \left(x\right):=[0,d(x\left)\right].$$

Then $\gamma \in \mathrm{MD}\left(A\right)$.

Proof. 

Since $d\in \mathrm{Der}\left(A\right)$, we obtain $\gamma (x\odot y)=[0,d(x\odot y\left)\right]=[0,(d\left(x\right)\odot y)\vee (x\odot d\left(y\right)\left)\right]$. Moreover, we have

$$\begin{array}{cccc}\hfill \left(\gamma \right(x)\odot y)\vee (x\odot \gamma (y\left)\right)& =\left(\right[0,d\left(x\right)]\odot y)\vee (x\odot [0,d\left(y\right)\left]\right)\hfill & \hfill & (\mathrm{Definition} 3)\hfill \\ & =[0,d(x)\odot y]\vee [0,x\odot d(y\left)\right]\hfill & \hfill & (\mathrm{Lemma} 5\left(3\right))\hfill \\ & =[0,(d\left(x\right)\odot y)\vee (x\odot d\left(y\right)\left)\right].\hfill & \hfill & (\mathrm{Lemma} 5\left(4\right))\hfill \end{array}$$

Hence, we conclude that $\gamma \in \mathrm{MD}\left(A\right)$. □

On the other hand, if there is a given $(\odot ,\vee )$-multiderivation $\sigma$ on A, then we can construct a corresponding $(\odot ,\vee )$-derivation d from $\sigma$. We need the following lemma to prepare.

Lemma 9. 

If A is an MV-algebra, and $M,N\in \mathrm{\Delta }\left(A\right)$, if both $\sup \left(M\right)$ and $\sup \left(N\right)$ exist, then

1. 

$\sup (M\odot N)$ exists and $\sup (M\odot N)=\sup \left(M\right)\odot \sup \left(N\right)$.

2. 

$\sup (M\vee N)$ exists and $\sup (M\vee N)=\sup \left(M\right)\vee \sup \left(N\right)$.

Proof. 

Denote $m_0=\sup \left(M\right)$ and $n_0=\sup \left(N\right)$.

(1) Firstly, we prove that $m_0\odot n_0$ is an upper bound of $M\odot N$. For any $m\in M$ and $n\in N$, we immediately have $m\odot n\le m_0\odot n_0$ by Lemma 1 (4). Hence, it is enough to show that $m_0\odot n_0$ is the least upper bound. Assume that $m\odot n\le x$ for all $m\in M,n\in N$. It tells us that $m\le n^{\ast }\oplus x$ and so $m_0\le n^{\ast }\oplus x$ by Lemma 1 (8) and the definition of least upper bound. Then, we have $m_0\odot n\le x$. Similarly, we obtain $n\le m_0^{\ast }\oplus x$ and $n_0\le m_0^{\ast }\oplus x$. Thus, we can prove that $m_0\odot n_0\le x$. Finally, $\sup (M\odot N)=\sup \left(M\right)\odot \sup \left(N\right)$ holds.

(2) For any $m\in M$ and $n\in N$, we have $m\le m_0$ and $n\le n_0$. So, $m\vee n\le m_0\vee n_0$ and $\sup (M\vee N)\le \sup \left(M\right)\vee \sup \left(N\right)$. Conversely, since $M\vee N⪰M,N$, it implies that $\sup (M\vee N)\ge \sup \left(M\right),\sup \left(N\right)$ and thus $\sup (M\vee N)\ge \sup \left(M\right)\vee \sup \left(N\right)$. Therefore, $\sup (M\vee N)=\sup \left(M\right)\vee \sup \left(N\right)$. □

Proposition 7. 

If A is an MV-algebra, $\sigma \in \mathrm{MD}\left(A\right)$, and $\sup \left(\sigma \right(x\left)\right)$ exists for any $x\in A$, define $\sup \sigma :A\to A$ by $\left(\sup \sigma \right)\left(x\right)=\sup \left(\sigma \right(x\left)\right)$. Then, $\sup \sigma \in \mathrm{Der}\left(A\right)$.

Proof. 

For any $x,y\in A$, we have

$$\begin{array}{cccc}\hfill \left(\sup \sigma \right)(x\odot y)& =\sup \left(\sigma \right(x\odot y\left)\right)\hfill & \hfill & (\mathrm{Definition}\mathrm{of}\sup \sigma )\hfill \\ & =\sup \left(\right(\sigma \left(x\right)\odot y)\vee (x\odot \sigma \left(y\right)\left)\right)\hfill & \hfill & (\mathrm{Equation}(3\left)\right)\hfill \\ & =\sup \left(\sigma \right(x)\odot y)\vee \sup (x\odot \sigma (y\left)\right)\hfill & \hfill & (\mathrm{Lemma} 9\left(2\right))\hfill \\ & =\left(\sup \right(\sigma \left(x\right))\odot \sup \{y\left\}\right)\vee \left(\sup \right\{x\}\odot \sup (\sigma \left(y\right)\left)\right)\hfill & \hfill & (\mathrm{Lemma} 9\left(1\right))\hfill \\ & =\left(\right(\sup \sigma \left)\right(x)\odot y)\vee (x\odot (\sup \sigma \left)\right(y\left)\right).\hfill & \hfill & (\mathrm{Definition}\mathrm{of}\sup \sigma )\hfill \end{array}$$

Hence, $\sup \sigma \in \mathrm{Der}\left(A\right)$. □

Remark 7. 

$\left(1\right)$ If MV-algebra A is complete, then $\sup \sigma$ is always an $(\odot ,\vee )$-derivation on A for an arbitrary $(\odot ,\vee )$-multiderivation σ on A.

$\left(2\right)$ If $\sigma \in \mathrm{MD}\left(A\right)$ and the image $\sigma \left(x\right)$ is finite for any $x\in A$, then $\sup \sigma$ is always an $(\odot ,\vee )$-derivation on A.

Next, we construct $(\odot ,\vee )$-multiderivations on subalgebras and direct products of MV-algebras from a given $(\odot ,\vee )$-multiderivation.

Proposition 8. 

Let A be an MV-algebra and $\sigma \in \mathrm{MD}\left(A\right)$. If S is a subalgebra of A and $\sigma \left(x\right)\subseteq S$ for any $x\in S$, then ${\sigma |}_S\in \mathrm{MD}\left(S\right)$.

Proof. 

For any $x,y\in S$, we know that $\sigma \left(x\right),\sigma \left(y\right)\subseteq S$ and so $\sigma \left(x\right)\odot y,x\odot \sigma \left(y\right)\subseteq S$. Then,

$${\sigma |}_S(x\odot y)=(\sigma \left(x\right)\odot y)\vee (x\odot \sigma \left(y\right))=(\sigma {|}_S\left(x\right)\odot y)\vee (x\odot \sigma {|}_S\left(y\right))\subseteq S\vee S=S$$

by Lemma 6 (3). Thus, ${\sigma |}_S\in \mathrm{MD}\left(S\right)$. □

Definition 5. 

If $\mathrm{\Omega }$ is a nonempty set, for each $i\in \mathrm{\Omega }$, let ${\sigma }_i$ be a multifunction on $A_i.$ The direct product of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$${\prod }_{i\in \mathrm{\Omega }}{\sigma }_i:{\prod }_{i\in \mathrm{\Omega }}{A}_i\to \mathrm{\Delta }\left({\prod }_{i\in \mathrm{\Omega }}{A}_i\right)$ is defined by

$$\left(\prod _{i\in \mathrm{\Omega }}{\sigma }_i\right)\left(g\right)=\prod _{i\in \mathrm{\Omega }}{\sigma }_i\left(g\left(i\right)\right)=\{{\left(x_i\right)}_{i\in \mathrm{\Omega }}\mid x_i\in {\sigma }_i\left(g\left(i\right)\right)\}$$

for all $g\in {\prod }_{i\in \mathrm{\Omega }}{A}_i$.

Lemma 10. 

Let Ω be a nonempty set, ${\left\{A_i\right\}}_{i\in \mathrm{\Omega }}$ be a family of MV-algebras, and $M_i,N_i\in \mathrm{\Delta }\left(A_i\right)$. Then, ${\prod }_{i\in \mathrm{\Omega }}(M_i\vee N_i)={\prod }_{i\in \mathrm{\Omega }}{M}_i\vee {\prod }_{i\in \mathrm{\Omega }}{N}_i$.

Proof. 

We first show that ${\prod }_{i\in \mathrm{\Omega }}(M_i\vee N_i)\subseteq {\prod }_{i\in \mathrm{\Omega }}{M}_i\vee {\prod }_{i\in \mathrm{\Omega }}{N}_i$. For any $x\in {\prod }_{i\in \mathrm{\Omega }}(M_i\vee N_i)$, there are $m_i\in M_i,n_i\in N_i$ for any $i\in \mathrm{\Omega }$ such that $x={(m_i\vee n_i)}_{i\in \mathrm{\Omega }}$. Denote $m={\left(m_i\right)}_{i\in \mathrm{\Omega }},n={\left(n_i\right)}_{i\in \mathrm{\Omega }}$, we have $x={(m_i\vee n_i)}_{i\in \mathrm{\Omega }}={\left(m_i\right)}_{i\in \mathrm{\Omega }}\vee {\left(n_i\right)}_{i\in \mathrm{\Omega }}=m\vee n\in {\prod }_{i\in \mathrm{\Omega }}{M}_i\vee {\prod }_{i\in \mathrm{\Omega }}{N}_i$. And vice versa. Therefore, ${\prod }_{i\in \mathrm{\Omega }}(M_i\vee N_i)={\prod }_{i\in \mathrm{\Omega }}{M}_i\vee {\prod }_{i\in \mathrm{\Omega }}{N}_i$. □

Proposition 9. 

Assume that Ω is a nonempty set and ${\left\{A_i\right\}}_{i\in \mathrm{\Omega }}$ is a family of MV-algebras. Then, ${\sigma }_i\in \mathrm{MD}\left(A_i\right)$ for any $i\in \mathrm{\Omega }$ iff ${\prod }_{i\in \mathrm{\Omega }}{\sigma }_i\in \mathrm{MD}\left({\prod }_{i\in \mathrm{\Omega }}{A}_i\right)$.

Proof. 

Denote $A={\prod }_{i\in \mathrm{\Omega }}{A}_i$ and $\sigma ={\prod }_{i\in \mathrm{\Omega }}{\sigma }_i$. For all $x={\left(x_i\right)}_{i\in \mathrm{\Omega }},y={\left(y_i\right)}_{i\in \mathrm{\Omega }}\in A$, we have

$$\sigma (x\odot y)=\sigma ({\left(x_i\right)}_{i\in \mathrm{\Omega }}\odot {\left(y_i\right)}_{i\in \mathrm{\Omega }})=\prod _{i\in \mathrm{\Omega }}{\sigma }_i(x_i\odot y_i),$$

$$\begin{array}{cc}\hfill \left(\sigma \right(x)\odot y)\vee (x\odot \sigma (y\left)\right)& =\left(\prod _{i\in \mathrm{\Omega }}{\sigma }_i\left(x_i\right)\odot {\left(y_i\right)}_{i\in \mathrm{\Omega }}\right)\vee \left({\left(x_i\right)}_{i\in \mathrm{\Omega }}\odot \prod _{i\in \mathrm{\Omega }}{\sigma }_i\left(y_i\right)\right)\hfill \\ & =\prod _{i\in \mathrm{\Omega }}({\sigma }_i\left(x_i\right)\odot y_i)\vee \prod _{i\in \mathrm{\Omega }}(x_i\odot {\sigma }_i\left(y_i\right))\hfill \\ & =\prod _{i\in \mathrm{\Omega }}(({\sigma }_i\left(x_i\right)\odot y_i)\vee (x_i\odot {\sigma }_i\left(y_i\right))).\hfill & \hfill & (\mathrm{Lemma} 10)\hfill \end{array}$$

We can immediately obtain ${\sigma }_i\in \mathrm{MD}\left(A_i\right)$ for all $i\in \mathrm{\Omega }$ iff $\sigma (x\odot y)=\left(\sigma \right(x)\odot y)\vee (x\odot \sigma (y\left)\right)$ by Equation (3). □

Finally, we investigate the condition when an $(\odot ,\vee )$-multiderivation $\sigma$ is isotone.

Definition 6. 

If A is an MV-algebra, and $\sigma \in \mathrm{MD}\left(A\right)$, we say σ is isotone if $\sigma \left(x\right)⪯\sigma \left(y\right)$ whenever $x\le y$.

Proposition 10. 

If A is an MV-algebra, and $\sigma \in \mathrm{MD}\left(A\right)$, then σ is isotone iff $\sigma (x\wedge y)⪯\sigma \left(x\right)\wedge y$ for all $x,y\in A$.

Proof. 

Assume $\sigma$ is isotone, then,

$$\sigma (x\wedge y)\subseteq \sigma (x\wedge y)\wedge \sigma (x\wedge y)⪯\sigma \left(x\right)\wedge \sigma \left(y\right)⪯\sigma \left(x\right)\wedge y$$

by Lemma 6 (3) and (2). Conversely, assume that $\sigma (x\wedge y)⪯\sigma \left(y\right)\wedge x$ for all $x,y\in A$. Let $x,y\in A$ with $x\le y$. Then, $\sigma \left(x\right)=\sigma (y\wedge x)⪯\sigma \left(y\right)\wedge x$. Thus, for every $a\in \sigma \left(x\right)$ there is $b\in \sigma \left(y\right)$ such that $a\le b\wedge x$. Hence, $a\le b$ and so $\sigma \left(x\right)⪯\sigma \left(y\right)$. □

Corollary 2. 

If A is an MV-algebra, and $S\subseteq A$ is a subalgebra of A, then the principal $(\odot ,\vee )$-multiderivation ${\sigma }_S$ is isotone.

Proof. 

Method 1: Let $x,y\in A$ and $x\le y$. For any $s\in S$, Lemma 1 (4) implies $x\odot s\le y\odot s$. Thus, ${\sigma }_S\left(x\right)⪯{\sigma }_S\left(y\right)$.

Method 2: It is enough to verify that ${\sigma }_S(x\wedge y)⪯{\sigma }_S\left(x\right)\wedge y$ for all $x,y\in A$ by Proposition 10. For any $s\in S$, Lemma 1 (6) implies

$$(x\wedge y)\odot s=(x\odot s)\wedge (y\odot s)\le (x\odot s)\wedge y.$$

Thus, ${\sigma }_S(x\wedge y)=(x\wedge y)\odot S⪯(x\odot S)\wedge y={\sigma }_S\left(x\right)\wedge y$. □

4. The Order Structure of $(\odot ,\vee )$-Multiderivations on a Finite MV-Chain

Let $\mathrm{MF}\left(A\right)$ be the set of multifunctions on an MV-algebra A. Define ≼ on $\mathrm{MF}\left(A\right)$ by:

$$(\forall \sigma ,{\sigma }^{\prime }\in \mathrm{MF}\left(A\right))\hspace{1em}\sigma \preccurlyeq {\sigma }^{\prime } \mathrm{if} \sigma \left(x\right)⪯{\sigma }^{\prime }\left(x\right),\forall x\in A.$$

Then, ≼ is a preorder on $\mathrm{MF}\left(A\right)$ and ${\mathbf{0}}_{\mathrm{MF}\left(A\right)}\preccurlyeq \sigma \preccurlyeq {\mathbf{1}}_{\mathrm{MF}\left(A\right)}$ for any $\sigma \in \mathrm{MF}\left(A\right)$, where ${\mathbf{0}}_{\mathrm{MF}\left(A\right)}$ and ${\mathbf{1}}_{\mathrm{MF}\left(A\right)}$ are defined by ${\mathbf{0}}_{\mathrm{MF}\left(A\right)}\left(x\right):=\left\{0\right\}$ and ${\mathbf{1}}_{\mathrm{MF}\left(A\right)}\left(x\right):=\left\{1\right\}$ for any $x\in A$, respectively. For any $\sigma \in \mathrm{MD}\left(A\right)$, we have ${\mathbf{0}}_{\mathrm{MF}\left(A\right)}\preccurlyeq \sigma \preccurlyeq I{d}_{\mathrm{MF}\left(A\right)}$, where $I{d}_{\mathrm{MF}\left(A\right)}\left(x\right)=\left\{x\right\}$, and it is plain that $\left\{0\right\}⪯\sigma \left(x\right)⪯\left\{x\right\}$, $\forall x\in A$.

For $\sigma ,{\sigma }^{\prime }\in \mathrm{MF}\left(A\right)$, set

$$\left(\sigma ⊠{\sigma }^{\prime }\right)\left(x\right):=\sigma \left(x\right)⊠{\sigma }^{\prime }\left(x\right),$$

(4)

for any $x\in A$ and $⊠\in \{\vee ,\wedge ,\cup ,\cap \}$.

Remark 8. 

1. 

Note that $\sigma \left(x\right)\vee {\sigma }^{\prime }\left(x\right)$ is meant in the sense of Definition 3, rather than the supremum of $\sigma \left(x\right)$ and ${\sigma }^{\prime }\left(x\right)$.

2. 

Note that $\sigma \vee {\sigma }^{\prime }$ is an upper bound of σ and ${\sigma }^{\prime }$ by Lemma 6 (1) but is not necessarily a least upper bound. For example, define $\sigma \in \mathrm{MF}\left(B_4\right)$ by $\sigma \left(a\right)=\sigma \left(b\right)=\{a,b\}$, $\sigma \left(0\right)=\left\{0\right\},\sigma \left(1\right)=\left\{1\right\}$. Then,

$$(\sigma \vee \sigma )\left(a\right)=(\sigma \vee \sigma )\left(b\right)=\{a,b,1\}.$$

It is clear that both σ and $\sigma \vee \sigma$ are upper bounds of σ and σ, but $\sigma \prec \sigma \vee \sigma$. In a word, $\sigma \vee \sigma$ is not a least upper bound of σ and σ.

More generally, let A be an MV-algebra which is not an MV-chain with two incomparable elements $a,b$. Define $\sigma \in \mathrm{MF}\left(A\right)$ as $\sigma \left(a\right)=\sigma \left(b\right)=\{a,b\}$, $\sigma \left(x\right)=\left\{x\right\}$ for $x\in A\setminus \{a,b\}$. $\sigma \vee \sigma$ is not a least upper bound of σ and σ.

In the sense of category theory, a preordered set P is called complete [27] (Section 8.5) if for every subset S of P both $\sup S$ and $\inf S$ exist (in P). Note that $\sup S$ and $\inf S$ need not be unique. For example, let $P=\{a,b\}$ and define a preorder ⪯ as follows: $a⪯b$, $b⪯a$. Take $S=\{a,b\}$. Then, both a and b are $\sup S$, also $\inf S$. Therefore, we use “a” rather than “the” concerning $\sup S$ and $\inf S$ in the following.

Let ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$ be a nonempty family of multifunctions on an MV-algebra A. Define a multifunction ${\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i$ on A, by

$$\left(\bigcup _{i\in \mathrm{\Omega }}{\sigma }_i\right)\left(x\right):=\bigcup _{i\in \mathrm{\Omega }}{\sigma }_i\left(x\right),$$

for any $x\in A$.

Analogue to [28] (Theorem I.4.2), we have the following.

Lemma 11. 

If A is an MV-algebra, then $(\mathrm{MF}\left(A\right),\preccurlyeq ,{\mathbf{0}}_{\mathrm{MF}\left(A\right)},{\mathbf{1}}_{\mathrm{MF}\left(A\right)})$ is a complete bounded preordered set, where ${\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i$ is a least upper bound of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$, and $\sigma \wedge {\sigma }^{\prime }$ is a greatest lower bound of σ and ${\sigma }^{\prime }$, respectively.

Proof. 

Note that ${\mathbf{0}}_{\mathrm{MF}\left(A\right)}\preccurlyeq \sigma \preccurlyeq {\mathbf{1}}_{\mathrm{MF}\left(A\right)}$ for any $\sigma \in \mathrm{MF}\left(A\right)$.

Let ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$ be a nonempty family of $\mathrm{MF}\left(A\right)$. Then, ${\sigma }_i\preccurlyeq {\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i$. Now we will prove that ${\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i$ is a least upper bound of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$. Assume that ${\sigma }_i\preccurlyeq \eta$ for every $i\in \mathrm{\Omega }$. For any $y\in \left({\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i\right)\left(x\right)$ where $x\in A$, there exists $k\in \mathrm{\Omega }$ such that $y\in {\sigma }_k\left(x\right)$. Since ${\sigma }_k\left(x\right)⪯\eta \left(x\right)$, there is $z\in \eta \left(x\right)$ such that $y\le z$, which shows ${\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i\preccurlyeq \eta$. Therefore, ${\bigcup }_{i\in \mathrm{\Omega }}{\sigma }_i$ is a least upper bound of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$.

Let

$$X^{\ell }=\{\lambda \in \mathrm{MF}\left(A\right)\mid \lambda \preccurlyeq {\sigma }_i,\forall i\in \mathrm{\Omega }\}$$

be the set of lower bounds of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$ in $\mathrm{MF}\left(A\right)$. Next, we verify that ${\bigcup }_{\lambda \in X^{\ell }}\lambda$ is indeed a greatest lower bound of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$. For any $i\in \mathrm{\Omega }$ and $\lambda \in X^{\ell }$, we have $\lambda \preccurlyeq {\sigma }_i$. Thus, ${\bigcup }_{\lambda \in X^{\ell }}\lambda \preccurlyeq {\sigma }_i$ and ${\bigcup }_{\lambda \in X^{\ell }}\lambda \in X^{\ell }$. Hence, ${\bigcup }_{\lambda \in X^{\ell }}\lambda$ is a greatest lower bound of ${\left\{{\sigma }_i\right\}}_{i\in \mathrm{\Omega }}$. Therefore, $\mathrm{MF}\left(A\right)$ is complete.

For any $\sigma ,{\sigma }^{\prime }\in \mathrm{MF}\left(A\right)$, since $\sigma \wedge {\sigma }^{\prime }\preccurlyeq \sigma ,{\sigma }^{\prime }$, it follows that $\sigma \wedge {\sigma }^{\prime }$ is a lower bound of $\sigma$ and ${\sigma }^{\prime }$. To verify that $\sigma \wedge {\sigma }^{\prime }$ is a greatest lower bound, let $\eta \preccurlyeq \sigma ,{\sigma }^{\prime }$. Then, for any $y\in \eta \left(x\right)$$(x\in A)$, there are $z\in \sigma \left(x\right)$ and $w\in {\sigma }^{\prime }\left(x\right)$ such that $y\le z$ and $y\le w$ by $\eta \left(x\right)⪯\sigma \left(x\right),{\sigma }^{\prime }\left(x\right)$. Hence,

$$y\le z\wedge w\in \sigma \left(x\right)\wedge {\sigma }^{\prime }\left(x\right).$$

Therefore, $\eta \left(x\right)⪯\sigma \left(x\right)\wedge {\sigma }^{\prime }\left(x\right)$. Thus, $\eta \preccurlyeq \sigma \wedge {\sigma }^{\prime }$. □

As already mentioned, ⪯ is not always a partial order on $\mathrm{\Delta }\left(A\right)$, where $M⪯N$ iff for each $m\in M$ there exists $n\in N$ such that $m\le n$. The binary relation ∼ on $\mathrm{\Delta }\left(A\right)$ defined by $M\sim N$ iff $M⪯N$ and $N⪯M$ is an equivalence relation. Given $M\in \mathrm{\Delta }\left(A\right)$, the equivalence class of M with respect to ∼ will be denoted by $\overline{M}$. If $M=\left\{x\right\}$ is a singleton, then we abbreviate $\overline{\left\{x\right\}}$ by $\overline{x}$. Thus, we can obtain a partial order ⪯ on $\mathrm{\Delta }\left(A\right)/\sim$ defined by $\overline{M}⪯\overline{N}$ iff $M⪯N$. We claim that ⪯ is well defined. In fact, if $M\sim M^{\prime },N\sim N^{\prime }$ and $M⪯N$, then $M^{\prime }⪯M⪯N⪯N^{\prime }$.

Recall that for a subset M of A, the lower set generated by M [29] is the set

$$\downarrow M=\{x\in A\mid \mathrm{there}\mathrm{exists}m\in M\mathrm{such}\mathrm{that}x\le m\}.$$

Lemma 12. 

Let $M,N\in \mathrm{\Delta }\left(A\right)$. Then, $\overline{M}=\overline{N}$ iff $\downarrow M=\downarrow N$.

Proof. 

It is sufficient to show that $M⪯N$ iff $\downarrow M\subseteq \downarrow N$.

Let $M⪯N$. For every $x\in \downarrow M$, there is $m\in M$ such that $x\le m$. Then, $M⪯N$ gives $m\le n$ for some $n\in N$. Hence, $x\le n$ and $x\in \downarrow N$. Therefore, $\downarrow M\subseteq \downarrow N$.

Conversely, assume that $\downarrow M\subseteq \downarrow N$. For any $m\in M$, we have $m\in \downarrow M\subseteq \downarrow N$. Thus, there exists $n\in N$ such that $m\le n$. Hence, $M⪯N$.

Similarly, $N⪯M$ iff $\downarrow N\subseteq \downarrow M$. □

Corollary 3. 

In general, let A be an MV-algebra, $M\in \mathrm{\Delta }\left(A\right)$, and $a\in A$. Then, $\overline{M}=\overline{a}$ iff $\sup M$ exists and $\sup M=a\in M$.

Assume $\overline{M}=\overline{a}$. Then a is an upper bound of M since $M⪯\left\{a\right\}$. To prove a is a least upper bound of M, let b be an upper bound of M. Since $\left\{a\right\}⪯M$, there exists $m\in M$ such that $a\le m$. Hence, $a\le m\le b$, which shows $\sup M=a\in M$.

Conversely, let $\sup M=a\in M$. It suffices to verify that $\downarrow M=\downarrow a$ by Lemma 12. If $x\in \downarrow M$, then there is $m\in M$ such that $x\le m\le a$. It follows that $x\in \downarrow a$ and $\downarrow M\subseteq \downarrow a$. If $x\in \downarrow a$, then $x\le a\in M$. Thus, $x\in \downarrow M$ and $\downarrow a\subseteq \downarrow M$. Therefore, $\downarrow M=\downarrow a$.

Corollary 4. 

Let $L_n$ with $n\ge 2$ and $M\in \mathrm{\Delta }\left(L_n\right)$. Then, $\overline{M}=\overline{\sup M}$.

Proof. 

Observe that $\sup M$ is exactly $\frac{i}{n-1}$ for a certain $0\le i\le n-1$. It suffices to verify that $\downarrow M=\downarrow \sup M$ by Lemma 12. Suppose $x\in \downarrow M$, there is $m\in M$ such that $x\le m$. Since $m\le \sup M$, it follows that $x\le \sup M$. Hence, $x\in \downarrow \sup M$. Conversely, assume $x\in \downarrow \sup M$, which means $x\le \sup M=\frac{i}{n-1}$. Since $\sup M\in M$, it follows that $x\in \downarrow M$. Therefore, $\downarrow M=\downarrow \sup M$ and $\overline{M}=\overline{\sup M}$. □

Note that the family of all lower sets of a poset A is a complete lattice by [30] (Example O-2.8). We will prove that the family of all nonempty lower sets of A is also a complete lattice, denoted by $(L_0\left(A\right),\subseteq )$.

Corollary 5. 

Let A be an MV-algebra, then $\mathrm{\Delta }\left(A\right)/\sim$ is isomorphic to the complete lattice $(L_0\left(A\right),\subseteq )$.

Proof. 

Since A has a least element 0, the intersection of a family of nonempty lower sets of A is still a nonempty lower set. Therefore, $L_0\left(A\right)$ is a complete lattice.

Define $\phi :\mathrm{\Delta }\left(A\right)/\sim \to L_0\left(A\right)$ by $\overline{M}\mapsto \downarrow M$. Lemma 12 shows that $\phi$ is well defined and injective, and $\phi$ is also surjective since $M=\downarrow M$ if $M\in L_0\left(A\right)$. As discussed in the proof of Lemma 12, $\overline{M}⪯\overline{N}$ iff $\downarrow M\subseteq \downarrow N$ for all M, $N\in \mathrm{\Delta }\left(A\right)$, which gives both $\phi$ and ${\phi }^{-1}$ are order preserving. Hence, $\phi$ is an isomorphism. □

Next, we study the order structure on $\mathrm{\Delta }\left(L_n\right)/\sim$. First, we need

Lemma 13. 

Let A be an MV-chain, $M,N\in \mathrm{\Delta }\left(A\right)$, and $\sup M,\sup N$ exist.

1. 

If $\overline{M}⪯\overline{N}$, then $\sup M\le \sup N$.

2. 

If $\sup M<\sup N$, then $\overline{M}⪯\overline{N}$.

3. 

$\overline{M}=\overline{N}$ iff the following conditions hold:

(a) 

$\sup M=\sup N$.

(b) 

$\sup M\in M\iff \sup N\in N$.

In particular, if A is a finite MV-chain, then $\overline{M}=\overline{N}$ iff (a) holds.

Proof. 

(1) Suppose $\overline{M}⪯\overline{N}$, then $M⪯N$. For any $m\in M$ there is $n\in N$ such that $m\le n\le \sup N$. According to the definition of $\sup M$, we have $\sup M\le \sup N$.

(2) Let $\sup M<\sup N$. Assume on the contrary $MN$. Then, there is $m\in M$ such that $m>n$ for any $n\in N$. The definition of $\sup N$ implies $m\ge \sup N$. Thus, $\sup N\le m\le \sup M$, which contradicts the fact that $\sup M<\sup N$.

(3) Assume that $\overline{M}=\overline{N}$. (a) follows from (1).

To prove that $\sup M\in M\iff \sup N\in N$, we assume $\sup M\in M$. Then, there exists $n_0\in N$ such that $\sup M\le n_0$ by $M⪯N$. Since $N⪯M$, we have $n_0\le \sup M$. Hence, $n_0=\sup M$. Therefore, $\sup N=\sup M=n_0\in N$ by (a). Symmetrically, $\sup N\in N\Rightarrow \sup M\in M$.

Conversely, assume that (a) and (b) hold, it suffices to show that $\downarrow M=\downarrow N$ by Lemma 12. Assume that $\downarrow M\ne \downarrow N$; without loss of generality, there is $y\in \downarrow M$ but $y\notin \downarrow N$. That is to say, for arbitrary $n\in N$ we have $n<y$. So, $\sup N\in N$ implies $\sup N<y$. Since $y\in \downarrow M$, there is $m\in M$ such that $y\le m$. It follows $\sup N<y\le m<\sup M$ by the definition of $\sup N$, which is contrary to $\sup M=\sup N$. Thus, $\overline{M}=\overline{N}$.

Assume A is a finite MV-chain, and (b) always holds. Hence, $\overline{M}=\overline{N}$ iff (a) holds. □

Remark 9. 

Note that $\sup M=\sup N$ may not imply $\overline{M}⪯\overline{N}$. For example, let $A=[0,1]$ be the standard MV-algebra and $\frac{1}{2}\in A$. Define $M=\downarrow \frac{1}{2}$ and $N=\{a\in A\mid 0\le a<\frac{1}{2}\}$. Then, $\sup M=\sup N=\frac{1}{2}$, but $\overline{M}\overline{N}$, since $\frac{1}{2}\in M$, there is no $y\in N$ such that $\frac{1}{2}\le y$.

Example 4. 

Consider the MV-chain $L_n$ with $n\ge 2$. Then, $\mathrm{\Delta }\left(L_n\right)/\sim$ is order isomorphic to $L_n$.

Proof. 

Define $f:L_n\to \mathrm{\Delta }\left(L_n\right)/\sim$ by $f\left(x\right)=\overline{x}$ for any $x\in L_n$. If $\overline{x}=\overline{y}$, then $x=\sup \left\{x\right\}=\sup \left\{y\right\}=y$ by Lemma 13 (3). Thus, f is injective. To prove f is surjective, assume $\overline{M}\in \mathrm{\Delta }\left(L_n\right)/\sim$, then $f\left(\sup M\right)=\overline{\sup M}=\overline{M}$ by Corollary 4.

It is enough to verify that f and $f^{-1}$ are order preserving. If $x\le y$, then $f\left(x\right)=\overline{x}⪯\overline{y}=f\left(y\right)$ since $\left\{x\right\}⪯\left\{y\right\}$ and Corollary 4. Conversely, suppose $\overline{x}⪯\overline{y}$, we have $x=\sup \left\{x\right\}\le \sup \left\{y\right\}=y$ by Lemma 13 (1). Therefore, f is an isomorphism. □

We next investigate the preorder on the set of $(\odot ,\vee )$-multiderivations.

Similar to $\mathrm{\Delta }\left(A\right)$, we can define an equivalence relation on $\mathrm{MD}\left(A\right)$ by $\sigma \sim {\sigma }^{\prime }$ iff $\sigma \preccurlyeq {\sigma }^{\prime }$ and ${\sigma }^{\prime }\preccurlyeq \sigma$, and define $\overline{\sigma }\preccurlyeq \overline{{\sigma }^{\prime }}$ in $\mathrm{MD}\left(A\right)/\sim$ iff $\sigma \preccurlyeq {\sigma }^{\prime }$. Observe that ≼ in $\mathrm{MD}\left(A\right)/\sim$ is a well-defined partial order by the hereditary order of ⪯. Clearly, $\left(\mathrm{MD}\right(A)/\sim ,\preccurlyeq )$ is a poset. By the definition of ⪯, we know $\overline{\sigma }=\overline{{\sigma }^{\prime }}$ iff $\overline{\sigma \left(x\right)}=\overline{{\sigma }^{\prime }\left(x\right)}$ for any $x\in A$.

For any $\sigma \in \mathrm{MD}\left(A\right)$, $\downarrow \sigma :A\to \mathrm{\Delta }\left(A\right)$ is defined as $(\downarrow \sigma )\left(x\right)=\downarrow \sigma \left(x\right)$. We claim that $\overline{\sigma }=\overline{\downarrow \sigma }$. In fact, $\sigma \preccurlyeq \downarrow \sigma$ is trivial. For any $y\in \downarrow \sigma \left(x\right)$, there exists $z\in \sigma \left(x\right)$ such that $y\le z$ by the definition of $\downarrow \sigma \left(x\right)$. Therefore, $\downarrow \sigma \left(x\right)⪯\sigma \left(x\right)$ for any $x\in A$ and $\downarrow \sigma \preccurlyeq \sigma$.

Lemma 14. 

If A is an MV-algebra, then:

1. 

$\sigma \vee {\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$ for all $\sigma ,{\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$.

2. 

$\downarrow \sigma \in \mathrm{MD}\left(A\right)$ for any $\sigma \in \mathrm{MD}\left(A\right)$.

Proof. 

(1) Let $\sigma ,{\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$ and $x,y\in A$. Then, we have

$$\begin{array}{cccc}\hfill (\sigma \vee {\sigma }^{\prime })(x\odot y)& =\sigma (x\odot y)\vee {\sigma }^{\prime }(x\odot y)\hfill & \hfill & (\mathrm{Definition}\mathrm{of}\sigma \vee {\sigma }^{\prime })\hfill \\ & =((\sigma \left(x\right)\odot y)\vee (x\odot \sigma \left(y\right)))\vee (({\sigma }^{\prime }\left(x\right)\odot y)\vee (x\odot {\sigma }^{\prime }\left(y\right)))\hfill & \hfill & \left(\sigma ,{\sigma }^{\prime }\in \mathrm{MD}\left(A\right)\right)\hfill \\ & =((\sigma \left(x\right)\odot y)\vee ({\sigma }^{\prime }\left(x\right)\odot y))\vee ((x\odot \sigma \left(y\right))\vee (x\odot {\sigma }^{\prime }\left(y\right)))\hfill & \hfill & (\mathrm{Lemma} 6\left(4\right)\mathrm{and}\left(5\right))\hfill \\ & =((\sigma \left(x\right)\vee {\sigma }^{\prime }\left(x\right))\odot y)\vee (x\odot (\sigma \left(y\right)\vee {\sigma }^{\prime }\left(y\right)))\hfill & \hfill & (\mathrm{Lemma} 7\left(2\right))\hfill \\ & =((\sigma \vee {\sigma }^{\prime })\left(x\right)\odot y)\vee (x\odot (\sigma \vee {\sigma }^{\prime })\left(y\right))\hfill & \hfill & (\mathrm{Definition}\mathrm{of}\sigma \vee {\sigma }^{\prime })\hfill \end{array}$$

and so $\sigma \vee {\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$.

(2) Assume $\sigma \in \mathrm{MD}\left(A\right)$. Let $a\in (\downarrow \sigma )(x\odot y)=\downarrow \sigma (x\odot y)=\downarrow \left(\right(\sigma \left(x\right)\odot y)\vee (x\odot \sigma \left(y\right)\left)\right)$. There exist $x_1\in \sigma \left(x\right)$ and $y_1\in \sigma \left(y\right)$ such that $a\le (x_1\odot y)\vee (x\odot y_1)$. It follows that

$$\begin{array}{cc}\hfill a& =a\wedge ((x_1\odot y)\vee (x\odot y_1))\hfill \\ & =(a\wedge (x_1\odot y))\vee (a\wedge (x\odot y_1))\hfill & \hfill & (\mathrm{Distributivity}\mathrm{of}A)\hfill \\ & =(b\odot y)\vee (x\odot c),\hfill & \hfill & (\mathrm{Lemma} 5\left(1\right))\hfill \end{array}$$

where $b\le x_1$ and $c\le y_1$. Hence, $a\in \left(\right(\downarrow \sigma \left)\right(x)\odot y)\vee (x\odot (\downarrow \sigma \left)\right(y\left)\right)$.

Conversely, let $a\in \left(\right(\downarrow \sigma \left)\right(x)\odot y)\vee (x\odot (\downarrow \sigma \left)\right(y\left)\right)$. There exist $x_1\in \sigma \left(x\right)$ and $y_1\in \sigma \left(y\right)$ such that

$$a=(b\odot y)\vee (x\odot c)\le (x_1\odot y)\vee (x\odot y_1),$$

where $b\le x_1$ and $c\le y_1$. Thus, $a\in (\downarrow \sigma )(x\odot y)$.

Therefore, $\downarrow \sigma \in \mathrm{MD}\left(A\right)$. □

Remark 10. 

When A is an MV-chain, $\sigma \vee {\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$ is a least upper bound of σ and ${\sigma }^{\prime }$ in $\mathrm{MD}\left(A\right)$. We know $\sigma \cup {\sigma }^{\prime }$ is a least upper bound of σ and ${\sigma }^{\prime }$ in $\mathrm{MF}\left(A\right)$. Note that $\mathrm{MD}\left(A\right)\subseteq \mathrm{MF}\left(A\right)$ and the preordered on $\mathrm{MF}\left(A\right)$. It suffices to verify that $\sigma \vee {\sigma }^{\prime }\sim \sigma \cup {\sigma }^{\prime }$. For all $x\in A$, $(\sigma \cup {\sigma }^{\prime })\left(x\right)⪯(\sigma \vee {\sigma }^{\prime })\left(x\right)$ is trivial. For any $y\in (\sigma \vee {\sigma }^{\prime })\left(x\right)$, there exist $z\in \sigma \left(x\right)$ and $z^{\prime }\in {\sigma }^{\prime }\left(x\right)$ such that $y=z\vee z^{\prime }$. Since A is an MV-chain, $y=z$ or $y=z^{\prime }$. Hence, $y\in (\sigma \cup {\sigma }^{\prime })\left(x\right)$, which implies $(\sigma \vee {\sigma }^{\prime })\left(x\right)⪯(\sigma \cup {\sigma }^{\prime })\left(x\right)$. Therefore, $(\sigma \cup {\sigma }^{\prime })\left(x\right)\sim (\sigma \vee {\sigma }^{\prime })\left(x\right)$ for all $x\in A$, and hence, $\sigma \vee {\sigma }^{\prime }\in \mathrm{MD}\left(A\right)$ is a least upper bound of σ and ${\sigma }^{\prime }$ in $\mathrm{MD}\left(A\right)$.

At the end of this section, we characterize the lattice $\mathrm{MD}\left(L_n\right)/\sim$$(n\ge 2)$.

Theorem 1. 

If $L_n$ is the n-element MV-chain with $n\ge 2$, then the lattices $\mathrm{MD}\left(L_n\right)/\sim$ and $\mathrm{Der}\left(L_n\right)$ are isomorphic.

Proof. 

Define a map $f:\mathrm{MD}\left(L_n\right)/\sim \to \mathrm{Der}\left(L_n\right)$ by

$$f\left(\overline{\sigma }\right)=\sup \sigma .$$

By Proposition 7 we know $\sup \sigma \in \mathrm{Der}\left(L_n\right)$. The order ≦ on $\mathrm{Der}\left(L_n\right)$ is defined as $d\leqq d^{\prime }$ iff $d\left(x\right)\le d^{\prime }\left(x\right)$, $\forall x\in L_n$.

Firstly, we prove that f is well defined. Suppose $\overline{\sigma }=\overline{{\sigma }^{\prime }}$, that is, $\overline{\sigma \left(x\right)}=\overline{{\sigma }^{\prime }\left(x\right)}$ for any $x\in L_n$. We get

$$\left(\sup \sigma \right)\left(x\right)=\sup \left(\sigma \left(x\right)\right)=\sup \left({\sigma }^{\prime }\left(x\right)\right)=\left(\sup {\sigma }^{\prime }\right)\left(x\right)$$

for any $x\in L_n$ by Lemma 13 (3). Thus, $f\left(\overline{\sigma }\right)=\sup \left(\sigma \right)=\sup \left({\sigma }^{\prime }\right)=f\left(\overline{{\sigma }^{\prime }}\right)$.

If $f\left(\overline{\sigma }\right)=f\left(\overline{{\sigma }^{\prime }}\right)$, that is, $\sup \left(\sigma \right)=\sup \left({\sigma }^{\prime }\right)$, then $\sup \left(\sigma \left(x\right)\right)=\sup \left({\sigma }^{\prime }\left(x\right)\right)$ for any $x\in L_n$. Lemma 13 (3) implies $\overline{\sigma \left(x\right)}=\overline{{\sigma }^{\prime }\left(x\right)}$ for any $x\in L_n$ and thus $\overline{\sigma }=\overline{{\sigma }^{\prime }}$. Hence, f is injective. For any $d\in \mathrm{Der}\left(L_n\right)$, there is ${\gamma }_d\in \mathrm{MD}\left(L_n\right)$ where ${\gamma }_d\left(x\right):=[0,d\left(x\right)]$ such that

$$f\left(\overline{{\gamma }_d}\right)\left(x\right)=\left(\sup {\gamma }_d\right)\left(x\right)=\sup \left({\gamma }_d\left(x\right)\right)=\sup [0,d\left(x\right)]=d\left(x\right)$$

for all $x\in L_n$ by Propositions 6 and 7. Thus, $f\left(\overline{{\gamma }_d}\right)=d$ and f is surjective.

To prove that f is an order-isomorphism, let $\overline{\sigma }\preccurlyeq \overline{{\sigma }^{\prime }}$, that is, for any $x\in L_n$, $\overline{\sigma \left(x\right)}⪯\overline{{\sigma }^{\prime }\left(x\right)}$. Corollary 4 implies that $\overline{\sigma \left(x\right)}=\overline{\sup \left(\sigma \right(x\left)\right)}$ for any $x\in L_n$. It follows that

$$\overline{\left(\sup \sigma \right)\left(x\right)}=\overline{\sup \left(\sigma \right(x\left)\right)}⪯\overline{\sup \left({\sigma }^{\prime }\left(x\right)\right)}=\overline{\left(\sup {\sigma }^{\prime }\right)\left(x\right)}$$

and thus $\left(\sup \sigma \right)\left(x\right)\le \left(\sup {\sigma }^{\prime }\right)\left(x\right)$ for any $x\in L_n$ since $\left(\sup \sigma \right)\left(x\right)$ is a singleton. Hence, $f\left(\overline{\sigma }\right)=\sup \sigma \leqq \sup {\sigma }^{\prime }=f\left(\overline{{\sigma }^{\prime }}\right)$. Conversely, assume $d,d^{\prime }\in \mathrm{Der}\left(L_n\right)$ and $d\leqq d^{\prime }$, which means $d\left(x\right)\le d^{\prime }\left(x\right)$ for all $x\in L_n$. Now the construction in Proposition 6 gives ${\gamma }_d=f^{-1}:A\to \mathrm{\Delta }\left(A\right)$, where ${\gamma }_d\left(x\right)=[0,d\left(x\right)]$. Furthermore, we have

$${\gamma }_d\left(x\right)=[0,d\left(x\right)]⪯[0,d^{\prime }\left(x\right)]={\gamma }_{d^{\prime }}\left(x\right)$$

for any $x\in L_n$ by the definition of ⪯. Thus, ${\gamma }_d\preccurlyeq {\gamma }_{d^{\prime }}$ and $f^{-1}\left(d\right)=\overline{{\gamma }_d}\preccurlyeq \overline{{\gamma }_{d^{\prime }}}=f^{-1}\left(d^{\prime }\right)$. □

Proposition 11. 

If $L_n$ is the n-element MV-chain with $n\ge 2$, then the lattices $\mathrm{\Delta }(L_n\times L_2)/\sim$ and $\mathrm{Der}\left(L_{n+1}\right)$ are isomorphic.

Proof. 

Recall that $\mathrm{Der}\left(L_{n+1}\right)$ is isomorphic to the lattice $(\mathcal{A}\left(L_{n+1}\right),\leqq )$ where $\mathcal{A}\left(L_{n+1}\right)=\{(x,y)\in L_{n+1}\times L_{n+1}\mid y\le x\}\setminus \left\{(0,0)\right\}$ [16], Theorem 5.6 and ≦ is defined by: for any $(x_1,y_1),(x_2,y_2)\in L_{n+1}\times L_{n+1}$, $(x_1,y_1)\leqq (x_2,y_2)$ iff $x_1\le x_2$ and $y_1\le y_2$. Moreover, $\mathrm{\Delta }(L_n\times L_2)/\sim$ is isomorphic to the lattice $L_0(L_n\times L_2)$ by Corollary 5.

Define a map $f:\mathcal{A}\left(L_{n+1}\right)\to L_0(L_n\times L_2)$ by:

$$f\left(\frac{k}{n},\frac{\ell }{n}\right)=\left\{\begin{array}{cc}\downarrow (\frac{k-1}{n-1},0),\hfill & \mathrm{if}\ell =0;\hfill \\ \downarrow (\frac{k-1}{n-1},0)\cup \downarrow (\frac{\ell -1}{n-1},1),\hfill & \mathrm{if}\ell >0,\hfill \end{array}\right.$$

where $0\le k,\ell \le n-1$. It is easy to see that f is injective. Now we show that f is surjective. For any $M\in L_0(L_n\times L_2)$, we claim M has at most two maximal elements. By way of contradiction, assume M has three different maximal elements denoted by $(a_n,b_n)$, $n=1,2,3$; then, there exist $1\le i<j\le 3$ such that $b_i=b_j$ since $b_n\in L_2$. Thus, $(a_i,b_i)$ and $(a_j,b_j)$ are comparable, which contradicts the fact that $(a_i,b_i)$ and $(a_j,b_j)$ are different maximal elements. If M has only one maximal element denoted by $(\frac{k}{n-1},a)$, then

$$M=\downarrow (\frac{k}{n-1},a)=\left\{\begin{array}{cc}f\left(\frac{k+1}{n},0\right),\hfill & \mathrm{if}a=0;\hfill \\ f\left(\frac{k+1}{n},\frac{k+1}{n}\right),\hfill & \mathrm{if}a=1.\hfill \end{array}\right.$$

If M has exactly two maximal elements denoted by $(\frac{k}{n-1},0)$ and $(\frac{\ell }{n-1},1)$, then

$$M=\downarrow (\frac{k}{n-1},0)\cup \downarrow (\frac{\ell }{n-1},1)=f\left(\frac{k+1}{n},\frac{\ell +1}{n}\right).$$

Therefore, f is surjective.

Since a bijection with supremum preserving is an order isomorphism, it suffices to verify that f preserves the supremum, that is,

$$f\left((\frac{k}{n},\frac{\ell }{n})\vee (\frac{p}{n},\frac{q}{n})\right)=f\left(\frac{k}{n},\frac{\ell }{n}\right)\cup f\left(\frac{p}{n},\frac{q}{n}\right)$$

for all $(\frac{k}{n},\frac{\ell }{n}),(\frac{p}{n},\frac{q}{n})\in \mathcal{A}\left(L_{n+1}\right)$.

Case 1. If $\ell =q=0$, then

$$\begin{array}{cc}\hfill f\left(\frac{k}{n},0\right)\cup f\left(\frac{p}{n},0\right)& =\downarrow (\frac{k-1}{n-1},0)\cup \downarrow (\frac{p-1}{n-1},0)\hfill \\ \hfill & =\downarrow (\max \{\frac{k-1}{n-1},\frac{p-1}{n-1}\},0)\hfill \\ \hfill & =\downarrow (\frac{\max \{k,p\}-1}{n-1},0)\hfill \\ \hfill & =f\left((\frac{k}{n},0)\vee (\frac{p}{n},0)\right).\hfill \end{array}$$

Case 2. If $\ell =0$, $q>0$, then

$$\begin{array}{cc}\hfill f\left(\frac{k}{n},0\right)\cup f\left(\frac{p}{n},\frac{q}{n}\right)& =\downarrow (\frac{k-1}{n-1},0)\cup \left(\downarrow (\frac{p-1}{n-1},0)\cup \downarrow (\frac{q-1}{n-1},1)\right)\hfill \\ \hfill & =\downarrow (\frac{\max \{k,p\}-1}{n-1},0)\cup \downarrow (\frac{q-1}{n-1},1)\hfill \\ \hfill & =f\left((\frac{k}{n},0)\vee (\frac{p}{n},\frac{q}{n})\right).\hfill \end{array}$$

The case $\ell >0$, $q=0$ is similar.

Case 3. If $\ell >0$, $q>0$, then

$$\begin{array}{cc}\hfill f\left(\frac{k}{n},\frac{\ell }{n}\right)\cup f\left(\frac{p}{n},\frac{q}{n}\right)& =\left(\downarrow (\frac{k-1}{n-1},0)\cup \downarrow (\frac{\ell -1}{n-1},1)\right)\cup \left(\downarrow (\frac{p-1}{n-1},0)\cup \downarrow (\frac{q-1}{n-1},1)\right)\hfill \\ \hfill & =\downarrow (\frac{\max \{k,p\}-1}{n-1},0)\cup \downarrow (\frac{\max \{\ell ,q\}-1}{n-1},1)\hfill \\ \hfill & =f\left(\frac{\max \{k,p\}}{n},\frac{\max \{\ell ,q\}}{n}\right)\hfill \\ \hfill & =f\left((\frac{k}{n},\frac{\ell }{n})\vee (\frac{p}{n},\frac{q}{n})\right).\hfill \end{array}$$

Now we verify that f is an isomorphism of posets and hence an isomorphism of lattices. For all $x,y\in \mathcal{A}\left(L_{n+1}\right)$,

$$x\leqq y\iff x\vee y=y\iff f\left(x\right)\cup f\left(y\right)=f(x\vee y)=f\left(y\right)\iff f\left(x\right)\subseteq f\left(y\right).$$

Hence, f is an isomorphism of lattices.

Therefore, $\mathcal{A}\left(L_{n+1}\right)\cong L_0(L_n\times L_2)$ and then $\mathrm{\Delta }(L_n\times L_2)/\sim \cong \mathrm{Der}\left(L_{n+1}\right)$. □

Corollary 6. 

If $L_n$ is the n-element MV-chain with $n\ge 2$, then $\mathrm{MD}\left(L_{n+1}\right)/\sim$ is isomorphic to the lattice $\mathrm{\Delta }(L_n\times L_2)/\sim$.

Proof. 

It follows from Theorem 1 and Proposition 11. □

Note that according to the isomorphism in Theorem 1, $|\mathrm{MD}\left(L_n\right)/\sim |=|\mathrm{Der}\left(L_n\right)|=\frac{(n-1)(n+2)}{2}$ by [16] (Theorem 3.11). However, the following Example 5 shows that the cardinalities of different equivalence classes with respect to the equivalence relation ∼ are different in general.

Example 5. 

Let $n=2$ and define $\delta \in \mathrm{MF}\left(L_2\right)$ by $\delta \left(0\right)=\left\{0\right\},\delta \left(1\right)=\{0,1\}$. Then, it is easy to check that

$$\mathrm{MD}\left(L_2\right)=\{{\mathbf{0}}_{\mathrm{MF}\left(L_2\right)},\mathrm{I}{\mathrm{d}}_{\mathrm{MF}\left(L_2\right)},\delta \},$$

$$\mathrm{MD}\left(L_2\right)/\sim =\{\left\{{\mathbf{0}}_{\mathrm{MF}\left(L_2\right)}\right\},\{\mathrm{I}{\mathrm{d}}_{\mathrm{MF}\left(L_2\right)},\delta \}\}.$$

It is clear that $\left|\overline{{\mathbf{0}}_{\mathrm{MF}\left(L_2\right)}}\right|=1$ but $\left|\overline{I{d}_{\mathrm{MF}\left(L_2\right)}}\right|=2$. Hence, $2=|\mathrm{MD}\left(L_2\right)/\sim |\nmid |\mathrm{MD}\left(L_2\right)|=3$.

So, the cardinality of $\mathrm{MD}\left(L_n\right)$ is not easy to deduce from Theorem 1. In the next section, we will investigate the enumeration of the set of $(\odot ,\vee )$-multiderivations on $L_n$ by constructing a counting principle (Theorem 3).

5. The Enumeration of $(\odot ,\vee )$-Multiderivations on a Finite MV-Chain

In this section, we determine the cardinality of $\mathrm{MD}\left(L_n\right)$. For small values of n, this can be performed with calculations using Python (see the Appendix A Figure A1) in Table 1:

Table 1. The cardinality of $\mathrm{MD}\left(L_n\right)$.

n	2	3	4	5	6
$\left|\mathrm{MD}\right(L_n\left)\right|$	3	16	63	220	723

The result cannot be obtained after $n\ge 6$ due to the limitation of computing resources. But we have shown the following general formula.

Theorem 2. 

Let $n\ge 2$ be a positive integer. Then, $|\mathrm{MD}\left(L_n\right)|={\displaystyle \frac{7·3^{n-1}-2^{n+2}+1}{2}}$.

In order to prove Theorem 2, we need the following Lemmas.

Lemma 15. 

Assume that A is an MV-chain and $\sigma \in \mathrm{MD}\left(A\right)$; then, the following results hold:

1. 

If $M\subseteq A$, then $M=M\vee M$.

2. 

For any $x\in A,n\in {\mathbb{N}}_{+}$, we have $\sigma \left(x^n\right)=x^{n-1}\odot \sigma \left(x\right)$, where $x^0=1$, $x^n={\underbrace{x\odot x\odot \cdots \odot x}}_n$.

Proof. 

(1) It follows immediately from Lemma 6 (3), as M is a sublattice.

(2) We prove $\sigma \left(x^n\right)=x^{n-1}\odot \sigma \left(x\right)$ by induction on n. Obviously, $\sigma \left(x^1\right)=\sigma \left(x\right)=1\odot \sigma \left(x\right)=x^{1-1}\odot \sigma \left(x\right)$.

Now, assume that $\sigma \left(x^n\right)=x^{n-1}\odot \sigma \left(x\right)$. By Equation (3), we have

$$\begin{array}{cc}\hfill \sigma \left(x^{n+1}\right)& =\sigma (x^n\odot x)\hfill \\ \hfill & =(\sigma \left(x^n\right)\odot x)\vee (x^n\odot \sigma \left(x\right))\hfill \\ & =(x^{n-1}\odot \sigma \left(x\right)\odot x)\vee (x^n\odot \sigma \left(x\right))\hfill \\ \hfill & =x^n\odot \sigma \left(x\right),\hfill \end{array}$$

so $\left(2\right)$ holds. □

Note that an MV-chain can be completely characterized by $\left(1\right)$. That is, if A is an MV-algebra, then A is an MV-chain iff $M=M\vee M$ for every $M\subseteq A$. In fact, by way of contraposition, assume that $x,y\in A$ and $x,y$ are incomparable, denote $z=x\vee y$. Let $M=\{x,y\}$. Then, $z=x\vee y\in M\vee M$ but $z\notin M$. This leads to a contradiction.

Let $n\in {\mathbb{N}}_{+}$ and $n\ge 2$. In $L_n$, we know $\frac{n-m-1}{n-1}={\left(\frac{n-2}{n-1}\right)}^m$ for every $m\in \{1,2,\cdots ,n-1\}$. So, any $x\in L_n\setminus \left\{1\right\}$ has a representation as a power of $\frac{n-2}{n-1}$.

Next, we give a counting principle for $(\odot ,\vee )$-multiderivations on a finite MV-chain $L_n$.

Theorem 3. 

Let σ be a multifunction on $L_n$ and $v=\frac{n-2}{n-1}$. Then, $\sigma \in \mathrm{MD}\left(L_n\right)$ iff σ satisfies the following conditions:

1. 

$\sigma \left(v^m\right)=v^{m-1}\odot \sigma \left(v\right)$, $\forall m\in \{1,2,\cdots ,n-1\}$.

2. 

$\sigma \left(v\right)=\sigma \left(v\right)\vee (v\odot \sigma (1\left)\right)$.

3. 

$\sigma \left(v\right)⪯\left\{v\right\}$.

Proof. 

Assume $\sigma \in \mathrm{MD}\left(L_n\right)$; then, for each $m\in \{1,2,\cdots ,n-1\}$, we have $\sigma \left(v^m\right)=v^{m-1}\odot \sigma \left(v\right)$ by Lemma 15 $\left(2\right)$, and $\sigma \left(v\right)=\sigma (v\odot 1)=\sigma \left(v\right)\vee (v\odot \sigma (1\left)\right)$ by Equation (3). Thus, $\sigma$ satisfies $\left(1\right)$ and $\left(2\right)$. Furthermore, $\left(3\right)$ holds by Proposition 4 (2).

Conversely, suppose that $\sigma$ satisfies $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$. Let $x,y\in L_n$. There are four cases:

If $x=y=1$, then it is easy to see that $\sigma (1\odot 1)=\sigma \left(1\right)=\sigma \left(1\right)\vee \sigma \left(1\right)$ by Lemma 15 (1).

If $x=1$ or $y=1$, and $x\ne y$. With out loss of generality, suppose that $x\ne 1$ and $y=1$, then $x=v^k$ for some $k\in \{1,2,\cdots ,n-1\}$. By (1), we have $\sigma (x\odot 1)=\sigma \left(x\right)=\sigma \left(v^k\right)=v^{k-1}\odot \sigma \left(v\right)$. Also, we have

$$\begin{array}{cc}\hfill \sigma \left(x\right)\vee (x\odot \sigma (1\left)\right)& =(v^{k-1}\odot \sigma \left(v\right))\vee (v^k\odot \sigma \left(1\right))\hfill \\ & =(v^{k-1}\odot \sigma \left(v\right))\vee (v^{k-1}\odot (v\odot \sigma \left(1\right)))\hfill \\ & =v^{k-1}\odot (\sigma \left(v\right)\vee (v\odot \sigma \left(1\right))\hfill & \hfill & (\mathrm{Lemma} 7\left(2\right))\hfill \\ & =v^{k-1}\odot \sigma \left(v\right).\hfill & \hfill & \left(\right(2)\mathrm{of}\mathrm{Theorem} 3)\hfill \end{array}$$

Hence, $\sigma (x\odot 1)=\sigma \left(x\right)=\left(\sigma \right(x)\odot 1)\vee (x\odot \sigma (1\left)\right)$.

If $x\ne 1$ and $y\ne 1$, then assume that $x=v^k$ and $y=v^{\ell }$ for some $k,\ell \in \{1,2,\cdots ,n-1\}$. We have

$$\sigma (x\odot y)=\sigma (v^k\odot v^{\ell })=\sigma \left(v^{k+\ell }\right)$$

and

$$(\sigma \left(x\right)\odot y)\vee (x\odot \sigma \left(y\right))=((v^{k-1}\odot \sigma \left(v\right))\odot v^{\ell })\vee (v^k\odot (v^{\ell -1}\odot \sigma \left(v\right)))=v^{k+\ell -1}\odot \sigma \left(v\right)$$

by Lemma 15 (1). Then, there are three cases:

For $k+\ell <n-1$, by (1) we obtain $\sigma \left(v^{k+\ell }\right)=v^{k+\ell -1}\odot \sigma \left(v\right)$.

For $k+\ell =n-1$, by $\left(3\right)$ we have $\sigma \left(v^{k+\ell }\right)=\sigma \left(v^{n-1}\right)=\sigma \left(0\right)⪯\left\{0\right\}$ and so $\sigma \left(0\right)=\left\{0\right\}$. And $v^{k+\ell -1}\odot \sigma \left(v\right)=v^{n-2}\odot \sigma \left(v\right)=v^{\ast }\odot \sigma \left(v\right)=\left\{0\right\}$. Thus, $\sigma (x\odot y)=\left(\sigma \right(x)\odot y)\vee (x\odot \sigma (y\left)\right)$.

For $n-1<k+\ell \le 2n-2$, we have $\sigma \left(v^{k+\ell }\right)=\sigma \left(0\right)=\left\{0\right\}=0\odot \sigma \left(v\right)=v^{k+\ell -1}\odot \sigma \left(v\right)$ by (3) and thus Equation (3) holds.

Therefore, we conclude that $\sigma \in \mathrm{MD}\left(L_n\right)$. □

Lemma 16. 

Let $P,Q\in \mathrm{\Delta }\left(L_n\right)$. Then, the following results hold:

1. 

$P\subseteq P\vee Q$ iff $\min Q\le \min P$.

2. 

$P\vee Q\subseteq P$ iff $[\min P,1]\cap Q\subseteq P$.

Proof. 

Denote $p_0=\min P$, $q_0=\min Q$.

(1) Assume $P\subseteq P\vee Q$, then there exist $p\in P$, $q\in Q$ such that $p_0=p\vee q\ge q$. Thus, $q_0\le q\le p_0$.

Conversely, suppose $q_0\le p_0$, then $p=p\vee q_0$ for any $p\in P$ since $p_0\le p$. Hence, $P\subseteq P\vee Q$.

(2) Assume $P\vee Q\subseteq P$; then, for all $q\in [p_0,1]\cap Q$, we have $q=p_0\vee q\in P\vee Q\subseteq P$. Thus, $[p_0,1]\cap Q\subseteq P$.

Conversely, assume $[p_0,1]\cap Q\subseteq P$ and $p\in P$, $q\in Q$. If $q\le p$, then $p\vee q=p\in P$. If $q>p$, then $p\vee q=q\in [p_0,1]\cap Q\subseteq P$. In either case, $p\vee q\in P$ and so $P\vee Q\subseteq P$. □

Lemma 17. 

Let $Q,Q^{\prime }\in \mathrm{\Delta }\left(L_n\right)$ and $1\notin Q$. Denote $v=\frac{n-2}{n-1}$. Then, the following results hold:

1. 

If $0\notin Q$, then $Q=v\odot Q^{\prime }$ iff $Q^{\prime }=Q\oplus v^{\ast }$.

2. 

If $0\in Q$, denote $Q_1=Q\setminus \left\{0\right\}$. Then, $Q=v\odot Q^{\prime }$ iff $Q^{\prime }=\left\{0\right\}\bigsqcup (Q_1\oplus v^{\ast })$, $\left\{v^{\ast }\right\}\bigsqcup (Q_1\oplus v^{\ast })$ or $\{0,v^{\ast }\}\bigsqcup (Q_1\oplus v^{\ast })$.

Proof. 

(1) Let $0\notin Q$ and $Q=v\odot Q^{\prime }$. Then, $0\notin Q^{\prime }$, otherwise, $0=v\odot 0\in v\odot Q^{\prime }=Q$, a contradiction. Thus, $0\notin Q^{\prime }$, which implies $\left\{v^{\ast }\right\}⪯Q^{\prime }$. Hence, we have

$$\begin{array}{cc}\hfill Q^{\prime }=Q^{\prime }\vee v^{\ast }& =\{q^{\prime }\vee v^{\ast }\mid q^{\prime }\in Q^{\prime }\}\hfill \\ \hfill & =\{(q^{\prime }\odot v)\oplus v^{\ast }\mid q^{\prime }\in Q^{\prime }\}\hfill \\ \hfill & =(Q^{\prime }\odot v)\oplus v^{\ast }\hfill \\ \hfill & =Q\oplus v^{\ast }.\hfill \end{array}$$

Conversely, assume $Q^{\prime }=Q\oplus v^{\ast }$. Since $1\notin Q$, we have $Q⪯\left\{v\right\}$. Hence,

$$\begin{array}{cc}\hfill Q=Q\wedge v& =\{q\wedge v\mid n\in Q\}\hfill \\ \hfill & =\{v\odot (q\oplus v^{\ast })\mid n\in Q\}\hfill \\ \hfill & =v\odot (Q\oplus v^{\ast })\hfill \\ \hfill & =v\odot Q^{\prime }.\hfill \end{array}$$

(2) Assume $0\in Q$ and $Q=v\odot Q^{\prime }$; then, $0=v\odot q^{\prime }$ for some $q^{\prime }\in Q^{\prime }$. Thus, $0\in Q^{\prime }$ or $v^{\ast }\in Q^{\prime }$. Denote $Q_0^{\prime }=\{0,v^{\ast }\}\cap Q^{\prime }$ and $Q_1^{\prime }=Q^{\prime }\setminus Q_0^{\prime }$. By $v\odot Q_0^{\prime }=\left\{0\right\}$ and $\left\{v^{\ast }\right\}⪯v\odot Q_1^{\prime }$, we have

$$\begin{array}{cc}\hfill Q_1=Q\setminus \left\{0\right\}& =(v\odot Q^{\prime })\setminus \left\{0\right\}\hfill \\ & =(v\odot (Q_0^{\prime }\cup Q_1^{\prime }))\setminus \left\{0\right\}\hfill \\ & =((v\odot Q_0^{\prime })\cup (v\odot Q_1^{\prime }))\setminus \left\{0\right\}\hfill & \hfill & (\mathrm{Lemma} 7\left(3\right))\hfill \\ & =(\left\{0\right\}\cup (v\odot Q_1^{\prime }))\setminus \left\{0\right\}\hfill \\ & =v\odot Q_1^{\prime }.\hfill \end{array}$$

Since $0\notin Q_1$, we obtain $Q_1^{\prime }=Q_1\oplus v^{\ast }$ by (1). Therefore,

$$Q^{\prime }=Q_0^{\prime }\bigsqcup Q_1^{\prime }=Q_0^{\prime }\bigsqcup (Q_1\oplus v^{\ast }),$$

where $Q_0^{\prime }=\left\{0\right\}$, $\left\{v^{\ast }\right\}$ or $\{0,v^{\ast }\}$.

Conversely, assume $0\in Q$ and $Q^{\prime }=Q_0^{\prime }\bigsqcup (Q_1\oplus v^{\ast })$, where $Q_0^{\prime }=\left\{0\right\}$, $\left\{v^{\ast }\right\}$ or $\{0,v^{\ast }\}$. From $1\notin Q_1$, it follows that $Q_1⪯\left\{v\right\}$ and

$$\begin{array}{cc}\hfill v\odot Q^{\prime }& =v\odot (Q_0^{\prime }\cup (Q_1\oplus v^{\ast }))\hfill \\ & =(v\odot Q_0^{\prime })\cup (v\odot (Q_1\oplus v^{\ast }))\hfill & \hfill & (\mathrm{Lemma} 7\left(3\right))\hfill \\ & =\left\{0\right\}\cup (Q_1\wedge v)=\left\{0\right\}\cup Q_1=Q.\hfill \end{array}$$

Hence, we complete the proof. □

We are now in a position to prove Theorem 2:

Proof of Theorem 2.

Assume that $\sigma$ is a multifunction on $L_n$ and denote $\frac{n-2}{n-1}$ by v. According to Theorem 3, $\sigma$ is uniquely determined by $\sigma \left(v\right)$ and $\sigma \left(1\right)$ if $\sigma \in \mathrm{MD}\left(L_n\right)$. Hence, it is enough to consider the values of $\sigma \left(v\right)$ and $\sigma \left(1\right)$. By Theorem 3, $\sigma \in \mathrm{MD}\left(L_n\right)$ iff

$$\sigma \left(v\right)⪯\left\{v\right\},$$

(5)

and

$$\sigma \left(v\right)=\sigma \left(v\right)\vee (v\odot \sigma (1\left)\right).$$

(6)

For convenience, we denote $P=\sigma \left(v\right)$, $Q^{\prime }=\sigma \left(1\right)$, $Q=v\odot \sigma \left(1\right)$, $p_0=\min P$ and $q_0=\min Q$. Equation (5) implies $1\notin P$. By Lemma 16, we know Equation (6) implies that $q_0\le p_0$ and $[p_0,1]\cap Q\subseteq P$. Assume that $p_0=\frac{k}{n-1}$ and $\left|P\right|=\ell$, where $0\le k\le n-2$ and $1\le \ell \le n-k-1$. Then, $P\setminus \left\{p_0\right\}\subseteq \left[\frac{k+1}{n-1},\frac{n-2}{n-1}\right]$. Thus, P has $C_{n-k-2}^{\ell -1}$ choices with respect to k and ℓ. Now, we will determine all choices of Q and $Q^{\prime }$.

Case 1. If $q_0=p_0$, then $Q=[q_0,1]\cap Q=[p_0,1]\cap Q\subseteq P$. Hence, $Q\setminus \left\{q_0\right\}$ can take any subset of $P\setminus \left\{p_0\right\}$ and so Q has $2^{\ell -1}$ choices.

If $q_0>0$, then $0\notin Q$, and by Lemma 17 (1) and $Q=v\odot Q^{\prime }$ we know $Q^{\prime }=Q\oplus v^{\ast }$. Hence, $Q^{\prime }$ has $2^{\ell -1}$ choices.

If $q_0=0$, then $0\in Q$, by Lemma 17 (2) and $Q=v\odot Q^{\prime }$ we have $Q^{\prime }=\left\{0\right\}\bigsqcup (Q_1\oplus v^{\ast })$, $\left\{v^{\ast }\right\}\bigsqcup (Q_1\oplus v^{\ast })$ or $\{0,v^{\ast }\}\bigsqcup (Q_1\oplus v^{\ast })$. Thus, $Q^{\prime }$ has $3·2^{\ell -1}$ choices.

Case 2. If $0<q_0<p_0$, denote $Q_1=(0,p_0)\cap Q$ and $Q_2=[p_0,1]\cap Q$. Since $0\notin Q$, we have $Q=Q_1\bigsqcup Q_2$. Notice that $Q_1\ne ⌀$, so there are $2^{k-1}-1$ choices of $Q_1$. Furthermore, since $Q_2=[p_0,1]\cap Q\subseteq P$, $Q_2$ can take any subset of P and so has $2^{\ell }$ choices. Thus, there are $(2^{k-1}-1)·2^{\ell }$ choices of Q in this case. Since $0\notin Q$, it follows that $Q^{\prime }$ has also $(2^{k-1}-1)·2^{\ell }$ choices by Lemma 17 (1).

Case 3. If $0=q_0<p_0$, denote $Q_1=(0,p_0)\cap Q$ and $Q_2=[p_0,1]\cap Q$, so we have $Q=\left\{0\right\}\bigsqcup Q_1\bigsqcup Q_2$. Since $Q_1\subset (0,p_0)$, there are $2^{k-1}$ choices of $Q_1$. Moreover, $Q_2$ has $2^{\ell }$ choices as in Case 2. Thus, there are $2^{k+\ell -1}$ choices of Q in this case. Since $0\in Q$, it follows that $Q^{\prime }$ has $3·2^{k+\ell -1}$ choices by Lemma 17 (2).

According to Theorem 3, we can determine the unique $(\odot ,\vee )$-multiderivation for each choices of $\sigma \left(1\right)$ and $\sigma \left(v\right)$.

Therefore, it follows

$$\begin{array}{cc}\hfill \left|\mathrm{MD}\right(L_n\left)\right|& =\sum _{k=1}^{n-2}\sum _{\ell =1}^{n-k-1}\left(\genfrac{}{}{0pt}{}{n-k-2}{\ell -1}\right)(2^{\ell -1}+(2^{k-1}-1)·2^{\ell }+3·2^{k-1}·2^{\ell })+\sum _{\ell =1}^{n-1}\left(\genfrac{}{}{0pt}{}{n-2}{\ell -1}\right)(3·2^{\ell -1})\hfill \\ \hfill & =\sum _{k=0}^{n-2}\sum _{\ell =1}^{n-k-1}\left(\genfrac{}{}{0pt}{}{n-k-2}{\ell -1}\right)(2^{k+\ell +1}-2^{\ell -1})\hfill \\ \hfill & =\sum _{k=0}^{n-2}\left((2^{k+2}-1)\sum _{\ell =1}^{n-k-1}\left(\genfrac{}{}{0pt}{}{n-k-2}{\ell -1}\right)·2^{\ell -1}\right)\hfill \\ \hfill & =\sum _{k=0}^{n-2}(2^{k+2}-1){(2+1)}^{n-k-2}\hfill \\ \hfill & =3^n\sum _{k=0}^{n-2}\left({\left(\frac{2}{3}\right)}^{k+2}-{\left(\frac{1}{3}\right)}^{k+2}\right)\hfill \\ \hfill & ={\displaystyle \frac{7·3^{n-1}-2^{n+2}+1}{2}}.\hfill \end{array}$$

□

6. Conclusions and Questions

In this paper, the point-to-point $(\odot ,\vee )$-derivations on MV-algebras have been extended to point-to-set $(\odot ,\vee )$-multiderivations. We show that $(\mathrm{MD}\left(L_n\right)/\sim ,\preccurlyeq )$ is isomorphic to the complete lattice $\mathrm{Der}\left(L_n\right)$, the underlying set of $(\odot ,\vee )$-derivations on $L_n$. This unveils a certain relevance between $(\odot ,\vee )$-multiderivations and $(\odot ,\vee )$-derivations. Moreover, by building a counting principle, we obtain the enumeration of $\mathrm{MD}\left(L_n\right)$.

This general study of $(\odot ,\vee )$-multiderivations has the advantage of developing into a system theory of sets and has potential wide applications: other logical algebras, control theory, interval analysis, and artificial intelligence.

We list three questions to be considered in the future:

(1) We have found two ways to construct $(\odot ,\vee )$-multiderivations from $(\odot ,\vee )$-derivations in Propositions 5 and 6. Are there other ways?

(2) We ask whether the equivalent characterization and enumeration of $(\odot ,\vee )$- multiderivations on finite MV-chains can be extended to finite MV-algebras.

(3) We ask whether MV-algebras A and $A^{\prime }$ are isomorphic if $\left(\mathrm{MD}\right(A),\preccurlyeq )$ and $(\mathrm{MD}\left(A^{\prime }\right),\preccurlyeq )$ are order isomorphic.