Derivations of MV-Algebras

Abstract

We introduce the notion of derivation for an MV-algebra and discuss some related properties. Using the notion of an isotone derivation, we give some characterizations of a derivation of an MV-algebra. Moreover, we define an additive derivation of an MV-algebra and investigate some of its properties. Also, we prove that an additive derivation of a linearly ordered MV-algebral is an isotone.

1. Introduction

In his classical paper [1], Chang invented the notion of MV-algebra in order to provide an algebraic proof of the completeness theorem of infinite valued Lukasiewicz propositional calculus. Recently, the algebraic theory of MV-algebras is intensively studied, see [2–5].

The notion of derivation, introduced from the analytic theory, is helpful to the research of structure and property in algebraic system. Several authors [6–9] studied derivations in rings and near rings. Jun and Xin [10] applied the notion of derivation in ring and near-ring theory to BCI-algebras. In [11], Szász introduced the concept of derivation for lattices and investigated some of its properties, for more details, the reader is referred to [9, 12–19].

In this paper, we apply the notion of derivation in ring and near-ring theory to MV-algebras and investigate some of its properties. Using the notion of an isotone derivation, we characterize a derivation of MV-algebra. We introduce a new concept, called an additive derivation of MV-algebras, and then we investigate several properties. Finally, we prove that an additive derivation of a linearly ordered MV-algebra is an isotone.

2. Preliminaries

Definition 2.1 (see [5]). An MV-algebra is a structure (𝑀,⊕,∗,0) where ⊕ is a binary operation, ∗ is a unary operation, and 0 is a constant such that the following axioms are satisfied for any 𝑎,𝑏∈𝑀: (MV1) (𝑀,⊕,0) is a commutative monoid,(MV2) (𝑎^∗)^∗=𝑎,(MV3) 0^∗⊕𝑎=0^∗,(MV4) (𝑎^∗⊕𝑏)^∗⊕𝑏=(𝑏^∗⊕𝑎)^∗⊕𝑎.

If we define the constant 1=0^∗ and the auxiliary operations ⊙,∨, and ∧ by 𝑎⊙𝑏=𝑎^∗⊕𝑏^∗∗,𝑎∨𝑏=𝑎⊕𝑏⊙𝑎^∗,𝑎∧𝑏=𝑎⊙𝑏⊕𝑎^∗, (2.1) then (𝑀,⊙,1) is a commutative monoid and the structure (𝑀,∨,∧,0,1) is a bounded distributive lattice. Also, we define the binary operation ⊝ by 𝑥⊝𝑦=𝑥⊙𝑦^∗. A subset 𝑋 of an MV-algebra 𝑀 is called subalgebra of 𝑀 if and only if 𝑋 is closed under the MV-operations defined in 𝑀. In any MV-algebras, one can define a partial order ≤ by putting 𝑥≤𝑦 if and only if 𝑥∧𝑦=𝑥 for each 𝑥,𝑦∈𝑀. If the order relation ≤, defined over 𝑀, is total, then we say that 𝑀 is linearly ordered. For an MV-algebra 𝑀, if we define 𝐵(𝑀)={𝑥∈𝑀∶𝑥⊕𝑥=𝑥}={𝑥∈𝑀∶𝑥⊙𝑥=𝑥}. Then, (𝐵(𝑀),⊕,∗,0) is both a largest subalgebra of 𝑀 and a Boolean algebra.

An MV-algebra 𝑀 has the following properties for all 𝑥,𝑦,𝑧∈𝑀(1)𝑥⊕1=1,(2)𝑥⊕𝑥^∗=1,(3)𝑥⊙𝑥^∗=0, (4)If 𝑥⊕𝑦=0, then 𝑥=𝑦=0,(5)If 𝑥⊙𝑦=1, then 𝑥=𝑦=1,(6)If 𝑥≤𝑦, then 𝑥∨𝑧≤𝑦∨𝑧 and 𝑥∧𝑧≤𝑦∧𝑧,(7)If 𝑥≤𝑦, then 𝑥⊕𝑧≤𝑦⊕𝑧 and 𝑥⊙𝑧≤𝑦⊙𝑧,(8)𝑥≤𝑦 if and only if 𝑦^∗≤𝑥^∗,(9)𝑥⊕𝑦=𝑦 if and only if 𝑥⊙𝑦=𝑥.

Theorem 2.2 (see [1]). The following conditions are equivalent for all 𝑥,𝑦∈𝑀(i)𝑥≤𝑦, (ii)𝑦⊕𝑥^∗=1, (iii)𝑥⊙𝑦^∗=0.

Definition 2.3 (see [1]). Let 𝑀 be an MV-algebra and 𝐼 be a nonempty subset of 𝑀. Then, we say that 𝐼 is an ideal if the following conditions are satisfied: (i)0∈𝐼, (ii)𝑥,𝑦∈𝐼 imply 𝑥⊕𝑦∈𝐼,(iii)𝑥∈𝐼 and 𝑦≤𝑥 imply 𝑦∈𝐼.

Proposition 2.4 (see [1]). Let 𝑀 be a linearly ordered MV-algebra, then 𝑥⊕𝑦=𝑥⊕𝑧 and 𝑥⊕𝑧≠1 implies that 𝑦=𝑧.

3. Derivations of MV-Algebras

Definition 3.1. Let 𝑀 be an MV-algebra, and let 𝑑∶𝑀→𝑀 be a function. We call 𝑑 a derivation of 𝑀, if it satisfies the following condition for all 𝑥,𝑦∈𝑀𝑑𝑥⊙𝑦=𝑑𝑥⊙𝑦⊕𝑥⊙𝑑𝑦. (3.1)
We often abbreviate 𝑑(𝑥) to 𝑑𝑥.

Example 3.2. Let 𝑀={0,𝑎,𝑏,1}. Consider Tables 1 and 2.
Then (𝑀,⊕,∗,0) is an MV-algebra. Define a map 𝑑∶𝑀→𝑀 by 𝑑𝑥=0if𝑥=0,𝑎,1,𝑎if𝑥=𝑏. (3.2)
Since 𝑑(𝑎⊙𝑏)=0 and (𝑑𝑎⊙𝑏)⊕(𝑎⊙𝑑𝑏)=(0⊙𝑏)⊕(𝑎⊙𝑎)=0⊕𝑎=𝑎,𝑑 is not derivation.

Table 1 

Table 2 

Example 3.3. Let 𝑀={0,𝑥₁,𝑥₂,𝑥₃,𝑥₄,1}. Consider Tables 3 and 4.
Then, (𝑀,⊕,∗,0) is an MV-algebra. Define a map 𝑑∶𝑀→𝑀 by 𝑑𝑥=0if𝑥=0,𝑥₁,𝑥₃,𝑥₂if𝑥=𝑥₂,𝑥₄,1. (3.3)
Then, it is easily checked that 𝑑 is a derivation of 𝑀.

Table 3 

Table 4 

Proposition 3.4. Let 𝑀 be an MV-algebra, and let 𝑑 be a derivation on 𝑀. Then, the following hold for every 𝑥∈𝑀: (i)𝑑0=0, (ii)𝑑𝑥⊙𝑥^∗=𝑥⊙𝑑𝑥^∗=0, (iii)𝑑𝑥=𝑑𝑥⊕(𝑥⊙𝑑1), (iv)𝑑𝑥≤𝑥,(v)If 𝐼 is an ideal of an MV-algebra 𝑀, then 𝑑(𝐼)⊆𝐼.

Proof. (i) 𝑑0=𝑑(𝑥⊙0)=(𝑑𝑥⊙0)⊕(𝑥⊙𝑑0)=𝑥⊙𝑑0.
Putting 𝑥=0, we get 𝑑0=0.
(ii) Let 𝑥∈𝑀, then 0=𝑑0=𝑑𝑥⊙𝑥^∗=𝑑𝑥⊙𝑥^∗⊕𝑥⊙𝑑𝑥^∗, (3.4) and so (ii) follows from (4). (iii)It is clear.(iv)Let 𝑥∈𝑀, from (ii), we have 1=0^∗=𝑑𝑥⊙𝑥^∗∗=(𝑑𝑥)^∗⊕𝑥, (3.5) from Theorem 2.2 we get 𝑑𝑥≤𝑥. (v)Let 𝑦∈𝑑(𝐼), then 𝑦=𝑑(𝑥) for some 𝑥∈𝐼. Since 𝑦=𝑑(𝑥)≤𝑥∈𝐼, thus 𝑦∈𝐼 and so 𝑑(𝐼)⊆𝐼.

Proposition 3.5. Let 𝑑 be a derivation of an MV-algebra 𝑀, and let 𝑥,𝑦∈𝑀. If x≤𝑦. Then, the following hold: (i)𝑑(𝑥⊙𝑦^∗)=0, (ii)𝑑𝑦^∗≤𝑥^∗, (iii)𝑑𝑥⊙𝑑𝑦^∗=0.

Proof. (i) Let 𝑥≤𝑦, then Theorem 2.2 implies that 𝑥⊙𝑦^∗=0, and so 𝑑(𝑥⊙𝑦^∗)=𝑑0=0.
(ii)From (i), we get 0=𝑑𝑥⊙𝑦^∗=𝑑𝑥⊙𝑦^∗⊕𝑥⊙𝑑𝑦^∗, (3.6) and by (4), we have 𝑥⊙𝑑𝑦^∗=0. Therefore, 𝑑𝑦^∗≤𝑥^∗.(iii)If 𝑥≤𝑦, then 𝑑𝑥≤𝑦, thus 𝑑𝑥⊙𝑑𝑦^∗≤𝑦⊙𝑑𝑦^∗, also 𝑑𝑦^∗≤𝑦^∗, and so 𝑦⊙𝑑𝑦^∗≤𝑦⊙𝑦^∗=0. Hence, 𝑑𝑥⊙𝑑𝑦^∗=0.

Proposition 3.6. Let 𝑀 be an MV-algebra, and let 𝑑 be a derivation on 𝑀. Then, the following hold: (i)𝑑𝑥⊙𝑑𝑥^∗=0, (ii)𝑑𝑥^∗=(𝑑𝑥)^∗ if and only if 𝑑 is the identity on 𝑀.

Proof. (i) It follows directly from Proposition 3.5(iii).
(ii)It is sufficient to show that if 𝑑𝑥^∗=(𝑑𝑥)^∗, then 𝑑 is the identity on 𝑀.
Assume that 𝑑𝑥^∗=(𝑑𝑥), from Proposition 3.4(ii), we have 𝑥⊙(𝑑𝑥)^∗=0, which implies that 𝑥≤𝑑𝑥. Therefore, 𝑑𝑥=𝑥.

Definition 3.7. Let 𝑀 be an MV-algebra and 𝑑 be a derivation on 𝑀. If 𝑥≤𝑦 implies 𝑑𝑥≤𝑑𝑦 for all 𝑥,𝑦∈𝑀, 𝑑 is called an isotone derivation.

Example 3.8. Let 𝑀 be an MV-algebra as in Example 3.3. It is easily checked that 𝑑 is an isotone derivation of 𝑀.

Proposition 3.9. Let 𝑀 be an MV-algebra, and let 𝑑 be aderivation of 𝑀. If 𝑑𝑥^∗=𝑑𝑥 for all 𝑥∈𝑀, then the following hold:(i)𝑑1=0,(ii)𝑑𝑥⊙𝑑𝑥=0, (iii)If 𝑑 is an isotone derivation of 𝑀, then 𝑑 is zero.

Proof. (i) It follows by putting 𝑥=0.
(ii)It follows from Proposition 3.6(i).(iii)Since 𝑑 is an isotone, hence 𝑑𝑥≤𝑑1 for all 𝑥∈𝑀. By (i), we have 𝑑𝑥≤0, and so 𝑑 is zero.

Definition 3.10. Let 𝑀 be an MV-algebra, and let 𝑑 be a derivation on 𝑀. If 𝑑(𝑥⊕𝑦)=𝑑𝑥⊕𝑑𝑦 for all 𝑥,𝑦∈𝑀,𝑑 is called an additive derivation.

Example 3.11. Let 𝑀 be an MV-algebra as in Example 3.3. It is easily checked that 𝑑 is an additive derivation of 𝑀.

Theorem 3.12. Let M be an MV-algebra, and let 𝑑 be a nonzero additive derivation of 𝑀. Then, 𝑑(𝐵(𝑀))⊆𝐵(𝑀).

Proof. Let 𝑦∈𝑑(𝐵(𝑀)), thus 𝑦=𝑑(𝑥) for some 𝑥∈𝐵(𝑀). Then, 𝑦⊕𝑦=𝑑𝑥⊕𝑑𝑥=𝑑(𝑥⊕𝑥)=𝑑𝑥=𝑦. (3.7)
Therefore 𝑦∈𝐵(𝑀), this complete the proof.

Theorem 3.13. Let 𝑑 be an additive derivation of a linearly ordered MV-algebra M. Then, either 𝑑=0 or 𝑑1=1.

Proof. Let 𝑑 be an additive derivation of a linearly ordered MV-algebra 𝑀. Hence, 𝑑1=𝑑𝑥⊕𝑥^∗=𝑑𝑥⊕𝑑𝑥^∗, (3.8) also, 𝑑1=𝑑(𝑥⊕1)=𝑑𝑥⊕𝑑1, (3.9) for all 𝑥∈𝑀. If 𝑑1≠1, then Proposition 2.4 implies that 𝑑𝑥^∗=𝑑1. Putting 𝑥=1, we get that 𝑑1=0. Therefore, 0=𝑑1=𝑑𝑥⊕𝑑1=𝑑𝑥, (3.10) for all 𝑥∈𝑀, and so 𝑑 is zero.

Proposition 3.14. Let 𝑀 be a linearly ordered MV-algebra, and let 𝑑₁,𝑑₂ additive derivations of 𝑀. Define 𝑑₁𝑑₂(𝑥)=𝑑₁(𝑑₂𝑥) for all 𝑥∈𝑀. If 𝑑₁𝑑₂=0, then 𝑑₁=0 or 𝑑₂=0.

Proof. Let 𝑑₁𝑑₂=0, 𝑥∈𝑀, and suppose that 𝑑₂≠0. Then, 0=𝑑₁𝑑₂𝑥=𝑑₁𝑑₂𝑥⊕𝑥⊙𝑑₂1=𝑑₁𝑑₂𝑥⊕𝑑₁𝑥=𝑑₁𝑥, (3.11) thus 𝑑₁=0. Similarly, we can prove that 𝑑₂=0.

Proposition 3.15. Let 𝑀 be a linearly ordered MV-algebra, and let 𝑑 be a nonzero additive derivation of 𝑀. Then, 𝑑(𝑥⊙𝑥)=𝑥⊕𝑥,∀𝑥∈𝑀. (3.12)

Proof. From Proposition 3.4(iii) and Theorem 3.13, we get that 𝑑𝑥=𝑑𝑥⊕𝑥; applying (9), we have 𝑑𝑥⊙𝑥=𝑥. Thus, 𝑑(𝑥⊕𝑥)=(𝑑𝑥⊙𝑥)⊕(𝑑𝑥⊙𝑥)=𝑥⊕𝑥. (3.13)

Theorem 3.16. Every nonzero additive derivation of a linearly ordered MV-algebra 𝑀 is an isotone derivation.

Proof. Assume that 𝑑 is an additive derivation of 𝑀, and 𝑥,𝑦∈𝑀. If 𝑥≤𝑦, then 𝑥^∗⊕𝑦=1, hence 1=𝑑1=𝑑𝑥^∗⊕𝑦=𝑑𝑥^∗⊕𝑑𝑦, (3.14) and so, (𝑑𝑦)^∗≤𝑑𝑥^∗, from (8), we have (𝑑𝑥^∗)^∗≤𝑑𝑦. Otherwise, 𝑑𝑥^∗≤𝑥^∗, again by (8) 𝑥≤(𝑑𝑥^∗)^∗. Since 𝑑𝑥≤𝑥, we get 𝑑𝑥≤𝑑𝑦.

Theorem 3.17. Let 𝑀 be a linearly ordered MV-algebra, and let 𝑑 be a nonzero additive deriviation of 𝑀. Then, 𝑑⁻¹(0)={𝑥∈𝑀∣𝑑𝑥=0} is an ideal of 𝑀.

Proof. From Proposition 3.4(i), we get that 0∈𝑑⁻¹(0). Let 𝑥,𝑦∈𝑑⁻¹(0); this implies that 𝑑(𝑥⊕𝑦)=0. And so 𝑥⊕𝑦∈𝑑⁻¹(0).
Now, let 𝑥∈𝑑⁻¹(0) and 𝑦≤𝑥. Using Theorem 3.16, we have that 𝑑𝑦≤𝑑𝑥, and so 𝑑𝑦=0.