EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 3, 2020, 498-512 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global The Structure of Pseudo-BF/BF ∗-algebra Hessah M. Al-Malki1, Deena S. Al-Kadi1,∗ 1 Department of Mathematics and Statistics, Faculty of Science, Taif University, Taif, Saudi Arabia Abstract. In this paper, we study the structure of pseudo-BF/BF ∗-algebra as a generalization of BF -algebra. We show how pseudo-BF/BF ∗-algebra and pseudo-BCK-algebra are related. We study some elementary properties related to pseudo-BF -algebra and pseudo-BF ∗-algebra. 2020 Mathematics Subject Classifications: 06F35, 03G25 Key Words and Phrases: pseudo-BF -algebra, pseudo-BF ∗-algebra, pseudo-ideal, pseudo- atoms. 1. Introduction Through the work of the Japanese mathematicians Imai and Iseki the notions of BCK/BCI-algebra were introduced (see [7] and [8]). Neggers and Sik introduced the concept of B-algebra, and obtained several results (we refer the reader to [13] for more details). In [17], Walendziak introduced a generalization of B-algebra named BF -algebra and investigated some properties of ideals and normal-ideals in BF -algebra and gave some characterization of them. In [6], Georgescu and Iorgulescu introduced an exten- sion of BCK-algebra called pseudo-BCK-algebra. Moreover, they gave the connection of pseudo-BCK-algebra with pseudo-MV -algebra and with pseudo-BL-algebra. Dudek and Jun introduced the notion pseudo-BCI-algebra as a natural generalization of BCI- algebra and of pseudo-BCK-algebra and investigated some of their properties. They gave some conditions for a pseudo-BCI-algebra to be a pseudo-BCK-algebra (see [4] for more details). In [10], Jun, Kim and Neggers studied pseudo-atoms, pseudo-ideals and pseudo-homomorphisms in pseudo-BCI-algebra. In [12], Kim and So discussed minimality on elements in pseudo-BCI-algebra and concluded some of the properties in B-algebra. Walendziak in [18] introduced the notion of pseudo-BCH-algebra and investigated some properties and gave conditions to when a pseudo-BCH-algebra be a pseudo-BCI-algebra. The authors G. Georgescu and A. Iorgulescu in [5], and independently Rachunek in [15], ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i3.3735 Email addresses: dak12le@hotmail.co.uk (Deena S. Al-Kadi), hhhmmm9999@hotmail.com (Hessah M. Al-Malki) https://www.ejpam.com 498 c© 2020 EJPAM All rights reserved. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 499 studied a non-commutative generalization of the MV -algebra named pseudo-MV -algebra. In [16], pseudo-BL-algebra was introduced as a generalization of BL-algebra and pseudo- MV -algebra and basic properties, filters, normal-filters and congruences were given. Di Nola, Georgescu and Iorgulescu, in [14], investigated pseudo-BL-algebra including def- inition, basic properties, filters, normal-filters and congruences. Moreover, they gave some important classes of pseudo-BL-algebra and some results concerning the pseudo- BL-chains. In [11], Jun, Kim and Neggers introduced the notion of pseudo-d-algebra as an extension of d-algebra and they showed that the class of pseudo-d-algebra can be in- cluded in the class of coupled d-algebra. In [1], the authors, introduced the concept of pseudo-BE-algebra. They studied the concepts of pseudo-subalgebra, pseudo-filter and pseudo-upper-set and proved that every pseudo-filter is a union of pseudo-upper-sets. In [9], Jun and Ahn studied some properties of pseudo-BH-algebra. Furthermore, they in- troduced the concept of pseudo-complicated-BH-algebra and got some related properties. In [3], Ciungu introduced and investigated pointed-pseudo-BE-algebra and commutative- pseudo-BE-algebra and proved that the class of commutative-pseudo-BE-algebra and the class of commutative-pseudo-BCK-algebra are equivalent. In this paper, we study the structure of pseudo-BF/BF ∗-algebra. We introduce,in the second section, the notion of pseudo-BF/BF ∗-algebra and find the relation between pseudo-BF/BF ∗-algebra with pseudo-BCK-algebra. In the third section, we study pseudo- subalgebra, pseudo-ideal and pseudo-normal-ideal of pseudo-BF -algebra. We study pseudo- atoms of pseudo-BF/BF ∗-algebra in the last section. We start by recalling the definitions and elementary properties related to the paper. Definition 1. [17, Definition 2.1] An algebra (E; •, 0) of type (2, 0) is called a BF -algebra if the following axioms are satisfies the following axiom, for all a, b ∈ E: (BF (1)) a • a = 0, (BF (2)) a • 0 = a, (BF (3)) 0 • (a • b) = b • a. Definition 2. [2, Definition 2.3] In BF -algebra (E; •, 0), we can define a binary relation ”≤” on E as follows: a ≤ b if and only if a • b = 0 for all a, b ∈ E. Any BF -algebra, satisfies the properties given in the following Proposition. Proposition 1. [17, Proposition 2.5] Let (E; •, 0) be a BF -algebra, then, (1) 0 • (0 • a) = a for all a ∈ E, (2) if 0 • a = 0 • b, then a = b for all a, b ∈ E, (3) if a • b = 0, then b • a = 0 for all a, b ∈ E. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 500 We give next the definition of pseudo-BCK-algebra. Definition 3. [6, Definition 3] An algebra (E;≤, •, ?, 0) of type (2, 2, 0), where ”≤” is a binary relation on a set E, ”•” and ”?” are binary operations on E and ”0” is a constant of E, is called a pseudo-BCK-algebra if the following are satisfied: ∀a, b, c ∈ E, (pBCK(1)) (a • b) ? (a • c) ≤ c • b and (a ? b) • (a ? c) ≤ c ? b, (pBCK(2)) a ? (a • b) ≤ b and a • (a ? b) ≤ b, (pBCK(3)) a ≤ a, (pBCK(4)) 0 ≤ a, (pBCK(5)) a ≤ b and b ≤ a then a = b, (pBCK(6)) a ≤ b ⇔ a • b = 0 if and only if a ? b = 0. Theorem 1. [6, Theorem 7] In a pseudo-BCK-algebra (E;≤, •, ?, 0), for all a, b, c ∈ E we have (a • b) ? c = (a ? c) • b. Theorem 2. [6, Theorem 8] In any pseudo-BCK-algebra (E;≤, •, ?, 0) we have, for all a, b, c ∈ E: (1) a • b ≤ c if and only if a ? c ≤ b, (2) a • b ≤ a and a ? b ≤ a. 2. Pseudo-BF/BF ∗-algebra In this section, we give a generalization of BF -algebra named pseudo-BF -algebra and study its structure. Also, we will introduce pseudo-BF ∗-algebra and find the relation between pseudo-BF/BF ∗-algebra and pseudo-BCK-algebra. Definition 4. An algebra (E; •, ?, 0) of type (2, 2, 0) is said to be a pseudo-BF -algebra, if the following axioms are satisfied for all a, b ∈ E : (pBF (1)) a • a = 0 and a ? a = 0, (pBF (2)) a • 0 = a and a ? 0 = a, (pBF (3)) 0 • (a ? b) = b ? a and 0 ? (a • b) = b • a. The following examples illustrates the definition. Example 1. Consider the group (G; +, 0), where ”+” is the usual addition. Define the operations ”•” and ”?” on G by: H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 501 a • b = (−b) + a and a ? b = (−b) + a for all a, b ∈ G then (G; •, ?, 0) is a pseudo-BF -algebra. Note: It is obvious that in any pseudo-BF -algebra E if a • b = a ? b for all a, b ∈ E then E is a BF -algebra. Example 2. Define the operations ”•” and ”?” on E = {0, 1, 2, 3}, by the following Cay- ley tables: Table 1 Table 2 • 0 1 2 3 0 0 1 2 3 1 1 0 3 0 2 2 3 0 2 3 3 0 2 0 ? 0 1 2 3 0 0 1 2 3 1 1 0 1 1 2 2 1 0 1 3 3 1 1 0 Then (E; •, 0) and (E; ?, 0) are BF -algebras (shown in [17]). It is obvious that a•a = 0 and a?a = 0. Moreover, a • 0 = a and a? 0 = a. It is direct to check that 0 • (a? b) = b ? a and 0 ? (a • b) = b • a is satisfied for all a, b ∈ E. Thus (E; •, ?, 0) is a pseudo-BF -algebra. Corollary 1. Any two BF -algebras does not necessarily construct a pseudo-BF -algebra. Moreover, if (R; •, ?, 0) is a pseudo-BF -algebra then it is not necessary for both (R; •, 0) and (R; ?, 0) to be a BF -algebra. The following two examples proves the Corollary. Example 3. Define the operations ”•” and ”?” on E = {0, 1, 2, 3, 4, 5}, by the following Cayley tables: Table 3 Table 4 • 0 1 2 3 4 5 0 0 2 1 3 4 5 1 1 0 2 4 5 3 2 2 1 0 5 3 4 3 3 4 5 0 2 1 4 4 5 3 1 0 2 5 5 3 4 2 1 0 ? 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 0 3 2 1 0 2 2 3 0 0 0 2 3 3 2 0 0 3 1 4 4 1 0 3 0 0 5 5 0 2 1 0 0 Then (E; •, 0),(E; ?, 0) are BF -algebras but (E; •, ?, 0) is not since 0 • (0 ? 1) = 0 • 1 = 2 6= 1 ? 0 = 1. Example 4. Let R be the set of real numbers. Define the operations ”•” and ”?” on R for all a, b ∈ R by: a • b =  a if b = 0, b if a = 0, 0 otherwise. a ? b =  a if b = 0, 0 if a = 0, a = b, b ? a otherwise. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 502 Then (R; •, ?, 0) is a pseudo-BF -algebra. The algebra (R; •, 0) is BF -algebra [17], but the algebra (R; ?, 0) is not. Proposition 2. If (E; •, ?, 0) is a pseudo-BF -algebra for all a, b ∈ E then (1) 0 • (0 • a) = a and 0 ? (0 ? a) = a, (2) 0 ? (0 • a) = a and 0 • (0 ? a) = a, (3) 0 • a = 0 ? b, implies a = b. Proof. (1) By (pBF (2)), (pBF (3)) and let a ∈ E then 0 • (0 • a) = 0 • [0 ? (a • 0)] = 0 • (0 ? a) = a ? 0 = a and 0 ? (0 ? a) = 0 ? [0 • (a ? 0)] = 0 ? (0 • a) = a • 0 = a. (2) Let a ∈ E. By (pBF (2)) and (pBF (3)) we obtain 0 ? (0 • a) = a • 0 = a and 0 • (0 ? a) = a ? 0 = a, that is (2) holds. (3) Let 0 • a = 0 ? b, then it follows from (1) and (2) that a = 0 ? (0 • a) = 0 ? (0 ? b) = b. Corollary 2. In a pseudo-BF -algebra (E; •, ?, 0), a • b = 0 does not imply b ? a = 0 and similarly a ? b = 0 does not imply b • a = 0. ∀a, b ∈ E. Proof. Let a, b ∈ E and a • b = 0. Then 0 = 0 ? 0 = 0 ? (a • b) = b • a. Then it is not necessary that b ? a = 0. Similarly, if a ? b = 0 then it is not necessary that b • a = 0. Note: From the proof of (Corollary 2) we see that if a • b = 0, then b • a = 0 and if a ? b = 0, then b ? a = 0, for all a, b ∈ E. As in BF -algebra, a binary relation ”≤” could be defined in pseudo-BF -algebra as follows: a ≤ b ⇔ a • b = 0 ⇔ a ? b = 0 ∀a, b ∈ E. Therefore we can rewrite the definition of a pseudo-BF -algebra with a binary relation ”≤” as follows: Definition 5. The algebra (E;≤, •, ?, 0) where ”≤” is a binary relation on a set E, ”•” and ”?” are binary operations on E and ”0” is an element of E, is said to be a pseudo- BF -algebra if for all a, b, c ∈ E the following axioms are satisfied: (pBF (1’)) a ≤ a, (pBF (2’)) a • 0 ≤ a and a ? 0 ≤ a, (pBF (3’)) 0 • (a ? b) ≤ b ? a and 0 ? (a • b) ≤ b • a, (pBF (4’)) a ≤ b ⇔ a • b = 0 ⇔ a ? b = 0. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 503 Proposition 3. The following proposition holds in any pseudo-BF -algebra (E;≤, •, ?, 0),: 0 ≤ a implies a = 0 ∀a ∈ E. Proof. Since 0 ≤ a, we have 0 • a = 0 ? a = 0 from (pBF (4’)). Using (Proposition 2 (1)), (pBF (1’)) and (pBF (4’)) we get a = 0 • (0 • a) = 0 • 0 = 0. Next we introduce pseudo-BF ∗-algebra and we find some results. Definition 6. A pseudo-BF -algebra (E; •, ?, 0) is called a pseudo-BF ∗-algebra, for all a, b, c ∈ E if it satisfies the following identity: (pBF ∗) (a • b) ? c = (a ? c) • b. We can see that any pseudo-BF ∗-algebra is a pseudo-BF -algebra and any pseudo-BF - algebra satisfying (pBF ∗) is a pseudo-BF ∗-algebra. Example 5. In Example 1, it is straight forward to see that (G; •, ?, 0) is a pseudo-BF ∗- algebra. Example 6. In Example 2, (E; •, ?, 0) is not a pseudo-BF ∗-algebra, as (1•1)?2 = 0?2 = 2 6= (1 ? 2) • 1 = 1 • 1 = 0. Proposition 4. Let (E;≤, •, ?, 0) be a pseudo-BF ∗-algebra. The following axioms are satisfied for any a, b, c ∈ E: (1) a ≤ 0 implies a = 0, (2) a • (a ? b) ≤ b and a ? (a • b) ≤ b, (3) a • b ≤ c if and only if a ? c ≤ b, (4) 0 • (a • b) = (0 ? a) ? (0 • b), (5) 0 ? (a ? b) = (0 • a) • (0 ? b), (6) 0 • a = 0 ? a. Proof. (1) Let a ≤ 0. Then a•0 = a?0 = 0 by (pBF (4’)). Multiplying by ”a” from the right we have 0?a = (a•0)?a = (a?a)•0 = 0•0 = 0 and 0•a = (a?0)•a = (a•a)?0 = 0?0 = 0, using (pBF ∗) and (pBF (1’)). Now, using (Proposition 2 (1)) and (pBF (1’)), we get a = 0 • (0 • a) = 0 • 0 = 0. (2) From (pBF ∗), (pBF (1’)) and (pBF (4’)), we have [a • (a ? b)] ? b = (a ? b) • (a ? b) = 0 and [a ? (a • b)] • b = (a • b) ? (a • b) = 0. Thus a • (a ? b) ≤ b and a ? (a • b) ≤ b. (3) By (pBF ∗) and (pBF (4’)) we have a • b ≤ c ⇔ (a • b) ? c = 0 ⇔ (a ? c) • b = 0 ⇔ a ? c ≤ b. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 504 (4) Let a, b ∈ E. Then by using (pBF (1’)), (pBF (4’)) and (pBF ∗) when needed we have (0 ? a) ? (0 • b) = ([(a • b) • (a • b)] ? a) ? (0 • b) = ([(a • b) ? a] • (a • b)) ? (0 • b) = ([(a?a)•b]•(a•b))?(0•b) = ((0•b)•(a•b))?(0•b) = ((0•b)?(0•b))•(a•b) = 0•(a•b). (5) Can be proved as (4). (6) Let a ∈ E. From (pBF (1’)), (pBF (4’)) and (pBF ∗) we have 0 • a = (a ? a) • a = (a • a) ? a = 0 ? a. Theorem 3. In a pseudo-BF ∗-algebra (E;≤, •, ?, 0), we have: a ≤ b and b ≤ a imply a = b, for all a, b ∈ E. Proof. Let a ≤ b and b ≤ a then a • b = 0, a ? b = 0 and b • a = 0, b ? a = 0. By (Proposition 2 (2)), we have a = 0 ? (0 • a) = 0 ? [(a ? b) • a]. By using (pBF ∗), (pBF (1’)) and (pBF (4’)) we get 0 ? [(a ? b) • a] = 0 ? [(a • a) ? b] = 0 ? (0 ? b). By (Proposition 2 (1)), we get 0 ? (0 ? b) = b. The proof is complete. The relation between pseudo-BCK-algebra and pseudo-BF/BF ∗-algebra is given in the following theorems. Theorem 4. Any pseudo-BCK-algebra is a pseudo-BF -algebra. Proof. Let (E;≤, •, ?, 0) be a pseudo-BCK-algebra. The axioms (pBF (1’)), (pBF (4’)) are clearly the axioms (pBCK(3)), (pBCK(6)). Put b = 0 in (Theorem 2 (2)) we get a•0 ≤ a and a?0 ≤ a. Then the axiom (pBF (2’)) holds. Now, we will show (pBF (3’)). By (pBCK(4)) and (pBCK(6)) we get [0•(a?b)]•(b?a) = 0•(b?a) = 0 and [0?(a•b)]?(b•a) = 0?(b•a) = 0 and so 0•(a?b) ≤ b?a and 0?(a•b) ≤ b•a. Thus E is a pseudo-BF -algebra. Theorem 5. Any pseudo-BCK-algebra is a pseudo-BF ∗-algebra. Proof. It is obvious from (Theorem 4) above and by using (Theorem 1) that (a•b)?c = (a?c)•b (that is (pBF ∗)). Therefore every pseudo-BCK-algebra is a pseudo-BF ∗-algebra. 3. Pseudo-Ideal of Pseudo-BF -algebra In this section, we start with the definition of pseudo-subalgebra of pseudo-BF -algebra. Then we study pseudo-ideal and pseudo-normal-ideal. We start with the following defini- tion. Definition 7. In a pseudo-BF -algebra (E; •, ?, 0), let φ 6= S ⊆ E. Then S is said to be a pseudo-subalgebra of E if: a • b ∈ S and a ? b ∈ S for all a, b ∈ S. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 505 Note: It is easy to see that if S is a pseudo-subalgebra of E, then 0 ∈ S. Lemma 1. In a pseudo-BF -algebra (E; •, ?, 0), let S be a pseudo-subalgebra of E. Then for a, b ∈ E we have: (1) If a • b ∈ S, then b • a ∈ S, (2) If a ? b ∈ S, then b ? a ∈ S. Proof. For a, b ∈ S, let a • b ∈ S and a ? b ∈ S. By (pBF (3)), b • a = 0 ? (a • b). Since 0 ∈ S and a • b ∈ S, we see that 0 ? (a • b) ∈ S and so b • a ∈ S and b ? a = 0 • (a ? b). Since 0 ∈ S and a ? b ∈ S, we see that 0 • (a ? b) ∈ S and so b ? a ∈ S. Definition 8. In a pseudo-BF -algebra (E; •, ?, 0), let φ 6= I ⊆ E. Then we say that I is a pseudo-ideal of E if it satisfies for all a, b ∈ E: (pI1) 0 ∈ I, (pI2) a • b ∈ I, a ? b ∈ I and b ∈ I implies a ∈ I. Example 7. In Example 2, let C = {0, 1}, A = {0, 3} and F = {0, 1, 2} be subsets of E. Then C is a pseudo-subalgebra of E, whereas F is not, as 1 • 2 = 3 /∈ F . Also, A is a pseudo-ideal of E, but C is not, because 3 • 1 = 0, 3 ? 1 = 1 ∈ C, 1 ∈ C, but 3 /∈ C. Definition 9. In a pseudo-BF -algebra (E; •, ?, 0), let I be a pseudo-ideal. We say that I is a pseudo-normal, if for any a, b, c ∈ E: a • b, a ? b ∈ I implies (c • a) ? (c • b) and (c ? a) • (c ? b) ∈ I. Note: {0} and E are always pseudo-ideals of E. Whereas if E is a pseudo-normal, {0} is not a pseudo-normal in general. Lemma 2. Let I be a pseudo-normal-ideal of a pseudo-BF -algebra (E; •, ?, 0) and a, b ∈ E. Then, (1) a ∈ I ⇒ 0 • a ∈ I and 0 ? a ∈ I, (2) a • b , a ? b ∈ I ⇒ b • a ∈ I and b ? a ∈ I. Proof. (1) Let a ∈ I. Then by (pBF (2)) we have a = a • 0 ∈ I and so a = a ? 0 ∈ I. Since I is a pseudo-normal-ideal, we get (0 • a) ? (0 • 0) and (0 ? a) • (0 ? 0) ∈ I. By (pBF (1)) then (0 • a) ? 0 and (0 ? a) • 0 ∈ I and 0 ∈ I from (pI1). By (pI2) we get (0 • a) , (0 ? a) ∈ I. (2) Let a • b , a ? b ∈ I. By (1) we get 0 ? (a • b) , 0 • (a ? b) ∈ I. Applying (pBF (3)) we have b • a , b ? a ∈ I. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 506 Proposition 5. In a pseudo-BF -algebra (E; •, ?, 0), let I be a pseudo-normal-ideal. Then I is a pseudo-subalgebra that satisfies the following condition: (pNI) If a ∈ E and b ∈ I, then a ? (a • b), a • (a ? b) ∈ I. Proof. Let a ∈ E and b ∈ I. By (Lemma 2 (1)), 0• b , 0?b ∈ I. We have (a•0)? (a• b) and (a ? 0) • (a ? b) ∈ I as I is a pseudo-normal-ideal. By (pBF (2)), a ? (a • b) and a • (a ? b) ∈ I. Thus (pNI) holds. Now let a, b ∈ I. Therefore a ? (a • b), a • (a ? b) ∈ I. By (Lemma 2 (2)), (a • b) ? a, (a?b)•a ∈ I ;a ∈ I. From (pI2) we have (a•b) , (a?b) ∈ I. Thus I is a pseudo-subalgebra satisfying (pNI). Proposition 6. In a pseudo-BF -algebra (E; •, ?, 0), let I be a pseudo-ideal. Then for a, b ∈ E where b ≤ a, if a ∈ I , we have b ∈ I. Proof. Let a ∈ I and b ≤ a. Thus b • a = 0, b ? a = 0. By (pI1) and (pI2), we have 0 ∈ I and so having b • a, b ? a ∈ I , a ∈ I we get b ∈ I. Theorem 6. In a pseudo-BF -algebra (E; •, ?, 0), let φ 6= I ⊆ E. Then I is a pseudo-ideal of E if and only if the following hold: (1) For all a, b, c ∈ E , a, b ∈ I and c • b ≤ a =⇒ c ∈ I. (2) For all a, b, c ∈ E , a, b ∈ I and c ? b ≤ a =⇒ c ∈ I. Proof. Let I be a pseudo-ideal of E. Let a, b, c ∈ E , a, b ∈ I and c • b ≤ a we have (c • b) ? a = 0 ∈ I from (pI1). Since a ∈ I then c • b ∈ I by (pI2). Since b ∈ I then c ∈ I by (pI2). Thus (1) is valid. Now, let a, b, c ∈ E , a, b ∈ I and c ? b ≤ a we have (c ? b) • a = 0 ∈ I from (pI1). Since a ∈ I then c ? b ∈ I by (pI2). Since b ∈ I then c ∈ I by (pI2). Thus (2) is true. Conversely, suppose that (1), (2) hold. Suppose that b ∈ I. By using (1), (2) we have 0 • b ≤ b and 0 ? b ≤ b, then 0 ∈ I. Now, let a • b, a ? b ∈ I and b ∈ I. By using (1), (2) we have a • b ≤ a • b and a ? b ≤ a ? b, then a ∈ I. Therefore I is a pseudo-ideal of E. Theorem 7. In a pseudo-BF -algebra (E; •, ?, 0), let I be a pseudo-subalgebra. Then I is a pseudo-ideal of E if and only if for a, b ∈ E if a ∈ I and b /∈ I then b • a and b ? a /∈ I. Proof. Let a, b ∈ E and let I be a pseudo-ideal of E where a ∈ I and b ∈ E − I. We prove by contradiction. Let b • a , b ? a /∈ E − I, we have b • a , b ? a ∈ I. Since a ∈ I then b ∈ I by (pI2). This contradicts the hypothesis (b ∈ E − I). Hence b • a , b ? a ∈ E − I. Conversely, let a ∈ I and b ∈ E− I ⇒ b •a , b ? a ∈ E− I. Since I is a pseudo-subalgebra, we have 0 ∈ I (by Definition 7). Now, assume that a, b ∈ E, a ∈ I and b • a , b ? a ∈ I. We prove by contradiction. Let b /∈ I ,i.e. b ∈ E − I. Then b • a , b ? a ∈ E − I by hypothesis. This contradicts the hypothesis (b • a , b ? a ∈ I). Hence b ∈ I. Therefore I is a pseudo-ideal of E. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 507 Proposition 7. In a pseudo-BF -algebra (E; •, ?, 0), let I be a pseudo-ideal. If J is a pseudo-ideal of I, then J is a pseudo-ideal of E as well. Proof. Assume that J is a pseudo-ideal of I, then 0 ∈ J . Let b ∈ J and a • b, a ? b ∈ J for any a ∈ E. If a ∈ I, then a ∈ J since J is a pseudo-ideal of I. If a /∈ I ,i.e. a ∈ E − I, then b, a • b, a ? b ∈ J ⊆ I and so a ∈ I. Hence a ∈ J . Thus J is a pseudo-ideal. Proposition 8. In a pseudo-BF -algebra (E; •, ?, 0), let I be a pseudo-ideal. Then ∀a ∈ E , a ∈ I we have 0 • (0 ? a), 0 ? (0 • a) ∈ I. Proof. Let a ∈ I and 0 ? a , 0 • a ∈ I, then 0 ∈ I from (pI1) and (pI2). Since a ∈ I and 0 ∈ I , by using (pBF (1)) we have 0 = a ? a, 0 = a • a ∈ I. (By Proposition 2 (2)) we obtain a ? a = [0 • (0 ? a)] ? a, a • a = [0 ? (0 • a)] • a ∈ I. Thus 0 • (0 ? a), 0 ? (0 • a) ∈ I from (pI2). 4. Pseudo-Atoms of Pseudo-BF/BF ∗-algebra In this section we introduce pseudo-atoms of pseudo-BF/BF ∗-algebra and prove re- lated properties. We start with the following definition. Definition 10. In a pseudo-BF -algebra (E; •, ?, 0), let τ be an element in E. If a ≤ τ implies a = τ ∀a ∈ E then we call τ a pseudo-atom of E and the collection of all pseudo-atoms of E is called the center of E and denoted by Lp(E). Theorem 8. In a pseudo-BF ∗-algebra (E; •, ?, 0) the following are equivalent for all a, b, c, d, τ ∈ E: (1) there exists a pseudo-atom τ , (2) τ = a ? (a • τ) and τ = a • (a ? τ); (3) (a • b) ? (a • τ) = τ • b and (a ? b) • (a ? τ) = τ ? b; (4) τ • (a ? b) = b ? (a • τ) and τ ? (a • b) = b • (a ? τ), (5) 0 ? (b • τ) = τ • b and 0 • (b ? τ) = τ ? b, (6) 0 ? (0 • τ) = τ and 0 • (0 ? τ) = τ , (7) 0 ? (0 • (τ ? c)) = τ ? c and 0 • (0 ? (τ • c) = τ • c, (8) c ? (c • (τ ? d)) = τ ? d and c • (c ? (τ • d)) = τ • d. Proof. (1) ⇒ (2). Assume that τ is a pseudo-atom of E. As a ? (a • τ) ≤ τ and a • (a ? τ) ≤ τ by (Proposition 4 (2)), we have τ = a ? (a • τ) and τ = a • (a ? τ). H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 508 (2) ⇒ (3). For all a ∈ E. By (pBF ∗) and (2), we have (a•b)?(a•τ) = [a?(a•τ)]•b = τ •b and (a ? b) • (a ? τ) = [a • (a ? τ)] ? b = τ ? b. (3) ⇒ (4). Replacing b by a? b in (3), we get τ • (a? b) = [a • (a? b)] ? (a • τ). By (pBF ∗) and (3), we have [a• (a?b)]? (a•τ) = [a? (a•τ)]• (a?b) = b? (a•τ). Also, replacing b by a • b in (3), we get τ ? (a • b) = [a ? (a • b)] • (a ? τ). By (pBF ∗) and (3), we have [a ? (a • b)] • (a ? τ) = [a • (a ? τ)] ? (a • b) = b • (a ? τ). (4) ⇒ (5). Put b = 0 and a = b in (4). Hence τ •(b?0) = 0?(b•τ) and τ ?(b•0) = 0•(b?τ). From (pBF (3)), then 0 ? (b • τ) = τ • b and 0 • (b ? τ) = τ ? b. (5) ⇒ (6). Put b = 0 in (5). Then it is straightforward that 0 ? (0 • τ) = τ • 0 = τ and 0 • (0 ? τ) = τ ? 0 = τ by (pBF (2)). (6) ⇒ (7). For any τ, c ∈ E. By (Proposition 4 (6)), we have 0?[0•(τ?c)] = 0•[0•(τ?c)] = 0 • [0 ? (τ ? c)]. By (Proposition 4 (5)), then 0 • [0 ? (τ ? c)] = 0 • [(0 • τ) • (0 ? c)]. By (Proposition 4 (4)), we get 0 • [(0 • τ) • (0 ? c)] = [0 ? (0 • τ)] ? [0 • (0 ? c)]. By (6), then [0 ? (0 • τ)] ? [0 • (0 ? c)] = τ ? c. Also, by (Proposition 4 (6),(4) and (5), respectively) and (6) we have 0 • [0 ? (τ • c)] = 0 ? [0 ? (τ • c)] = 0 ? [0 • (τ • c)] = 0 ? [(0 ? τ) ? (0 • c)] = [0 • (0 ? τ)] • [0 ? (0 • c)] = τ • c. Thus (7) holds. (7) ⇒ (8). For any c, d, τ ∈ E, we have τ ? d = 0 ? [0 • (τ ? d)] = 0 ? [(c ? c) • (τ ? d)] = 0 ? ([c • (τ ? d)] ? c) from (7), (pBF (1)) and (pBF ∗). By (Proposition 4 (5) and (6), respectively) then 0?([c•(τ ?d)]?c) = (0• [c•(τ ?d)])•(0?c) = (0? [c•(τ ?d)])•(0?c). Using (pBF ∗), (0 ? [c • (τ ? d)]) • (0 ? c) = (0 • (0 ? c)) ? [c • (τ ? d)]. By (Proposition 4 (6)), we get (0 • (0 ? c)) ? [c • (τ ? d)] = (0 ? (0 ? c)) ? [c • (τ ? d)]. Using (pBF (3)), the hypothesis and (pBF (2)), respectively we have (0 ? (0 ? c)) ? [c • (τ ? d)] = (0 ? [0 • (c ? 0)]) ? [c • (τ ? d)] = (c ? 0) ? [c • (τ ? d)] = c ? [c • (τ ? d)]. Similarly c • [c ? (τ • d)] = τ • d is proved. (8) ⇒ (1). Let c ≤ τ we have c • τ = c ? τ = 0. By (pBF (2)) we have τ = τ • 0. Then by (8) with d = 0 we obtain τ • 0 = c • [c ? (τ • 0)]. Using (pBF (2)) we have c • [c ? (τ • 0)] = c • [c ? τ ] = c • 0 = c. Thus τ is a pseudo-atom of E. Corollary 3. In a pseudo-BF ∗-algebra (E; •, ?, 0), let τ be a pseudo-atom of E. Then τ • a and τ ? a are pseudo-atoms, for all a ∈ E. Hence Lp(E) is a pseudo-subalgebra of E. Proof. For a, b ∈ E, let b ≤ τ • a and b ≤ τ ? a then b ? (τ • a) = 0 and b • (τ ? a) = 0. Multiplying by ”b” from the right we have (τ • a) ? b = 0 ? (0 • [(τ • a) ? b]) and (τ ? a) • b = 0•(0?[(τ ?a)•b]) from (Theorem 8 (7)). By (pBF (3)) we get 0?(0•[(τ•a)?b]) = 0?[b?(τ•a)] and 0•(0? [(τ ?a)•b]) = 0• [b•(τ ?a)]. By the hypothesis (b?(τ •a) = 0 and b•(τ ?a) = 0) and (BF (1)) we have 0 ? [b ? (τ • a)] = 0 ? 0 = 0 and 0 • [b • (τ ? a)] = 0 • 0 = 0. Then τ • a ≤ b and τ ? a ≤ b and so b = τ • a and b = τ ? a, thus τ • a and τ ? a are pseudo- atoms. By (Definition 10) we have Lp(E) is the set of all pseudo-atoms of E then τ • a and τ ? a ∈ L(E). Therefore Lp(E) is a pseudo-subalgebra of E. H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 509 Corollary 4. If pseudo-BF -algebra (E; •, ?, 0) is generated by an element g then g is a pseudo-atom. Proof. For g ∈ E, suppose that g generates E and let τ be a pseudo-atom of E. Thus we have g ≤ τ . Then g • τ = 0 and g ? τ = 0. By (Corollary 2) we get τ • g = 0 and τ ? g = 0. Therefore τ ≤ g and so τ = g. Hence g is a pseudo-atom. Proposition 9. In a pseudo-BF -algebra (E; •, ?, 0), let τ ∈ E. If {0, τ} is a pseudo-ideal then 0 6= τ is a pseudo-atom. Proof. Let {0, τ} be a pseudo-ideal of E and for all a ∈ E let a ≤ τ we have a • τ = a ? τ = 0 ∈ {0, τ} from (pI1). By (pI2) we have a ∈ {0, τ}, then a = 0 or a = τ . Since τ 6= 0 and (pBF (1)), we get a = τ . Thus τ is a pseudo-atom of E. Proposition 10. In a pseudo-BF ∗-algebra (E; •, ?, 0), if a non-zero element is a pseudo- atom of E, then any pseudo-subalgebra is a pseudo-ideal. Proof. We prove (pI1) and (pI2). Let S be a pseudo-subalgebra of E, then 0 ∈ S from (Definition 7). For (pI2), let b • a , b ? a ∈ S and a ∈ S. By (Theorem 8 (2) and (5), respectively) we have b = a • (a ? b) = a • [0 • (b ? a)]. Since 0, b ? a ∈ S and S is a pseudo-subalgebra of E, we obtain 0 • (b ? a) ∈ S. So a • [0 • (b ? a)] ∈ S. Also, similarly we can show it if b • a ∈ S. Then b ∈ S. Hence the proposition is proved. For any pseudo-BF -algebra (E; •, ?, 0), define the subsets K(E), V (τ) of E as follows: K(E) = {a ∈ E : 0 ≤ a} and V (τ) = {a ∈ E : τ ≤ a}. Theorem 9. In a pseudo-BF ∗-algebra (E; •, ?, 0) if τ and ω is pseudo-atoms then the following hold: (1) a ∈ V (τ), b ∈ V (ω), imply a • b ∈ V (τ • ω) and a ? b ∈ V (τ ? ω), (2) a, b ∈ V (τ), implies a ? b , a • b ∈ K(E), (3) If τ 6= ω, then we have a • b , a ? b ∈ K(E),for all a ∈ V (τ) , b ∈ V (ω), (4) a ∈ V (ω), implies τ • a = τ • ω and τ ? a = τ ? ω, (5) If τ 6= ω, then V (τ) ∩ V (ω) = φ. Proof. (1) Let a ∈ V (τ), b ∈ V (ω). Then τ ≤ a we have τ • a = τ ? a = 0 and ω ≤ b we have ω•b = ω?b = 0. From (Theorem 8 (7)) we obtain (τ•ω)?(a•b) = [0•(0?(τ•ω))]?(a•b). Using (pBF ∗), [0 • (0 ? (τ • ω))] ? (a • b) = [0 ? (a • b)] • [0 ? (τ • ω)]. By (Proposition 4 (6) and (4), respectively) then [0 ? (a • b)] • [0 ? (τ •ω)] = [0 • (a • b)] • [0 ? (τ •ω)] = [(0?a)?(0•b)]• [0?(τ •ω)]. By applying (pBF ∗), we get [(0?a)?(0•b)]• [0?(τ •ω)] = H. M. Al-Malki, D. S. Al-Kadi / Eur. J. Pure Appl. Math, 13 (3) (2020), 498-512 510 [(0 ? a) • [0 ? (τ • ω)]] ? (0 • b) = [(0 • [0 ? (τ • ω)]) ? a] ? (0 • b). By (Theorem 8 (7)) we have [(0 • [0 ? (τ • ω)]) ? a] ? (0 • b) = [(τ • ω) ? a] ? (0 • b). Using (pBF ∗) we get [(τ • ω) ? a] ? (0 • b) = [(τ ? a) • ω] ? (0 • b). From the hypothesis we have [(τ ? a) •ω] ? (0 • b) = (0 •ω) ? (0 • b). By (Proposition 4 (6) and (4), respectively) we get (0•ω)? (0• b) = (0?ω)? (0• b) = 0• (ω • b). Using the hypothesis and (pBF (1)), respectively we get 0 • (ω • b) = 0 • 0 = 0, and so τ •ω ≤ a • b. Thus a • b ∈ V (τ •ω) and similarly a ? b ∈ V (τ ? ω). (2) Let a, b ∈ V (τ), by (1) we have a • b ∈ V (τ • τ), a ? b ∈ V (τ ? τ). Using (pBF (1)) then a • b ∈ V (0), a ? b ∈ V (0). We get 0 ≤ a • b, 0 ≤ a ? b. Then a • b, a ? b ∈ K(E). (3) Let 0 be a pseudo-atom from (Definition 10) we get a • b ≤ 0 then a • b = 0. By (Corollary 2) we get b • a = 0. Using (pBF (3)) then 0 ? (a • b) = 0 and so 0 ≤ a • b. Therefore a•b ∈ V (0) and so a•b ∈ K(E). Similarly we can show that a?b ∈ K(E). (4) Let a ∈ V (ω), then ω ≤ a we have ω • a = 0 and ω ? a = 0. By (Theorem 8 (3)) we get (τ • a) ? (τ • ω) = ω • a = 0. So τ • a ≤ τ • ω. Moreover, τ • ω is a pseudo-atom by (Corollary 3). Therefore τ • a = τ • ω. Similarly τ ? a = τ ? ω. (5) We prove by contradiction. Let τ 6= ω and let V (τ) ∩ V (ω) 6= φ then there exists c ∈ V (τ) ∩ V (ω). From (1), we have c • c ∈ V (τ • ω), c ? c ∈ V (τ ? ω). Using (pBF (1)) then c • c = 0 = c ? c and so 0 ∈ V (τ • ω), V (τ ? ω). Hence τ • ω ≤ 0 and τ ? ω ≤ 0. That is τ • ω, τ ? ω are pseudo-atoms from (1), then τ • ω = 0 = τ ? ω we have τ ≤ ω. That is ω is a pseudo-atom then τ = ω this is a contradiction with hypothesis (τ 6= ω). Thus V (τ) ∩ V (ω) = φ. Proposition 11. In a pseudo-BF ∗-algebra (E; •, ?, 0), let τ ∈ E. Then τ is a pseudo- atom if and only if there is a ∈ E such that τ = 0 • a. Proof. Let τ be a pseudo-atom of E. Then τ = 0 • (0 ? τ), from (Theorem 8 (6)). Set a = 0 ? τ , we get τ = 0 • a. Conversely, let τ = 0 • a for some a ∈ E. We use (Proposition 2 (2)) to have 0 • (0 ? τ) = 0•(0?(0•a)) = 0•a = τ . By (Theorem 8 (6) and (1)) we conclude that τ is a pseudo-atom. Proposition 12. In a pseudo-BF ∗-algebra (E; •, ?, 0), the following properties hold for any a, b, c ∈ E: (1) if a ≤ b then c • b ≤ c • a and c ? b ≤ c ? a, (2) if a ≤ b , b ≤ c then a ≤ c, (3) if a • b = c = a ? b then c • a = c ? a, (4) (a • b) • (c • b) ≤ a • c and (a ? b) ? (c ? b) ≤ a ? c, (5) if a ≤ b then a • c ≤ b • c and a ? c ≤ b ? c. REFERENCES 511 Proof. (1) Let a, b ∈ E, a ≤ b then a • b = 0 and a ? b = 0. By (Theorem 8 (3)) then (c • b) ? (c • a) = a • b = 0 and (c ? b) • (c ? a) = a ? b = 0 we get c • b ≤ c • a and c ? b ≤ c ? a. (2) Let a, b, c ∈ E, a ≤ b and b ≤ c we have a • b = 0, a ? b = 0 and b • c = 0, b ? c = 0. Also, by (1) since b ≤ c then a • c ≤ a • b ⇒ a • c ≤ 0. By (Proposition 4 (1)) we get a • c = 0 and so a ≤ c. (3) Let a • b = c = a ? b. By using (pBF (1)) and (pBF ∗) we obtain c ? a = (a • b) ? a = (a ? a) • b = 0 • b. By (Proposition 4 (6)), 0 • b = 0 ? b. Using (pBF (1)) and (pBF ∗) we have 0 ? b = (a • a) ? b = (a ? b) • a = c • a. (4) By (pBF ∗), (Theorem 8 (3)) and (pBF (1)), respectively we have [(a • b) • (c • b)] ? (a • c) = [(a • b) ? (a • c)] • (c • b) = (c • b) • (c • b) = 0. Then (a • b) • (c • b) ≤ a • c. Similarly, (a ? b) ? (c ? b) ≤ a ? c. (5) Suppose that a, b ∈ E, a ≤ b we have a • b = 0, a ? b = 0. Using (4), we have (a • c) • (b • c) ≤ a • b but a • b = 0. By (Proposition 4 (1)) then (a • c) • (b • c) = 0 and so a • c ≤ b • c. By a similar way, we can show that a ? c ≤ b ? c. Theorem 10. In a pseudo-BF ∗-algebra (E; •, ?, 0), the set K(E) is a pseudo-subalgebra. Proof. For a, b ∈ K(E), we have 0 ≤ a, 0 ≤ b, then 0 • a = 0, 0 ? a = 0 and 0 • b = 0, 0 ? b = 0. By using (Proposition 12 (5)) since 0 ≤ a we get 0 • b ≤ a • b and 0 ? b ≤ a ? b. Hence 0 ≤ a•b and 0 ≤ a?b and so a•b, a?b ∈ K(E). 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