Commutators in BCI -algebras

Abstract

In this paper, the notions of commutators and pseudo commutators of elements (subsets) of a BCI-algebra are introduced and some properties are given. The concept of solvable BCI-algebras are also discussed and their properties are investigated. It is proved that the class of solvable BCI-algebras is closed under sub-algebra, cartesian product and inverse image operations.

Keywords

(pseudo) commutators derived sub-algebra

1 Introduction

Various types of BCK/BCI-algebras as algebras are strongly connected with non-classical propositional calculi are studied by many authors. A short survey of basic results on BCK/BCI-algebras can be found in the book [7]. The class of BCI-algebras introduced by Iseki and Tanaka [5] is an important class of logical algebras. One of the motivations for their study is based on set theory and another one classical and non-classical propositional calculus. By definition, the notion of BCI-algebras generalizes the algebra of sets with the set subtraction as well as the notion of implication algebras. BCI-algebras are generalizations of BCK-algebras. Hence most of the algebras related to the t-norm based logic, such as MV and Boolean algebras, etc. are extensions of BCK-algebras. This shows that BCK/BCI-algebras are considerably general structures.

The concepts of commutators and solvability in BCK-algebras are defined and their properties are studied [8, 9].

In this paper, we give a new definition of the notions of commutators and solvability in BCI-algebra which have more properties.

Definition 1.1. [3] By a BCI-algebra, we mean an algebra (X, * , 0) of type (2, 0) satisfying the following axioms: for all x, y, z ∈ X,

(BCI 1) ((x * y) * (x * z)) * (z * y) =0,

(BCI 2) (x * (x * y)) * y = 0,

(BCI 3) x * x = 0,

(BCI 4) x * y = y * x = 0 implies x = y .

On any BCI-algebra (X, * , 0) the natural order can be defined by putting x ≤ y if and only if x * y = 0, for all x, y ∈ X. If in a BCI-algebra (X, * , 0) the condition 0 * x = 0, for all x ∈ X hold, then it is a BCK-algebra [2]. In any BCK-algebra X for all x, y ∈ X, the following axioms hold [1 –7].

(1.1) x * y ≤ x.

(1.2) (x ∧ y) * x = (x ∧ y) * y,

where x ∧ y = y * (y * x) .

The set {x ∈ X : 0 * x = 0} is called BCK-part of BCI-algebra X and is denoted by BCK (X). If X = BCK (X) , then X is a BCK-algebra.

The following statements are true in any BCI-algebra X, for all x, y, z ∈ X [2].

(1.3) (x * y) * z = (x * z) * y .

(1.4) x * 0 = x.

(1.5) 0 * (x * y) = (0 * x) * (0 * y).

(1.6) x * (x * (x * y)) = x * y.

(1.7) x ≤ y implies x * z ≤ y * z and z * y ≤ z * x .

A non-empty subset S of a BCI-algebra X is called a BCI-sub-algebra of X, if x * y ∈ S whenever x, y ∈ S. Also a non-empty subset I of a BCI-algebra X is called a BCI-ideal if

(i) 0 ∈ I,

(ii) x * y ∈ I and y ∈ I imply x ∈ I for all x, y ∈ X.

Let I be an ideal of a BCI-algebra X. Then the relation θ defined by (x, y) ∈ θ if and only if x * y ∈ I and y * x ∈ I is a congruence relation on X. Let C_x denote the class of x ∈ X, then C₀ = I. In fact, it is the greatest closed ideal contained in I. Assume that X/I = {C_x : x ∈ X}. Then (X/I, * , C₀) is a BCI-algebra, where C_x * C_y = C_x*y for all x, y ∈ X, also C_x ≤ C_y if and only if x ≤ y. Let (X, * , 0) and (Y, . , 0) be two BCI-algebras. A mapping f : X ⟶ Y is called a homomorphism from X to Y if for any x, y ∈ X, f (x * y) = f (x) . f (y) holds. Assume that {X_i : i ∈ I} is a BCI-algebraic family. Denote $\prod_{i \in I} X_{i}$ for the Cartesian product of {X_i : i ∈ I}. Define operation * on $\prod_{i \in I} X_{i}$ by {x_i} _i∈I * {y_i} _i∈I = {x_i * y_i} _i∈I. It is easily seen that $(\prod_{i \in I} X_{i}, *, 0)$ is a BCI-algebra where 0 = {0_i} _i∈I.

Definition 1.2. [3 , 5–7] Let X be a BCK-algebra. Then

i) X is called bounded, if there exists the greatest element of X, which be denoted by 1, we denote 1 * x by Nx, In a bounded BCK-algebra X, we have NNx ≤ x.

ii) For a bounded BCK-algebra X, if an element x satisfies NNx = x, then x is called an involution.

iii) X is called involutory BCK-algebra, if NNx = x, for all x ∈ X .

Definition 1.3. [1 , 7] Let X be a BCI-algebra. Then

i) X is called p-semisimple BCI-algebra if $0 * (0 * x) = x,$ ii) X is called an implicative BCI-algebra if $(x * (x * y)) * (y * x) = y * (y * x),$ iii) X is called an associative BCI-algebra if $(x * y) * z = x * (y * z),$ iv) X is called commutative BCI-algebra if $x \leq y imply x = x \land y,$ v) X is called positive implicative BCI-algebra if it satisfies in property $(x * (x * y)) * (y * x) = x * (x * (y * (y * x))),$ vi) X is called weakly commutative BCI-algebra if it satisfies the following identity $(x * (x * y)) * (0 * (x * y)) = y * (y * x) .$ for all x, y, z ∈ X.

A BCI-algebra X is implicative if and only if it is both commutative and positive implicative.

In any commutative BCI-algebra X, for all x, y ∈ X the following statements hold.

(1.8) (y ∧ x) = y * (y * (y ∧ x)),

(1.9) (y ∧ x) * (x ∧ y) =0 * (x * y),

In any associative BCI-algebra X, for all x, y ∈ X the following statements hold:

(1.10) 0 * x = x,

(1.11) x * y = y * x,

In any p-semisimple BCI-algebra X, for all x, y ∈ X we have:

(1.12) x * (0 * y) = y * (0 * x),

(1.13) x * (x * y) = y,

(1.14) 0 * (y * x) = x * y.

2 Commutators in BCI-algebras

From now on, (X, * , 0) or simply X is a BCI-algebra, unless otherwise specified.

Definition 2.1. Let x₁, x₂ be elements of X . The element ((x₂ ∧ x₁) * (x₁ ∧ x₂)) * (0 * (x₁ * x₂) of X is called a pseudo-commutator of x₁ and x₂ of weight 2 and denoted by [x₁, x₂], i.e.,

(2.1) [x₁, x₂] = ((x₂ ∧ x₁) * (x₁ ∧ x₂)) * (0 * (x₁ * x₂))

If x₁ * x₂ = [x₁, x₂] * (x₂ * x₁) or x₂ * x₁ = [x₁, x₂] * (x₁ * x₂), then [x₁, x₂] is called the commutator of x₁ and x₂.

If x₁, x₂, x₃, . . . , x_n (n ≥ 2) be elements of X, we define

[x₁, x₂, . . . . . . , x_n] = [[x₁, . . . . , x_n-1] , x_n] as a pseudo-commutator of weight n.

Example 2.2. Let X = {0, 1, 2, . . . , n}. Define a binary operation “*” on X by $x * y = {\begin{matrix} 0 & if x \leq y \\ x & otherwise \end{matrix}$

Then (X, * , 0) is a BCI-algebra. In this algebra we have [1, 2] = ((2 ∧ 1) * (1 ∧ 2)) * (0 * (1 * 2)) = (1 * 0) *0 = 1 and [2, 1] = ((1 ∧ 2) * (2 ∧ 1)) * (0 * (2 * 1)) = ((2 * 2) * (1 * 0)) * (0 * 2) = (0 * 1) *0 = 0 *0 = 0. So [1, 2] ≠ [2, 1]. Also [1, 2, 4] = [[1, 2] , 4] = [1, 4] = ((4 ∧ 1) * (1 ∧ 4)) * (0 * (1 * 4)) = (1 * 0) *0 = 1 *0 = 1 and [n, n - 1, . . . , 1] = [[. . . [n, n - 1] , n - 2] , . . . , ] , 2] , 1] = [[. . . [0, n - 2] , . . . , ] , 2] , 1] = . . . = [[0, 2] , 1] = [0, 1] =0.

Example 2.3. Let X = {0, 1, 2, 3, 4}. Define “*” by the following table

$\begin{matrix} \begin{matrix} * & 0 & 1 & 2 & 3 & 4 \end{matrix} \\ \begin{matrix} 0 & 0 & 0 & 0 & 3 & 3 \end{matrix} \\ \begin{matrix} 1 & 1 & 0 & 1 & 4 & 3 \end{matrix} \\ \begin{matrix} 2 & 2 & 2 & 0 & 3 & 3 \end{matrix} \\ \begin{matrix} 3 & 3 & 3 & 3 & 0 & 0 \end{matrix} \\ \begin{matrix} 4 & 4 & 3 & 4 & 1 & 0 \end{matrix} \end{matrix}$

Then (X, * , 0) is a BCI-algebra. We have [x, y] =0 for all x, y ∈ X and it is not difficult to verify that 1 * 3 ≠ [1, 3] * (3 * 1) and 4 * 3 ≠ [4, 3] * (3 * 4) but 3 * 1 = [3, 1] * (1 * 3) and 3 * 4 = [3, 4] * (4 * 3), also x * y = [x, y] * (y * x), for another x, y ∈ X. So for all x, y ∈ X, [x, y] are commutators of x and y.

Obviously, any commutator is a pseudo-commutator but the converse is not always true. By Definition 2.1 and Example 2.2, there exist x and y such that [x, y] ≠ [y, x] in BCI-algebras, so this definition of commutators is not the commutators in the sense of usual commutators of groups theory. Thus the commutators defined in this paper essentially directed commutators.

Now, let us give serval useful properties of commutators in BCI-algebras.

Lemma 2.4. For any x, y ∈ X, we have

i) [x, x] =0,

ii) [0, x] =0,

iii) [x, 0] =0.

Proof. $\begin{matrix} i) [x, x] \\ = ((x \land x) * (x \land x)) * (0 * (x * x)) \\ = ((x * (x * x)) * (x * (x * x))) * (0 * 0) \\ = ((x * 0) * (x * 0)) * 0 = (x * x) * 0 \\ = 0 * 0 = 0 \\ ii) [0, x] \\ = ((x \land 0) * (0 \land x)) * (0 * (0 * x)) \\ = ((0 * (0 * x)) * (x * (x * 0))) * (0 * (0 * x)) \\ = ((0 * (0 * x)) * (x * x)) * (0 * (0 * x)) \\ = ((0 * (0 * x)) * 0) * (0 * (0 * x)) \\ = (0 * (0 * x)) * (0 * (0 * x)) = 0 \end{matrix}$ $\begin{matrix} iii) [x, 0] \\ = ((0 \land x) * (x \land 0)) * (0 * (x * 0)) \\ = ((x * (x * 0)) * (0 * (0 * x))) * (0 * (x * 0)) \\ = ((x * x) * (0 * (0 * x))) * (0 * (x * 0)) \\ = (0 * (0 * (0 * x))) * (0 * x) \\ = (0 * x) * (0 * x) = 0 □ \end{matrix}$

Lemma 2.5. For any x, y ∈ BCK (X), we have

i) if y ≤ x, then [x, y] =0,

ii) [x, x * y] =0,

iii) [x * y, x] =0,

iv) [x, y] ≤ x,

v) [x, y] ≤ y,

vi) [x, y] = y if and only if y = 0.

Proof. i) Let y ≤ x. Then y * x = 0, therefore $\begin{matrix} [x, y] & = & ((y \land x) * (x \land y)) * (0 * (x * y)) \\ = & ((x * (x * y)) * (y * (y * x))) * (0 * (x * y)) \\ = & ((x * (x * y)) * (y * 0)) * 0 \\ = & (x * (x * y)) * y = 0 \end{matrix}$ $\begin{matrix} ii) [x, x * y] & = & (((x * y) \land x) * (x \land (x * y))) \\ * (0 * (x * (x * y))) \\ = & ((x * (x * (x * y))) * ((x * y) \\ * ((x * y) * x))) * 0 \\ = & ((x * y) * ((x * y) * 0)) * 0 \\ = & (x * y) * (x * y) = 0 \\ iii) [x * y, x] & = & ((x \land (x * y)) * ((x * y) \land x)) \\ * (0 * ((x * y) * x)) \\ = & (((x * y) * ((x * y) * x)) \\ * (x * (x * (x * y)))) * (0 * 0) \\ = & ((x * y) * 0) * (x * y) \\ = & (x * y) * (x * y) = 0 \\ iv) [x, y] * x & = & (((y \land x) * (x \land y)) * (0 * (x * y))) * x \\ = & ((y \land x) * x) * (x \land y) \\ = & ((y \land x) * y) * (x \land y) \\ = & 0 * (x \land y) = 0 \end{matrix}$

Therefore, [x, y] ≤ x. $\begin{matrix} v) [x, y] * y \\ = (((y \land x) * (x \land y)) * (0 * (x * y))) * y \\ = ((y \land x) * y) * (x \land y) \\ = 0 * (x \land y) = 0 \end{matrix}$

Therefore, [x, y] ≤ y.

vi) If [x, y] = y, then [x, y] = y ≤ y ∧ x ≤ y. Thus y ∧ x = y, whence y * x = (y ∧ x) * x = (y ∧ x) * y = 0. So, y ≤ x. But in this case [x, y] =0. Hence y = 0. The converse statement is obvious. □

Lemma 2.6. Let X be a bounded BCK-algebra. Then for any x ∈ X

i) [1, x] =0,

ii) [x, 1] = x * NNx.

Proof. i) Since x ≤ 1, for all x ∈ X, then [1, x] =0 by use part (iii) Lemma 2.5. $\begin{matrix} ii) [x, 1] & = & ((1 \land x) * (x \land 1)) * (0 * (x * 1)) \\ = & ((x * (x * 1)) * (1 * (1 * x))) * 0 \\ = & (x * 0) * (1 * (1 * x)) \\ = & x * NNx □ \end{matrix}$

Theorem 2.7. X is involutory BCK-algebra if and only if [x, 1] =0, for all x ∈ X.

Proof. Let X be a involutory BCK-algebra. Then for any x ∈ X, we have [x, 1] = x * NNx = x * x = 0.

Conversely, let [x, 1] =0, for all x ∈ X . Since [x, 1] = x * NNx = 0, therefore x ≤ NNx, but NNx ≤ x. Thus NNx = x, for all x ∈ X. Hence X is a involutory BCK-algebra. □

3 Commutator of sub-algebras

In this section, we introduce the notion of the commutator of two subsets of BCK-algebras.

Definition 3.1. Let X₁, X₂ be non-empty subsets of X. Define the pseudo-commutator of X₁ and X₂ to be ${[x_{1}, x_{2}] : x_{1} \in X_{1}, x_{2} \in X_{2}}$

Also, the subset {[x₁, x₂] |x₁, x₂ ∈ X} of X is called the derived subset of X.

Example 3.2. Let X = {0, a, b, c, d} and (*) operation be given by the following table

$\begin{matrix} * & 0 & a & b & c & d \\ 0 & 0 & 0 & 0 & 0 & 0 \\ a & a & 0 & 0 & 0 & 0 \\ b & b & b & 0 & 0 & 0 \\ c & c & b & a & 0 & 0 \\ d & d & d & d & d & 0 \end{matrix}$

Then (X, * , 0) is a BCI-algebra. Consider X₁ = {a, b, c} and X₂ = {d}. It is easy to check that a pseudo-commutator of subsets of X₁ and X₂ is {a, b, c}, which is not a sub-algebra of X.

In general, the pseudo commutators of the subsets of X are not a sub-algebra of X. Because the product of two commutators need not be a commutator. To resolve this problem we define the commutator as follows.

Definition 3.3. Let X₁, X₂, . . . , X_n be non-empty subsets of X. Define a commutator of X₁ and X₂ to be $[X_{1}, X_{2}] = {\prod [x_{1}, x_{2}] : x_{1} \in X_{1}, x_{2} \in X_{2}}$

More generally, for n ≥ 2 $[X_{1}, . . ., X_{n}] = [[X_{1}, . . ., X_{n - 1}], X_{n}]$ [X, X] is a product of a finite number of elements [x, y] such that x, y ∈ X and is called derived sub-algebra of X and is denoted by X′.

X′ = [X, X] = {∏ [x_i, y_i] : x_i, y_i ∈ X}

Generally, [X₁, X₂] ≠ [X₂, X₁]. Since for any x ∈ X, [x, 0] = [0, x] = [x, x] =0, for any two non-empty sub-algebras X₁, X₂ of X, 0 ∈ [X₁, X₂], that is, 0 ∈ X′.

Example 3.4. Let X = {0, a, b, c, d} and (*) operation be given by the following table

$\begin{matrix} * & 0 & a & b & c & d \\ 0 & 0 & 0 & 0 & 0 & 0 \\ a & a & 0 & a & 0 & a \\ b & b & b & 0 & 0 & 0 \\ c & c & c & c & 0 & c \\ d & d & d & b & b & 0 \end{matrix}$

It is not difficult to verify that (X, * , 0) is a solvable BCI-algebra. Consider A = {0, a} and B = {0, c}. Then A and B are sub-algebras of X. It is easy to check that [A, B] = {0, a} and [B, A] = {0}, therefore [A, B] ≠ [B, A]. Now X′ = {0, a, b} is a sub-algebra of X, but X′ is not an ideal of X, because d * b = b ∈ X′ and b ∈ X′ but d ∉ X′.

Remark 3.5. In Example 3.4 if we put A = {0, a} and B = {0, b, c} we see that A and B are two sub-algebras of X such that [A, B] = {0, a} which is not sub-algebra of B.

Theorem 3.6. X is commutative if and only ifX′ = {0}.

Proof. Let X be a commutative BCI-algebraand x, y ∈ X. Then for any x, y ∈ X, [x, y] = ((y ∧ x) * (x ∧ y)) * (0 * (x * y)) = (0 * (x * y)) * (0 * (x * y)) =0. Thus, X′ is generated by 0. Therefore, X′ = {0}.

Conversely, let X′ = [X, X] = {0} and x ≤ y. Then for all x, y ∈ X, the commutator of x and y is [x, y] =0. Since x ≤ y, x * y = 0 and $\begin{matrix} [x, y] & = & ((y \land x) * (x \land y)) * (0 * (x * y)) \\ = & ((x * (x * y)) * (y * (y * x))) * (0 * (x * y)) \\ = & ((x * 0) * (y * (y * x))) * (0 * 0) \\ = & x * (y * (y * x)) = 0 \end{matrix}$

So, x ≤ y * (y * x), also we have (y * (y * x)) ≤ x. Thus, if x ≤ y, then x = y * (y * x) = x ∧ y . Hence X is a commutative BCI-algebra. □

Theorem 3.7.X′ is a sub-algebra ofX.

Proof. Let x, y ∈ X′. Then a_i, b_i, c_i, d_i, ∈ X exist for i = 1, 2, . . . , n such that x = ∏ [a_i, b_i] and y = ∏ [c_i, d_i]. Therefore, for e_i, f_i ∈ X (i = 1, 2, . . . , n), x ∗ y = (∏ [a_i, b_i]) ∗ (∏ [c_i, d_i]) = ∏ [e_i, f_i] ∈ X′, then x * y ∈ X′ and X′ is a subalgebra of X. □

Corollary 3.8. If X ⊆ Y, then X′ ⊆ Y′.

Proof. Let t ∈ X′. Then there exist a_i, b_i ∈ X such that t = ∏ [a_i, b_i], since a_i, b_i ∈ X ⊆ Y, therefore a_i, b_i ∈ Y and t = ∏ [a_i, b_i], that is, t ∈ Y′. Thus, X′ ⊆ Y′. □

4 Solvable BCI-algebras

Definition 4.1. Let C¹ (X) = X, C² (X) = [X, X] , . . . , C^k (X) = [C^k-1 (X) , X] and C₁ (X) = X, C₂ (X) = [X, X] , . . . , C_k (X) = [C_k-1 (X) , C_k-1 (X)]. X is called to be solvable if their exists m ∈ N such that C_m (X) = {0}.

Example 4.2. Let X = [0, 1]. Define a binary operation “*” on X by $x * y = {\begin{matrix} 0 & if x \leq y \\ x & otherwise \end{matrix}$

Then (X, * , 0) is a BCI-algebra. Let X₁ = [0, 1], X₂ = [0, 1/2] , X₃ = [0, 1/3] , . . . , X_n = [0, 1/n]. Then [X₁, X₂] = [0, 1/2] and [X₂, X₁] = [0, 1/2).Therefore [X₁, X₂] ≠ [X₂, X₁]. [X₁, X₂, . . . , X_n] = [0, 1/n]. Also C¹ (X) = X = [0, 1] , C² (X) = [X, X] = [0, 1] , . . . , C^k (X) = [C^k-1 (X) , X] = [0, 1] , for any k ∈ N and C₁ (X) = X = C₂ (X) = [X, X] = . . . = C_k (X) = [C_k-1 (X) , C_k-1 (X)] = [0, 1] .

Theorem 4.3. a) Let X be an associative BCI-algebra. Then X′ = {0}, i.e, X is solvable.

b) Let X be a p-semisimple BCI-algebra. Then X′ = {0}, i.e, X is solvable.

Proof. a) Let X be an associative BCI-algebra. Then for all x, y ∈ X, we have $\begin{matrix} [x, y] & = & ((x * (x * y)) * (y * (y * x))) * (0 * (x * y)) \\ = & (((x * x) * y) * ((y * y) * x)) * (0 * (x * y)) \\ = & ((0 * y) * (0 * x)) * (0 * (x * y)) \\ = & (y * x) * (y * x) = 0 \end{matrix}$

Thus X′ = {0}. b)Let X be a p-semisimple BCI-algebra. Then for all x, y ∈ X we have $\begin{matrix} [x, y] & = & ((x * (x * y)) * (y * (y * x))) * (0 * (x * y)) \\ = & (y * x) * (0 * (x * y)) \\ = & (y * x) * (y * x) = 0 \end{matrix}$

Thus [x, y] =0. So X′ = {0}. □

Theorem 4.4. Let X be a weakly commutative BCI-algebra. Then X′ = {0}, i.e, X is solvable.

Proof. Let X be a weakly BCI-algebra. Then for all x, y ∈ X we have $\begin{matrix} [x, y] & = & ((x * (x * y)) * (y * (y * x))) * (0 * (x * y)) \\ = & ((x * (x * y)) * (((x * (x * y)) \\ * (0 * (x * y)))) * (0 * (x * y)) \end{matrix}$

So [x, y] =0, for all x, y ∈ X. Hence X′ = {0}. □

By the following examples, it could be shown that the converse of Theorems 4.3 and 4.4 are not true.

Example 4.5. Let X = {0, 1, 2, 3, 4}. We define a binary operation (*) on X by the following table

$\begin{matrix} * & 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 4 & 3 \\ 1 & 1 & 0 & 0 & 4 & 0 \\ 2 & 2 & 2 & 0 & 4 & 0 \\ 3 & 3 & 3 & 3 & 0 & 0 \\ 4 & 4 & 4 & 4 & 3 & 0 \end{matrix}$

It is not difficult to verify that (X, * , 0) is a solvable BCI-algebra. However it is not an associative BCI-algebra, since 2 * (1 * 4) =2 * 3 =4 ≠ (2 * 1) *4 = 2 *4 = 3. Moreover, X is not a p-semisimple BCI-algebra, because 0 * (0 * 2) =0 * 0 =0 ≠ 2.

Example 4.6. Let X = {0, 1, a, b, c}. Define (*) by the following table

$\begin{matrix} * & 0 & 1 & c & c & a \\ 0 & 0 & 0 & c & c & a \\ 1 & 1 & 0 & c & c & a \\ a & a & a & 0 & 0 & c \\ b & b & a & a & 0 & c \\ c & c & c & a & a & 0 \end{matrix}$

Then (X, * , 0) is a BCI-algebra. Also X is commutative, therefore X′ = {0}. So X is a solvable BCI-algebra, but X is not weakly commutative, since (1 * (1 * b)) * (0 * (1 * b)) =0 ≠ b * (b * 1) = b * a = 1.

Theorem 4.7. If Y is a sub-algebra of X, then Y⁽ⁿ⁾ is a sub-algebra of X⁽ⁿ⁾.

Proof. Let Y be a sub-algebra of X. We have C₁ (Y) = Y is a sub-algebra of C₁ (X) = X. Since Y is a sub-algebra of X, Y ⊆ X and by Corollary 3.8 Y′ ⊆ X′. But Y′ and X′ are two sub-algebras of X, so C₂ (Y) = Y′ is a sub-algebra of C₂ (X) = X′. Let C_k (Y) = Y^(k) be a sub-algebra of C_k (X) = X^(k). Therefore we have C_k+1 (Y) = Y^(k+1) = (Y^(k)) ′, i.e., a sub-algebra of (X^(k)) ′ = C_k+1 (X). □

Corollary 4.8. For any non-negative integer number n, X⁽ⁿ⁾ is a sub-algebra of X^(n-1).

Proof. Since X′ is sub-algebra of X and by Theorems 3.7 and 4.7, X″ is a sub-algebra of X′ and X^″′ is a sub-algebra of X″. By induction we show that X⁽ⁿ⁾ is a sub-algebra of X^(n-1). □

Example 4.9. Let X = {0, 1, 2, . . . , n - 1} and “*” is given by $x * y = {\begin{matrix} 0 & if x \leq y \\ x & otherwise \end{matrix}$

Then (X, * , 0) is a positive implicative BCI-algebra of order n.

X′ = {0, 1, 2, . . . , n - 2}, therefore X′ is a sub-algebra of X of order n - 1. Also X″ = {0, 1, 2,..., n − 3} is a sub-algebra of X′ and X⁽³⁾ = {0, 1, 2, . . . , n - 4} is a sub-algebra of X″,...,X^(n-1) = {0} is a sub-algebra of X^(n-2).

Lemma 4.10. Let A, B ⊆ X. Then

a) A′ ∪ B′ ⊆ (A ∪ B) ′ .

b) (A ∩ B) ′ ⊆ A′ ∩ B′ .

Proof. a) Since A, B ⊆ A ∪ B, by Corollary 3.8, we have A′ ⊆ (A ∪ B) ′ and B′ ⊆ (A ∪ B) ′. So A′ ∪ B′ ⊆ (A ∪ B) ′

b) Since A ∩ B ⊆ A, A ∩ B ⊆ B, (A ∩ B) ′ ⊆ A′ and (A ∩ B) ′ ⊆ B′. So (A ∩ B) ′ ⊆ A′ ∩ B′. □

Corollary 4.11. The intersection of any arbitrary family of sub-algebras of a solvable BCI-algebra is again a solvable BCI-algebra.

Proof. Let {X_i} _i∈I be the family of solvable sub-algebras of X. Since ∩X_i ⊆ X_i and X_i is solvable, there exists n ∈ N such that $X_{i}^{(n)} = {0}$ . Therefore, (∩ X_i) ⁽ⁿ⁾ = {0}. Hence ∩_i∈IX_i is a solvable BCI-algebra. □

We defined X as a solvable BCI-algebra, if there exists a non-negative integer n such that X⁽ⁿ⁾ = {0} . Note that, if X is a commutative BCI-algebra, then X is a solvable BCI-algebra. Since any implicative BCI-algebra X is a commutative BCI-algebra, X′ = {0} for any implicative BCI-algebra X, so any implicative BCI-algebra X is a solvable BCI-algebra.

Example 4.12. Let X = [0, 1]. Define a binary operation “*” on X by

$x * y = {\begin{matrix} 0 & if x \leq y \\ x - y & otherwise \end{matrix}$

Then (X, * , 0) is a BCI-algebra. We see that X′ = {0}. Therefore X is a solvable BCI-algebra. If we define $x * y = {\begin{matrix} 0 & if x \leq y \\ x & otherwise \end{matrix}$

Then (X, * , 0) is a BCI-algebra and X⁽ⁿ⁾ = [0, 1), for any n ≥ 1, so X is not a solvable BCI-algebra.

Example 4.13. Let X = {0, a, b, c, d}. Define a binary operation “*” on X by the following table

$\begin{matrix} * & 0 & a & b & c & d \\ 0 & 0 & 0 & 0 & d & c \\ a & a & 0 & 0 & d & c \\ b & b & b & 0 & d & c \\ c & c & c & c & 0 & d \\ d & d & d & d & c & 0 \end{matrix}$

It is not difficult to verify that (X, * , 0) is a BCI-algebra. It is easy to check that X′ = {0, a} and X″ = {0}. So X is a solvable BCI-algebra.

5 Connection between quotient BCI-algebras and commutators

Let f be a homomorphism from a BCI-algebra (X, * , 0) to a BCI-algebra (Y, * ′, 0′). Then $\begin{matrix} f ([x, y]) \\ = f (((y \land x) * (x \land y)) * (0 * (x * y))) \\ = (f (y \land x) *^{'} f (x \land y)) *^{'} f (0 * (x * y)) \\ = (f (x * (x * y)) *^{'} f (y * (y * x))) \\ *^{'} f (0 * (x * y)) \\ = ((f (x) *^{'} f (x * y)) *^{'} (f (y) *^{'} f (y * x))) \\ *^{'} (f (0) *^{'} f (x * y)) \\ = ((f (x) *^{'} (f (x) *^{'} f (y))) *^{'} (f (y) \\ *^{'} (f (y) *^{'} f (x)))) *^{'} (f (0) *^{'} (f (x) *^{'} f (y))) \\ = ((f (x) \land^{'} f (y)) *^{'} (f (y) \land^{'} f (x))) \\ *^{'} (f (0) *^{'} (f (x) *^{'} f (y))) \\ = [f (x), f (y)] \end{matrix}$ i.e., for any x, y ∈ X if [x, y] ∈ X′, then f ([x, y]) = [f (x) , f (y)] ∈ Y′.

Theorem 5.1. Let I be an ideal of X. Then X/I is a commutative BCI-algebra if and only if X′ ⊆ I.

Proof. Let X/I be a commutative BCI-algebra and x ∈ X′. Then there exist a_i, b_i ∈ X such that x = ∏ [a_i, b_i] = [a, b]. Therefore $\begin{matrix} C_{x} & = & C_{[a, b]} \\ = & C_{((b \land a) * (a \land b)) * (0 * (a * b))} \\ = & (C_{(b \land a)} * C_{(a \land b)}) * C_{0 * (a * b)} \\ = & (C_{a * (a * b)} * C_{b * (b * a)}) * C_{0 * (a * b)} \\ = & ((C_{a} * (C_{a} * C_{b})) * (C_{b} * (C_{b} * C_{a}))) \\ * (C_{0} * (C_{a} * C_{b})) \\ = & C_{0} = I \end{matrix}$

Thus x ∼ 0 and so x * 0, 0 * x ∈ I, hence x ∈ I . Thus, X′ ⊆ I.

Conversely, let X′ ⊆ I and C_x, C_y ∈ X/I such that C_x ≤ C_y. We show that C_x = C_x ∧ C_y = C_x∧y. Since C_x ≤ C_y, x ≤ y, then x * y = 0. X′ ⊆ I implies that [x, y] ∈ I, for all x, y ∈ X. Hence $\begin{matrix} C_{[x, y]} & = & C_{0} = C_{((y \land x) * (x \land y)) * (0 * (x * y))} \\ = & (C_{y \land x} * C_{x \land y}) * C_{0 * (x * y)} \\ = & (C_{x * (x * y)} * C_{x \land y}) * (C_{0 * (x * y)}) \\ = & (C_{x * 0} * C_{x \land y}) * (C_{0 * 0}) \\ = & (C_{x} * C_{x \land y}) * C_{0} = C_{0} \end{matrix}$

Then C_x * C_x∧y = C₀, that is, C_x ≤ C_x∧y. Since (X/I, * , C₀) is a BCI-algebra, C_x∧y = C_y*(y*x) = C_y * (C_y * C_x) ≤ C_x. Therefore C_x∧y = C_x. Hence X/I is a commutative BCI-algebra. □

Theorem 5.2. Let f be an isomorphism from a BCI-algebra X to a BCI-algebra Y . X is a solvable BCI-algebra if and only if Y is a solvable BCI-algebra.

Proof. Since f (X) = Y, f (X′) = Y′, because if y ∈ f (X′) , then there exists x ∈ X′ such that f (x) = y. Since x ∈ X′, there exist a_i, b_i ∈ X such that x = ∏ [a_i, b_i]. Consequently f (x) = f (∏ [a_i, b_i]) = prodf [a_i, b_i] = ∏ [f (a_i) , f (b_i)] = y ∈ Y′. So f (X′) ⊆ Y′.

Conversely, let h ∈ Y′. Then there exist h_i, k_i ∈ Y such that h = ∏ [h_i, k_i]. Since f is an isomorphism, there exist a_i, b_i ∈ X such that h_i = f (a_i) , k_i = f (b_i). Hence h = ∏ [h_i, k_i] = ∏ [f (a_i) , f (b_i)] = ∏f [a_i, b_i] = f (∏ [a_i, b_i]) ∈ f (X′) . By induction, we can show that f (X⁽ⁿ⁾) = Y⁽ⁿ⁾ . Since X is a solvable BCI-algebra, there exists n ∈ N such that X⁽ⁿ⁾ = {0}. Therefore {0} = f ({0}) = f (X⁽ⁿ⁾) = Y⁽ⁿ⁾ . Hence Y is a solvable BCI-algebra.

Conversely, let Y be a solvable BCI-algebra. Since f (X) = Y, f (X⁽ⁿ⁾) = Y⁽ⁿ⁾ = {0}, therefore f (X⁽ⁿ⁾) = {0} = f ({0}). So X⁽ⁿ⁾ = {0}, that is X is a solvable BCI-algebra. □

In the above theorem if f is an epimorphism from X to BCI-algebra Y and X is a solvable BCI-algebra, then Y is a solvable BCI-algebra. Indeed, f (X) = Y implies that f (X⁽ⁿ⁾) = Y⁽ⁿ⁾ . Since X is solvable, there exists m ∈ N such that X^(m) = {0}. Therefore Y^(m) = f (X^(m)) = f ({0}) = {0} . Hence Y is a solvable BCI-algebra.

Theorem 5.3. If I is an ideal of X and A, B ⊆ X, then [A/I, B/I] = [A, B]/I.

Proof. If t ∈ [A, B]/I, then there exists x ∈ [A, B] such that t = C_x. Therefore there exist a_i ∈ A, b_i ∈ B such that x = ∏ [a_i, b_i]. it shows that $\begin{matrix} t & = & C_{x} = C_{\prod [a_{i}, b_{i}]} = \prod C_{[a_{i}, b_{i}]} \\ = & \prod C_{((b_{i} \land a_{i}) * (a_{i} \land b_{i})) * (0 * (a_{i} * b_{i}))} \\ = & \prod ((C_{(b_{i} \land a_{i})} * C_{(a_{i} \land b_{i})}) * C_{0 * (a_{i} * b_{i})}) \\ = & \prod ((C_{a_{i} * (a_{i} * b_{i})} * C_{b_{i} * (b_{i} * a_{i})}) * C_{0 * (a_{i} * b_{i})}) \\ = & \prod (((C_{a_{i}} * (C_{a_{i}} * C_{b_{i}})) * (C_{b_{i}} * (C_{b_{i}} \\ * C_{a_{i}}))) * (C_{0} * (C_{a} * C_{b}))) \\ = & \prod ((C_{b_{i}} \land C_{a_{i}}) * (C_{a_{i}} \land C_{b_{i}})) \\ * (C_{0} * (C_{a i} * C_{b i}))) \\ = & \prod [C_{a_{i}}, C_{b_{i}}] \in [A / I, B / I] \end{matrix}$

Therefore t ∈ [A/I, B/I], hence [A, B]/I ⊆ [A/I, B/I]. If t ∈ [A/I, B/I], then there exist C_{a
_i} ∈ A/I and C_{b
_i} ∈ B/I such that a_i ∈ A, b_i ∈ B and $\begin{matrix} t & = & \prod [C_{a_{i}}, C_{b_{i}}] \\ = & \prod (((C_{b_{i}} \land C_{a_{i}}) * (C_{a_{i}} \land C_{b_{i}})) \\ * (C_{0} * (C_{a_{i}} * C_{b_{i}}))) \\ = & \prod (((C_{a_{i}} * (C_{a_{i}} * C_{b_{i}})) * (C_{b_{i}} * (C_{b_{i}} * C_{a_{i}}))) \\ * (C_{0} * (C_{a_{i}} * C_{b_{i}}))) \\ = & \prod ((C_{(a_{i} * (a_{i} * b_{i})} * C_{b_{i} * (b_{i} * a_{i}))}) * C_{(0 * (a_{i} * b_{i}))}) \\ = & \prod (C_{((a_{i} * (a_{i} * b_{i})) * (b_{i} * (b_{i} * a_{i}))) * (0 * (a_{i} * b_{i}))}) \\ = & \prod C_{[a_{i}, b_{i}]} \\ = & C_{\prod [a_{i}, b_{i}]} \in [A, B] / I \end{matrix}$

Thus, [A/I, B/I] ⊆ [A, B]/I. So [A/I, B/I] = [A, B]/I. □

Theorem 5.4. The subalgebras of a solvable BCI-algebra also are solvable.

Proof. Let Y be a subalgebra of X. Then Y⁽ⁿ⁾ is a subalgebra of X⁽ⁿ⁾ for all n ∈ N. Since X is a solvable BCI-algebra, there exists m ∈ N such that X^(m) = {0} and so Y^(m) = {0} . Thus, Y is a solvable BCI-algebra. □

Theorem 5.5. Let I be an ideal of X. If I and X/I are solvable BCI-algebras, then X is a solvable BCI-algebra.

Proof. Let f be the natural homomorphism from X onto X/I. Since X/I is solvable, for some n ∈ N, f (X⁽ⁿ⁾) = (X/I) ⁽ⁿ⁾ = {I}. Hence X⁽ⁿ⁾ is a sub-algebra of Ker (f) = I. Since I is a solvable BCI-algebra, then by Theorem 5.4, X⁽ⁿ⁾ is solvable. Therefore there exists a positive integer k such that X^(n+k) = (X⁽ⁿ⁾) ^(k) = {0}. Hence X is solvable. □

Remark 5.6. Let (X, * , 0) and (Y, * ′, 0′) be two algebras of type (2, 0) and X be a BCI-algebra. If f : X ⟶ Y be a homomorphism from X to Y, then f (X′) = (f (X)) ′ . Because if y ∈ f (X′) , then there exists x ∈ X′ such that f (x) = y. Since x ∈ X′, there exist a_i, b_i ∈ X such that x = ∏ [a_i, b_i]. Consequently y = f (x) = f (∏ [a_i, b_i]) = ∏f [a_i, b_i] = ∏ [f (a_i) , f (b_i)] ∈ (f (X)) ′ . So f (X′) ⊆ (f (X)) ′ .

Conversely, let y ∈ (f (X)) ′ Then there exist h_i, k_i ∈ f (X) such that y = ∏ [h_i, k_i]. Since h_i, k_i ∈ f (X), there exist a_i, b_i ∈ X such that h_i = f (a_i) , k_i = f (b_i). Hence y = ∏ [h_i, k_i] = ∏ [f (a_i) , f (b_i)] = ∏f [a_i, b_i] = f (∏ [a_i, b_i]) ∈ f (X′) . So (f (X)) ′ ⊆ f (X′) . Hence (f (X)) ′ = f (X′) .

Also, (f(X))″ = ((f(X))′)′ = (f(X′))′ = f(X″). By induction, we can show that f (X⁽ⁿ⁾) = (f (X)) ⁽ⁿ⁾ .

Theorem 5.7. Let (X_i, * _i, 0_i) be an indexed family of BCI-algebras. Then $\prod_{i \in I} X_{i}$ is solvable if and only if X_i is solvable, for all i ∈ I.

Proof. If $\prod_{i \in I} X_{i}$ is solvable, then there exists n ∈ N such that $(\prod_{i \in I} X_{i})^{(n)} = {0} .$ The project mapping $P : \prod_{i \in I} X_{i} \to X_{i}$ given by P ((x_i) _i∈I) = x_i is an epimorphism. So, $P (\prod_{i \in I} X_{i}) = X_{i}$ and by use Remark 5.6 $X_{i}^{(n)} = (P (\prod_{i \in I} X_{i}))^{(n)} = (P (\prod_{i \in I} X_{i})^{(n)}) = P ({0}) = {0_{i}}$ . Hence X_i is solvable.

Conversely, let X_i be solvable. If $X_{i}^{'} = {0}$ , then we show that $(\prod_{i \in I} X_{i})^{'} = {0}$ . For proof this problem, let $X_{i}^{'} = {0}$ and $y \in (\prod_{i \in I} X_{i})^{'}$ . Then there exist sequences a = (a_i) _i∈I = (a₁, a₂, . . . .) and b = (b_i) _i∈I = (b₁, b₂, . . . .) in $\prod_{i \in I} X_{i}$ such that $\begin{matrix} t & = & [(a_{i}), (b_{i})] \\ = & ((((a_{i}) * ((a_{i}) * (b_{i}))) * ((b_{i}) * ((b_{i}) * (a_{i})))) \\ * ((0_{i}) * ((a_{i}) * (b_{i})))) \\ = & ((((a_{1}, a_{2}, . . .) * ((a_{1}, a_{2}, . . .) * (b_{1}, b_{2}, . . .))) \\ * ((b_{1}, b_{2}, . . .) * ((b_{1}, b_{2}, . . .) * (a_{1}, a_{2}, . . .)))) \\ * ((0_{1}, 0_{2}, . . .) * ((a_{1}, a_{2}, . . .) * (b_{1}, b_{2}, . . .)))) \\ = & (((((a_{1}) * (a_{1} * b_{1})) * ((b_{1}) * (b_{1} * a_{1}))) \\ * ((0_{1}) * (a_{1} * b_{1}))), (((a_{2}) * (a_{2} * b_{2})) \\ * ((b_{2}) * (b_{2} * a_{2}))) * ((0_{2}) * (a_{2} * b_{2}))), . . .) \\ = & ([a_{1}, b_{1}], [a_{2}, b_{2}], . . ., [a_{i}, b_{i}], . . .) \\ = & (0_{1}, 0_{2}, . . .) = (0)_{i} = 0 \end{matrix}$

Thus, y = 0 and hence $(\prod_{i \in I} X_{i})^{'} = {0}$ . By induction we show that if $X_{i}^{(n)} = {0}$ , then $(\prod_{i \in I} X_{i})^{(n)} = {0}$ . So $\prod_{i \in I} X_{i}$ is solvable. □

By the following example we show that the homomorphic image of a solvable BCI-algebra it may not be a BCI-algebra.

Example 5.8. [2] For the Wronskis algebra (X, * , 0), X = N ∪ A ∪ B in which N is the set of nonnegative integers, A = {a_n : n ∈ N} and B = {b_n : n ∈ N} and each pair of the sets N, A, B are disjoint, we define a binary operation “*” on X as follows: for any m, n ∈ N, $\begin{matrix} m * n = \max {0, m - n}, \\ m * a_{n} = m * b_{n} = 0, \\ a_{m} * n = a_{m + n}, \\ b_{m} * n = b_{m + n}, \\ a_{m} * a_{n} = b_{m} * b_{n} = n * m, \\ a_{m} * b_{n} = b_{m} * a_{n} = (n + 1) * m . \end{matrix}$

Then (X, * , 0) is a BCI-algebra with BCI-ordering $\begin{matrix} 0 \leq 1 \leq 2 \leq . . . \leq a_{2} \leq a_{1} \leq a_{0} \\ 0 \leq 1 \leq 2 \leq . . . \leq b_{2} \leq b_{1} \leq b_{0} \end{matrix}$

By ordinary calculations we see that X′ = {0, 1, 2}, and X″ = {0}. Therefore, X is a solvable BCI-algebra. Then the collection {N, A, B} determines a partition of X, thus we have a congruence relation θ on X where x ∼ _θy if and only if both of x and y are in one of the there sets N, A and B. We obtain a quotient algebra $(\frac{X}{θ}, *, N)$ , where $\frac{X}{θ} = {N, A, B}$ . Now, it is easy to calculate that the “*” multiplication table on $\frac{X}{θ}$ is

$\begin{matrix} * & N & A & B \\ N & N & N & N \\ A & A & N & N \\ B & B & N & N \end{matrix}$

Since A * B = B * A = N and A ≠ B, BCI3 does not hold for the quotient algebra $\frac{X}{θ}$ . The mapping $f : X \to \frac{X}{θ}$ , x ↦ θ_x is an BCI-epimorphism. However, as we have seen, $\frac{X}{θ} = f (X)$ is not a BCI-algebra. Thus, homomorphic image of solvable BCI-algebra X is not a solvable BCI-algebra.

Theorem 5.9. Solvable BCI-algebras do not form a variety.

6 Conclusion

For elements x and y of X, x ∧ y = y * (y * x), it is not difficult to see that x ∧ y is a lower bound of x and y. A BCI-algebra which is a lower semi-lattice with respect to ∧ is said to be commutative. We define a commutator of x and y by [x, y] = ((y ∧ x) * (x ∧ y)) * (0 * (x * y)). A BCI-algebra X is called solvable if there exists a natural number n such that X⁽ⁿ⁾ = {0}, where X′ = [X, X] is a product of a finite number of elements [x, y] such that x, y ∈ X and is called a derived subalgebra of X and X^(k+1) = [X^(k), X^(k)].

The results of this paper show that:

(1) In general, X′ is a sub-algebra of X but is not an ideal of X.

(2) the intersection of any arbitrary family of sub-algebras of a solvable BCI-algebra, is again a solvable BCI-algebra. But in general, the union of two solvable BCI-sub-algebras of a BCI-algebra may not be a solvable BCI-algebra.

(3) Sub-algebras and inverse images of solvable BCI-algebras also are solvable.

(4) X is commutative if and only if X′ = {0}.

(5) If Y is a subalgebra of X. Then Y⁽ⁿ⁾ is a sub-algebra of X⁽ⁿ⁾.

(6) if I, $\frac{X}{I}$ are solvable BCI-algebra, then X is also solvable BCI-algebra.

(7) [A/I, B/I] = [A, B]/I.

(8) X/I is a commutative BCI-algebra if and only if X′ ⊆ I.

(9) $\prod_{i \in I} X_{i}$ is solvable if and only if X_i is solvable for all i ∈ I.

Footnotes

Acknowledgments

The authors would like to express their gratitude to anonymous referees for their comments and suggestions which improved the paper.

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