Characterizations of non-associative rings by their intuitionistic fuzzy bi-ideals EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 1, 2019, 226-250 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Characterizations of non-associative rings by their intuitionistic fuzzy bi-ideals Nasreen Kausar1,∗, Muhammad Azam Waqar2 1 Department of Mathematics, Quaid-i-Azam University Islamabad, Pakistan 2 Department of School of Business Mangment, NFC IEFR FSD, Pakistan Abstract. The purpose of this paper is to initiate and study on the generalization of the fuzzifica- tion of ideals in a class of non-associative and non-commutative algebraic structures (LA-ring). We characterize different classes of LA-ring in terms of intuitionistic fuzzy left (resp. right, bi-, generalized bi-, (1, 2)-) ideals. 2010 Mathematics Subject Classifications: 17D05, 17D99 Key Words and Phrases: Intuitionistic fuzzy left (right, bi-, generalized bi-, (1, 2)-) ideals In 1972, a generalization of abelian semigroups initiated by Kazim et al [11]. In ternary commutative (abelian) law: abc = cba, they introduced braces on the left side of this law and explored a new pseudo associative law, that is (ab)c = (cb)a. This law (ab)c = (cb)a is called the left invertive law. A groupoid S is said to be a left almost semigroup (abbreviated as LA-semigroup), if it satisfies the left invertive law: (ab)c = (cb)a. An LA-semigroup is a midway structure between an abelian semigroup and a groupoid. Ideals in LA-semigroup have been investigated by [16]. In [9] (resp. [4]), a groupoid S is said to be medial (resp. paramedial) if (ab)(cd) = (ac)(bd) (resp. (ab)(cd) = (db)(ca)). In [11], an LA-semigroup is medial, but in general an LA-semigroup needs not to be paramedial. Every LA-semigroup with left identity is paramedial by Protic et al [16] and also satisfies a(bc) = b(ac), (ab)(cd) = (dc)(ba). Kamran [10], extended the notion of LA-semigroup to the left almost group (LA- group). An LA-semigroup S is said to be a left almost group, if there exists left identity e ∈ S such that ea = a for all a ∈ S and for every a ∈ S, there exists b ∈ S such that ba = e. Shah et al [20], initiated the concept of left almost ring (abbreviated as LA-ring) of finitely nonzero functions, which is a generalization of a commutative semigroup ring. By a left almost ring, we mean a non-empty set R with at least two elements such that (R,+) is an LA-group, (R, ·) is an LA-semigroup, both left and right distributive laws hold. For example, from a commutative ring (R,+, ·) , we can always obtain an LA-ring (R,⊕, ·) by ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i1.3344 Email addresses: kausar.nasreen57@gmail.com (N. Kausar), azamwaqar4@gmail.com (M. A. Waqar) http://www.ejpam.com 226 c© 2019 EJPAM All rights reserved. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 227 defining for all a, b ∈ R, a ⊕ b = b − a and a · b is same as in the ring. Despite the fact that the structure is non-associative and non-commutative, however it possesses properties which usually come across in associative and commutative algebraic structures. A non-empty subset A of an LA-ring R is called an LA-subring of R if a − b and ab ∈ A for all a, b ∈ A. A is called a left (resp. right) ideal of R if (A,+) is an LA-group and RA ⊆ A (resp. AR ⊆ A). A is called an ideal of R if it is both a left ideal and a right ideal of R. An LA-subring A of R is called a bi-ideal of R if (AR)A ⊆ A. A non-empty subset A of R is called a generalized bi-ideal of R if (A,+) is an LA-group and (AR)A ⊆ A. Every bi-ideal of R is a generalized bi-ideal of R. An LA-subring A of R is called (1, 2)-ideal of R if (AR)A2 ⊆ A. We will initiate the concept of regular (resp. left regular, right regular, (2, 2)-regular, left weakly regular, right weakly regular, intra-regular) LA-rings. We will also define the concept of intuitionistic fuzzy left (resp. right, bi-,generalized bi-, (1, 2)-) ideals. We will describe a study of regular (resp. left regular, right regular, (2, 2)-regular, left weakly regular, right weakly regular, intra-regular) LA-rings by the properties of intuitionistic fuzzy left (right, bi-, generalized bi-) ideals. In this regard, we will prove that in regular (resp. left weakly regular) LA-rings, the concept of intuitionistic fuzzy (right, two-sided) ideals coincides. We will also show that in right regular (resp. (2, 2)- regular, right weakly regular, intra-regular) LA-rings, the concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides. Also in left regular LA-rings with left identity, the concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides. We will also characterize left weakly regular LA-rings in terms of intuitionistic fuzzy right (two-sided, bi-, generalize bi-) ideals. 1. Basic Definitions and Preliminary Results After the introduction of fuzzy set by Zadeh [22], several researchers explored on the generalization of the notion of fuzzy set. The concept of intuitionistic fuzzy set was introduced by Atanassov [1, 2], as a generalization of the notion of fuzzy set. Liu [13], introduced the concept of fuzzy subrings and fuzzy ideals of a ring. Many authors have explored the theory of fuzzy rings (for example [6, 12, 14, 15, 21]). Gupta et al [6], gave the idea of intrinsic product of fuzzy subsets of a ring. Kuroki [12], characterized regular (intra-regular, both regular and intra-regular) rings in terms of fuzzy left (right, quasi, bi-) ideals. An intuitionistic fuzzy set (briefly, IFS) A in a non-empty set X is an object having the form A = {(x, µA(x), γA(x)) : x ∈ X}, where the functions µA : X → [0, 1] and γA : X → [0, 1] denote the degree of membership and the degree of nonmembership, respectively and 0 ≤ µA(x) + γA(x) ≤ 1 for all x ∈ X [1, 2]. An intuitionistic fuzzy set A = {(x, µA(x), γA(x)) : x ∈ X} in X can be identified to be an ordered pair (µA, γA) in IX × IX , where IX is the set of all functions from X to [0, 1]. For the sake of simplicity, we shall use the symbol A = (µA, γA) for the IFS A = {(x, µA(x), γA(x)) : x ∈ X}. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 228 Banerjee et al [3] and Hur et al [7], initiated the notion of intuitionistic fuzzy subrings and intuitionistic fuzzy ideals of a ring. Subsequently many authors studied the intu- itionistic fuzzy subrings and intuitionistic fuzzy ideals of a ring by describing the different properties (see [8]). Shah et al [18], have initiated the concept of intuitionistic fuzzy normal LA-subrings of an LA-ring. We initiate the notion of intuitionistic fuzzy left (resp. right, bi-, generalized bi-,(1, 2)- ) ideals of an LA-ring R. [18] An intuitionistic fuzzy set (IFS) A = (µA, γA) of an LA-ring R is called an intu- itionistic fuzzy LA-subring of R if (1) µA (x− y) ≥ min{µA (x) , µA(y)}, (2) γA (x− y) ≤ max{γA (x) , γA(y)}, (3) µA (xy) ≥ min{µA (x) , µA (y)}, (4) γA (xy) ≤ max{γA (x) , γA (y)} for all x, y ∈ R. An IFS A = (µA, γA) of an LA-ring R is called an intuitionistic fuzzy left ideal of R if (1) µA (x− y) ≥ min{µA (x) , µA(y)}, (2) γA (x− y) ≤ max{γA (x) , γA(y)}, (3) µA (xy) ≥ µA (y) , (4) γA (xy) ≤ γA (y) for all x, y ∈ R. An IFS A = (µA, γA) of an LA-ring R is called an intuitionistic fuzzy right ideal of R if (1) µA (x− y) ≥ min{µA (x) , µA(y)}, (2) γA (x− y) ≤ max{γA (x) , γA(y)}, (3) µA (xy) ≥ µA (x) , (4) γA (xy) ≤ γA (x) for all x, y ∈ R. An IFS A = (µA, γA) of an LA-ring R is called an intuitionistic fuzzy ideal of R if it is both an intuitionistic fuzzy left ideal and an intuitionistic fuzzy right ideal of R. Example 1. Let R = {a, b, c, d}. Define + and · in R as follows : + a b c d a a b c d b d a b c c c d a b d b c d a and · a b c d a a a a a b a b a b c a a c c d a b c d Then R is an LA-ring and A = (µA, γA) be an IFS of R. We define µA(a) = µA(c) = 0.7, µA(b) = µA(d) = 0 and γA(a) = γA(c) = 0, γA(b) = γA(d) = 0.7. Then A = (µA, γA) is an intuitionistic fuzzy ideal of R. Every intuitionistic fuzzy left (resp. right, two-sided) ideal of an LA-ring R is an intuitionistic fuzzy LA-subring of R, but the converse is not true. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 229 Example 2. R = {0, 1, 2, 3, 4, 5, 6, 7} is an LA-ring. + 0 1 2 3 4 5 6 7 0 0 1 2 3 4 5 6 7 1 2 0 3 1 6 4 7 5 2 1 3 0 2 5 7 4 6 3 3 2 1 0 7 6 5 4 4 4 5 6 7 0 1 2 3 5 6 4 7 5 2 0 3 1 6 5 7 4 6 1 3 0 2 7 7 6 5 4 3 2 1 0 and · 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 1 0 4 4 0 0 4 4 0 2 0 4 4 0 0 4 4 0 3 0 0 0 0 0 0 0 0 4 0 3 3 0 0 3 3 0 5 0 7 7 0 0 7 7 0 6 0 7 7 0 0 7 7 0 7 0 3 3 0 0 3 3 0 Let A = (µA, γA) be an IFS of an LA-ring R. We define µA(0) = µA(4) = 0.7, µA(1) = µA(2) = µA(3) = µA(5) = µA(6) = µA(7) = 0 and γA(0) = γA(4) = 0, γA(1) = γA(2) = γA(3) = γA(5) = γA(6) = γA(7) = 0.7. Then A = (µA, γA) is an intuitionistic fuzzy LA-subring of R, but not an intuitionistic fuzzy right ideal of R, because µA(41) = µA(3) = 0. µA(4) = 0.7. ⇒ µA(41) � µA(4). and γA(41) = γA(3) = 0.7. γA(4) = 0. ⇒ γA(41) � γA(4). An Intuitionistic fuzzy LA-subring A = (µA, γA) of an LA-ring R is called an intu- itionistic fuzzy bi-ideal of R if (1) µA ((xy)z) ≥ min {µA(x), µA(z)} , (2) γA ((xy)z) ≤ max {γA(x), γA(y)} for all x, y, z ∈ R. An IFS A = (µA, γA) of an LA-ring R is called an intuitionistic fuzzy generalized bi-ideal of R if (1) µA (x− y) ≥ min {µA(x), µA(y)} , (2) γA (x− y) ≤ max {γA(x), γA(y)} , (3) µA ((xy)z) ≥ min {µA(x), µA(z)} , (4) γA ((xy)z) ≤ max {γA(x), γA(z)} for all x, y, z ∈ R. An intuitionistic fuzzy LA-subring A = (µA, γA) of an LA-ring R is called an intuition- istic fuzzy (1, 2)-ideal of R if (1) µA((xw)(yz)) ≥ min {µA(x), µA(y), µA(z)} , (2) γA((xw)(yz)) ≤ max {γA(x), γA(y), γA(z)} for all x, y, z, w ∈ R. We note that an LA-ring R can be considered an intuitionistic fuzzy set of itself and we write R = IR, i.e., R = (µR, γR) = (1, 0) for all x ∈ R. Let A and B be two intuitionistic fuzzy sets of an LA-ring R. Then (1) A ⊆ B ⇔ µA ⊆ µB and γA ⊇ γB, (2) A = B ⇔ A ⊆ B and B ⊆ A, N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 230 (3) Ac = (γA, µA) , (4) A ∩B = (µA ∧ µB, γA ∨ γB) = (µA∧B, γA∨B) , (5) A ∪B = (µA ∨ µB, γA ∧ γB) = (µA∨B, γA∧B) , (6) 0 ∼= (0, 1), 1 ∼= (1, 0) . [18] Let A be a non-empty subset of an LA-ring R. Then the intuitionistic characteristic of A is denoted by χA = 〈µχA , γχA〉 and defined by µχA (x) = { 1 if x ∈ A 0 if x /∈ A and γχA (x) = { 0 if x ∈ A 1 if x /∈ A The product of A = (µA, γA) and B = (µB, γB) is denoted by A◦B = (µA◦µB, γA◦γB) and defined by: (µA ◦ µB)(x) =  ∨ x= n∑ i=1 aibi {∧ni=1{µA(ai) ∧ µB(bi)}} if x = n∑ i=1 aibi, ai, bi ∈ R 0 if x 6= n∑ i=1 aibi and (γA ◦ γB)(x) =  ∧ x= n∑ i=1 aibi {∨ni=1{γA(ai) ∨ γB(bi)}} if x = n∑ i=1 aibi, ai, bi ∈ R 1 if x 6= n∑ i=1 aibi Now we are giving the some fundamental properties, which will be very helpful for next section. Theorem 1. Let A and B be two non-empty subsets of an LA-ring R. Then the following conditions hold. (1) If A ⊆ B then χA ⊆ χB. (2) χA ◦ χB = χAB. (3) χA ∪ χB = χA∪B. (4) χA ∩ χB = χA∩B. Proof. Straight forward. Let A = (µA, γA) and B = (µB, γB) be two intuitionistic fuzzy sets of an LA-ring R. The sum of A and B is denoted by A+B = (µA + µB, γA + γB) and defined by (µA + µB)(x) = ∨x=y+z(µA(y) ∧ µB(z)) and (γA + γB)(x) = ∧x=y+z(γA(y) ∨ γB(z)), for all x ∈ R. Lemma 1. Let A = (µA, γA) and B = (µB, γB) be two intuitionistic fuzzy sets of an LA-ring R. Then A+B is also an intuitionistic fuzzy set of R. Proof. It is sufficient to show that 0 ≤ (µA + µB)(x) + (γA + γB)(x) ≤ 1 for all x ∈ R. Now (µA + µB)(x) = ∨x=y+z(µA(y) ∧ µB(z)) ≤ ∨x=y+z((1− γA(y)) ∧ (1− γB(z))) N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 231 = 1− ∧x=y+z(γA(y) ∨ γB(z)) = 1− (γA + γB)(x). Since µA(y) ≤ 1 − γA(y) and µA(z) ≤ 1 − γA(z) for all y, z ∈ R. Hence A + B is an intuitionistic fuzzy set of R. Lemma 2. Every intuitionistic fuzzy left (resp. right, two-sided) ideal of an LA-ring R is an intuitionistic fuzzy bi-ideal of R. Proof. Straight forward. Lemma 3. Every intuitionistic fuzzy bi-ideal of an LA-ring R is an intuitionistic fuzzy (1, 2)-ideal of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy bi-ideal of R and a, x, y, z ∈ R. Thus µA((xa)(yz)) ≥ min{µA(x), µA(yz)} ≥ min{µA(x), µA(y), µA(z)} and γA((xa)(yz)) ≤ max{γA(x), γA(yz)} ≤ max{γA(x), γA(y), γA(z)}. Hence A = (µA, γA) is an intuitionistic fuzzy (1, 2)-ideal of R. Remark 1. Every intuitionistic fuzzy left (resp. right, two-sided) ideal of an LA-ring R is an intuitionistic fuzzy (1, 2)-ideal of R. Proposition 1. Let R be an LA-ring having the property a = a2 for every a ∈ R. Then every intuitionistic fuzzy (1, 2)-ideal of R is an intuitionistic fuzzy bi-ideal of R. Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy (1, 2)-ideal of R and a, x, y ∈ R. Thus µA((xa)y) = µA((xa)(yy)) ≥ min{µA(x), µA(y), µA(y)} = min{µA(x), µA(y)} and γA((xa)y) = γA((xa)(yy)) ≤ max{γA(x), γA(y), γA(y)} = max{γA(x), γA(y)}. Therefore A = (µA, γA) is an intuitionistic fuzzy bi-ideal of R. Theorem 2. If {Ai}i∈I is a family of intuitionistic fuzzy (1, 2)-ideals of an LA-ring R, then ∩Ai is also an intuitionistic fuzzy (1, 2)-ideal of R, where ∩Ai = (∧µAi ,∨γAi) and ∧µAi(x) = inf {µAi(x) | i ∈ I, x ∈ R} and ∨ γAi(x) = sup {γAi(x) | i ∈ I, x ∈ R} . Proof. Straight forward. Remark 2. Intersection of a family of intuitionistic fuzzy bi-ideals of an LA-ring R, is also an intuitionistic fuzzy bi-ideal of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 232 Lemma 4. [18] Let R be an LA-ring and ∅ 6= A ⊆ R. Then A is an LA-subring of R if and only if the intuitionistic characteristic function χA = 〈µχA , γχA〉 of A is an intuitionistic fuzzy LA-subring of R. Proposition 2. Let R be an LA-ring and ∅ 6= A ⊆ R. Then A is a left (resp. right) ideal of R if and only if the intuitionistic characteristic function χA = 〈µχA , γχA〉 of A is an intuitionistic fuzzy left (resp. right) ideal of R. Proof. Straight forward. Theorem 3. Let R be an LA-ring and ∅ 6= A ⊆ R. Then A is a (1, 2)-ideal of R if and only if the intuitionistic characteristic function χA = 〈µχA , γχA〉 of A is an intuitionistic fuzzy (1, 2)-ideal of R. Proof. Let A be a (1, 2)-ideal of R, this implies that A is an LA-subring of R. Then χA is an intuitionistic fuzzy LA-subring of R by the Lemma 4. Let a, x, y, z ∈ R. If x, y, z ∈ A, then by definition of intuitionistic characteristic function µχA(x) = 1 = µχA(y) = µχA(z) and γχA(x) = 0 = γχA(y) = µχA(z). Since (xa)(yz) ∈ A, A being a (1, 2)-ideal of R, so µχA((xa)(yz)) = 1 and γχA((xa)(yz)) = 0. Thus µχA((xa)(yz)) ≥ min{µχA(x), µχA(y), µχA(z)} and γχA((xa)(yz)) ≤ max{γχA(x), γχA(y), γχA(z)}. Similarly, we have µχA((xa)(yz)) ≥ min{µχA(x), µχA(y), µχA(z)} and γχA((xa)(yz)) ≤ max{γχA(x), γχA(y), γχA(z)}, when x, y, z /∈ A. Hence the intuitionistic characteristic function χA = 〈µχA , γχA〉 of A is an intuitionistic fuzzy (1, 2)-ideal of R. Conversely, suppose that the intuitionistic characteristic function χA = 〈µχA , γχA〉 of A is an intuitionistic fuzzy (1, 2)-ideal of R, this means that χA is an intuitionistic fuzzy LA-subring of R. Then A is an LA-subring of R by the Lemma 4. Let t ∈ (AR)A2, this implies that t = (xa)(yz), where x, y, z ∈ A and a ∈ R. Then by definition µχA(x) = 1 = µχA(y) = µχA(z) and γχA(x) = 0 = γχA(y) = γχA(z). Now µχA((xa)(yz)) ≥ µχA(x) ∧ µχA(y) ∧ µχA(z) = 1 and γχA((xa)(yz)) ≤ γχA(x) ∨ γχA(y) ∨ γχA(z) = 0, χA being an intuitionistic fuzzy (1, 2)-ideal ofR. Thus µχA((xa)(yz)) = 1 and γχA((xa)(yz)) = 0, i.e., (xa)(yz) ∈ A. Hence A is a (1, 2)-ideal of R. Remark 3. Let R be an LA-ring and ∅ 6= A ⊆ R. Then A is a bi-ideal of R if and only if the intuitionistic characteristic function χA = 〈µχA , γχA〉 of A is an intuitionistic fuzzy bi-ideal of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 233 Zadeh [22], introduced the concept of level set. Das [5], studied the fuzzy groups, level subgroups and gave the proper definition of a level set such that: let µ be a fuzzy subset of a non-empty set S, for t ∈ [0, 1], the set µt = {x ∈ S | µ(x) ≥ t}, is called a level subset of the fuzzy subset µ. Now we give the definition of strong level set. Let A = (µA, γA) be an intuitionistic fuzzy set of an LA-ring R, then for all r, t ∈ (0, 1], we define a set A(r,t) = {x ∈ R | µA(x) ≥ r and γA(x) ≤ t}, which is called the (r, t)-strong level set of A. It is clear that A(r,t) = U(µA; r) ∩ L(γA; t) for all r, t ∈ (0, 1]. Lemma 5. Let A = (µA, γA) be an IFS of an LA-ring R. Then A is an intuitionistic fuzzy LA-subring of R if and only if A(r,t) is an LA-subring of R for all r, t ∈ (0, 1]. Proof. Straight forward. Proposition 3. Let A = (µA, γA) be an IFS of an LA-ring R. Then A is an intuitionistic fuzzy left (resp. right) ideal of R if and only if A(r,t) is a left (resp. right) ideal of R for all r, t ∈ (0, 1]. Proof. Straight forward. Theorem 4. Let A = (µA, γA) be an IFS of an LA-ring R. Then A is an intuitionistic fuzzy (1, 2)-ideal of R if and only if A(r,t) is a (1, 2)-ideal of R for all r, t ∈ (0, 1]. Proof. Let A = (µA, γA) be an intuitionistic fuzzy (1, 2)-ideal of R, this implies that A is an intuitionistic fuzzy LA-subring of R. Then A(r,t) is an LA-subring of R by the Lemma 5. Let x, y, z ∈ A(r,t) and a ∈ R, so µA(y), µA(y), µA(z) ≥ r and γA(y), γA(y), γA(z) ≤ t. By our assumption µA((xy)(az)) ≥ µA(y) ∧ µA(y) ∧ µA(z) ≥ r and γA((xy)(az)) ≤ µA(x) ∨ µA(y) ∨ µA(z) ≤ t. Thus µA((xy)(az)) ≥ r and γA((xy)(az)) ≤ t, i.e., (xy)(az) ∈ A(r,t). So A(r,t) is a (1, 2)-ideal of R. Conversely, suppose that A(r,t) is a (1, 2)-ideal of R, this means that A(r,t) is an LA- subring of R. Then A is an intuitionistic fuzzy LA-subring of R by the Lemma 5. Let x, y, z, a ∈ R. We have to show that µA((xy)(az)) ≥ µA(x) ∧ µA(y) ∧ µA(y) and γA((xy)(az)) ≤ γA(x) ∨ γA(y) ∨ γA(y). We assume a contradiction µA((xy)(az)) ≤ µA(x) ∨ µA(y) ∨ µA(y) and γA((xy)(az)) ≥ γA(x) ∧ γA(y) ∧ γA(y). N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 234 Let µA(x) = r = µA(y) = µA(z) and γA(x) = t = γA(y) = γA(z), this implies that µA(x), µA(y), µA(z) ≥ r and γA(x), γA(y), γA(z) ≤ t, i.e., x, y, z ∈ A(r,t). But µA((xy)(az)) ≤ r and γA((xy)(az)) ≥ t, i.e., (xy)(az) /∈ A(r,t), which is a contradiction. So µA((xy)(az)) ≥ µA(x) ∧ µA(y) ∧ µA(y) and γA((xy)(az)) ≤ γA(x) ∨ γA(y) ∨ γA(y). Remark 4. Let A = (µA, γA) be an IFS of an LA-ring R. Then A is an intuitionistic fuzzy bi-ideal of R if and only if A(r,t) is a bi-ideal of R for all r, t ∈ (0, 1]. 2. Characterizations of LA-rings In this section, we characterize different classes of LA-ring in terms of intuitionistic fuzzy left (right, bi-, generalized bi-) ideals. An LA-ring R is called regular, if for every element x ∈ R, there exists an element a ∈ R such that x = (xa)x. An LA-ring R is called intra-regular, if for every element x ∈ R, there exist elements ai, bi ∈ R such that x = ∑n i=1(aix 2)bi. An LA-ring R is called left (resp. right) regular, if for every element x ∈ R, there exists an element a ∈ R such that x = ax2 (resp. x2a). An LA-ring R is called completely regular, if it is regular, left regular and right regular. An LA-ring R is called (2, 2)-regular, if for every element x ∈ R, there exists an element a ∈ R such that x = (x2a)x2. An LA-ring R is called locally associative LA-ring if (a.a).a = a.(a.a) for all a ∈ R. A ring R is called left (resp. right) weakly regular if I2 = I, for every left (resp. right) ideal I of R, equivalently x ∈ RxRx(x ∈ xRxR) for every x ∈ R. An LA-ring R is called weakly regular if it is both left weakly regular and right weakly regular [17]. Now we define this notion in a class of non-associative and non-commutative rings (LA-ring). An LA-ring R is called left (resp. right) weakly regular, if for every element x ∈ R, there exist elements a, b ∈ R such that x = (ax)(bx) (resp. x = (xa)(xb)) . An LA-ring R is called weakly regular if it is both left weakly regular and right weakly regular. Lemma 6. Every intuitionistic fuzzy right ideal of an LA-ring R with left identity e, is an intuitionistic fuzzy ideal of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy right ideal of R and x, y ∈ R. Thus µA (xy) = µA ((ex) y) = µA ((yx) e) ≥ µA (yx) ≥ µA (y) and γA (xy) = γA ((ex) y) = γA ((yx) e) ≤ γA (yx) ≤ γA (y) . Hence A is an intuitionistic fuzzy ideal of R. Lemma 7. Every intuitionistic fuzzy right ideal of a regular LA-ring R, is an intuitionistic fuzzy ideal of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 235 Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy right ideal of R. Let x, y ∈ R, this implies that there exists a ∈ R, such that x = (xa)x. Thus µA(xy) = µA(((xa)x)y) = µA((yx)(xa)) ≥ µA(yx) ≥ µA(y) and γA(xy) = γA(((xa)x)y) = γA((yx)(xa)) ≤ γA(yx) ≤ γA(y). Therefore A is an intuitionistic fuzzy ideal of R. Proposition 4. Let R be a regular LA-ring having the property a = a2 for every a ∈ R, with left identity e. Then every intuitionistic fuzzy generalized bi-ideal of R is an intu- itionistic fuzzy bi-ideal of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, this implies that there exists a ∈ R such that x = (xa)x. We have to show that A is an intuitionistic fuzzy LA-subring of R. Thus µA(xy) = µA(((xa)x)y) = µA(((xa)x2)y) = µA(((xa)(xx))y) = µA((x((xa)x))y) ≥ min{µA(x), µA(y)} and γA(xy) = γA(((xa)x)y) = γA(((xa)x2)y) = γA(((xa)(xx))y) = γA((x((xa)x))y) ≤ max{γA(x), γA(y)}. Hence A is an intuitionistic fuzzy LA-subring of R. Lemma 8. Let R be an LA-ring with left identity e. Then Ra is the smallest left ideal of R containing a. Proof. Let x, y ∈ Ra and r ∈ R. This implies that x = r1a and y = r2a, where r1, r2 ∈ R. Now x− y = r1a− r2a = (r1 − r2)a ∈ Ra and rx = r(r1a) = (er)(r1a) = ((r1a)r)e = ((r1a)(er))e = ((r1e)(ar))e = (e(ar))(r1e) = (ar)(r1e) = ((r1e)r)a ∈ Ra. Since a = ea ∈ Ra. Thus Ra is a left ideal of R containing a. Let I be another left ideal of R containing a. So ra ∈ I, where ra ∈ Ra, i.e., Ra ⊆ I. Hence Ra is the smallest left ideal of R containing a. Lemma 9. Let R be an LA-ring with left identity e. Then aR is a left ideal of R. Proof. Straight forward. Proposition 5. Let R be an LA-ring with left identity e. Then aR ∪ Ra is the smallest right ideal of R containing a. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 236 Proof. Let x, y ∈ aR ∪ Ra, this means that x, y ∈ aR or Ra. Since aR and Ra both are left ideals of R, so x − y ∈ aR and Ra, i.e., x − y ∈ aR ∪ Ra. We have to show that (aR ∪Ra)R ⊆ (aR ∪Ra). Now (aR ∪Ra)R = (aR)R ∪ (Ra)R = (RR)a ∪ (Ra)(eR) ⊆ Ra ∪ (Re)(aR) = Ra ∪R(aR) = Ra ∪ a(RR) ⊆ Ra ∪ aR = aR ∪Ra. ⇒ (aR ∪Ra)R ⊆ aR ∪Ra. As a ∈ Ra, i.e., a ∈ aR ∪ Ra. Let I be another right ideal of R containing a. Since aR ∈ IR ⊆ I and Ra = (RR)a = (aR)R ∈ (IR)R ⊆ IR ⊆ I, i.e., aR ∪Ra ⊆ I. Therefore aR ∪Ra is the smallest right ideal of R containing a. Lemma 10. Let R be an LA-ring. Then A◦B ⊆ A∩B for every intuitionistic fuzzy right ideal A and every intuitionistic fuzzy left ideal B of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy right and B = (µB, γB) be an intuitionistic fuzzy left ideal of R and x ∈ R. If x cannot be expressible as x = ∑n i=1 aibi, where ai, bi ∈ R and n is any positive integer, then obvious A ◦B ⊆ A ∩B, otherwise we have (µA ◦ µB) (x) = ∨x=∑n i=1 aibi {∧ni=1 {µA (ai) ∧ µB (bi)}} ≤ ∨x=∑n i=1 aibi {∧ni=1 {µA (aibi) ∧ µB (aibi)}} = ∨x=∑n i=1 aibi {∧ni=1(µA ∧ µB) (aibi)} = (µA ∧ µB) (x) = (µA ∩ µB) (x) ⇒ µA ◦ µB ⊆ µA ∩ µB Similarly, we have γA ◦ γB ⊇ γA ∪ γB. Hence A ◦ B ⊆ A ∩ B for every intuitionistic fuzzy right ideal A and every intuitionistic fuzzy left ideal B of R. Theorem 5. Let R be an LA-ring with left identity e, such that (xe)R = xR for all x ∈ R. Then the following conditions are equivalent. (1) R is a regular. (2) A ∩ B = A ◦ B for every intuitionistic fuzzy right ideal A and every intuitionistic fuzzy left ideal B of R. Proof. Suppose that (1) holds. Since A◦B ⊆ A∩B, for every intuitionistic fuzzy right ideal A and every intuitionistic fuzzy left ideal B of R by the Lemma 10. Let x ∈ R, this implies that there exists an element a ∈ R such that x = (xa)x. Thus (µA ◦ µB)(x) = ∨x=∑n i=1 aibi {∧ni=1 {µA (ai) ∧ µB (bi)}} ≥ min{µA(xa), µB(x)} ≥ min{µA(x), µB(x)} = (µA ∧ µB)(x) = (µA ∩ µB)(x). N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 237 ⇒ µA ∩ µB ⊆ µA ◦ µB. Similarly, we have γA ∪ γB ⊇ γA ◦ γB. Hence A ∩ B = A ◦ B, i.e., (1)⇒ (2) . Assume that (2) is true and a ∈ R. Then Ra is a left ideal of R containing a by the Lemma 8 and aR∪Ra is a right ideal of R containing a by the Proposition 5. So χRa is an intuitionistic fuzzy left ideal and χaR∪Ra is an intuitionistic fuzzy right ideal of R, by the Lemma 2. By our assumption χaR∪Ra ∩ χRa = χaR∪Ra ◦ χRa, i.e., χ(aR∪Ra)∩Ra = χ(aR∪Ra)Ra. Thus (aR ∪ Ra) ∩ Ra = (aR ∪ Ra)Ra. Since a ∈ (aR ∪ Ra) ∩ Ra, i.e., a ∈ (aR ∪ Ra)Ra, so a ∈ (aR)(Ra) ∪ (Ra)(Ra). Now (Ra)(Ra) = ((Re)a)(Ra) = ((ae)R)(Ra) = (aR)(Ra). This implies that (aR)(Ra) ∪ (Ra)(Ra) = (aR)(Ra) ∪ (aR)(Ra) = (aR)(Ra). Thus a ∈ (aR)(Ra). Then a = (ax)(ya) = ((ya)x)a = (((ey)a)x)a = (((ay)e)x)a = ((xe)(ay))a = (a((xe)y))a ∈ (aR)a, for any x, y ∈ R. This means that a ∈ (aR)a, i.e., a is regular. Hence R is a regular, i.e., (2)⇒ (1) . Theorem 6. Let R be a regular locally associative LA-ring having the property a = a2 for every a ∈ R. Then for every intuitionistic fuzzy bi-ideal A = (µA, γA) of R, A(an) = A(a2n) for all a ∈ R, where n is any positive integer. Proof. For n = 1. Let a ∈ R, this implies that there exists an element x ∈ R such that a = (ax)a. Now a = (ax)a = (a2x)a2, because a = a2. Thus µA (a) = µA ( (a2x)a2 ) ≥ min{µA ( a2 ) , µA ( a2 ) } = µA ( a2 ) = µA (aa) ≥ min{µA (a) , µA (a)} = µA(a). Similarly, γA(a) = γA ( a2 ) , therefore A(a) = A ( a2 ) . Now a2 = aa = ((a2x)a2)((a2x)a2) = (a4x2)a4, then the result is true for n = 2. Suppose that the result is true for n = k, i.e., A(ak) = A ( a2k ) . Now ak+1 = aka = ((a2kxk)a2k)((a2x)a2) = (a2(k+1)xk+1)a2(k+1). Thus µA ( ak+1 ) = µA ( (a2(k+1)xk+1)a2(k+1) ) ≥ min{µA(a2(k+1)), µA ( a2(k+1) ) } = µA ( a2(k+1) ) = µA ( ak+1ak+1) ) ≥ min{µA ( ak+1) ) , µA ( ak+1) ) } = µA(ak+1)). Similarly, γA(ak+1) = γA ( a2(k+1) ) , therefore A(ak+1) = A(a2(k+1)). Hence by induc- tion method, the result is true for all positive integers. Lemma 11. Every intuitionistic fuzzy left (right) ideal of (2, 2)-regular LA-ring R, is an intuitionistic fuzzy ideal of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 238 Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy right ideal of R and x, y ∈ R, this means that there exists a ∈ R such that x = (x2a)x2. Thus µA(xy) = µA(((x2a)x2)y) = µA((yx2)(x2a)) ≥ µA(yx2) ≥ µA(y) and γA(xy) = γA(((x2a)x2)y) = γA((yx2)(x2a)) ≤ γA(yx2) ≤ γA(y). Therefore A is an intuitionistic fuzzy ideal of R. Similarly, for left ideal. Remark 5. The concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides in (2, 2)-regular LA-rings. Proposition 6. Every intuitionistic fuzzy generalized bi-ideal of (2, 2)-regular LA-ring R with left identity e, is an intuitionistic fuzzy bi-ideal of R. Proof. Assume that A = (µA, γA) is an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, then there exists an element a ∈ R such that x = (x2a)x2. We have to show that A is an intuitionistic fuzzy LA-subring of R. Thus µA(xy) = µA(((x2a)x2)y) = µA(((x2a)(xx))y) = µA((x((x2a)x))y) ≥ min{µA(x), µA(y)} and γA(xy) = γA(((x2a)x2)y) = γA(((x2a)(xx))y) = γA((x((x2a)x))y) ≤ max{γA(x), γA(y)}. So A is an intuitionistic fuzzy LA-subring of R. Theorem 7. Let R be a (2, 2)-regular locally associative LA-ring. Then for every intu- itionistic fuzzy bi-ideal A = (µA, γA) of R, A(an) = A(a2n) for all a ∈ R, where n is any positive integer. Proof. Same as Theorem 6. Lemma 12. Let R be a right regular LA-ring. Then every intuitionistic fuzzy left (right) ideal of R is an intuitionistic fuzzy ideal of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy right ideal of R and x, y ∈ R, this implies that there exists a ∈ R such that x = x2a. Thus µA(xy) = µA((x2a)y) = µA(((xx)a)y) = µA(((ax)x)y) = µA((yx)(ax)) ≥ µA(yx) ≥ µA(y) and γA(xy) = γA((x2a)y) = γA(((xx)a)y) = γA(((ax)x)y) = γA((yx)(ax)) ≤ γA(yx) ≤ γA(y). Hence A is an intuitionistic fuzzy ideal of R. Similarly, for left ideal. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 239 Remark 6. The concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides in right regular LA-rings. Proposition 7. Let R be a right regular LA-ring with left identity e. Then every intu- itionistic fuzzy generalized bi-ideal of R is an intuitionistic fuzzy bi-ideal of R. Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, this means that there exists a ∈ R such that x = x2a. We have to show that A is an intuitionistic fuzzy LA-subring of R. Thus µA(xy) = µA((x2a)y) = µA(((xx)(ea))y) = µA(((ae)(xx))y) = µA((x((ae)x))y) ≥ min{µA(x), µA(y)} and γA(xy) = γA((x2a)y) = γA(((xx)(ea))y) = γA(((ae)(xx))y) = γA((x((ae)x))y) ≤ max{γA(x), γA(y)}. Therefore A is an intuitionistic fuzzy LA-subring of R. Lemma 13. Let R be a left regular LA-ring with left identity e. Then every intuitionistic fuzzy left (right) ideal of R is an intuitionistic fuzzy ideal of R. Proof. Assume that A = (µA, γA) is an intuitionistic fuzzy right ideal of R and x, y ∈ R, then there exists an element a ∈ R such that x = ax2. Thus µA(xy) = µA((ax2)y) = µA((a(xx))y) = µA((x(ax))y) = µA((y(ax))x) ≥ µA(y(ax)) ≥ µA(y) and γA(xy) = γA((ax2)y) = γA((a(xx))y) = γA((x(ax))y) = γA((y(ax))x) ≤ γA(y(ax)) ≤ γA(y). So A is an intuitionistic fuzzy ideal of R. Similarly, for left ideal. Remark 7. The concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides in left regular LA-rings with left identity. Proposition 8. Every intuitionistic fuzzy generalized bi-ideal of a left regular LA-ring R with left identity e, is an intuitionistic fuzzy bi-ideal of R. Proof. Let A be an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, this im- plies that there exists a ∈ R such that x = ax2. We have to show that A is an intuitionistic fuzzy LA-subring of R. Thus µA(xy) = µA((ax2)y) = µA((a(xx))y) = µA((x(ax))y) ≥ min{µA(x), µA(y)} and γA(xy) = γA((ax2)y) = γA((a(xx))y) = γA((x(ax))y) ≤ max{γA(x), γA(y)}. Hence A is an intuitionistic fuzzy LA-subring of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 240 Theorem 8. Let R be a regular and right regular locally associative LA-ring. Then for every intuitionistic fuzzy right ideal A = (µA, γA) of R, A(an) = A(a3n) for all a ∈ R, where n is any positive integer. Proof. For n = 1. Let a ∈ R, this means that there exists an element x ∈ R such that a = (ax)a and a = a2x. Now a = (ax)a = (ax)(a2x) = a3x2. Thus µA(a) = µA(a3x2) ≥ µA(a3) = µA(aa2) ≥ min{µA (a) , µA ( a2 ) } ≥ min{µA (a) , µA (a) , µA (a)} = µA (a) . Similarly, γA (a) = γA ( a3 ) , so A (a) = A ( a3 ) . Here a2 = aa = (a3x2)(a3x2) = a6x4, then the result is true for n = 2. Assume that the result is true for n = k, i.e., A(ak) = A(a3k). Now ak+1 = aka = (a3kx2k)(a3x2) = a3(k+1)x2(k+1). Thus µA(ak+1) = µA(a3(k+1)x2(k+1)) ≥ µA(a3(k+1)) = µA(a3k+3) = µA(ak+1a2k+2) ≥ min{µA ( ak+1 ) , µA ( a2k+2 ) } ≥ min{µA ( ak+1 ) , µA ( ak+1 ) , µA ( ak+1 ) } = µA ( ak+1 ) . Similarly, γA ( ak+1 ) = γA ( a3(k+1) ) , so A(ak+1) = A(a3(k+1)). Hence by induction method, the result is true for all positive integers. Theorem 9. Let R be a right regular locally associative LA-ring. Then for every intu- itionistic fuzzy right ideal A = (µA, γA) of R, A(an) = A(a2n) for all a ∈ R, where n is any positive integer. Proof. For n = 1. Let a ∈ R, then there exists an element x ∈ R such that a = a2x. Thus µA(a) = µA(a2x) ≥ µA(a2) = µA(aa) ≥ min{µA (a) , µA (a)} = µA (a) . Similarly, γA(a) = γA(a2), therefore A (a) = A ( a2 ) . Now a2 = aa = (a2x)(a2x) = a4x2, then the result is true for n = 2. Suppose that the result is true for n = k, i.e., A(ak) = A(a2k). Now ak+1 = aka = (a2kxk)(a2x) = a2(k+1)x(k+1). Thus µA(ak+1) = µA(a2(k+1)x(k+1)) ≥ µA(a2(k+1)) = µA(a2k+2) = µA(ak+1ak+1) ≥ min{µA ( ak+1 ) , µA ( ak+1 ) } = µA ( ak+1 ) . Similarly, γA(ak+1) = γA(a2(k+1)), therefore A(ak+1) = A(a2(k+1)). Hence by induction method, the result is true for all positive integers. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 241 Lemma 14. Let R be a right regular locally associative LA-ring with left identity e. Then for every intuitionistic fuzzy right ideal A = (µA, γA) of R, A(ab) = A(ba) for all a, b ∈ R. Proof. Let a, b ∈ R. By using Theorem 9 (for n = 1) . Now µA(ab) = µA((ab)2) = µA((ab)(ab)) = µA((ba)(ba)) = µA((ba)2) = µA(ba) and γA(ab) = γA((ab)2) = γA((ab)(ab)) = γA((ba)(ba)) = γA((ba)2) = γA(ba). Thus A(ab) = A(ba). Remark 8. It is easy to see that, if R is a left regular locally associative LA-ring with left identity e. Then for every intuitionistic fuzzy left ideal A = (µA, γA) of R, A(an) = A(a2n) for all a ∈ R, where n is any positive integer. And also for every intuitionistic fuzzy left ideal A = (µA, γA) of R, A(ab) = A(ba) for all a, b ∈ R. Lemma 15. Let R be a right weakly regular LA-ring. Then every intuitionistic fuzzy left (right) ideal of R is an intuitionistic fuzzy ideal of R. Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy right ideal of R and x, y ∈ R, this means that there exist a, b ∈ R such that x = (xa)(xb). Thus µA(xy) = µA(((xa)(xb))y) = µA((((xb)a)x)y) = µA((((ab)x)x)y) = µA((yx)((ab)x)) = µA((yx)(nx)) say ab = n ≥ µA(yx) ≥ µA(y) and γA(xy) = γA(((xa)(xb))y) = γA((((xb)a)x)y) = γA((((ab)x)x)y) = γA((yx)((ab)x)) = γA((yx)(nx)) ≤ γA(yx) ≤ γ(y). Therefore A is an intuitionistic fuzzy ideal of R. Similarly, for left ideal. Remark 9. The concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides in right weakly regular LA-rings. Proposition 9. Every intuitionistic fuzzy generalized bi-ideal of a right weakly regular LA-ring R with left identity e, is an intuitionistic fuzzy bi-ideal of R. Proof. Assume that A = (µA, γA) is an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, then there exist elements a, b ∈ R such that x = (xa)(xb). We have to show that A is an intuitionistic fuzzy LA-subring of R. Thus µA(xy) = µA(((xa)(xb))y) = µA((x((xa)b))y) ≥ min{µA(x), µA(y)} and γA(xy) = γA(((xa)(xb))y) = γA((x((xa)b))y) ≤ max{γA(x), γA(y)}. So A is an intuitionistic fuzzy LA-subring of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 242 Lemma 16. Let R be a left weakly regular LA-ring. Then every intuitionistic fuzzy right ideal of R is an intuitionistic fuzzy ideal of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy right ideal of R and x, y ∈ R, this implies that there exist a, b ∈ R such that x = (ax)(bx). Thus µA(xy) = µA(((ax)(bx))y) = µA(y(bx))(ax) ≥ µA(y(bx)) ≥ µA(y) and γA(xy) = γA(((ax)(bx))y) = γA(y(bx))(ax) ≤ γA(y(bx)) ≤ γA(y). Hence A is an intuitionistic fuzzy ideal of R. Remark 10. The concept of intuitionistic fuzzy (right, two-sided) ideals coincides in left weakly regular LA-rings. Lemma 17. Let R be a left weakly regular LA-ring with left identity e. Then every intu- itionistic fuzzy left ideal of R is an intuitionistic fuzzy ideal of R. Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy left ideal of R and x, y ∈ R, this means that there exist a, b ∈ R such that x = (ax)(bx). Thus µA(xy) = µA(((ax)(bx))y) = µA(((ab)(xx))y) = µA((x((ab)x))y) = µA((y((ab)x))x) ≥ µA(x) and γA(xy) = γA(((ax)(bx))y) = γA(((ab)(xx))y) = γA((x((ab)x))y) = γA((y((ab)x))x) ≤ γA(x). Therefore A is an intuitionistic fuzzy ideal of R. Remark 11. The concept of intuitionistic fuzzy (left, two-sided) ideals coincides in left weakly regular LA-rings with left identity e. Proposition 10. Every intuitionistic fuzzy generalized bi-ideal of a left weakly regular LA-ring R with left identity e, is an intuitionistic fuzzy bi-ideal of R. Proof. Assume that A = (µA, γA) is an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, then there exist elements a, b ∈ R such that x = (ax)(bx). We have to show that A is an intuitionistic fuzzy LA-subring of R. Thus µA(xy) = µA(((ax)(bx))y) = µA(((ab)(xx))y) = µA((x((ab)x))y) ≥ min{µA(x), µA(y)} and γA(xy) = γA(((ax)(bx))y) = γA(((ab)(xx))y) = γA((x((ab)x))y) ≤ max{γA(x), γA(y)}. So A is an intuitionistic fuzzy LA-subring of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 243 Remark 12. It is easy to see that, if R is a left (right) weakly regular locally associative LA-ring. Then for every intuitionistic fuzzy left (right) ideal A = (µA, γA) of R, A(an) = A(a2n) for all a ∈ R, where n is any positive integer. Theorem 10. Let R be an LA-ring with left identity e, such that (xe)R = xR for all x ∈ R. Then the following conditions are equivalent. (1) R is a left weakly regular. (2) A ∩ B = A ◦ B for every intuitionistic fuzzy right ideal A and every intuitionistic fuzzy left ideal B of R. Proof. Suppose that (1) holds. Since A ◦ B ⊆ A ∩ B for every intuitionistic fuzzy right ideal A = (µA, γA) and every intuitionistic fuzzy left ideal B = (µB, γB) of R by the Lemma 10. Let x ∈ R, this implies that there exist a, b ∈ R such that x = (ax)(bx) = (ab)(xx) = x((ab)x). Now (µA ◦ µB)(x) = ∨x=∑n i=1 aibi {∧ni=1 {µA (ai) ∧ µB (bi)}} ≥ µA(x) ∧ µB((ab)x) ≥ µA(x) ∧ µB(x) = (µA ∩ µB)(x) and (γA ◦ γB)(x) = ∧x=∑n i=1 aibi {∨ni=1 {γA (ai) ∨ γB (bi)}} ≤ γA(x) ∨ γB((ab)x) ≤ γA(x) ∨ γB(x) = (γA ∪ γB)(x). Thus µA∩µB ⊆ µA◦µB and γA∪γB ⊇ γA◦γB, i.e., A∩B ⊆ A◦B. Hence A∩B = A◦B, i.e., (1)⇒ (2) . Assume that (2) is true and a ∈ R. Then Ra is a left ideal of R containing a by the Lemma 8 and aR ∪ Ra is a right ideal of R containing a by the Proposition 5. So χRa is an intuitionistic fuzzy left ideal and χaR∪Ra is an intuitionistic fuzzy right ideal of R, by the Proposition 2. Then by our assumption χaR∪Ra ∩ χRa = χaR∪Ra ◦ χRa, i.e., χ(aR∪Ra)∩Ra = χ(aR∪Ra)Ra by the Theorem 1. Thus (aR∪Ra)∩Ra = (aR∪Ra)Ra. Since a ∈ (aR ∪ Ra) ∩ Ra, i.e., a ∈ (aR ∪ Ra)Ra, so a ∈ (aR)(Ra) ∪ (Ra)(Ra). This implies that a ∈ (aR)(Ra) or a ∈ (Ra)(Ra). If a ∈ (Ra)(Ra), then R is a left weakly regular. If a ∈ (aR)(Ra), then (aR)(Ra) = ((ea)(RR))(Ra) = ((RR)(ae))(Ra) = (((ae)R)R)(Ra) = ((aR)R)(Ra) = ((RR)a)(Ra) = (Ra)(Ra). Hence R is a left weakly regular, i.e., (2)⇒ (1) . Theorem 11. Let R be an LA-ring with left identity e, such that (xe)R = xR for all x ∈ R. Then the following conditions are equivalent. (1) R is a left weakly regular. (2) A∩ I ⊆ A ◦ I for every intuitionistic fuzzy bi-ideal A and every intuitionistic fuzzy ideal I of R. (3) B ∩ I ⊆ B ◦ I for every intuitionistic fuzzy generalized bi-ideal B and every intu- itionistic fuzzy ideal I of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 244 Proof. Assume that (1) holds. Let B = (µB, γB) be an intuitionistic fuzzy generalized bi-ideal and I = (µI , γI) be an intuitionistic fuzzy ideal of R. Let x ∈ R, this means that there exist a, b ∈ R such that x = (ax)(bx) = (ab)(xx) = x((ab)x). Now (µB ◦ µI)(x) = ∨x=∑n i=1 aibi {∧ni=1 {µB (ai) ∧ µI (bi)}} ≥ µB(x) ∧ µI((ab)x) ≥ µB(x) ∧ µI(x) = (µB ∩ µI)(x). ⇒ µB ∩ µI ⊆ µB ◦ µI . Similarly, γA ∪ γB ⊇ γA ◦ γB. Hence A ∩ B ⊆ A ◦ B, i.e., (1) ⇒ (3) . It is clear that (3)⇒ (2) . Suppose that (2) holds. Then A∩ I ⊆ A ◦ I, where A is an intuitionistic fuzzy right ideal of R. Since A ◦ I ⊆ A∩ I, so A ◦ I = A∩ I. Therefore R is a left weakly regular by the Theorem 10, i.e., (2)⇒ (1) . Theorem 12. Let R be an LA-ring with left identity e, such that (xe)R = xR for all x ∈ R. Then the following conditions are equivalent. (1) R is a left weakly regular. (2) A∩ I ∩C ⊆ (A ◦ I) ◦C for every intuitionistic fuzzy bi-ideal A, every intuitionistic fuzzy ideal I and every intuitionistic fuzzy right ideal C of R. (3) B ∩ I ∩C ⊆ (B ◦ I) ◦C for every intuitionistic fuzzy generalized bi-ideal B, every intuitionistic fuzzy ideal I and every intuitionistic fuzzy right ideal C of R. Proof. Suppose that (1) holds. Let B = (µB, γB) be an intuitionistic fuzzy generalized bi-ideal, I = (µI , γI) be an intuitionistic fuzzy ideal and C = (µC , γC) be an intuitionistic fuzzy right ideal of R. Let x ∈ R, then there exist elements a, b ∈ R such that x = (ax)(bx). Here x = (ax)(bx) = (xb)(xa) xb = ((ax)(bx))b = ((xx)(ba))b = (b(ba))(xx) = c(xx) = x(cx) say c = b(ba) Now ((µB ◦ µI) ◦ µC)(x) = ∨x=∑n i=1 aibi {∧ni=1 {(µB ◦ µI) (ai) ∧ µC (bi)}} ≥ (µB ◦ µI)(xb) ∧ µC(xa) ≥ (µB ◦ µI)(xb) ∧ µC(x) = ∨xb=∑n i=1 piqi {∧ni=1 {µB (pi) ∧ µI (qi)}} ∧ µC(x) ≥ µB(x) ∧ µI(cx) ∧ µC(x) ≥ µB(x) ∧ µI(x) ∧ µC(x) = (µB ∩ µI ∩ µC)(x). ⇒ µB ∩ µI ∩ µC ⊆ (µB ◦ µI) ◦ µC . Similarly, γB ∪ γI ∪ γC ⊇ (γB ◦ γI) ◦ γC , i.e., B ∩ I ∩C ⊆ (B ◦ I) ◦C. Hence (1)⇒ (3) . It is clear that (3)⇒ (2) , every intuitionistic fuzzy bi-ideal of R is an intuitionistic fuzzy N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 245 generalized bi-ideal of R. Assume that (2) is true. Then A∩ I ∩R ⊆ (A ◦ I) ◦R, where A is an intuitionistic right ideal of R, i.e., A∩ I ⊆ A◦ I. Since A◦ I ⊆ A∩ I, so A◦ I = A∩ I. Therefore R is a left weakly regular by the Theorem 10, i.e., (2)⇒ (1) . Lemma 18. Every intuitionistic fuzzy left (right) ideal of an intra-regular LA-ring R, is an intuitionistic fuzzy ideal of R. Proof. Suppose that A = (µA, γA) is an intuitionistic fuzzy right ideal of R. Let x, y ∈ R, this implies that there exist ai, bi ∈ R, such that x = ∑n i=1(aix 2)bi. Thus µA(xy) = µA(((aix 2)bi)y) = µA((ybi)(aix 2)) ≥ µA(ybi) ≥ µA(y) and γA(xy) = γA(((aix 2)bi)y) = γA((ybi)(aix 2)) ≤ γA(ybi) ≤ γA(y). Hence A is an intuitionistic fuzzy ideal of R. Similarly, for left ideal. Remark 13. The concept of intuitionistic fuzzy (left, right, two-sided) ideals coincides in inta-regular LA-rings. Proposition 11. Every intuitionistic fuzzy generalized bi-ideal of an intra-regular LA-ring R with left identity e, is an intuitionistic fuzzy bi-ideal of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy generalized bi-ideal of R and x, y ∈ R, this implies that there exist ai, bi ∈ R such that x = ∑n i=1(aix 2)bi. We have to show that A is an intuitionistic fuzzy LA-subring of R. Now x = (aix 2)bi = (aix 2)(ebi) = (aie)(x 2bi) = (aie)((xx)bi) = (aie)((bix)x) = (x(bix))(eai) = (x(bix))ai = (ai(bix))x = (ai(bix))(ex) = (xe)((bix)ai) = (bix)((xe)ai) = (bix)((aie)x) = (x(aie))(xbi) = x((x(aie))bi) = xn, say n = (x(aie))bi Thus µA(xy) = µA((xn)y) ≥ min{µA(x), µA(y)} and γA(xy) = γA((xn)y) ≤ max{γA(x), γA(y)}. Hence A is an intuitionistic fuzzy LA-subring of R. Theorem 13. Let R be an LA-ring with left identity e, such that (xe)R = xR for all x ∈ R. Then the following conditions are equivalent. (1) R is an intra-regular. (2) A ∩ B ⊆ A ◦ B for every intuitionistic fuzzy right ideal B and every intuitionistic fuzzy left ideal A of R. N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 246 Proof. Assume that (1) holds. Let x ∈ R, then there exist elements ai, bi ∈ R such that x = ∑n i=1(aix 2)bi. Now x = (aix 2)bi = (ai(xx))bi = (x(aix))(ebi) = (xe)((aix)bi) = (aix)((xe)bi). Thus (µA ◦ µB)(x) = ∨x=∑n i=1 aibi {∧ni=1µA (ai) ∧ µB (bi)} ≥ min{µA(aix), µB((xe)bi)} ≥ min{µA(x), µB(x)} = (µA ∧ µB)(x) = (µA ∩ µB)(x). ⇒ µA ∩ µB ⊆ µA ◦ µB. Similarly, we have γA ∪ γB ⊇ µA ◦ µB. Hence A ∩B ⊆ A ◦B, i.e., (1)⇒ (2) . Suppose that (2) is true and a ∈ R, then Ra is a left ideal of R containing a by the Lemma 8 and aR ∪ Ra is a right ideal of R containing a by the Proposition 5. This means that χRa is an intuitionistic fuzzy left ideal and χaR∪Ra is an intuitionistic fuzzy right ideal of R, by the Lemma 2. By our supposition χaR∪Ra ∩ χRa ⊆ χRa ◦ χaR∪Ra, i.e., χ(aR∪Ra)∩Ra ⊆ χ(Ra)(aR∪Ra). Thus (aR∪Ra)∩Ra ⊆ Ra(aR∪Ra). Since a ∈ (aR∪Ra)∩Ra, i.e., a ∈ Ra(aR ∪Ra) = (Ra)(aR) ∪ (Ra)(Ra). Now (Ra)(aR) = (Ra)((ea)(RR)) = (Ra)((RR)(ae)) = (Ra)(((ae)R)R) = (Ra)((aR)R) = (Ra)((RR)a) = (Ra)(Ra). This implies that (Ra)(aR) ∪ (Ra)(Ra) = (Ra)(Ra) ∪ (Ra)(Ra) = (Ra)(Ra) = ((Ra)a)R = ((Ra)(ea))R = ((Re)(aa))R = (Ra2)R. Thus a ∈ (Ra2)R, i.e., a is an intra regular. Therefore R is an intra-regular, i.e., (2)⇒ (1) . Theorem 14. Let R be an intra-regular locally associative LA-ring. Then for every intu- itionistic fuzzy right ideal A = (µA, γA) of R, A(an) = A(a2n) for all a ∈ R, where n is a positive integer. Proof. For n = 1. Let a ∈ R, this implies that there exist elements xi, yi ∈ R such that a = ∑n i=1(xia 2)yi. Thus µA (a) = µA((xia 2)yi) ≥ µA(xia 2) ≥ µA(a2) N. Kausar, M. A. Waqar / Eur. J. Pure Appl. Math, 12 (1) (2019), 226-250 247 = µA(aa) ≥ min{µA (a) , µA (a)} = µA (a) . Similarly, γA (a) = γA(a2), so A(a) = A(a2). Result is also true for n = 2, as a2 = aa = ((xia 2)yi)((xia 2)yi) = (x2i a 4)y2i . Assume that the result is true for n = k, i.e., A(ak) = A(a2k). Now ak+1 = aka = ((xki a 2k)yki )((xia 2)yi) = (xk+1 i a2(k+1))yk+1 i . Thus µA(ak+1) = µA((xk+1 i a2(k+1))yk+1 i ) ≥ µA(xk+1 i a2(k+1)) ≥ µA(a2(k+1)) = µA(ak+1ak+1) ≥ min{µA ( a(k+1) ) , µA ( a(k+1) ) } = µA ( a(k+1) ) . Similarly, γA (a) = γA(a2(k+1)), so A(ak+1) = A(a2(k+1)). Hence by induction method, the result is true for all positive integers. Proposition 12. Let R be an intra-regular locally associative LA-ring with left identity e. Then for every intuitionistic fuzzy right ideal A = (µA, γA) of R, A(ab) = A(ba) for all a, b ∈ R. Proof. Same as Lemma 14. Theorem 15. If an IFS A = (µA, γA) of an LA-ring R is an intuitionistic fuzzy (1, 2)- ideal of R, then so is �A = (µA, µA) ( resp. ♦A = (γA, γA)). Proof. Let A = (µA, γA) be an intuitionistic fuzzy (1, 2)-ideal of R. We have to show that �A = (µA, µA) is also an intuitionistic fuzzy (1, 2)-ideal of R. Now µA(x− y) = 1− µA(x− y) ≤ 1−min {µA(x), µA(y)} = max {1− µA(x), 1− µA(y)} = max{µA(x), µA(y)}. µA(xy) = 1− µA(xy) ≤ 1−min {µA(x), µA(y)} = max {1− µA(x), 1− µA(y)} = max{µA(x), µA(y)}. µA((xa)(yz)) = 1− µA((xa)(yz)) ≤ 1−min {µA(x), µA(y), µA(z)} = max {1− µA(x), 1− µA(y), 1− µA(z)} = max{µA(x), µA(y), µA(z)}. Hence �A is an intuitionistic fuzzy (1, 2)-ideal of R. Similarly, for ♦A = (γA, γA) . Remark 14. 1. An IFS A = (µA, γA) of an LA-ring R is an intuitionistic fuzzy (1, 2)- ideal of R if and only if �A = (µA, µA) (resp. ♦A = (γA, γA)) is an intuitionistic fuzzy (1, 2)-ideal of R. 2. If an IFS A = (µA, γA) of an LA-ring R is an intuitionistic fuzzy bi-ideal of R, then so is �A = (µA, µA) (resp. ♦A = (γA, γA)). 3. An IFS A = (µA, γA) of an LA-ring R is an intuitionistic fuzzy bi-ideal of R if and only if �A = (µA, µA) (resp. ♦A = (γA, γA)) is an intuitionistic fuzzy bi-ideal of R. PSEUDO-INTEGRALITY RELATIVE TO A RATIONAL CYCLIC MONOID 248 Theorem 16. An IFS A = (µA, γA) of an LA-ring R is an intuitionistic fuzzy (1, 2)-ideal of R if and only if the fuzzy subsets µA and γA are fuzzy (1, 2)-ideals of R. Proof. Let A = (µA, γA) be an intuitionistic fuzzy (1, 2)-ideal of R, this implies that µA is a fuzzy (1, 2)-ideal of R. We have to show that γA is also a fuzzy (1, 2)-ideal of R. Now γA(x− y) = 1− γA(x− y) ≥ 1−max{γA(x), γA(y)} = min{1− γA(x), 1− γA(y)} = min{γA(x), γA(y)}. γA(xy) = 1− γA(xy) ≥ 1−max{γA(x), γA(y)} = min{1− γA(x), 1− γA(y)} = min{γA(x), γA(y)}. γA((xa)(yz)) = 1− γA((xa)(yz)) ≥ 1−max{γA(x), γA(y), γA(z)} = min{1− γA(x), 1− γA(y), 1− γA(z)} = min{γA(x), γA(y), γA(z)}. Therefore γA is a fuzzy (1, 2)-ideal of R. Conversely, suppose that µA and γA are fuzzy (1, 2)-ideals of R. We have to show that A = (µA, γA) is an intuitionistic fuzzy (1, 2)-ideal of R. Now 1− γA(x− y) = γA(x− y) ≥ min{γA(x), γA(y)} = min{1− γA(x), 1− γA(y)} = 1−max{γA(x), γA(y)}. 1− γA(xy) = γA(xy) ≥ min{γA(x), γA(y)} = min{1− γA(x), 1− γA(y)} = 1−max{γA(x), γA(y)}. 1− γA((xa)(yz)) = γA((xa)(yz)) ≥ min{γA(x), γA(y), γA(z)} = min{1− γA(x), 1− γA(y), 1− γA(z)} = 1−max{γA(x), γA(y), γA(z)}. Therefore A is an intuitionistic fuzzy (1, 2)-ideal of R. Remark 15. An IFS A = (µA, γA) of an LA-ring R is an intuitionistic fuzzy bi-ideal of R if and only if the fuzzy subsets µA and γA are fuzzy bi-ideals of R. References [1] K. T. Atanassov, Intuitionistic fuzzy sets, Fuzzy sets and systems, 20(1986) 87-96. [2] K. T. Atanassov, New operations defined over the intuitionistic fuzzy sets, Fuzzy sets and systems, 61(1994) 137-142. REFERENCES 249 [3] B. Banerjee and D. K. Basnet, Intuitionistic fuzzy subrings and ideals, J. Fuzzy Math., 11(2003) 139-155. [4] R. J. Cho, J. Jezek and T. Kepka, Paramedial Groupoids, Czechoslovak Math. J., 49(1999) 277-290. [5] P. S. Das, Fuzzy groups and level subgroups, J. Math. Anal. Appli., 84(1981) 264-269. [6] K. C. Gupta and M. K. Kantroo, The intrinsic product of fuzzy subsets of a ring, Fuzzy sets and systems, 57(1993) 103-110. [7] K. Hur, H. w. Kang and H. k. Song, Intuitionistic Fuzzy subgroups and subrings, Honam Math. J., 25(2003) 19-41. [8] K. Hur, S. Y. Jang and H. W. Kang, Intuitionistic fuzzy ideals of a ring, J. Korea Soc. Math. Educ. Ser. B: Pure Appl. Math., 12 (2005) 193-209. [9] J. Jezek and T. Kepka, Medial groupoids, Rozpravy CSAV Rada mat. a prir. ved 93/2, 1983, 93 pp. [10] M. S. Kamran, Conditions for LA-semigroups to resemble associative structures, Ph.D. Thesis, Quaid-i-Azam University, Islamabad, 1993. [11] M. A. Kazim and M. Naseerdin, On almost semigroups, Alig. Bull. Math., 2(1972) 1-7. [12] N. Kuroki, Regular fuzzy duo rings, Inform. Sci., 94(1996) 119-139. [13] W. J. Liu, Fuzzy invariant subgroups and ideals, Fuzzy sets and systems, 8(1982) 133-139. [14] T. K. Mukherjee and M. K. Sen, On fuzzy ideals of a ring 1, Fuzzy sets and systems, 21(1987) 99-104. [15] T. K. Mukherjee and M. K. Sen, Prime fuzzy ideals in rings, Fuzzy sets and systems, 32(1989) 337-341. [16] P. V. Protic and N. Stevanovic, AG-test and some general properties of Abel- Grassmann’s groupoids, Pure Math. Appli., 6(1995) 371-383. [17] V. S. Ramamurthi, Weakly regular rings, Canad. Math. Bull., 16 (1973) 317-321. [18] T. Shah, N. Kausar and I. Rehman, Intuitionistic fuzzy normal subrings over a non- associative ring, An. St. Univ. Ovidius Constanta, 1(2012) 369-386. [19] Shah, Kausar, “Characterizations of non-associative ordered semigroups by their fuzzy bi-ideals”, Theoretical Computer Science, Vol. 529 (2014) 96-110. REFERENCES 250 [20] T. Shah and I. Rehman, On LA-rings of finitely non-zero functions, Int. J. Contempt. Math. Sci., 5(2010) 209-222. [21] U. M. Swamy and K. L. N. Swamy, Fuzzy prime ideals of rings, J. Math. Anal. Appl., 134(1988) 94-103. [22] L. A. Zadeh, Fuzzy sets, Information and control, 8(1965) 338-363.