Approximations in left almost polygroups

Abstract

In this paper, we study the idea of rough approximations in LA-polygroups, and present some examples in this respect. In particular, we study some properties of rough LA-subpolygroup which is an extension of LA-subpolygroup. We prove that the lower (upper) approximation of an LA-subpolygroup is an LA-subpolygroup. We consider the factor LA-polygroup P/H and interpret the lower and upper approximations as subsets of the factor LA-polygroup P/H. Then, we introduce the concept of factor rough LA-subpolygroups. Also, we provided some results on homomorphisms of rough LA-subpolygroups.

Keywords

LA-polygroups LA-subpolygroups Lower approximations Upper approximations

1 Introduction

The notion of rough set was introduced by Pawlak in his paper [27]. The theory of rough sets has emerged as another major mathematical approach for managing uncertainty that arises from inexact, noisy, or incomplete information. It is turning out to be methodologically significant to the domains of artificial intelligence and cognitive sciences, especially in the representation of and reasoning with vague and/or imprecise knowledge, data classification, data analysis, machine learning and knowledge discovery. Bonikowaski [2] studied algebraic properties of rough sets. Biswas and Nanda [1] gave the notion of rough subgroups. Goldhaber and Enrich [12] studied the homomorphism and isomorphism of rough groups. Dudek et al. [10] applied roughness to BCI-algebras. Dudek and Jun [11] studied rough subalgebras of some binary algebras connected with logics. Qurashi and Shabir [28, 29] introduced the concept of generalized roughness of ideals in quantales. Shabir et al. considered roughness in ordered semigroups [30] and ternary semigroups [31]. Wang and Chen [33] presented a short note on some properties of rough groups. Yaqoob et al. [35 –37] applied rough set theory to Γ-hyperideals in left almost Γ-semihypergroups and to bi-Γ-hyperideals of Γ-semihypergroups.

The concept of a hypergroup is a generalization of group, first was introduced by Marty [20]. Nowadays hyperstructures are widely studied from theoretical point of view and for their applications in many subjects of pure and applied mathematics. In a classical algebraic structure, the composition of two elements is an element, while in an algebraic hyperstructure, the composition of two elements is a set. The principal notions of hypergroup theory can be found in [5, 32]. Polygroup is a special class of hypergroup which have many applications. For example, polygroups which are certain subclasses of hypergroups are studied by Ioulidis in [15] and are used to study colour algebra [3, 4]. Davvaz and Alp [6] introduced the idea of Cat¹-polygroup and pullback cat¹-polygroup. The concept of topological polygroups is a generalization of the concept of topological groups which have been introduced in [13]. In [21], the concept of thin n-subpolygroups was given and the notion of wreath product of n-polygroups was studied. A book written by Davvaz [7] is devoted especially to the study of polygroups.

A left almost semigroup (LA-semigroup) or an Abel-Grassmann’s groupoid (AG-groupoid) has been investigated in several papers. The left almost semigroup, was first introduced by Kazim and Naseerudin [16]. They introduced braces on the left of the ternary commutative law abc = cba, to get a new pseudo associative law, that is (ab) c = (cb) a, and named it as left invertive law. Later, Mushtaq and others [22 –25] investigated the structure and added many useful results to the theory of LA-semigroups. Mushtaq and Kamran [26] introduced the concept of left almost groups. The work on left almost hyperstructures starts with the notion of left almost semihypergroups (abbreviated as LA-semihypergroups) defined by Hila and Dine [14] in 2011. In [34], the authors defined the concept of left almost polygroups and provided several examples. They also discussed the quotient structure and isomorphism theorems for left almost polygroups.

There are several authors who applied the rough set theory to algebraic hyperstructures, for instance, Davvaz applied roughness to polygroups [8, 9]. Fotea [17, 18] applied roughness to hypergroups.

In this paper, we combine the rough set theory and LA-polygroups. We provide some properties of rough LA-subpolygroups which are an extension of LA-subpolygroups. We proved some results on rough LA-subpolygroups in a factor LA-polygroup and homomorphisms of rough LA-subpolygroups.

2 Left almost polygroups

In this section, we shall recall some central definitions and useful elementary results concerning LA-polygroups, which will be helpful for our further studies.

Definition 2.1. [34] A multivalued system 〈P, ∘, e, ⁻¹〉, where e ∈ P, ^{- 1}: P ⟶ P and $\circ : P \times P ⟶ P^{*} (P)$ is called an LA-polygroup, if, for all x, y, z ∈ P, the following axioms hold:

there exists a left identity e ∈ P such that e ∘ x = x,

left invertive law: (x ∘ y) ∘ z = (z ∘ y) ∘ x,

e ∈ x ∘ x⁻¹ ∩ x⁻¹ ∘ x (we call x⁻¹ the inverse of x),

x ∈ y ∘ z ⇒ y ∈ x ∘ z⁻¹.

The following elementary facts about LA-polygroups follow easily from the above axioms: $e^{- 1} = e and (x^{- 1})^{- 1} = x .$

Example 2.2. Let P = {e, a, b, c} and the binary hyperoperation “∘” be defined as in the following table:

∘	e	a	b	c
e	e	a	b	c
a	b	{a, b, c}	{e, a, c}	{a, b}
b	a	{e, b, c}	{a, b, c}	{a, b}
c	c	{a, b}	{a, b}	{e, c}

Then 〈P, ∘, e, ⁻¹〉 is an LA-polygroup.

Lemma 2.3. In an LA-polygroup 〈P, ∘, e, ⁻¹〉 the following laws holds for all a, b, c, d ∈ P.

Medial law: (a ∘ b) ∘ (c ∘ d) = (a ∘ c) ∘ (b ∘ d),

a ∘ (b ∘ c) = b ∘ (a ∘ c),

Paramedial law: (a ∘ b) ∘ (c ∘ d) = (d ∘ c) ∘ (b ∘ a).

Definition 2.4. [34] A non-empty subset K of an LA-polygroup 〈P, ∘, e, ⁻¹〉 is called an LA-subpolygroup of P if, under the hyperoperation in P, K itself forms an LA-polygroup.

Lemma 2.5. [34] A non-empty subset K of the LA-polygroup 〈P, ∘, e, ⁻¹〉 is an LA-subpolygroup of it, if and only if the following relations are satisfied.

For all a, b ∈ K ⇒ a ∘ b ⊆ K.

For all a ∈ K ⇒ a⁻¹ ∈K.

Lemma 2.6. [34] If K is an LA-subpolygroup of the LA-polygroup 〈P, ∘, e, ⁻¹〉, then, for every a, b ∈ P, we have:

K = K ∘ K.

e ∘ K = K ∘ e = K.

a ∘ K = (K ∘ a) ∘ e,

(a ∘ b) ∘ K = K ∘ (b ∘ a).

It is important to note that there is no concept of polygroup theoretic normality in LA-polygroups, meaning that we can factor an LA-polygroup by any of its LA-subpolygroups. We know that if 〈P, ∘, e, ⁻¹〉 is a polygroup and K is its subpolygroup, then (K ∘ a) ∘ (K ∘ b) ≠ K ∘ (a ∘ b), unless K is normal in P. But for LA-polygroups we have no such condition because of the medial property, that is, if K ∘ a, K ∘ b belong to P/K, then by using the medial law $\begin{matrix} (K \circ a) \circ (K \circ b) & = & (K \circ K) \circ (a \circ b) \\ = & K \circ (a \circ b), \end{matrix}$ without having an extra condition on P.

Remark 2.7. [34] An LA-polygroup can be partitioned only into right cosets (or left cosets) and we do not require the two side decomposition.

Theorem 2.8. [34] If 〈P, ∘, e, ⁻¹〉 is an LA-polygroup and K is an LA-subpolygroup of P, then $P / K = {K \circ a ∣ a \in P}$ is an LA-polygroup, too.

3 Rough approximations in LA-polygroups

In this section, we will present some results related to roughness in LA-polygroups.

Definition 3.1. Let 〈P, ∘, e, ⁻¹〉 be an LA-polygroup, and let H be an LA-subpolygroup of P. Let A be a non-empty subset of P. Then the sets $\begin{matrix} \underline{Apr} (H, A) & = & {x \in P : H \circ x \subseteq A} \\ and \bar{Apr} (H, A) & = & {x \in P : (H \circ x) \cap A \neq \emptyset}, \end{matrix}$ are called, respectively, lower and upper approximations of a set A with respect to the LA-subpolygroup H.

For a non-empty subset A of P, $Apr (H, A) = (\underline{Apr} (H, A), \bar{Apr} (H, A))$ is called a rough set with respect to H or simply a rough subset of ℘ (P) × ℘ (P) if $\underline{Apr} (H, A) \neq \bar{Apr} (H, A),$ where ℘ (P) denotes the power set of P.

Example 3.2. Consider Example 1, and let A = {a, b, c} and H = {e, c}. Then $\underline{Apr} (H, A) = {a, b} and \bar{Apr} (H, A) = {e, a, b, c} .$

Proposition 3.3. Let H and K be LA-subpolygroups of an LA-polygroup P. Let A and B be any non-empty subsets of P. Then

$\underline{Apr} (H, A) \subseteq A \subseteq \bar{Apr} (H, A);$

$\bar{Apr} (H, A \cup B) = \bar{Apr} (H, A) \cup \bar{Apr} (H, B);$

$\underline{Apr} (H, A \cap B) = \underline{Apr} (H, A) \cap \underline{Apr} (H, B);$

A ⊆ B implies $\underline{Apr} (H, A) \subseteq \underline{Apr} (H, B);$

A ⊆ B implies $\bar{Apr} (H, A) \subseteq \bar{Apr} (H, B);$

$\underline{Apr} (H, A \cup B) \supseteq \underline{Apr} (H, A) \cup \underline{Apr} (H, B);$

$\bar{Apr} (H, A \cap B) \subseteq \bar{Apr} (H, A) \cap \bar{Apr} (H, B);$

H ⊆ K implies $\bar{Apr} (H, A) \subseteq \bar{Apr} (K, A) .$

Proof. The proof is straightforward. □

Proposition 3.4. Let H be an LA-subpolygroup of an LA-polygroup P. Let A and B be non-empty subsets of P. Then $\bar{Apr} (H, A) \circ \bar{Apr} (H, B) = \bar{Apr} (H, A \circ B) .$

Proof. Let c be any element of $\bar{Apr} (H, A \circ B)$ . Then H∘ c ∩ A ∘ B ≠ ∅. Thus there exists an element x in P such that x ∈ H ∘ c ∩ A ∘ B, and so x ∈ H ∘ c and x ∈ A ∘ B. Then x ∈ a ∘ b with a ∈ A and b ∈ B. Since $\begin{matrix} c & \in & H \circ x \subseteq H \circ (a \circ b) \\ = & (H \circ a) \circ (H \circ b), \end{matrix}$ we have c ∈ y ∘ z with y ∈ H ∘ a and z ∈ H ∘ b. Then a ∈ H ∘ y, and so a ∈ H ∘ y ∩ A. Thus $y \in \bar{Apr} (H, A)$ . Similarly we have $z \in \bar{Apr} (H, B)$ . Thus $c \in y \circ z \subseteq \bar{Apr} (H, A) \circ \bar{Apr} (H, B),$ and so we have $\bar{Apr} (H, A \circ B) \subseteq \bar{Apr} (H, A) \circ \bar{Apr} (H, B)$ .

Conversely, let c be any element of $\bar{Apr} (H, A) \circ \bar{Apr} (H, B)$ , then c ∈ a ∘ b with $a \in \bar{Apr} (H, A)$ and $b \in \bar{Apr} (H, B)$ . Thus there exist elements x and y in P such that x ∈ H ∘ a ∩ A and y ∈ H ∘ b ∩ B, and so x ∈ H ∘ a, x ∈ A, y ∈ H ∘ b, and y ∈ B. Since H is an LA-subpolygroup, x ∘ y ⊆ (H ∘ a) ∘ (H ∘ b) = H ∘ (a ∘ b), and x ∘ y ⊆ A ∘ B. Thus x ∘ y ⊆ H ∘ (a ∘ b) ∩ A ∘ B, which yields that $c \in a \circ b \subseteq \bar{Apr} (H, A \circ B),$ and so $\bar{Apr} (H, A) \circ \bar{Apr} (H, B) \subseteq \bar{Apr} (H, A \circ B)$ . Therefore we have $\bar{Apr} (H, A) \circ \bar{Apr} (H, B) = \bar{Apr} (H, A \circ B) .$ □

Proposition 3.5. Let H be an LA-subpolygroup of an LA-polygroup P. Let A and B be non-empty subsets of P. Then $\underline{Apr} (H, A) \circ \underline{Apr} (H, B) \subseteq \underline{Apr} (H, A \circ B) .$

Proof. Let c be any element of $\underline{Apr} (H, A) \circ \underline{Apr} (H, B)$ . Then c ∈ a ∘ b with $a \in \underline{Apr} (H, A)$ and $b \in \underline{Apr} (H, B)$ . Thus H ∘ a ⊆ A and H ∘ b ⊆ B. Since H is an LA-subpolygroup, so $\begin{matrix} H \circ c & \subseteq & H \circ (a \circ b) = (H \circ a) \circ (H \circ b) \\ \subseteq & A \circ B, \end{matrix}$ and so $c \in \underline{Apr} (H, A \circ B)$ . Thus $\underline{Apr} (H, A) \circ \underline{Apr} (H, B) \subseteq \underline{Apr} (H, A \circ B)$ .

The following example shows that $\underline{Apr} (H, A \circ B) ⊈ \underline{Apr} (H, A) \circ \underline{Apr} (H, B) .$ □

Example 3.6. Let P ={ e, a, b, c, d } and (∘) be the binary hyperoperation in P defined by the following table:

∘	e	a	b	c	d
e	e	a	b	c	d
a	a	{e, a}	b	c	d
b	c	c	{e, a}	d	b
c	b	b	d	{e, a}	c
d	d	d	c	b	{e, a}

Clearly (∘) is non-associative as {b} = (b ∘ e) ∘ e ≠ b ∘ (e ∘ e) = {c}. Then P is an LA-polygroup with left identity e and the elements of P satisfy the left invertive law. Let H = {e, a} be an LA-subpolygroup of P. Let A = {e, b} and B = {c, d} be non-empty subsets of P. Now $\underline{Apr} (H, A) = {b}$ and $\underline{Apr} (H, B) = {c, d},$ thus $\underline{Apr} (H, A) \circ \underline{Apr} (H, B) = {b, d} .$ Now A ∘ B = {b, c, d} and $\underline{Apr} (H, A \circ B) = {b, c, d} .$ Hence it is clear that $\underline{Apr} (H, A) \circ \underline{Apr} (H, B) \subseteq \underline{Apr} (H, A \circ B)$ but $\underline{Apr} (H, A \circ B) ⊈ \underline{Apr} (H, A) \circ \underline{Apr} (H, B) .$

Remark 3.7. Let H and K be LA-subpolygroups of an LA-polygroup P. Then, as is well known and easily seen, H ∩ K is also an LA-subpolygroup of P.

Proposition 3.8. Let H and K be LA-subpolygroups of an LA-polygroup P. Let A be a non-empty subset of P. Then

$\bar{Apr} (H \cap K, A) = \bar{Apr} (H, A) \cap \bar{Apr} (K, A);$

$\underline{Apr} (H \cap K, A) = \underline{Apr} (H, A) \cap \underline{Apr} (K, A) .$

Proof. (1) Suppose that $\begin{matrix} \forall a \in \bar{Apr} (H \cap K, A) \\ \Leftrightarrow (H \cap K) \circ a \cap A \neq \emptyset \\ \Leftrightarrow \exists x \in (H \cap K) \circ a \cap A \\ \Leftrightarrow x \in (H \cap K) \circ a and x \in A \\ \Leftrightarrow x \in H \circ a, x \in A and x \in K \circ a, x \in A \\ \Leftrightarrow x \in H \circ a \cap A and x \in K \circ a \cap A \\ \Leftrightarrow x \in \bar{Apr} (H, A) and x \in \bar{Apr} (K, A) \\ \Leftrightarrow x \in \bar{Apr} (H, A) \cap \bar{Apr} (K, A) . \end{matrix}$ Thus $\bar{Apr} (H \cap K, A) = \bar{Apr} (H, A) \cap \bar{Apr} (K, A) .$

(2) Suppose that $\begin{matrix} \forall a \in \underline{Apr} (H \cap K, A) \\ \Leftrightarrow (H \cap K) \circ a \subseteq A \\ \Leftrightarrow H \circ a \subseteq A and K \circ a \subseteq A \\ \Leftrightarrow a \in \underline{Apr} (H, A) and a \in \underline{Apr} (K, A) \\ \Leftrightarrow a \in \underline{Apr} (H, A) \cap \underline{Apr} (K, A) . \end{matrix}$ Thus $\underline{Apr} (H \cap K, A) = \underline{Apr} (H, A) \cap \underline{Apr} (K, A)$ . □

4 Lower and upper rough LA-subpolygroups

Definition 4.1. A non-empty subset A of an LA-polygroup P is called an upper rough LA-subpolygroup of P if the upper approximation $\bar{Apr} (H, A)$ of A is an LA-subpolygroup of P.

Definition 4.2. A non-empty subset A of an LA-polygroup P is called a lower rough LA-subpolygroup of P if the lower approximation $\underline{Apr} (H, A)$ of A is an LA-subpolygroup of P.

Theorem 4.3. Let H be an LA-subpolygroup of an LA-polygroup P. If A is an LA-subpolygroup of P, then $\bar{Apr} (H, A)$ is an LA-subpolygroup of P.

Proof. Let a and b be any elements of $\bar{Apr} (H, A)$ . Then there exist elements x and y in P such that x ∈ H ∘ a ∩ A and y ∈ H ∘ b ∩ A. Thus x ∈ H ∘ a, y ∈ H ∘ b and x ∈ A, y ∈ A. Since A is an LA-subpolygroup of P, x ∘ y ⊆ A. And since H is an LA-subpolygroup of P, so $x \circ y \subseteq (H \circ a) \circ (H \circ b) = H \circ (a \circ b) .$

Thus x ∘ y ⊆ H ∘ (a ∘ b) ∩ A, and so $a \circ b \subseteq \bar{Apr} (H, A)$ . Therefore the first condition of Lemma 2.5 is satisfied.

Now we have to prove the second condition of Lemma 2.5. Let a be any element of $\bar{Apr} (H, A)$ . Then there exist x ∈ H ∘ a ∩ A for some x ∈ P, that is, x ∈ H ∘ a and x ∈ A. Then since A is an LA-subpolygroup of P, x⁻¹ ∈ A. On the other hand, since x ∈ h ∘ a for some h ∈ H, and since H is an LA-subpolygroup of P and h⁻¹ ∈ H, we have $x^{- 1} \in (h \circ a)^{- 1} = h^{- 1} \circ a^{- 1} \subseteq H \circ a^{- 1} .$

Thus x⁻¹ ∈ H ∘ a⁻¹ ∩ A, and so $a^{- 1} \in \bar{Apr} (H, A)$ . This implies that $\bar{Apr} (H, A)$ is an LA-subpolygroup of P.

Theorem 4.4. Let H be an LA-subpolygroup of an LA-polygroup P. If A is an LA-subpolygroup of P such that H ⊆ A, then $\underline{Apr} (H, A)$ is an LA-subpolygroup of P.

Proof. Let H be an LA-subpolygroup of an LA-polygroup P. Let a and b be any elements of $\underline{Apr} (H, A)$ . Then H ∘ a ⊆ A and H ∘ b ⊆ A. Since H and A are LA-subpolygroup of P such that H ⊆ A, so $\begin{matrix} H \circ (a \circ b) & = & (H \circ a) \circ (H \circ b) \\ \subseteq & A \circ A \subseteq A . \end{matrix}$ This implies that $a \circ b \subseteq \underline{Apr} (H, A)$ . Therefore the first condition of Lemma 2.5 is satisfied.

Now we have to prove the second condition of Lemma 2.5. Let a be any element of $\underline{Apr} (H, A)$ . Then a = e ∘ a ∈ H ∘ a ⊆ A. Since A is an LA-subpolygroup of P, a⁻¹ ∈ A. Thus we have H ∘ a⁻¹ ⊆ A ∘ A ⊆ A. This implies that $a^{- 1} \in \underline{Apr} (H, A)$ . Therefore $\underline{Apr} (H, A)$ is an LA-subpolygroup of P. □

The following example shows that the converse of Theorem 4.3 and Theorem 4.4 does not hold in general, i.e. $\bar{Apr} (H, A)$ and $\underline{Apr} (H, A)$ are LA-subpolygroup of P but A is not an LA-subpolygroup.

Example 4.5. Let P ={ e, a, b, c } and (∘) be the binary hyperoperation in P defined by the following table:

∘	e	a	b	c
e	e	a	b	c
a	b	{e, a, b}	{a, b, c}	a
b	a	{a, b, c}	{e, a, b}	b
c	c	b	a	e

Here P is an LA-polygroup. Let H = {e, c} be an LA-subpolygroup of P. Let A = {a, c} be a non-empty subset of P. Now

\bar{Apr} (H, A) = {e, a, b, c}

which is an LA-subpolygroup of P, but A is not an LA-subpolygroup of P. Similarly, let B = {e, a, c} be a non-empty subset of P. Then

\underline{Apr} (H, A) = {e, c}

which is an LA-subpolygroup of P, but A is not an LA-subpolygroup of P.

Remark 4.6. The product H ∘ K of LA-subpolygroups H and K of an LA-polygroup P is also an LA-subpolygroup of P.

Proposition 4.7. Let H and K be LA-subpolygroups of an LA-polygroup P. If A is an LA-subpolygroup of P, then $\bar{Apr} (H, A) \circ \bar{Apr} (K, A) \supseteq \bar{Apr} (H \circ K, A) .$

Proof. Let c be any element of $\bar{Apr} (H \circ K, A) .$ Then there exists element x in P such that x ∈ (H ∘ K) ∘ c ∩ A and so x ∈ (H ∘ K) ∘ c and x ∈ A. Now x ∈ (h ∘ k) ∘ c for h ∈ H and k ∈ K. Now by left invertive law and by Lemma 2.6(iv), we have $\begin{matrix} x & \in & (h \circ k) \circ c = (c \circ k) \circ h \\ \subseteq & (c \circ k) \circ H = H \circ (k \circ c) . \end{matrix}$

This implies that x ∈ H ∘ (k ∘ c) ∩ A. Thus $k \circ c \subseteq \bar{Apr} (H, A) .$ As K is an LA-subpolygroup so for any k ∈ K, we have $k^{- 1} \in K \subseteq \bar{Apr} (K, A) .$ So $c \in k^{- 1} \circ \bar{Apr} (H, A) \subseteq \bar{Apr} (K, A) \circ \bar{Apr} (H, A) .$ Therefore, we get $\bar{Apr} (H \circ K, A) \subseteq \bar{Apr} (H, A) \circ \bar{Apr} (K, A) .$ □

Corollary 4.8. Let H and K be LA-subpolygroups of an LA-polygroup P. If A is a non-empty subset of P, then $\bar{Apr} (H \circ K, A) \subseteq K \circ \bar{Apr} (H, A) .$

Proposition 4.9. Let H and K be LA-subpolygroups of an LA-polygroup P. If A is an LA-subpolygroup of P, then $\underline{Apr} (H, A) \circ \underline{Apr} (K, A) \subseteq \underline{Apr} (H \circ K, A) .$

Proof. Let c be any element of $\underline{Apr} (H, A) \circ \underline{Apr} (K, A)$ . Then c ∈ a ∘ b with $a \in \underline{Apr} (H, A)$ and $b \in \underline{Apr} (K, A)$ . Thus H ∘ a ⊆ A and K ∘ b ⊆ A. Since H and K are LA-subpolygroups of P, so by medial law, we have $\begin{matrix} (H \circ K) \circ (a \circ b) & = & (H \circ a) \circ (K \circ b) \\ \subseteq & A \circ A \subseteq A . \end{matrix}$ Therefore, we have $c \in a \circ b \subseteq \underline{Apr} (H \circ K, A)$ . Thus $\underline{Apr} (H, A) \circ \underline{Apr} (K, A) \subseteq \underline{Apr} (H \circ K, A) .$ □

Theorem 4.10. Let H and A be LA-subpolygroups of an LA-polygroup P.

If H ⊈ A, then $\underline{Apr} (H, A) = \emptyset .$

If H ⊆ A, then $\underline{Apr} (H, A) = A .$

Proof. (1) If H ⊈ A, then we have to show that H ∘ x ⊈ A for any x ∈ P. Assume that there exist an element x in P such that H ∘ x ⊆ A. Then x = e ∘ x ∈ H ∘ x ⊆ A, where e is the identity of H. Since A is an LA-subpolygroup of P, it follows that x⁻¹ ∈ A. Thus $\begin{matrix} H & = & e \circ H \subseteq (x^{- 1} \circ x) \circ H \\ = & (H \circ x) \circ x^{- 1} \\ \subseteq & A \circ A = A . \end{matrix}$ Thus H ⊆ A. This is a contradiction. Hence for all x ∈ G, H ∘ x ⊈ A. So we get $\underline{Apr} (H, A) = \emptyset .$

(2) Let x ∈ A. If H ⊆ A, then H ∘ x ⊆ A ∘ A ⊆ A. This implies that $x \in \underline{Apr} (H, A) .$ Thus $A \subseteq \underline{Apr} (H, A) .$ Using Proposition 3.3(1), we get $\underline{Apr} (H, A) = A .$ □

5 Rough LA-subpolygroups in a factor LA-polygroup

The lower and upper approximations in a factor LA-polygroup can be presented in an equivalent form as shown below. Let H be an LA-subpolygroup of an LA-polygroup P, and A a non-empty subset of P. Then $\underline{\underline{Apr}} (H, A) = {H \circ a \in P / H : H \circ a \subseteq A}$ and $\bar{\bar{Apr}} (H, A) = {H \circ a \in P / H : H \circ a \cap A \neq \emptyset} .$

Example 5.1. Consider P = {e, a, b, c, d, f, g, h} be an LA-polygroup defined by the table "†".

Now consider H = {e, b} is an LA-subpolygroup of P. Then the factor LA-polygroup P/H = {H, H_b, H_d, H_g} is given by the table:

⊙	H	H _b	H _d	H _g
H	H	H _b	H _d	H _g
H _b	H _b	H, H_b	H _d	H _g
H _d	H _g	H _g	H_d, H_g	H, H_b, H_d
H _g	H _d	H _d	H, H_b, H_g	H_d, H_g

By routine calculations one can see that P/H = {H, H_b, H_d, H_g} is an LA-polygroup. Now, let A = {e, a, b, c, h}. Then $\underline{\underline{Apr}} (H, A) = {H, H_{b}}$ and $\bar{\bar{Apr}} (H, A) = {H, H_{b}, H_{g}} .$

Proposition 5.2. Let H and K be LA-subpolygroups of an LA-polygroup P. Let A and B be any non-empty subsets of P such that H ⊆ A and H ⊆ B. Then

$\underline{\underline{Apr}} (H, A) \subseteq A / H \subseteq \bar{\bar{Apr}} (H, A);$

$\bar{\bar{Apr}} (H, A \cup B) = \bar{\bar{Apr}} (H, A) \cup \bar{\bar{Apr}} (H, B);$

$\underline{\underline{Apr}} (H, A \cap B) = \underline{\underline{Apr}} (H, A) \cap \underline{\underline{Apr}} (H, B);$

A ⊆ B implies $\underline{\underline{Apr}} (H, A) \subseteq \underline{\underline{Apr}} (H, B);$

A ⊆ B implies $\bar{\bar{Apr}} (H, A) \subseteq \bar{\bar{Apr}} (H, B);$

$\underline{\underline{Apr}} (H, A \cup B) \supseteq \underline{\underline{Apr}} (H, A) \cup \underline{\underline{Apr}} (H, B);$

$\bar{\bar{Apr}} (H, A \cap B) \subseteq \bar{\bar{Apr}} (H, A) \cap \bar{\bar{Apr}} (H, B);$

H ⊆ K implies $\bar{\bar{Apr}} (H, A) \subseteq \bar{\bar{Apr}} (K, A) .$

Proof. The proof follows from the definitions.□

Proposition 5.3. Let H be an LA-subpolygroup of an LA-polygroup P. Let A and B be non-empty subsets of P. Then $\bar{\bar{Apr}} (H, A) ⊙ \bar{\bar{Apr}} (H, B) \supseteq \bar{\bar{Apr}} (H, A \circ B) .$

Proof. Let H ∘ c be any element of $\bar{\bar{Apr}} (H, A \circ B)$ . Then H∘ c ∩ A ∘ B ≠ ∅. Thus there exists an element x in P such that x ∈ H ∘ c ∩ A ∘ B, and so x ∈ H ∘ c and x ∈ A ∘ B. Then x ∈ a ∘ b with a ∈ A and b ∈ B. Since $\begin{matrix} c & \in & H \circ x \subseteq H \circ (a \circ b) \\ = & (H \circ a) \circ (H \circ b), \end{matrix}$ we have c ∈ y ∘ z with y ∈ H ∘ a and z ∈ H ∘ b. Then a ∈ H ∘ y, and so H∘ y ∩ A ≠ ∅. Thus $H \circ y \in \bar{\bar{Apr}} (H, A)$ . Similarly we have $H \circ z \in \bar{\bar{Apr}} (H, B)$ . Thus $\begin{matrix} H \circ c & \subseteq & (H \circ y) ⊙ (H \circ z) \\ \subseteq & \bar{\bar{Apr}} (H, A) ⊙ \bar{\bar{Apr}} (H, B), \end{matrix}$ and so we have $\bar{\bar{Apr}} (H, A \circ B) \subseteq \bar{\bar{Apr}} (H, A) ⊙ \bar{\bar{Apr}} (H, B)$ .

Proposition 5.4. Let H be an LA-subpolygroup of an LA-polygroup P. Let A and B be non-empty subsets of P. Then $\underline{\underline{Apr}} (H, A) ⊙ \underline{\underline{Apr}} (H, B) \subseteq \underline{\underline{Apr}} (H, A \circ B) .$

Proof. Let H ∘ c be any element of

\underline{\underline{Apr}} (H, A) ⊙ \underline{\underline{Apr}} (H, B)

. Then H ∘ c = H ∘ x ⊙ H ∘ y with

H \circ x \in \underline{\underline{Apr}} (H, A)

and

H \circ y \in \underline{\underline{Apr}} (H, B)

. Thus H ∘ x ⊆ A and H ∘ y ⊆ B. Since H is an LA-subpolygroup, so

∘	e	a	b	c	d	f	g	h
e	e	a	b	c	d	f	g	h
a	a	{e, a}	c	{b, c}	f	{d, f}	h	{g, h}
b	b	c	{e, b}	{a, c}	d	f	g	h
c	c	{b, c}	{a, c}	{e, a, b, c}	f	{d, f}	h	{g, h}
d	g	h	g	h	{d, g}	{f, h}	{e, b, d}	{a, c, f}
f	h	{g, h}	h	{g, h}	{f, h}	{d, f, g, h}	{a, c, f}	{e, a, b, c, d, f}
g	d	f	d	f	{e, b, g}	{a, c, h}	{d, g}	{f, h}
h	f	{d, f}	f	{d, f}	{a, c, h}	{e, a, b, c, g, h}	{f, h}	{d, f, g, h}

Table† Cayley table for P = {e, a, b, c, d, f, g, h} in Example 5.1.

H \circ c = (H \circ x) \circ (H \circ y) \subseteq A \circ B,

and so

H \circ c \in \underline{\underline{Apr}} (H, A \circ B)

. Thus

\underline{\underline{Apr}} (H, A) ⊙ \underline{\underline{Apr}} (H, B) \subseteq \underline{\underline{Apr}} (H, A \circ B)

The following example shows that $\underline{\underline{Apr}} (H, A \circ B) ⊈ \underline{\underline{Apr}} (H, A) \circ \underline{\underline{Apr}} (H, B) .$

Example 5.5. Consider Example 5.1, and let A = {e, a, c, f} and B = {b, c, d, g, h}. Then $\underline{\underline{Apr}} (H, A) = {H}$ and $\underline{\underline{Apr}} (H, B) = {H_{b}, H_{g}} .$ Now A ∘ B = P, so $\underline{\underline{Apr}} (A \circ B) = {H, H_{b}, H_{d}, H_{g}}$ which shows that $\begin{matrix} {H, H_{b}, H_{d}, H_{g}} & = & \underline{\underline{Apr}} (H, A \circ B) \\ ⊈ & \underline{\underline{Apr}} (H, A) ⊙ \underline{\underline{Apr}} (H, B) \\ = & {H_{b}, H_{g}} . \end{matrix}$

Theorem 5.6. Let H be an LA-subpolygroup of an LA-polygroup P. If A is an LA-subpolygroup of P, then $\bar{\bar{Apr}} (H, A)$ is an LA-subpolygroup of P/H.

Proof. Let H ∘ a and H ∘ b be any elements of $\bar{\bar{Apr}} (H, A)$ . Then there exist elements x and y in P such that x ∈ H ∘ a ∩ A and y ∈ H ∘ b ∩ A. Thus x ∈ H ∘ a, y ∈ H ∘ b, x ∈ A, and y ∈ A. Then, since H is an LA-subpolygroup, so $x \circ y \subseteq (H \circ a) \circ (H \circ b) = H \circ (a \circ b) .$ Since A is an LA-subpolygroup of P, x ∘ y ⊆ A. Thus x ∘ y ⊆ (H ∘ (a ∘ b)) ∩ A. Therefore we have $(H \circ a) ⊙ (H \circ b) = H ⊙ (a \circ b) \in \bar{\bar{Apr}} (H, A) .$

Let H ∘ a be any element of $\bar{\bar{Apr}} (H, A)$ . Then there exists an element x in P such that x ∈ H ∘ a ∩ A. Thus x ∈ H ∘ a and x ∈ A. Thus x ∈ h ∘ a for some h ∈ H. Note that (H ∘ a) ⁻¹ = H ∘ a⁻¹. Then we have $x^{- 1} \in (h \circ a)^{- 1} = h^{- 1} \circ a^{- 1} \subseteq H \circ a^{- 1} .$

Since x⁻¹ ∈ A, x⁻¹ ∈ H ∘ a⁻¹ ∩ A, and so $H \circ a^{- 1} \in \bar{\bar{Apr}} (H, A)$ . Therefore, we obtain that $\bar{\bar{Apr}} (H, A)$ is an LA-subpolygroup of P/H.□

The following example shows that, some times $\bar{\bar{Apr}} (H, A)$ is an LA-subpolygroup of P/H, but A is not an LA-subpolygroup of P.

Example 5.7. Again, consider Example 5.1, and let A = {e, a, b, c, d, g}. Then $\bar{\bar{Apr}} (H, A) = {H, H_{b}, H_{d}, H_{g}},$ which is an LA-subpolygroup of P/H but A is not an LA-subpolygroup of P.

Theorem 5.8. Let H be an LA-subpolygroup of an LA-polygroup P. If A is an LA-subpolygroup of P such that H ⊆ A, then $\underline{\underline{Apr}} (H, A)$ is an LA-subpolygroup of P/H.

Proof. Let H ∘ a and H ∘ b be any elements of $\underline{\underline{Apr}} (H, A)$ , then H ∘ a ⊆ A and H ∘ b ⊆ A. Since H is an LA-subpolygroup, so $H \circ (a \circ b) = (H \circ a) \circ (H \circ b) \subseteq A \circ A \subseteq A,$ and so we have $(H \circ a) ⊙ (H \circ b) = H ⊙ (a \circ b) \in \underline{\underline{Apr}} (H, A) .$

Let H ∘ a be any element of $\underline{\underline{Apr}} (H, A)$ . Then, since a = e ∘ a ∈ H ∘ a ⊆ A, a⁻¹ ∈ A. Thus H ∘ a⁻¹ ⊆ A ∘ A ⊆ A. This means that $(H \circ a)^{- 1} = H \circ a^{- 1} \in \underline{\underline{Apr}} (H, A) .$

Thus $\underline{\underline{Apr}} (H, A)$ is an LA-subpolygroup of P/H.□

The following example shows that, some times $\underline{\underline{Apr}} (H, A)$ is an LA-subpolygroup of P/H, but A is not an LA-subpolygroup of P.

Example 5.9. Again, consider Example 5.1, and let A = {e, a, b, c, f}. Then $\underline{\underline{Apr}} (H, A) = {H, H_{b}},$ which is an LA-subpolygroup of P/H but A is not an LA-subpolygroup of P.

6 Homomorphism Problems

In this section, we will study some results on homomorphisms of rough LA-polygroups.

Definition 6.1. Let 〈P₁, ∘ ₁, e₁, ⁻¹〉 and 〈P₂, ∘ ₂, e₂, ⁻¹〉 be two LA-polygroups. Let f be a mapping from P₁ into P₂ such that f (e₁) = e₂. Then, f is called

an inclusion homomorphism if f (x ∘ ₁y) ⊆ f (x) ∘ ₂f (y).

a good homomorphism if f (x ∘ ₁y) = f (x) ∘ ₂f (y).

If f is surjective then f is called an epimorphism. We denote by e₂ the left identity of P₂. Then the set $ker (f) = {x \in P_{1} : f (x) = e_{2}}$ is called the Kernel of f. It is easy to show that, ker(f) is an LA-subpolygroup of P.

Proposition 6.2. Let f be an inclusion homomorphism of an LA-polygroup P₁ to an LA-polygroup P₂. If A is a non-empty subset of P₁, then $f (A) = f (ker (f) \circ_{1} A) .$

Proof. Let y be any element of f (A). Then f (a) = y for some a ∈ A. We note that e ∈ ker(f). Thus we have $y = f (a) = f (e \circ_{1} a) \in f (ker (f) \circ_{1} A),$ and so f (A) ⊆ f (ker(f) ∘ ₁A).

Conversely, let y be any element of f (ker(f) ∘ ₁A). Then f (a) = y for some a ∈ ker(f) ∘ ₁A. Thus a ∈ b ∘ ₁c with b ∈ ker(f) and c ∈ A. Then $\begin{matrix} y & = & f (a) \in f (b \circ_{1} c) \subseteq f (b) \circ_{2} f (c) \\ = & e_{2} \circ_{2} f (c) = f (c) \in f (A), \end{matrix}$ and so f (ker(f) ∘ ₁A) ⊆ f (A). Therefore, f (A) = f (ker(f) ∘ ₁A).□

Proposition 6.3. Let f be a good homomorphism of an LA-polygroup P₁ to an LA-polygroup P₂. If A is a non-empty subset of P₁, then $f (A) \subseteq f (\bar{Apr} (H, A)) \subseteq f (H \circ_{1} A) .$

Proof. By Proposition 3.3(1), $A \subseteq \bar{Apr} (H, A)$ , and since f is a homomorphism of P₁ to P₂, we have $f (A) \subseteq f (\bar{Apr} (H, A)) .$ To see $f (\bar{Apr} (H, A)) \subseteq f (H \circ_{1} A)$ , let y be any element of $f (\bar{Apr} (H, A))$ ; then f (a) = y for some $a \in \bar{Apr} (H, A)$ . Thus there exists an element x in P₁ such that x ∈ H ∘ ₁a ∩ A. Then x ∈ H ∘ ₁a and x ∈ A, that is, a ∈ H ∘ ₁x. Then we have $y = f (a) \in f (H \circ_{1} x) \subseteq f (H \circ_{1} A),$ and so $f (\bar{Apr} (H, A)) \subseteq f (H \circ_{1} A)$ , which completes the proof.□

The following example shows that $f (A) ⊉ f (\bar{Apr} (H, A)) .$

Example 6.4. Let P₁ ={ e, a, b, c } and P₂ ={ g, x, y, z } be two LA-polygroups with the hyperoperations defined in the following tables:

∘₁	e	a	b	c
e	e	a	b	c
a	a	e	b	c
b	c	c	{e, a}	b
c	b	b	c	{e, a}

∘₂	g	x	y	z
g	g	x	y	z
x	y	{g, z}	x	y
y	x	y	{g, z}	x
z	z	x	y	g

Let f: P₁ → P₂ be defined by f (e) = g, f (a) = z, f (b) = y and f (c) = x. Then, clearly f is a good homomorphism from P₁ to P₂. Now consider H = {e, a} and A = {a, b} contained in P₁. Then f (A) = {y, z} and $f (\bar{Apr} (H, A)) = {g, y, z} .$ This shows that $f (A) \subseteq f (\bar{Apr} (H, A))$ but $f (A) ⊉ f (\bar{Apr} (H, A)) .$

Proposition 6.5. Let f be a good homomorphism of an LA-polygroup P₁ to an LA-polygroup P₂. If A is a non-empty subset of P₁, then $f (A) = f (\bar{Apr} (ker (f), A)) .$

Proof. Since ker(f) is an LA-subpolygroup of P₁, it follows from Proposition 6.2 and Proposition 6.3 that $\begin{matrix} f (A) & \subseteq & f (\bar{Apr} (ker (f), A)) \\ \subseteq & f (ker (f) \circ_{1} A) \\ = & f (A) . \end{matrix}$ Therefore, we obtain that $f (A) = f (\bar{Apr} (ker (f), A))$ , which completes the proof.□

Theorem 6.6. Let f be an epimorphism of an LA-polygroup P₁ to an LA-polygroup P₂. Let H and A be LA-subpolygroups of P₁, such that H ⊆ A. Then

$f (\bar{Apr} (H, A)) \subseteq \bar{Apr} (f (H), f (A)) .$

$f (\underline{Apr} (H, A)) = \underline{Apr} (f (H), f (A)) .$

Proof. (1) Let $f (a) \in f (\bar{Apr} (H, A))$ with $a \in \bar{Apr} (H, A) .$ Then by the definition of $\bar{Apr} (H, A)$ we have H ∘ ₁a ∩ A ≠ ∅, so f (H ∘ ₁a ∩ A) ≠ ∅. It follows that f (H) ∘ ₂f (a) ∩ f (A) ≠ ∅. Thus $f (a) \in \bar{Apr} (f (H), f (A)) .$ Hence $f (\bar{Apr} (H, A)) \subseteq \bar{Apr} (f (H), f (A)) .$

(2) Since f is an epimorphism from P₁ to P₂, H and A are LA-subpolygroups of P₁, then f (H) and f (A) are LA-subpolygroups of P₂. Since H ⊆ A if and only if f (H) ⊆ f (A), by Theorem 4.10 we can get $\underline{Apr} (H, A) = A$ if and only if $\underline{Apr} (f (H), f (A)) = f (A) .$ Thus $f (\underline{Apr} (H, A)) = \underline{Apr} (f (H), f (A)) .$

7 Conclusion

Roughness of algebraic structures play an important role in mathematics with wide range of applications. Researchers apply roughness into the algebraic system and find interesting algebraic properties of them with some applications. In [19], the authors considered the application of rough soft hyperrings corresponding to decision making methods. In this paper, we combine the rough set theory and LA-polygroups. We provide some properties of rough LA-subpolygroups which is an extension of LA-subpolygroups. We proved some results on rough LA-subpolygroups in a factor LA-polygroup and homomorphisms of rough LA-subpolygroups.

Footnotes

Acknowledgement

The author is highly grateful to referees for their valuable comments and suggestions for improving the paper, and to the Assoc. Editor of the journal Professor Bijan Davvaz for editing and communicating the paper. The author would like to thank Deanship of Scientific Research at Majmaah University for supporting this work under Project Number 1440-3.

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