On derivations of EQ-algebras

1 Introduction

Recently, EQ-algebra has been introduced by Nov a ´ k and De Baets [9], which aims at develop an algebraic structure of truth values for fuzzy type theory. EQ-algebras are important and necessary algebras in many ways. First of all, logical algebras based residuated lattice are all special types of EQ-algebras. Next, an EQ-algebra has three binary operations (multiplication ⊙, meet ∧, and fuzzy equality ∼), and we notice that the typical operations the above algebras are multiplication ⊙ and implication →. In EQ-algebras, implication → is defined by binary operations fuzzy equality ∼ and meet ∧, we have x → y = (x ∧ y) ∼ x. However, fuzzy equality ∼ cannot be established from the implication →. Last but not the least, EQ-algebras start a pattern to develop a fuzzy logic with a single implication only without non-commutative conjunction [10]. In these senses, it is necessary to research EQ-algebras.

The notion of derivations, which proposed in the analytic theory, speeds up our research about properties and structures of algebraic systems. Posner [9] introduced the notion of derivations in prime rings (R, +, ·) in 1957, which is a map d: R → R satisfying two conditions: d (x · y) = d (x) · y + d (y) · x and d (x + y) = d (x) + d (y) for all x, y ∈ R. In what follows, many authors [1 , 7] further researched several properties about derivations in other algebras inspired by Posner. Such as Xin et al. [10] introduced the notion of derivations on lattice (L, ∧, ∨) in 2008, which is a map d: L → L satisfying the following condition: d (x ∧ y) = d (x) ∧ y + d (y) ∧ x for all x, y ∈ L. They discuss some related properties of some particular derivations, they also obtain distributive lattices by isotone derivations. In 2012, Xin [11] study the fixed points of derivations on lattices. Inspired by this, authors investigated the notion of derivations in residuated lattices and obtain that the fixed point set of good ideal derivations is a residuated lattice [6]. Hence, it is meaningful to introduce the derivations theory of EQ-algebras for studying the common properties of derivations in the above-mentioned algebras and characterizing some related properties of some particular derivations about EQ-algebras. It gives us a great deal of inspiration to discuss the derivations on EQ-algebras.

This paper is organized as follows: in Section 2, we review the basic notions, special types of EQ-algebras and their properties. In Section 3, we introduce derivations of EQ-algebras and investigate some properties. In particular, we discuss the fixed point set of useful isotone derivations is a good EQ-algebra. In Section 4 we show that Fix _{d t ^⊙} (E) and Fix _{g t ^→} (E) are order isomorphic.

2 Preliminaries

In this section, we present some results of EQ-algebras, which will be used in the following of the paper.

Definition 2.1. [9] An algebra structure (E, ∧, ⊙, ∼, 1) of type (2, 2, 2, 0) is called an EQ-algebra if it satisfies the following conditions, for all a, b, c, d ∈ E;

(E1) (E, ∧, 1) is a commutative idempotent monoid (i.e. ∧-semilattice with top element 1);

(E2) (E, ⊙, 1) is a commutative monoid and ⊙ is isotone w.r.t. ≤ (with a ≤ b defined as a ∧ b = a);

(E3) a ∼ a = 1;

(E4) ((a ∧ b) ∼ c) ⊙ (d ∼ a) ≤ c ∼ (d ∧ b);

(E5) (a ∼ b) ⊙ (c ∼ d) ≤ (a ∼ c) ∼ (b ∼ d);

(E6) (a ∧ b ∧ c) ∼ a ≤ (a ∧ b) ∼ a;

(E7) a ⊙ b ≤ a ∼ b.

The binary operations ∧, ⊙ and ∼ are called meet, multiplication and fuzzy equality, respectively. Clearly, (E, ≤) is a partial order. Now, we denote, for all a, b ∈ E,

a ˜ = a ∼ 1 and a → b = (a ∧ b) ∼ a

In what follows, we let E = (E, ∧, ∼, 1) be an equality algebra.

Proposition 2.2. [4, 9] Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. For all a, b, c ∈ E it holds that

(1) a ∼ b = b ∼ a, (symmetry)

(2) (a ∼ b) ⊙ (b ∼ c) ≤ a ∼ c, (transitivity)

(3) a ∼ b ≤ a → b and a → a = 1,

(4) (a → b) ⊙ (b → c) ≤ a → c and (b → c) ⊙ (a → b) ≤ a → c,

(5) if a ≤ b, then a → b = 1,

(6) a ⊙ b ≤ b ≤ b ˜ ≤ a → b ,

(7) a ⊙ ( a ∼ b ) ≤ b ˜ ,

(8) a ⊙ b ≤ a, a ⊙ b ≤ a ∧ b, c ⊙ (a ∧ b) = (c ⊙ a) ∧ (c ⊙ b),

(9) if a ≤ b → c (a ≤ b ∼ c), then a ⊙ b ≤ c ˜ .

If E is residuated, then we have

(10) a ⊙ (a → b) ≤ b,

(11) a → (b → c) = (a ⊙ b) → c = b → (a → c),

(12) a ⊙ (b  →  c)   ≤  b  →   (a  ⊙  c)   ≤   (a  ⊙ b)   →   (a ⊙ c).

Definition 2.3. [9] Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then E is called:

good if a ˜ = a for all a ∈ E.

residuated if (a ⊙ b) ∧ c = a ⊙ b iff a ∧ ((b ∧ c) ∼ b) = a for all a, b, c ∈ E.

Lemma 2.4. [9] Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then the following statements are equivalent:

(1) E is good,

(2) a ⊙ (a ∼ b) ≤ b holds for any a, b ∈ E,

(3) a ⊙ (a → b) ≤ b holds for any a, b ∈ E.

Theorem 2.5. [9] Let (E, ∧, ⊙, ∼, 1) be a good EQ-algebra. Then the following hold for all x, y, z ∈ E:

(1) x ≤ (x → y) → y,

(2) x → (y → z) = y → (x → z).

Definition 2.6. [5] Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra, we denote by B (E) the set of all idempotent elements, that is, B (E) = {a ∈ E ∣ a ⊙ a = a}.

Definition 2.7. [3] Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and ∅ ≠ F ⊆ E. F is called a prefilter in E if it satisfies for any x, y ∈ E,

(1) 1 ∈ F,

(2) if x ∈ F, x → y ∈ F, then y ∈ F.

A prefilter F is said to be a filter if it satisfies

(3) If x → y ∈ F, then (x ⊙ z) → (y ⊙ z) ∈ F for any x, y, z ∈ E.

Lemma 2.8. [9] Let F be a filter of an EQ-algebra. Then for all a, b ∈ E it holds that

(1)If a ∈ F and a ≤ b, then b ∈ F.

(2)If a, a ∼ b ∈ F, then b ∈ F.

3 Derivations on EQ-algebras

In this section, we introduce and investigate derivations on EQ-algebras and some basic properties of them.

Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then we define a binary operation on E as follows: for any x, y ∈ E, x ∨ ₁ y = ((x → y) → y) ∧ ((y → x) → x).

Proposition 3.1. Let (E, ∧, ⊙, ∼, 1) be a good EQ-algebra. For any x, y ∈ E,

(1) x, y ≤ x ∨ ₁ y,

(2) x ∨ ₁1 = 1,

(3) x ∨ ₁ x = x,

(4) x ≤ y if and only if x ∨ ₁ y = y.

Proof. (1) Since E is a good EQ-algebra, by Theorem 2.5 we can obtain x ≤ (x → y) → y and the Proposition proposition 2.2 (6) we have x ≤ (y → x) → x. Hence, x ≤ x ∨ ₁ y.

(2) x ∨ ₁1 = ((x → 1) →1) ∧ ((1 → x) → x) =1 ∧ 1 =1.

(3) x ∨ ₁ x = ((x → x) → x) ∧ ((x → x) → x) =1 ∧ x = x.

(4) If x ≤ y, then we have x → y = 1 according to Proposition 2.2 (5). Hence we can obtain that x ∨ ₁ y = ((x → y) → y) ∧ ((y → x) → x) = (1 → y) ∧ ((y → x) → x) = y ∧ ((y → x) → x) = y by Theorem 2.5. If x ∨ ₁ y = y, then we have x ≤ y from the statement (1).

Definition 3.2. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. A map d : E → E is called a derivation if it satisfies the following condition: for any x, y ∈ E,

d (x ⊙ y) = (d (x) ⊙ y) ∨ ₁ (x ⊙ d (y))

Note that if an EQ-algebra E becomes a residuated lattice, the derivations on residuated lattice introduced in [6] and the derivations on EQ-algebra coincide, which means that the notion of the derivations of EQ-algebras is a generalization of one of residuated lattices.

The following example shows that derivations on EQ-algebras are nontrivial generalization of the derivations on residuated lattices.

Example 3.3. Let E = {0, a, b, c, d, 1} be a chain and operations ⊙ and ∼ be defined as follows:

Then (E, ∧, ⊙, ∼, 1) is an EQ-algebra (see [9]). However, E is not a residuated lattice, because b ⊙ c = 0 ≤ a and c → a = a and a ≤ b. The derived operation → is described as the following Cayley table:

Now, we define a map d : E → E as follows:

It is easy to check that d is a derivation on E.

Proposition 3.4. Let (E, ∧, ⊙, ∼, 1) be a good EQ-algebra and d be a derivation on E. Then, for all x ∈ E we have:

(1) d (0) =0,

(2) d (x) ≥ x ⊙ d (1),

(3) d (x ⁿ ) = x ^n-1 ⊙ d (x), for n ≥ 1.

Proof. (1) Since d is a derivation on E, we conclude that d (0) = d (0 ⊙0) = (d (0) ⊙0) ∨ ₁ (0 ⊙ d (0)) =0 ⊙ d (0) =0

(2) For all x ∈ E, we have d (x) = d (x ⊙ 1) = (d (x) ⊙1) ∨ ₁ (x ⊙ d (1)),we can know that d (x) ≥ x ⊙ d (1) from Proposition 3.1 (1).

(3) According to Definition 3.2 and Proposition 3.1 (3), we have d (x ²) = d (x ⊙ x) = (x ⊙ d (x)) ∨ ₁ (d (x) ⊙ x) = x ⊙ d (x). Then by summary we get d (x ⁿ ) = x ^n-1 ⊙ d (x).

In the following, we will introduce and investigate some related properties of isotone and contractive derivations.

Definition 3.5. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a derivation on E.

(1) d is called an isotone derivation provided that x ≤ y implies d (x) ≤ d (y) for all x, y ∈ E;

(2) d is called a contractive derivation provided that d (x) ≤ x for all x ∈ E.

Example 3.6. Let E={0,a,b,1} be a chain with Cayley tables as follows:

Then (E, ∧, ⊙, ∼, 1) is an EQ-algebra [see [8]]. Now, we define a map d : E → E as follows:

We can easily check that d is a contractive derivation, however d is not an isotone derivation, since a ≤ b, d (a) = a, d (b) =0 and 0 ≤ a.

Example 3.7. Let E = {0, a, b, c, 1} be a chain with Cayley tables as follows:

Then (E, ∧, ⊙, ∼, 1) is an EQ-algebra. The derived operation → is described as the following Cayley table:

Now, we define a map d : E → E as follows:

One can check that d is an isotone derivation but is not a contractive derivation, because d (b) = c and c ≥ b.

Proposition 3.8. Let (E, ∧, ⊙, ∼, 1) be a good EQ-algebra and d be a contractive derivation on E. Then for all x, y ∈ E we have:

(1) d (x) ⊙ d (y) ≤ d (x ⊙ y),

(2) if d is isotone and E is a residuated EQ-algebra, then d (x → y) ≤ d (x) → d (y) ≤ d (x) → y,

(3) d (1) =1 if and only if d is an identity derivation.

Proof. (1) Since d is a contractive derivation on E, we have d (y) ≤ y, hence d (x) ⊙ d (y) ≤ d (x) ⊙ y. Similarly, d (x) ⊙ d (y) ≤ d (y) ⊙ x. Therefore d (x) ⊙ d (y) ≤ d (x ⊙ y) according to Proposition 3.1 (1).

(2) For all x, y ∈ E, since x ⊙ (x → y) ≤ y from Lemma 2.4, we deduce d (x ⊙ (x → y)) ≤ d (y). From the statement (1), we conclude that d (x → y) ⊙ d (x) ≤ d (x ⊙ (x → y)), it follows that d (x → y) ⊙ d (x) ≤ d (y). According to E is a residuated EQ-algebra, which implies d (x → y) ≤ d (x) → d (y). On the other hand, from d (y) ≤ y, we infer d (x) → d (y) ≤ d (x) → y by Proposition 2.2 (6). Therefore, we obtain d (x → y) ≤ d (x) → d (y) ≤ d (x) → y.

(3) By Proposition 3.4 (2), we have x ⊙ d (1) ≤ d (x) for all x ∈ E. Assume that d (1) =1, we get x = x ⊙ d (1) ≤ d (x) ≤ x showing that d (x) = x. Therefore, d is an identity derivation. If d is an identity derivation, then d (x) = x, we have d (1) =1.

In what follows, we give some characterizations of isotone derivations.

Theorem 3.9. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a contractive derivation on E satisfying: d (1) ∈ B (E). Then the following are equivalent:

(1) d is isotone,

(2) d (x) ≤ d (1), for all x ∈ E,

(3) d (x) = d (1) ⊙ x, for all x ∈ E,

(4) d (x ∧ y) = d (x) ∧ d (y), for all x, y ∈ E,

(5) d (x ⊙ y) = d (x) ⊙ d (y), for all x, y ∈ E.

Proof. (1) ⇒ (2) Since for all x ∈ E, x ≤ 1, we have d (x) ≤ d (1).

(2) ⇒ (3) Suppose that d (x) ≤ d (1) for all x ∈ E. Notice that d (1) ∈ B (E), we obtain d (x) = d (1) ∧ d (x) = d (1) ⊙ d (x) ≤ d (1) ⊙ x. On the other hand, we conclude from Proposition 3.4 (2) that d (x) ≥ x ⊙ d (1). Thus, d (x) = d (1) ⊙ x.

(3) ⇒ (4) Let d (x) = d (1) ⊙ x for all x ∈ E. It follows that d (x ∧ y) = d (1) ⊙ (x ∧ y) = d (1) ∧ (x ∧ y) = (d (1) ∧ x) ∧ (d (1) ∧ y)) = (d (1) ⊙ x) ∧ (d (1) ⊙ y)) = d (x) ∧ d (y).

(4) ⇒ (1) Assume that x ≤ y, we have x ∧ y = x. It follows from (4) that d (x) = d (x ∧ y) = d (x) ∧ d (y). Thus, we obtain d (x) ≤ d (y).

(3) ⇒ (5) By (3), we have d (x ⊙ y) = d (1) ⊙ (x ⊙ y) = (d (1) ⊙ x) ⊙ (d (1) ⊙ y) = d (x) ⊙ d (y).

(5) ⇒ (3) It follows from Proposition 3.4 (2) that d (x) ≥ x ⊙ d (1). From (5), we can get d (x) = d (1 ⊙ x) = d (1) ⊙ d (x) ≤ d (1) ⊙ x. An isotone and a contractive derivation d on E is said to be useful provided that d (1) ∈ B (E).

Example 3.10. Let E = {0, a, b, c, d, 1} be a set with Hasse diagram and Cayley tables as follows.

Then (E, ∧, ⊙, ∼, 1) is an EQ-algebra and B (E) = {b, c, 0, 1}. Now, we define a map d : E → E as follows: d ( x ) = { 0 , x = 0 , b d , x = a , d c , x = c , 1 .

Then we can check that d is a useful derivation on E since d (1) = c ∈ B (E).

The following results will be useful for researching the algebraic structure of the fixed point set for useful derivations.

Proposition 3.11. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a useful derivation on E. Then we have:

(1) for all x ∈ E, d ² (x) = d (x),

(2) for all x, y ∈ E, d (x) ≤ y if and only if d (x) ≤ d (y),

(3) for all x, y ∈ E, d (d (x) ∧ d (y)) = d (d (x) ∧ y) = d (x ∧ d (y)) = d (x ∧ y).

Proof.(1) Since d is an isotone and a contractive derivation on E and d (1) ∈ B (E), we have d (x) = d (1) ⊙ x by Theorem 3.9 (3). It follows that d (d (x)) = d (1) ⊙ d (x) = d (1) ⊙ (d (1) ⊙ x) = d (1) ⊙ x = d (x), that is, d ² (x) = d (x).

(2) For all x, y ∈ E, assume that d (x) ≤ y, we have d (d (x)) ≤ d (y). By the statement (1), we get d (d (x)) = d (x). Thus d (x) ≤ d (y). Conversely, suppose that d (x) ≤ d (y), we have d (x) ≤ d (y) ≤ y.

(3) From the statement (1), we conclude that d (d (x)) = d (x) for all x ∈ E. From d (y) ≤ y, we have x ∧ d (y) ≤ x ∧ y. It follows that d (x ∧ d (y)) ≤ d (x ∧ y). On the other hand, combining d (x ∧ y) ≤ d (y) and d (x ∧ y) ≤ d (x) ≤ x, we obtain d (x ∧ y) ≤ x ∧ d (y). Then d (x ∧ y) = d (d (x ∧ y)) ≤ d (x ∧ d (y)). Thus d (x ∧ y) = d (x ∧ d (y)) for all x, y ∈ E. It follows that d (d (x) ∧ y) = d (d (x) ∧ d (y)). In a similar way, we can prove d (x ∧ y) = d (d (x) ∧ y). Therefore, we conclude that d (d (x) ∧ d (y)) = d (d (x) ∧ y) = d (x ∧ d (y)) = d (x ∧ y).

Let d be a derivation on an EQ-algebra, we denote by Fix _d (E) the set of all fixed elements of E for d, that is, Fix _d (E) = {x ∈ E|d (x) = x}.

Proposition 3.12. Let (E, ∧, ⊙, ∼, 1) be a good EQ-algebra and d be an isotone and a contractive derivation on E. Then we have: for all x, y ∈ Fix _d (E), x ⊙ y, x ∼ ₁ y ∈ Fix _d (E), where x ∼ ₁ y = d (x ∼ y).

Proof. For all x, y ∈ Fix _d (E), we have d (x) = x and d (y) = y. It follows from Proposition 3.11 (1) that x ⊙ y = d (x) ⊙ d (y) ≤ d (x ⊙ y) ≤ x ⊙ y, which implies d (x ⊙ y) = x ⊙ y. Thus, we can get that x ⊙ y ∈ Fix _d (E). d (x ∼ ₁ y) = d (d (x ∼ y)) = d (x ∼ y) = x ∼ ₁ y combining Proposition 3.11 (1). Hence we obtain x ∼ ₁ y ∈ Fix _d (E).

As an application of the above propositions, we have the following result.

Theorem 3.13. Let (E, ∧, ⊙, ∼, 1) be a good EQ-algebra and d be a useful derivation on E, such that d (x ∼ y) ≤ d (x) ∼ d (y). Then (Fix _d (E), ∧ ₁, ⊙, ∼ ₁, 1′) is a good EQ-algebra, where x ∧ ₁ y = d (x ∧ y), x ∼ ₁ y = d (x ∼ y) and 1′ = d (1) for all x, y ∈ Fix _d (E).

Proof. First, since for all x, y ∈ Fix _d (E), d is an isotone and a contractive derivation on E, we have d (x ∧ y) ≤ d (x) = x and d (x ∧ y) ≤ d (y) = y. we can obtain that d (x ∧ y) is a lower bound of x, y ∈ Fix _d (E). If exists z ∈ Fix _d (E) such that z is a lower bound of x, y ∈ Fix _d (E), we know that z ≤ x ∧ y. Then we obtain z = d (z) ≤ d (x ∧ y) since d is an isotone. So, d (x ∧ y) is the greatest lower bound of x and y. For all x ∈ Fix _d (E), according to Theorem 3.9 we can obtain x = d (x) ≤ d (1), thus 1′ is the greatest element in Fix _d (E). So (Fix _d (E), ∧ ₁, 1′) is a ∧-semilattice with top element 1′.

Now we will prove (E, ⊙, 1′) is a commutative monoid. It is clearly that Fix _d (E) is closed under ⊙ by Proposition 3.12, and from Theorem 3.9 we can see that x ⊙ 1′ = d (x) ⊙ d (1) = d (x ⊙ 1) = d (x) = x.

For all a, b, c, d ∈ Fix _d (E), ((a ∧ b) ∼ c) ⊙ (d ∼ a) ≤ c ∼ (d ∧ b). Since d is isotone and by Theorem 3.9, we have d {((a ∧ b) ∼ c) ⊙ (d ∼ a)} = d ((a ∧ b) ∼ c) ⊙ d (d ∼ a) ≤ d (c ∼ (d ∧ b)). We conclude that ((a ∧ b) ∼ ₁ c) ⊙ (d ∼ ₁ a) ≤ c ∼ ₁ (d ∧ b). Similarly, we can prove (a ∧ b ∧ c) ∼ ₁ a ≤ (a ∧ b) ∼ ₁ a.

For all a, b, c, d ∈ Fix _d (E), (a ∼ b) ⊙ (c ∼ d) ≤ (a ∼ c) ∼ (b ∼ d). Since d is an isotone and by Theorem 3.9, we have d (x ∼ y) ≤ d (d (x) ∼ d (y)) and d ((a ∼ b) ⊙ (c ∼ d)) = d (a ∼ b) ⊙ d (c ∼ d) ≤ d ((a ∼ c) ∼ (b ∼ d)) ≤ d (d (a ∼ c) ∼ d (b ∼ d)). Hence, (a ∼ ₁ b) ⊙   (c ∼ ₁ d)   ≤   (a ∼ ₁ c) ∼ ₁ (b ∼ ₁ d).

For any a, b ∈ Fix _d (E), we know that a ⊙ b ≤ a ∼ b. a ⊙ b = d (a ⊙ b) ≤ d (a ∼ b) = a ∼ ₁ b according to d is isotone.

For all a  ∈  Fix _d (E), a ∼ ₁ a  =  d (a  ∼  a)   =  d (1) =1′.

Finally, we will prove a ˜ = a . For all a ∈ Fix _d (E), we have 1 ∼ ₁ a = d (1 ∼ a) = d (a) = a since E is a good EQ-algebra.

Therefore, we can conclude that (Fix _d (E), ∧, ⊙, ∼ ₁, 1′) is a good EQ-algebra.

4 Simple derivations and their fixed point sets

In this section, we study simple derivations and discuss some properties about them. In particular, we introduce and study simple ⊙-derivation and simple →-derivation and discuss relationship of their fixed point sets.

In what follows, let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then for t ∈ E, we define a map d _{t ^⊙} as follows: d _{t ^⊙} (x) = t ⊙ x for any x ∈ E.

Theorem 4.1. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then for all t ∈ E, the map d _{t ^⊙} is an isotone and a contractive derivation on E, which is called a simple ⊙-derivation.

Proof. For all x, y ∈ E, we obtain d _{t ^⊙} (x⊙ y) = t ⊙ (x ⊙ y) = (t ⊙ (x ⊙ y)) ∨ ₁ (t ⊙ (x ⊙ y)) = ((t ⊙ x) ⊙ y) ∨ ₁ (x ⊙ (t ⊙ y))   = (d _{t ^⊙} (x) ⊙ y) ∨ ₁ (x ⊙ d _{t ^⊙}   (y)). That is, d _{t ^⊙} is a derivation on E. For any x, y ∈ E, let x ≤ y, we know that d _{t ^⊙} (x) = t ⊙ x ≤ t ⊙ y = d _{t ^⊙} (y), hence, d _{t ^⊙} is isotone. And for all x ∈ E, d _{t ^⊙} (x) = t ⊙ x ≤ x by Proposition 2.2 (6), we have d _{t ^⊙} is contractive. Therefore, d _{t ^⊙} is a simple ⊙-derivation on E.

In what follows, we focus on algebraic structure the set of all simple ⊙-derivations. We denote by D (E) = {d _{t ^⊙} |t ∈ E} be the set of all simple ⊙-derivations of E.

Theorem 4.2. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then (D (E), ⊓, d _1^⊙ ) is a ∧-semilattice with top element d _1^⊙ , where (d _{a ^⊙} ⊓ d _{b ^⊙} ) (x) = (d _{a ^⊙} (x)) ∧ (d _{b ^⊙} (x)), for any d _a , d _b ∈ D (E).

Proof. For all d _a , d _b ∈ D (E) and x ∈ E, according to Proposition 2.2 (8) we have (d _{a ^⊙} ⊓ d _{b ^⊙} ) (x) = (d _{a ^⊙} (x)) ∧ (d _{b ^⊙} (x)) = (a ⊙ x) ∧ (b ⊙ x) = (a ∧b) ⊙ (x) = d _(a∧b)^⊙ (x), that is, d _{a ^⊙} ⊓ d _{b ^⊙} = d _{a ^⊙} ∧ d _{b ^⊙} . Moreover, for all d _a ∈ D (E) and x ∈ E, we have (d _{a ^⊙} ⊓ d _0^⊙ ) (x) = (d _{a ^⊙} (x)) ∧ (d _0^⊙ (x)) = (a ⊙ x) ∧ (0 ⊙ x) =0, (d _{a ^⊙} ⊓ d _1^⊙ ) (x) = (d _{a ^⊙} (x)) ∧ (d _1^⊙ (x))   =   (a ⊙ x) ∧ (1 ⊙ x)   =  1 ⊙ x  =  d _1^⊙ (x). Thus, we can conclude that (D (E), ⊓, d _0^⊙ , d _1^⊙ ) is a ∧-semilattice with top element d _1^⊙ .

Theorem 4.3. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra satisfying x ⊙ x = x for all x ∈ E. Then (D (E), ⊗, d _1^⊙ ) is a commutative monoid, where (d _{a ^⊙} ⊗ d _{b ^⊙} ) (x) = (d _{a ^⊙} (x)) ⊙ (d _{b ^⊙} (x)), for any d _a , d _b ∈ D (E).

Proof. For all d _a , d _b ∈ D (E) and x ∈ E, we obtain (d _{a ^⊙} ⊗ d _{b ^⊙} ) (x) = (d _{a ^⊙} (x)) ⊙ (d _{b ^⊙} (x))   =   (a ⊙ x) ⊙ (b   ⊙x) = a ⊙ b ⊙ x ⊙ x = (a ⊙ b) ⊙ x = d _(a⊙b)^⊙ (x), that is, (d _{a ^⊙} ⊗ d _{b ^⊙} ) = d _(a⊙b)^⊙ and hence d _{a ^⊙} ⊗ d _{b ^⊙} ∈ D (E). Furthermore, (d _{a ^⊙} ⊗ d _1^⊙ ) (x) = (d _{a ^⊙} (x)) ⊙ (d _1^⊙ (x)) = a ⊙ x = d _{a ^⊙} (x). Thus, (D (E), ⊗, d _1^⊙ ) is a commutative monoid.

The following we will introduce implicative derivations on EQ-algebras.

Definition 4.4. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. A map f : E → E is called an implicative derivation on E if it satisfies the following condition: for any x, y ∈ E, f (x → y) = (f (x) → y) ∨ ₁ (x → f (y)).

Now, we present an example for implicative derivations on EQ-algebras.

Example 4.5. Let E={0,a,b,1} be a chain and operations ⊙ and ∼ be defined as follows:

Then (E, ∧, ⊙, ∼, 1) is an EQ-algebra [see [8]]. The derived operation → is described as the following Cayley table:

Now, we define a map f : E → E as follows:

One can easily check that f is an implicative derivation on EQ-algebras.

From now on, we will discuss the adjoint derivation of simple ⊙-derivations, we firstly recall the notion of Galois connections.

Let S and T be two posets, d : S → T and f : T → S, a pair (d, f) of functions is called a Galois connection or an adjunction between S and T if

(1) d and f are isotone;

(2) d (s) ≤ t if and only if s ≤ f (t), for all s ∈ S and t ∈ T.

In an adjunction (d, f), the function d is called the upper adjoint and f the lower adjoint.

Definition 4.6. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a derivation on E. The derivation d is called to be residuated if there exists an implicative derivation f on E such that a pair (d, f) forms a Galois connections. The implicative derivation f is called the adjoint derivation of the derivation d.

From the notion of Galois connections, we have that if derivation d is residuated, then d and it is adjoint derivation must be isotone. In particular, if the derivation d has the adjoint derivation f, then the adjoint of d is unique. Therefore, we will denote this unique f by d ^*.

Proposition 4.7. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a residuted derivation on E, then for all x, y ∈ E we have:

(1) d ^* (1) =1,

(2) d ^* (x) ≥ x,

(3) d ^* (x) ∨ ₁ d ^* (y) ≤ d ^* (x → y),

(4) d ^* (x ∧ y) = d ^* (x) ∧ d ^* (y),

(5) If d is a contractive derivation, then d ^* (x) → d ^* (y) ≤ d ^* (x → y).

Proof. (1) Since d ^* is the adjoint derivation of d, we have d ^* (1) = d ^* (1 →1) = (d ^* (1) →1) ∨ ₁ (1 → d ^* (1)) =1 ∨ ₁ (1 → d ^* (1)) =1.

(2) For all x ∈ E, we have d ^* (x) = d ^* (1 → x) = (d ^* (1) → x) ∨ ₁ (1 → d ^* (x)) = (1 → x) ∨ ₁ (1 → d ^* (x)) = x ∨ ₁ d ^* (x). Thus, we can obtain d ^* (x) ≥ x by Proposition 3.1 (4).

(3) For all x, y ∈ E, we can get d ^* (x) ≤ y → d ^* (x) and d ^* (y) ≤ x → d ^* (y). It follows from Definition 4.4 that d ^* (x) ∨ ₁ d ^* (y) ≤ (x → d ^* (y)) ∨ ₁ (y → d ^* (x)) = d ^* (x → y).

(4) Since d ^* is the adjoint derivation of d, we know that d ^* is isotone. It follows that d ^* (x ∧ y) ≤ d ^* (x), d ^* (y) for all x, y ∈ E. Hence, d ^* (x ∧ y) is a lower bound of d ^* (x) and d ^* (y). Suppose now that t ∈ E is any lower bound of d ^* (x) and d ^* (y), we have t ≤ d ^* (x), d ^* (y). Since d ^* is the adjoint derivation of d, we obtain d (t) ≤ x, y, that is, d (t) ≤ x ∧ y. We can get t ≤ d ^* (x ∧ y). Thus, d ^* (x ∧ y) is the infimum of d ^* (x) and d ^* (y). Therefore, we conclude that d ^* (x ∧ y) = d ^* (x) ∧ d ^* (y).

(5) Since d is a contractive derivation, we have d (x) ≤ x, combining d ^* is the adjoint derivation of d, we have x ≤ d ^* (x). We obtain that d ^* (x) → d ^* (y) ≤ x → d ^* (y) from Proposition 2.2 (6). Combining (3), we obtain x → d ^* (y) ≤ d ^* (x → y). Therefore, d ^* (x) → d ^* (y) ≤ d ^* (x → y).

Next, let (E, ∧, ⊙, ∼, 1) be an EQ-algebra, t ∈ E, we define a map f _{t ^→} : E → E as follows: f _{t ^→} (x) = t → x for all x ∈ E.

Theorem 4.8. Let (E, ∧, ⊙, ∼, 1) be a residuated EQ-algebra. Then for all t ∈ E, f _{t ^→} is the adjoint derivation of d _{t ^⊙} on E, that is, d t ∗ = f t → . Therefore we call f _{t ^→} a simple →-derivation.

Proof. For all t, x, y ∈ E, we can get (t → x) → y ≤ x → y ≤ t → (x → y) from x ≤ t → x. Since (f _{t ^→} (x) → y) ∨ ₁ (x → f _{t ^→} (y))   =   ((t → x) → y) ∨ ₁ (x → (t → y))   =   ((t → x) → y) ∨ ₁ (t → (x → y))   = t → (x → y) = f _{t ^→} (x → y) follow from Proposition 2.2 (11), that is, f _{t ^→} (x → y) = (f _{t ^→} (x) →y) ∨ ₁ (x → f _{t ^→} (y)). Therefore, f _{t ^→} is an implicative derivation on E. Meanwhile, for all x, y ∈ E, let x ≤ y, we know that f _{t ^→} (x) = t→ x ≤ t → y = f _{t ^→} (y), hence f _{t ^→} is isotone. And for all x, y ∈ E, we have f _{t ^→} (x) = t ⊙ x ≤ y if and only if x ≤ t → y = f _{t ^→} (y). Thus, (d _{t ^⊙} , f _{t ^→} ) is a Galois connection. Therefore, we obtain that f _{t ^→} is the adjoint derivation of d _{t ^⊙} , that is, d t ∗ = f _{t ^→} .

In what follows, we denote by Fix _d (E) = {x ∈ E|d (x) = x}. In the following we will present the relationship between Fix _{d t ^⊙} (E) and Fix _{g t ^→} (E).

Theorem 4.9. Let (E, ∧, ⊙, ∼, 1) be a residuated EQ-algebra and t be an idempotent element of E. Then for all t ∈ E, Fix _{d t ^⊙} (E) and Fix _{f t ^→} (E) are order isomorphic.

Proof. For all t ∈ E, let h : Fix _{d t ^⊙} (E) ⟶ Fix _{f t ^→} (E) be defined by h (x) = t → x for all x ∈ Fix _{d t ^⊙} (E). Clearly, h is a map which is from Fix _{d t ^⊙} (E) to Fix _{f t ^→} (E), that is, for all x ∈ Fix _{d t ^⊙} (E), we have f _{t ^→} (t → x) = t → (t → x) = (t ⊙ t) → x = t → x from Proposition 2.2 (11), hence t → x ∈ Fix _{f t ^→} (E). Therefore h is well defined.

Let us check that h is injective. For all x, y ∈ Fix _{d t ^⊙} (E), that is, x = t ⊙ x and y = t ⊙ y, if h (x) = h (y), then t → x = t → y. We have x ≤ t → x from property of →, then x ≤ t → y. It follows that t ⊙ x ≤ y, which means x ≤ y. Similarly, we can prove y ≤ x. Then we obtain x = y. Consequently, h is injective.

Now, we prove that h is surjective. First, we can easily prove the fact (*) that x → (x ⊙ y) = y if and only if there exists z ∈ E such that x → z = y for all x, y ∈ E. For all x ∈ Fix _{f t ^→} (E), then x = f _{t ^→} (x) = t → x. Using the fact (*), we have that h (t ⊙ x) = t → (t ⊙ x) = t → (t ⊙ (t → x)) = t → x = x and d _{t ^⊙} (t ⊙ x) = t ⊙ t ⊙ x = t ⊙ x. Thus, we conclude that h is surjective.

Next, we show that h and h ^-1 are order-preserving, where the inverse map h ^-1 : Fix _{f t ^→} (E) ⟶Fix _{d t ^⊙} (E) for all x ∈ Fix _{f t ^→} (E), where h ^-1 (x) = t ⊙ x. Notice that for all x ∈ Fix _{d t ^⊙} (E), h [h ^-1 (x)] = h (t ⊙ x) = t  →   (t ⊙ x)   =  x since exists t ∈ E such that for all x ∈ Fix _{f t ^→} (E), f _t (x) = t →x = x from the fact (*). Therefore, h ∘ h ^-1 = I _{Fix f t ^→} (E). Similarly, h - 1 ∘ h = I Fix d t ⊙ ( E ) .

Combining them, we obtain that h is an order isomorphism from Fix _{d t ^⊙} (E) to Fix _{f t ^→} (E). Therefore, ordered sets Fix _{d t ^⊙} (E) and Fix _{f t ^→} (E) are isomorphic.

Definition 4.10. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then an EQ-algebra is said to be divisible provided that x ∧ y = x ⊙ (x → y), for all x, y ∈ E.

Theorem 4.11. Let (E, ∧, ⊙, ∼, 1) be a residuated EQ-algebra. Then the following assertions are equivalent:

(1) E is divisible,

(2) for all t ∈ E, Fix _{d t ^⊙} (E) = (t], where (t] = {x ∈ E ∣ x ≤ t}.

Proof. (1)⇒ (2) Assume that E satisfies x ⊙ y = x ∧ y = x ⊙ (x → y) for all x, y ∈ E. That is, for all x ∈ E, if x ≤ t, we have x = x ∧ t = x ⊙ t, hence, we can conclude that x ∈ Fix _{d t ^⊙} (E), which implies that (t] ⊆ Fix _{d t ^⊙} (E). Next, we will show that Fix _{d t ^⊙} (E) ⊆ (t]. For all x ∈ Fix _{d t ^⊙} (E), we have d _{t ^⊙} (x) = x. Thus, d _{t ^⊙} (x) = t ⊙ x ≤ t ∧ x, that is, x = t ∧ x which implies that x ≤ t, thus we can get x ∈ (t]. It follows that Fix _{d t ^⊙} (E) ⊆ (t]. Therefore, we obtain Fix _{d t ^⊙} (E) = (t].

(2)⇒ (1) Suppose that Fix _{d t ^⊙} (E) = (t] for all t ∈ E. Since t ∈ (t], we get t ∈ Fix _{d t ^⊙} (E). It follows that d _{t ^⊙} (t) = t, that is, t ⊙ t = t for all t ∈ E. Next, we will prove that x ⊙ y = x ⊙ (x → y). On the one hand, we can prove x ⊙ (x → y) ≤ x ⊙ y, that is, x → y ≤ x → (x ⊙ y). For all x, y ∈ E, by Proposition 2.2 (12), we can obtain x⊙ (x → y) ≤ (x ⊙ x) → (x ⊙ y), that is, x ⊙ (x → y) ≤ x → (x ⊙ y). Then we have x → y ≤ x → (x → (x ⊙ y)) = (x ⊙ x) → (x ⊙ y) = x → (x ⊙ y). It follows that x ⊙ (x → y) ≤ x ⊙ y. On the other hand, from y ≤ x → y, we obtain x ⊙ y ≤ x ⊙ (x → y). Thus, we get x ⊙ y = x ⊙ (x → y). Finally, we will show x ⊙ y = x ∧ y. For all x, y ∈ E, we have x ⊙ y ≤ x ∧ y by Proposition 2.2 (8). Now, for all m ∈ E, if m ≤ x and m ≤ y, we can obtain m ⊙ m ≤ x ⊙ y. Hence we have m ≤ x ⊙ y. It follows that x ⊙ y = x ∧ y = x ⊙ (x → y).

In what follows, we will give the definition of d-filter and investigate some properties.

Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a derivation of E. Then a filter F is called a d-filter if x ∈ F implies d (x) ∈ F for all x ∈ E. As is usually done, given a filter F, we can define an equivalence relation on E by x ≈ _F y iff x ∼ y ∈ F, for all x, y ∈ E. Then relation ≈ _F is a congruence relation on E. One can know that a quotient structure E/F is an EQ-algebra for a filter F [9].

In the following we will present an example of d-filter.

Example 4.12. Let E = {0, a, b, 1} be a chain. We define two binary operations ‘⊗’ and ‘∼’ by the following Cayley tables:

Then (E, ∧, ⊗, ∼, 1) is an EQ-algebra (see [8]). The derived operation ‘→’ is described as the following Cayley table:

It is routine to verify that F : = {a, 1} is a filter of E. Now, we define a map d : E → E as follows: d ( x ) = { 0 , x = 0 , b a , x = a , 1 .

One can easily check that d is a derivation and F is a d-filter of E.

Proposition 4.13. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra. Then for a useful derivation d of E, we have Fix _d (E) = d (E).

Proof. For all x ∈ Fix _d (E), we have d (x) = x, that is, x ∈ d (E). Conversely, for all x ∈ d (E), exists y ∈ E, such that d (y) = x. By Proposition 3.12 (1) and Theorem 3.9 (3), we have d (x) = d (d (y)) = d (y) ⊙ d (1) = d (1) ⊙ y = d (y) = x, we get x ∈ Fix _d (E).

Theorem 4.14. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a useful derivation and F be a d-filter of E. Then a map d/F : E/F → E/F defined by for all x/F ∈ E/F, (d/F) (x/F) = dx/F is a useful derivation of E/F.

Proof. Firstly, for all x/F, y/F ∈ E/F, if x/F = y/F, then we have dx/F = (d/F) (x/F) = (d/F) (y/F) = dx/F. Hence, d/F is well defined. Next, since d is a derivation, we have (d/F) ((x ⊙ y)/F) = d (x ⊙ y)/F = (d (x) ⊙ y) ∨ ₁ (x ⊙ d (y))/F. Thus, we can obtain d/F is a derivation. Now, we can prove d/F is isotone and contractive. For all x/F, y/F ∈ E/F, we can know that x/F ≤ y/F iff x/F ∧ y/F = x/F iff (x ∧ y)/F = x/F. Since F is a d-filter and by Theorem 3.9 (4), we have d (x ∧ y)/F = (dx ∧ dy)/F = dx/F ∧ dy/F = dx/F. This means that dx/F ≤ dy/F, that is, d/F is isotone. For all x/F ∈ E/F, since d is a contractive derivation, we have (d/F) (x/F) = d (x)/F ≤ x/F, hence, d/F is contractive. We can conclude that d/F is a useful derivation of E/F.

Proposition 4.15. Let (E, ∧, ⊙, ∼, 1) be an EQ-algebra and d be a useful derivation and F be a d-filter of E. Then d (F) is a filter of (d (E), ∧ ₁, ⊙, ∼ ₁, 1′).

Proof. Firstly, since 1 ∈ F, we have d (1) ∈ dF. Next, we will prove if dx, dx → ₁ dy ∈ dF, then we can get dy ∈ dF. Notice that dx → ₁ dy = (dx ∧ ₁ dy) ∼ ₁ dx = d (dx ∧ dy) ∼ ₁ dx = d (d (dx ∧ dy) ∼ dx) = d  (d (d (x ∧ y))   ∼  dx)   =  d (d (x ∧ y) ∼ dx)   =  d ((dx ∧ dy)   ∼dx) = d (dx → dy) ∈ dF from Theorem 3.9 (4) and Proposition 3.11 (1), which implies dx→ dy ∈ F. Since dx ∈ dF ⊆ F, we can obtain that dy ∈ F. Therefore, it holds d (dy) = dy ∈ dF according to Proposition 3.11 (1). Finally, for all dx, dy, dz ∈ dE, if dx → ₁ dy ∈ dF, then we have dx → ₁ dy = d (dx → dy) ∈ dF. Hence, we can know that dx → dy ∈ F, since F is a filter, there exists dz ∈ dF ⊆ F such that (dx ⊙ dz) → (dy ⊙ dz) ∈ F due to the Definition 2.7, that is, we have d ((dx ⊙ dz) → (dy ⊙ dz)) ∈ dF.

Note that the quotient structure (d/F) (E/F) = Fix _d/F (E/F) is an EQ-algebra by Theorem 3.13. By Proposition 4.15, we have that d (E)/d (F) is an EQ-algebra. It is natural to ask what the relation between EQ-algebra (d/F) (E/F) and d (E)/d (F) is. Now, we will give the answer.

Theorem 4.16. Let d be a useful derivation and F be a d-filter of E. Then we have (d/F) (E/F) = Fix _d/F (E/F) is order isomorphic to d (E)/d (F), in other words, (d/F) (E/F) = Fix _d/F (E/F) ≅ d (E)/d (F).

Proof. Define a map φ: Fix _d/F (E/F) → d (E)/d (F) as follows: φ (x/F) = dx/dF for all x/F ∈ Fix _d/F (E/F).

On the one hand, we show that φ is well defined. In fact, for all x/F, y/F ∈ E/F, if x/F = y/F, then we have x ∼ y ∈ F. Since d is a d-filter, we have d (x ∼ y) ∈ F. By Theorem 3.13 we get d (x ∼ y) ≤ dx ∼ dy, then we can obtain dx ∼ dy ∈ F due to F is a filter, that is, d (dx ∼ dy) = dx ∼ ₁ dy ∈ dF. Hence, we can know that dx/dF = dy/dF, so φ is well defined.

One the other hand, for all x/F, y/F ∈ Fix _d/F (E/F), we will check that x/F ≤ y/F only and only if φ (x/F) ≤ ₁ φ (y/F). In fact, for all x/F ∈ Fix _d/F (E/F), we have x/F = (d/F) (x/F) = dx/F from Theorem 4.14. Hence we can get dx/F ≤ dy/F iff dx/F ∧ dy/F = dx/F iff (dx ∧ dy)/F = d (x ∧ y)/F = dx/F from Theorem 3.9 iff d (x ∧ y) ∼ dx ∈ F. We observe that d (x ∧ y) ∼ dx = d (d (x ∧ y)) ∼ dx = d (dx ∧ dy) ∼ dx ∈ F from Theorem 3.9 (4) and Proposition 3.11 (1), that is, d (d (dx ∧ dy) ∼ dx) = d (dx ∧ dy) ∼ ₁ dx = (dx ∧ ₁ dy) ∼ ₁ dx ∈ dF, so we can obtain that dx/dF ≤ ₁ dy/dF. Conversely, if dx/dF ≤ ₁ dy/dF, then we have d (d (dx ∧ dy) ∼ dx) ∈ dF, it follows from Theorem 3.9 (4) and Proposition 3.11 (1) we get d (x ∧ y) ∼ dx ∈ F. Thus, we conclude that x/F = dx/F ≤ dy/F = y/F. Finally, we can easily see that φ is surjective. Hence, we can obtain Fix _d/F (E/F) ≅ d (E)/d (F).

5 Conclusions

In this paper, we had introduced the concept of derivations, which was helpful for studying properties of algebraic systems. Meanwhile, we discussed some properties of isotone and contractive derivations and studied some characterizes of useful derivations. Then we also showed that the fixed point set of useful derivations is a good EQ-algebra. Moreover, we gave simple ⊙-derivations and their adjoint derivations and proved that the fixed point set of simple ⊙-derivations and that of their adjoint derivations were order isomorphism. Finally, the quotient algebras induced by d-filter were studied.