Int. J. Anal. Appl. (2023), 21:85 On Prime E-ideals of Almost Distributive Lattices N. Rafi1, Y. Monikarchana2, Ravikumar Bandaru3, Aiyared Iampan4,∗ 1Department of Mathematics, Bapatla Engineering College, Bapatla, Andhra Pradesh 522 101, India 2Department of Mathematics, Mohan Babu University, A. Rangampet, Tirupati, Andhra Pradesh 517 102, India 3Department of Mathematics, GITAM(Deemed to be University), Hyderabad Campus, Telangana 502 329, India 4Fuzzy Algebras and Decision-Making Problems Research Unit, Department of Mathematics, School of Science, University of Phayao, Mae Ka, Mueang, Phayao 56000, Thailand ∗Corresponding author: aiyared.ia@up.ac.th Abstract. In an almost distributive lattice (ADL), the idea of E-ideals is introduced, and their properties are discussed. In terms of a congruence, an equivalence is established between the minimal prime E- ideals of an ADL and its quotient ADL. Finally, topological investigations are performed on prime E-ideals and minimal prime E-ideals. 1. Introduction In the article by Swamy and Rao [9], the concept of an Almost Distributive Lattice (ADL) was introduced as a generalization of Boolean algebras and distributive lattices. This allowed for the abstraction of various ring-theoretic generalizations. They also introduced the notion of an ideal in an ADL, noting that the set of principal ideals in an ADL forms a distributive lattice. This extension of lattice theory notions to ADLs was significant. The concept of normal lattices was initially introduced by Cornish [2]. Later, Rao and Ravi Kumar presented the concept of a minimal prime ideal belonging to an ideal in an ADL [6]. In another paper by Rao and Ravi Kumar [7], the notion of a normal ADL was defined, providing equivalent Received: Jun. 7, 2023. 2020 Mathematics Subject Classification. 06D99, 06D15. Key words and phrases. almost distributive lattice(ADL); prime filter; E-ideal; E-normal ADL; congruence; compact; Hausdorff space; closure. https://doi.org/10.28924/2291-8639-21-2023-85 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-85 2 Int. J. Anal. Appl. (2023), 21:85 conditions for an ADL to be considered normal in terms of its annulets. These papers contributed to the understanding of ADLs and their properties. The study of D-filters in lattices and their properties was carried out by Kumar et al. [4]. They investigated the properties of D-filters in lattices, providing valuable insights. In the same line of research, we investigated the notions of prime E-ideals and E-ideals in an ADL. The properties of these ideals are thoroughly examined, and it is established that every proper E-ideal must satisfy a set of equivalent conditions to become a prime E-ideal. It is also proven that every maximal E-ideal in an ADL is a prime E-ideal. Furthermore, the paper introduces the concept of OE(M) as the intersection of all minimal prime E-ideals contained in a prime E-ideal M in an ADL R. An ADL is defined as E-normal, characterized in terms of relative dual annihilators with respect to an ideal E. An equivalence between the minimal prime E-ideals of an ADL and its quotient ADL is derived with respect to a congruence. The topological properties of the space of all prime E-ideals and the space of all minimal prime E-ideals in an ADL are also investigated. 2. Preliminaries In this section, we recall certain definitions and important results from [5] and [9], those will be required in the text of the paper. Definition 2.1. [9] An algebra R = (R,∨,∧, 0) of type (2, 2, 0) is called an Almost Distributive Lattice (abbreviated as ADL), if it satisfies the following conditions: (1) (a∨b) ∧c = (a∧c) ∨ (b∧c) (2) a∧ (b∨c) = (a∧b) ∨ (a∧c) (3) (a∨b) ∧b = b (4) (a∨b) ∧a = a (5) a∨ (a∧b) = a (6) 0 ∧a = 0 (7) a∨ 0 = a, for all a,b,c ∈ R. Example 2.1. Every non-empty set X can be regarded as an ADL as follows. Let x0 ∈ X. Define the binary operations ∨,∧ on X by x ∨y =  x if x 6= x0 y if x = x0 x ∧y =  y if x 6= x0 x0 if x = x0. Then (X,∨,∧,x0) is an ADL (where x0 is the zero) and is called a discrete ADL. If (R,∨,∧, 0) is an ADL, for any a,b ∈ R, define a ≤ b if and only if a = a ∧b (or equivalently, a∨b = b), then ≤ is a partial ordering on R. Int. J. Anal. Appl. (2023), 21:85 3 Theorem 2.1. [9] If (R,∨,∧, 0) is an ADL, for any a,b,c ∈ R, we have the following: (1) a∨b = a ⇔ a∧b = b (2) a∨b = b ⇔ a∧b = a (3) ∧ is associative in R (4) a∧b∧c = b∧a∧c (5) (a∨b) ∧c = (b∨a) ∧c (6) a∨ (b∧c) = (a∨b) ∧ (a∨c) (7) a∧ (a∨b) = a, (a∧b) ∨b = b and a∨ (b∧a) = a (8) a∧a = a and a∨a = a. It can be observed that an ADL R satisfies almost all the properties of a distributive lattice except the right distributivity of ∨ over ∧, commutativity of ∨, commutativity of ∧. Any one of these properties make an ADL R a distributive lattice. As usual, an element m ∈ R is called maximal if it is a maximal element in the partially ordered set (R,≤). That is, for any a ∈ R, m ≤ a ⇒ m = a. The set of all maximal elements of an ADL R is denoted by M. As in distributive lattices [1,3], a non-empty subset I of an ADL R is called an ideal of R if a∨b ∈ I and a∧x ∈ I for any a,b ∈ I and x ∈ R. Also, a non-empty subset F of R is said to be a filter of R if a∧b ∈ F and x ∨a ∈ F for a,b ∈ F and x ∈ R. The set I(R) of all ideals of R is a bounded distributive lattice with least element {0} and greatest element R under set inclusion in which, for any I,J ∈ I(R), I ∩J is the infimum of I and J while the supremum is given by I ∨J := {a ∨b | a ∈ I,b ∈ J}. A proper ideal(filter) P of R is called a prime ideal(filter) if, for any x,y ∈ R, x∧y ∈ P (x∨y ∈ P ) ⇒ x ∈ P or y ∈ P. A proper ideal(filter) M of R is said to be maximal if it is not properly contained in any proper ideal(filter) of R. It can be observed that every maximal ideal(filter) of R is a prime ideal(filter). Every proper ideal(filter) of R is contained in a maximal ideal(filter). For any subset S of R the smallest ideal containing S is given by (S] := {( n∨ i=1 si ) ∧ x | si ∈ S,x ∈ R and n ∈ N}. If S = {s}, we write (s] instead of (S] and such an ideal is called the principal ideal of R. Similarly, for any S ⊆ R, [S) := {x∨( n∧ i=1 si ) | si ∈ S,x ∈ R and n ∈N}. If S = {s}, we write [s) instead of [S) and such a filter is called the principal filter of R. For any a,b ∈ R, it can be verified that (a] ∨ (b] = (a∨b] and (a] ∧ (b] = (a∧b]. Hence the set (IPI(R),∨,∩) of all principal ideals of R is a sublattice of the distributive lattice (I(R),∨,∩) of all ideals of R. Also, we have that the set (F(R),∨,∩) of all filters of R is a bounded distributive lattice. Theorem 2.2. [6] Let R be an ADL with maximal elements. Then P is a prime ideal of R if and only if R\P is a prime filter of R. Definition 2.2. [5] An ADL R is said to be an associate ADL, if the operation ∨ is associative on R. 4 Int. J. Anal. Appl. (2023), 21:85 Definition 2.3. [8] For any nonempty subset A of an ADL R, define A+ = {x ∈ R | a∨x is maximal, for all a ∈ A}. Here A+ is called the dual annihilator of A in R. For any a ∈ R, we have {a}+ = (a]+, where (a] is the principal filter generated by a. An element a of an ADL R is called dual dense element if (a]+ = M and the set E of all dual dense elements in an ADL R is an ideal if E is non-empty. 3. E-ideals of ADLs In this section, we present the concepts of prime E-ideals and E-ideals in an Abstract Distributive Lattice (ADL) and explore their properties. We observe that any proper E-ideal in an ADL can be transformed into a prime E-ideal based on a set of equivalent conditions. Additionally, we establish that the intersection of all minimal prime E-ideals contained in a prime E-ideal M is denoted as OE(M). Furthermore, we introduce the notion of E-normal ADLs, which are characterized in relation to the relative dual annihilators with respect to an ideal E. We establish an equivalence between the minimal prime E-ideals of an ADL and its quotient ADL with respect to a congruence. Definition 3.1. An ideal G of R is said to be an E-ideal of R if E ⊆ G. Now we have the example of an E-ideal of an ADL. Example 3.1. Let R = {0,a,b,c,d,e,f ,g} and define ∨, ∧ on R as follows: ∧ 0 a b c d e f g 0 0 0 0 0 0 0 0 0 a 0 a b c d e f g b 0 a b c d e f g c 0 c c c 0 0 c 0 d 0 d e 0 d e g g e 0 d e 0 d e g g f 0 f f c g g f g g 0 g g 0 g g g g ∨ 0 a b c d e f g 0 0 a b c d e f g a a a a a a a a a b b b b b b b b b c c a b c a b f f d d a a a d d a d e e b b b e e b e f f a b f a b f f g g a b f d e f g Then (R,∨, ∧) is an ADL. Clearly, we have that E = {0,g} and G = {0,c, f ,g} are ideals of R satisfying E ⊆ G. Therefore G is an E-ideal of R. Consider an ideal H = {0,c} of R, but not an E-ideal. It is easy to verify the proof of the following result. Lemma 3.1. For any non-empty subset A of an ADL R, (A]∨E is the smallest E-ideal of R containing A. Int. J. Anal. Appl. (2023), 21:85 5 We denote (A] ∨ E by AE, i.e., AE = (A] ∨ E. For, A = {a}, we denote simply (a)E for {a}E. Clearly, we have that (a)E is the smallest E-ideal containing a, which is known as the principal E-ideal generated by a. Lemma 3.2. For any two elements x,y of an ADL R with maximal element m, we have the following: (1) (0)E = E (2) (m)E = R (3) x ≤ y implies (x)E ⊆ (y)E (4) (x ∨y)E = (x)E ∨ (y)E (5) (x ∧y)E = (x)E ∩ (y)E (6) (x)E = E if and only if x ∈ E. Proof. (1) Now (0)E = (0] ∨E = E. (2) Now (m)E = (m] ∨E = R∨E = R. (3) Let x ≤ y. Then (x] ⊆ (y]. Now (x)E = (x] ∨E ⊆ (y] ∨E = (y)E. Therefore (x)E ⊆ (y)E. (4) Clearly, we have that (x ∨ y] = (x] ∨ (y]. Now, (x ∨ y)E = (x ∨ y] ∨ E = (x] ∨ (y] ∨ E = ((x] ∨E) ∨ ((y] ∨E)) = (x)E ∨ (y)E. Therefore (x ∨y)E = (x)E ∨ (y)E. (5) Since x ∧y ≤ y and y ∧x ≤ x and hence (x ∧y] ⊆ (x] and (y ∧x] ⊆ (y]. Since (x ∧y] = (y ∧x], we get that (x ∧ y] ⊆ (x] ∩ (y]. Let t ∈ (x] ∩ (y]. Then t ∈ (x] and t ∈ (y]. That implies x ∧ t = t and y ∧ t = t. Therefore x ∧ y ∧ t = t and hence t ∈ (x ∧ y]. Thus (x] ∩ (y] ⊆ (x ∧ y], which gives (x∧y] = (x]∩(y]. Now (x∧y)E = (x∧y]∨E = [(x]∩(y]]∨E = ((x]∨E)∩((y]∨E) = (x)E∩(y)E. Hence (x ∧y)E = (x)E ∩ (y)E. (6) Assume that (x)E = E. Then (x] ∨E = E. That implies (x] ⊆ E and hence x ∈ E. Conversely, assume that x ∈ E. Then (x] ⊆ E. This implies that (x] ∨E ⊆ E. Since E ⊆ (x] ∨E, we get that E = (x] ∨E. Therefore (x)E = E. � We denote I(R),IE(R)and IPEF (R) as the set of all ideals, E-ideals and principal E-ideals of an ADL R respectively. Theorem 3.1. IE(R) forms a distributive lattice contained in I(R), and IPEF (R) forms a sublattice of IE(R). Definition 3.2. An E-ideal Q is said to be proper if Q ( R. A proper E-ideal Q is said to be maximal if it is not properly contained in any proper E-ideal of R. A proper E-ideal Q of an ADL R is said to be a prime E-ideal if Q is a prime filter of R. 6 Int. J. Anal. Appl. (2023), 21:85 Example 3.2. Consider a distributive lattice L = {0,a,b,c, 1} and discrete ADL A = {0′,a′}. � � � @ @ @ @ @ @ � � � d d d d d a b c 1 0 Clearly, R = A × L = {(0′, 0), (0′,a), (0′,b), (0′,c), (0′, 1), (a′, 0), (a′,a), (a′,b), (a′,c), (a′, 1)} is an ADL with zero element (0, 0′). Clearly, the dense set E = {(0′, 0), (0′,a)}. Consider the E-ideals: I1 = {(0′, 0), (0′,a), (0′,b)} I2 = {(0′, 0), (0′,a), (0′,c)} I3 = {(0′, 0), (0′,a), (a′, 0), (a′,a)} I4 = {(0′, 0), (0′,a), (0′,c), (a′, 0), (a′,a)(a′,c)} I5 = {(0′, 0), (0′,a), (0′,b), (a′, 0), (a′,a), (a′,b)} I6 = {(0′, 0), (0′,a), (0′,b), (0′,c), (0′, 1)} Clearly, I4, I5 and I6 are prime E-ideal. But I1 is not a prime E-ideal, because (a′,b)∧(0′,c) = (0′,a) ∈ I1, but (a′,b) /∈ I1. and (0′,c) /∈ I1. And also, I2 is not a prime E-ideal, because (0′,b) ∧ (a′,c) = (0′,a) ∈ I2, but (0′,b) /∈ I2 and (a′,c) /∈ I2. Theorem 3.2. For any E-ideal Q of R, the following conditions are equivalent: (1) Q is a prime E-ideal (2) for any two E-ideals G,H of R,G ∩H ⊆ Q ⇒ G ⊆ Q or H ⊆ Q (3) for any x,y ∈ R, (x)E ∩ (y)E ⊆ Q ⇒ x ∈ Q or y ∈ Q. Proof. (1) ⇒ (2) Assume (1). Let G and H be two E-ideals of R such that G∩H ⊆ Q. We prove that G ⊆ Q or H ⊆ Q. Suppose G * Q and H * Q. Choose x,y ∈ R such that x ∈ G \Q and y ∈ H\Q. By our assumption we have that x ∧ y /∈ Q. Since x ∈ G,y ∈ H, which gives x ∧ y ∈ G ∩H ⊆ Q. Therefore x ∧y ∈ Q, we get a contradiction. Thus G ⊆ Q or H ⊆ Q. (2) ⇒ (3) Assume (2). Let x,y ∈ R with (x)E ∩ (y)E ⊆ Q. Since (x)E and (y)E are E-ideals of R, and by our assumption, we get that (x)E ⊆ Q or (y)E ⊆ Q. Hence x ∈ Q or y ∈ Q. (3) ⇒ (1) Assume (3). Let x,y ∈ R with x∧y ∈ Q. Since Q is an E-ideal, we have that (x)E∩(y)E = (x ∧y)E ⊆ Q. By our assumption, we get that x ∈ Q or y ∈ Q. Hence Q is prime. � Theorem 3.3. Every maximal E-ideal of an ADL R is a prime E-ideal. Int. J. Anal. Appl. (2023), 21:85 7 Proof. Let N be a maximal E-ideal of R. Let a,b ∈ R with a /∈ N and b /∈ N. Then N ∨ (a)E = R and N ∨ (b)E = R. That implies R = N ∨ ((a)E ∩ (b)E) = N ∨ (a∧b)E. If a∧b ∈ N then N = R, we get a contradiction. Therefore a∧b /∈ N and hence N is prime. � Corollary 3.1. Let N1,N2,N3, . . . ,Nn and N be maximal E-ideals of an ADL R with n⋂ i=1 Ni ⊆ N, then Nj ⊆ N, for some j ∈{1, 2, 3, . . . ,n}. Theorem 3.4. A proper E-ideal Q of an ADL R is a prime E-ideal if and only if R\Q is a prime filter such that (R\Q) ∩E = ∅. Proof. Assume that Q is a prime E-ideal of R. Clearly, R \Q is a prime filter of R. We prove that (R\Q) ∩E = ∅. If (R\Q) ∩E 6= ∅, choose x ∈ (R\Q) ∩E. That implies x ∈ E ⊆ Q, which gives a contradiction. Hence (R\Q) ∩E = ∅. Conversely, assume that R\Q is a prime filter of R such that (R \Q) ∩E = ∅. Clearly, Q is a prime ideal of R and E ⊆ R \ (R \Q) = Q. Therefore Q is a prime E-ideal of R. � Theorem 3.5. Let G be a E-ideal of an ADL R, and K be any non-empty subset of R, which is closed under the operation ∧ such that G ∩K = ∅. Then there exists a prime E-ideal Q of R containing G such that Q∩K = ∅. Proof. Let K be a non-empty subset of R, which is closed under the operation ∧ such that G∩K = ∅. Consider F = {H | H is an E−ideal of R,G ⊆ H and H∩K = ∅}. Clearly, it satisfies the hypothesis of the Zorn’s lemma and hence F has a maximal element say Q. That is, Q is an E-ideal of R such that G ⊆ Q and Q∩K = ∅. Let x,y ∈ R be such that x∧y ∈ Q. We prove that x ∈ Q or y ∈ Q. Suppose that x /∈ Q and y /∈ Q. Then clearly Q∨(x)E and Q∨(y)E are E-ideals of R such that Q ( Q∨(x)E and Q ( Q∨(y)E. Since Q is maximal in F, we get that (Q∨(x)E)∩K 6= ∅ and (Q∨(y)E)∩K 6= ∅. Choose s ∈ (Q ∨ (x)E) ∩ K and t ∈ (Q ∨ (y)E) ∩ K. Then s ∈ (Q ∨ (x)E),t ∈ (Q ∨ (y)E) and s,t ∈ K. Since K is closed under ∧, we get s ∧ t ∈ K. Now s ∧ t = {Q ∨ (x)E}∩{Q ∨ (y)E} = Q∨{(x)E ∩ (y)E} = Q∨ (x ∧y)E. Since x ∧y ∈ Q, we get that s ∧ t ∈ Q. Since s ∧ t ∈ K, we get that s ∧ t ∈ Q∩K, which is a contradiction to Q∩K = ∅. Therefore either x ∈ Q or y ∈ Q. Thus Q is a prime E-ideal of R. � Corollary 3.2. For any E-ideal G of an ADL R with x /∈ G, there exists a prime E-ideal Q of R such that G ⊆ Q and x /∈ Q. Corollary 3.3. For any E-ideal G of an ADL R, G = ⋂ {Q | Q is a prime E− ideal of R and G ⊆ Q}. Corollary 3.4. E is the intersection of all prime E-ideals of R. Proof. Let Q be any prime E-ideal of R. Clearly, we have that E ⊆ ⋂ Q. Let Q be any prime E-ideal of an ADL R and x ∈ ⋂ Q. Suppose x /∈ E. Then there exists prime filter N such that x ∈ N and 8 Int. J. Anal. Appl. (2023), 21:85 N ∩ E = ∅. That implies x /∈ R \ N and E ⊆ R \ N. Therefore R \ N is a prime E-ideal of R and x /∈ R\N, which is a contradiction. Therefore x ∈ E and hence ⋂ Q ⊆ E. Thus E = ⋂ Q. � Theorem 3.6. In an ADL the following are equivalent: (1) Every proper E-ideal is prime (2) IE(R) is a chain (3) IPEF (R) is a chain. Proof. (1) ⇒ (2) Assume (1). Clearly (IE(R),⊆) is a poset. Let S and T be two proper E-ideals of R. By (1), we have that S∩T is a prime E-ideal of R. Since S∩T ⊆ S∩T, we get S ⊆ S∩T ⊆ T or T ⊆ S ∩T ⊆ S. Hence IE(R) is a chain. (2) ⇒ (3) It is obvious. (3) ⇒ (1) Assume that (3). Let G be a proper E-ideal of R. We prove that G is prime. Let x,y ∈ R such that (x)E ∩ (y)E ⊆ G. By our assumption, we get that (x)E ⊆ (y)E or (y)E ⊆ (x)E. That implies x ∈ (x)E = (x)E ∩ (y)E ⊆ G or y ∈ (y)E = (x)E ∩ (y)E ⊆ G. Therefore G is a prime E-ideal of R. � Now we introduce the concept of a relative dual annihilator in the following definition. Definition 3.3. For any nonempty subset S of R, define (S,E) = {a ∈ R | s∧a ∈ E, f or all s ∈ S}. We call this set as relative dual annihilator of S with respect to the ideal E. For S = {s}, we denote ({s},E) by (s,E). Lemma 3.3. If S,T are nonempty subsets of an ADL R, then we have the following: (1) (R,E) = E = (M,E) (2) (E,E) = R (3) E ⊆ (S,E) (4) (S,E) is a E-ideal of R (5) S ⊆ E if and only if (S,E) = R (6) if S ⊆ T, then (T,E) ⊆ (S,E) and ((S,E),E) ⊆ ((T,E),E) (7) S ⊆ ((S,E),E) (8) (((S,E),E),E) = (S,E) (9) (S,E) = ([S),E) (10) ⋂ i∈4 (Si,E) = ( ⋃ i∈4 Si,E ) (11) (S,E) ⊆ (S ∩T, (T,E)) (12) if S ⊆ T, then (S, (T,E)) = (S,E) (13) (S ∪T,E) ⊆ (S, (T,E)) ⊆ (S ∩T,E) (14) (S, (S,E)) = (S,E). Int. J. Anal. Appl. (2023), 21:85 9 Proof. (1) Let x ∈ (R,E). Then a ∧ x ∈ E, for all a ∈ R. That implies x ∧ x ∈ E. So that x ∈ E. Hence (R,E) ⊆ E. Let x ∈ E. Then a∧x ∈ E, for all a ∈ R. Thus x ∈ (R,E). Therefore E ⊆ (R,E) and hence (R,E) = E. Clearly, we have that (M,E) = E. (2) Let x ∈ E. Then x ∧ a ∈ E, for all a ∈ R. Since x ∧ a ∈ E, for all x ∈ E, we get that a ∈ (E,E), for all a ∈ R. Therefore R ⊆ (E,E) and hence R = (E,E). (3) Let x ∈ E. Then y ∧ x ∈ E, for all y ∈ R. Then a ∧ x ∈ E, for all a ∈ S ⊆ R. That implies x ∈ (S,E). Therefore E ⊆ (S,E). (4) Let a,b ∈ (S,E). Then s∧a,s∧b ∈ E, for all s ∈ S. This implies (s∧a)∨(s∧b) ∈ E. Therefore s ∧ (a ∨b) ∈ E. Hence a ∨b ∈ (S,E). Let a ∈ (S,E) and b ∈ R with b ≤ a. Then s ∧ a ∈ E and s ∧b ≤ s ∧a, for all s ∈ S. Since s ∧a ∈ E and E is an ideal, we get s ∧b ∈ E. Hence b ∈ (S,E), for all s ∈ S. Thus (S,E) is an ideal of R. Since E ⊆ (S,E), we get that (S,E) is an E-ideal of R. (5) Suppose (S,E) = R. Let m ∈M. Then m ∈ (S,E). That implies a = m ∧a ∈ E, for all a ∈ S. Hence a ∈ E, for all a ∈ S. Therefore S ⊆ E. Conversely, assume that S ⊆ E. Let x ∈ R. Since E is an ideal, we get a∧x ∈ E, for all a ∈ S ⊆ E. Hence x ∈ (S,E). Therefore (S,E) = R. (6) Suppose S ⊆ T. Let a ∈ (T,E). Then t∧a ∈ E, for all t ∈ T. Since S ⊆ T, we get that s∧a ∈ E, for all s ∈ S. That implies a ∈ (S,E). Therefore (T,E) ⊆ (S,E) and hence ((S,E),E) ⊆ ((T,E),E). (7) Let x ∈ (S,E). Then s ∧ x ∈ E, for all s ∈ S. That implies x ∧ s ∈ E, for all x ∈ (S,E). That implies s ∈ ((S,E),E), for all s ∈ S. Thus S ⊆ ((S,E),E). (8) By (7), we have that (((S,E),E),E) ⊆ (S,E). Let x /∈ (((S,E),E),E). Then there exists an element a /∈ ((S,E),E) such that a ∧ x /∈ E. Since S ⊆ ((S,E),E), we have that a /∈ S. So that a ∧ x /∈ E and s /∈ S. Therefore x /∈ (S,E), it concludes that (S,E) ⊆ (((S,E),E),E). Thus (((S,E),E),E) = (S,E). (9) Since S ⊆ (S], we get that ((S],E) ⊆ (S,E). Let x ∈ (S,E). Then a∧x ∈ E, for all a ∈ S ⊆ (S]. That implies x ∈ ((S],E). Therefore (S,E) ⊆ ((S],E). Therefore (S,E) ⊆ ((S],E). Hence (S,E) = ((S],E). (10) Since Si ⊆ ⋃ i∈4 Si, for all i ∈ 4, we get that ( ⋃ i∈4 Si,E ) ⊆ (Si,E), for all i ∈ 4. That implies( ⋃ i∈4 Si,E ) ⊆ ⋂ i∈4 (Si,E). Let x ∈ ⋂ i∈4 (Si,E). Then x ∈ (Si,E), for all i ∈4. That implies a∧x ∈ E, for all a ∈ Si ⊆ ⋃ Si. That implies ⋂ i∈4 (Si,E) ⊆ ( ⋃ i∈4 Si,E ) . Therefore ⋂ i∈4 (Si,E) = ( ⋃ i∈4 Si,E). (11) Since E is an ideal in R, we have that E ⊆ (T,E) and hence we get that (S,E) ⊆ (S, (T,E)). Since S ∩T ⊆ S, we get that (S, (T,E)) ⊆ (S ∩T, (T,E)). Therefore (S,E) ⊆ (S ∩T, (T,E)). (12) Let S,T be two non empty subsets of R such that S ⊆ T. Since E ⊆ (T,E), we have that (S,E) ⊆ (S, (T,E)). Let x ∈ (S, (T,E)). Then a ∧ x ∈ (T,E), for all a ∈ S. That implies a ∧ x ∈ (S,E), for all a ∈ S. Since a ∧ x ∈ (S,E), we get that s ∧ (a ∧ x) ∈ E, for all s ∈ S and hence a ∧ x ∈ E, for all a ∈ S. Therefore x ∈ (S,E) and hence (S, (T,E)) ⊆ (S,E). Thus (S, (T,E)) = (S,E). (13) Clearly, we have that (S ∪T,E) ⊆ (S,E) and E ⊆ (T,E). So that (S,E) ⊆ (S, (T,E)). Also 10 Int. J. Anal. Appl. (2023), 21:85 S∩T ⊆ S. It follows that (S, (T,E)) ⊆ (S∩T,E). Therefore (S∪T,E) ⊆ (S, (T,E)) ⊆ (S∩T,E). (14) It is clear by (12). � Proposition 3.1. Let S and T be any two ideals of and ADL R. Then we have the following: (1) (S,E) ∩ ((S,E),E) = E (2) (S ∨T,E) = (S,E) ∩ (T,E) (3) ((S ∩T,E),E) ⊆ ((S,E),E) ∩ ((T,E),E). Proof. (1) We have that E ⊆ (S,E)∩((S,E),E). Let x ∈ (S,E)∩((S,E),E). Then x ∈ (S,E) and x ∈ ((S,E),E). Since x ∈ ((S,E),E)), we have that a∧x ∈ E, for all a ∈ (S,E). Since x ∈ (S,E), we get that x ∈ E and hence (S,E) ∩ ((S,E),E) ⊆ E. Thus (S,E) ∩ ((S,E),E) = E. (2) Clearly, S ⊆ S∨T and T ⊆ S∨T. Then ((S∨T ),E) ⊆ (S,E) and ((S∨T ),E) ⊆ (T,E). That implies ((S ∨ T ),E) ⊆ (S,E) ∩ (T,E). Let x ∈ (S,E) ∩ (T,E). Then x ∈ (S,E) and x ∈ (T,E). That implies s ∧x ∈ E, for all s ∈ S and t ∧x ∈ E, for all t ∈ T. That implies (s ∧x) ∨ (t ∧x) ∈ E and have (s ∨ t) ∧ x ∈ E. Since s ∈ S and t ∈ T, we get s ∨ t ∈ S ∨T. Therefore (s ∨ t) ∧ x ∈ E, for all s ∨ t ∈ S ∨T. That implies x ∈ (S ∨T,E). Therefore (S,E) ∩ (T,E) ⊆ (S ∨T,E). Hence (S,E) ∩ (T,E) = (S ∨T,E). (3) Since S∩T ⊆ S and S∩T ⊆ T, we get that (S,E) ⊆ (S∩T,E) and (T,E) ⊆ (S∩T,E). That implies ((S ∩ T,E),E) ⊆ ((S,E),E) and ((S ∩ T,E),E) ⊆ ((T,E),E). Hence ((S ∩ T,E),E) ⊆ ((S,E),E) ∩ ((T,E),E). � Theorem 3.7. For any non-empty subset S of an ADL R, (S,E) = ⋂ s∈S ((s],E). Proof. Let x ∈ ⋂ s∈S ((s],E). Then x ∈ ((s],E), for all s ∈ S. That implies t ∧ x ∈ E, for all t ∈ (s] and for all s ∈ S. It follows that s ∧ x ∈ E for all s ∈ S. Therefore x ∈ (S,E). Hence x ∈ ⋂ s∈S ((s],E) ⊆ (S,E). Let s be any element of S. Take t ∈ (s]. Then s ∧t = t. Now, x ∈ (S,E). That implies s ∧x ∈ E, for all s ∈ S. So that t ∧x = t ∧ s ∧x ∈ E, for all t ∈ (s] and for all s ∈ S. That implies x ∈ ((s],E), for all s ∈ S. Therefore x ∈ ⋂ s∈S ((s],E) and hence (S,E) ⊆ ⋂ s∈S ((s],E). Thus (S,E) = ⋂ s∈S ((s],E). � Corollary 3.5. Let x ∈ R and S be arbitrary subset of R. Then (S, (x]) = ⋂ a∈S (a, (x]). Corollary 3.6. For any x,y ∈ R we have the following: (1) ((x],E) = (x,E) (2) x ≤ y ⇒ (y,E) ⊆ (x,E) (3) (x ∨y,E) = (x,E) ∩ (y,E) (4) ((x ∧y,E),E) = ((x,E),E) ∩ ((y,E),E) (5) (x,E) = R ⇔ x ∈ E. Int. J. Anal. Appl. (2023), 21:85 11 Theorem 3.8. Let G be an E-ideal of an ADL R. Then (1) G ∩ (G,E) = E (2) ((G ∨ (G,E)),E) = E. Proof. (1) It is clear. (2) Clearly, ((G∨(G,E)),E) ⊆ (G,E)∩((G,E),E). Let a ∈ (G,E)∩((G,E),E). Let b ∈ G∨(G,E). Then b = c ∨ d, for some c ∈ G and d ∈ (G,E). That implies a ∧ c ∈ E and a ∧ d ∈ E. Now a∧b = a∧(c∨d) = (a∧c)∨(a∧d) ∈ E, for all b ∈ G∨(G,E). Therefore a ∈ ((G∨(G,E)),E) and hence (G,E)∩((G,E),E) ⊆ ((G∨(G,E)),E). Thus E = (G,E)∩((G,E),E) = ((G∨(G,E)),E). � Consider two ADLs R1 and R2 with zero elements 0 and 0′ respectively. Let M and M′ be denotes the set of all maximal elements of ADLs R1 and R2 respectively. Lemma 3.4. Let R1 and R2 be two ADLs with m ∈M and m′ ∈M′. Then for any (x,y) ∈ R1×R2, we have the following: (1) (x,y)+ = (a)+ × (y)+ (2) (x,y)+ = (m,m′) if and only if (x)+ = M and (y)+ = M′ (3) ((x,y),E) = (a,E) × (y,E). Let E1 and E2 be dual dense sets of R1 and R2 respectively. From the above result, it can be concluded that E = E1×E2 is a dual dense set of R1×R2. Further, every dual dense set of R1×R2 is form the form E1 ×E2. Theorem 3.9. Let Mi be a prime Ei−ideals of ADLs Ri, for i = 1, 2. Then M1 ×R2 and R1 ×M2 are prime E-ideals of R1 ×R2. Proof. Since E1 ⊆ M1 and E2 ⊆ M2, we get E1 × E2 ⊆ M1 × R2 and E1 × E2 ⊆ R1 × M2. That implies M1 × R2 and R1 × M2 are E-ideals of R1 × R2. Let (a,b), (c,d) ∈ R1 × R2 with (a,b) ∧ (c,d) ∈ M1 × R2. Then a ∧ c ∈ M1. Since M1 is a prime E1−ideal of R1, we get a ∈ M1 or c ∈ M1. Thus (a,b) ∈ M1 × R2 or (c,d) ∈ M1 × R2. Therefore M1 × R2 is a prime E-ideal of R1 ×R2. Similarly, we can prove that R1 ×M2 is also a prime E-ideal of R1 ×R2. � Theorem 3.10. Let R1 and R2 be two ADLs with zero elements 0 and 0′ respectively. For any prime E-ideal P of R1 ×R2, P is of the form P1 ×R2 or R1 ×P2, where Pi is a prime Ei−ideal of Ri, for i = 1, 2. Proof. Let P be a prime E-ideal of R1 × R2. Consider P1 = π1(P ) = {x1 ∈ R1 | (x1,x2) ∈ P, for some x2 ∈ R2} and P2 = π2(P ) = {x2 ∈ R2 | (x1,x2) ∈ P, for some x1 ∈ R1}. It is easy to verify that Pi is Ei−ideals of Ri, for i = 1, 2. We first show that Pi is prime Ei−ideals of Ri, for i = 1, 2. Suppose P1 = R1 and P2 = R2. Let (a,b) ∈ R1 ×R2. Then there exist x ∈ R1 and y ∈ R2 such that (a,y) ∈ P and (x,b) ∈ P. Since (a, 0′)∧(a,y) ∈ P and (0,b)∧(x,b) ∈ P, we get (a, 0′) ∈ P 12 Int. J. Anal. Appl. (2023), 21:85 and (0,b) ∈ P. Therefore (a,b) = (a, 0′) ∨ (0,b) ∈ P. Hence P = R1 ×R2, which is a contradiction to that P is proper. Next suppose that P1 6= R1 and P2 6= R2. Choose a ∈ R1 \P1 and b ∈ R2 \P2. Then (a,y) /∈ P for all y ∈ R2 and (x,b) /∈ P1 for all x ∈ R1. In particular, (a, 0′) /∈ P and (0,b) /∈ P. Since P is prime, we get (0, 0′) /∈ P, which is a contradiction. From the above observations, we get that either P1 = R1 and P2 6= R2 or P1 6= R1 and P2 = R2. Case (i): Suppose P1 = R1 and P2 6= R2. Let x2,y2 ∈ R2 be such that x2∧y2 ∈ P2. Then there exists a ∈ R1 = P1 such that (a,x2∧y2) ∈ P. Therefore (a,x2)∧(a,y2) = (a∧a, (x2∧y2)) = (a,x2∧y2) ∈ P. Since P is prime, we get (a,x2) ∈ P or (a,y2) ∈ P. Hence x2 ∈ P2 or y2 ∈ P2. Therefore P2 is a prime E2−ideal of R2. We now show that P = R1 ×P2. Clearly P ⊆ R1 ×P2. On the other hand, suppose (a,y) ∈ R1 ×P2. Since P1 = R1, there exists b ∈ R2 such that (a,b) ∈ P and there exists x ∈ R1 such that (x,y) ∈ P. Since (a, 0′)∧(a,b) = (a, 0′) and (0,y)∧(x,y) = (0,y), we get (a, 0′) ∈ P and (0,y) ∈ P. Since P is an ideal, it gives (a,y) = (a, 0′) ∨ (0,y) ∈ P. Hence R1 ×P2 ⊆ P. Therefore P = R1 ×P2. Case (ii): Suppose P1 6= R1 and P2 = R2. Similarly, we can prove that P1 is prime E1−ideal of R1 and P = P1 ×R2. � Theorem 3.11. Let S be a sub ADL of an ADL R and P is a prime E-ideal of S. Then there exists a prime E-ideal Q of R such that Q∩S = P. Proof. Let P be a prime E-ideal of S. Then S \ P is a prime filter of S. Consider I = (P ]. Then P ⊆ I ∩ S. Suppose I ∩ (S \ P ) 6= ∅. Choose x ∈ I ∩ (S \ P ). Then x ∈ I and x ∈ (S \ P ). Since x ∈ I = (P ], there exists a1 ∨a2 ∨ . . .∨an ∈ P such that x = y ∧ (a1 ∨a2 ∨ . . .∨an). Since P is an ideal of S, we get a1∨a2∨ . . .∨an ∈ P and hence x ∈ P. Since x ∈ (S\P ), we get a contradiction. Hence I∩(S\P ) = ∅. Then there exists a prime E-ideal Q of R such that I ⊆ Q and Q∩(S\P ) = ∅. Since I ⊆ Q, we get I ∩S ⊆ Q∩S. Since Q∩ (S \P ) = ∅, we get Q ⊆ P. Hence, both observations lead to P ⊆ I ∩S ⊆ Q∩S ⊆ P ∩S ⊆ P. Therefore P = Q∩S. � Now, we have the following definition. Definition 3.4. A prime E-ideal M of an ADL R containing an E-ideal G is said to be a minimal prime E-ideal belonging to G if there exists no prime E-ideal N such that G ⊆ N ⊆ M. Note that if we take E = G in the above definition then we say that M is a minimal prime E-ideal. Example 3.3. From the Example 3.2, we have that I6 is a prime E-ideal and I1 is a E-ideal of R. Clearly I1 ⊆ I6. Clearly there is no E-ideal N of R such that I1 ⊆ N ⊆ I6. Hence I6 is a minimal prime E-ideal belonging to I1. Proposition 3.2. Let G be an E-ideal and M, a prime E-ideal of R with G ⊆ M. Then M is a minimal prime E-ideal belonging to G if and only if R\M is a maximal filter with (R\M) ∩G = ∅. Int. J. Anal. Appl. (2023), 21:85 13 Proof. Clearly, R\M is a proper filter and we have (R\M)∩G = ∅. We prove that R\M is maximal. Let N be any proper filter of R such that N ∩ G = ∅ and R \ M ⊆ N. Then G ⊆ R \ N ⊆ M. By the minimality of M, we get R \ N = M. Therefore R \ M is a maximal filter with respect to the property (R \ M) ∩ G = ∅. Conversely, assume that R \ M be a maximal filter with respect to the property (R \ M) ∩ G = ∅. We prove that M is minimal. If N is any prime E-ideal of R such that E ⊆ G ⊆ N ⊆ M. Clearly, R\N is a filter such that R\M ⊆ R\N and (R\N) ∩G = ∅, which is a contradiction. Therefore M is a minimal prime E-ideal belonging to G. � Theorem 3.12. Let G be an E-ideal and M, a prime E-ideal of R with G ⊆ M. Then M is a minimal prime E-ideal belonging to G if and only if for any a ∈ M, there exists b /∈ M such that a∧b ∈ G. Proof. Assume that M is a minimal prime E-ideal belonging to G. Then R \ M is a maximal filter with respect to the property that (R \ M) ∩ G = ∅. Let a ∈ M. Then a /∈ R \ M. That implies R \M ⊂ (R \M) ∨ [a). By the maximality of R \M, we get that ((R \M) ∨ [a)) ∩G 6= ∅. Choose s ∈ ((R \ M) ∨ [a)) ∩ G. Then there exists b ∈ R \ M such that s = b ∧ a and s ∈ G. Therefore b∧a ∈ G. Conversely, assume that for any a ∈ M, there exists b /∈ M such that a∧b ∈ G. Suppose M is not a minimal prime E-ideal belonging to G. Then there exists a prime E-ideal N of R such that E ⊆ G ⊆ N ⊆ M. Choose a ∈ M \N. Then, by the our assumption, there exists b /∈ M such that a ∧b ∈ G ⊆ N. Since a /∈ N, we get that b ∈ N ⊆ M, which is a contradiction. Therefore M is a minimal prime E-ideal belonging to G. � Corollary 3.7. A prime E-ideal M of an ADL R is minimal if and only if for any a ∈ M there exists b /∈ M such that a∧b ∈ E. Definition 3.5. For any prime E-ideal M of R, define the set OE(M) as follows: OE(M) = {x ∈ R | x ∈ (y,E), for some y /∈ M}. Clearly, observe that OE(M) = ⋃ y /∈M (y,E). Lemma 3.5. Let M be prime E-ideal of an ADL R. Then OE(M) is an E-ideal such that OE(M) is contained in M. Proof. Let a,b ∈OE(M). There exist elements s /∈ M and t /∈ M such that a ∈ (s,E) and b ∈ (t,E). That implies ((s,E),E) ⊆ (a,E) and ((t,E),E) ⊆ (b,E). So that ((s ∧ t,E),E) = ((s,E),E) ∩ ((t,E),E) ⊆ (a,E) ∩ (b,E) = (a ∨ b,E). Hence a ∨ b ∈ ((a ∨ b,E),E) ⊆ (((s ∧ t,E),E),E) = (s ∧ t,E). Since s ∧ t /∈ M, we get that a ∨b ∈ OE(M). Let a ∈ OE(M) and b ≤ a. There exists s /∈ M such that a ∈ (s,E). Since (s,E) is an ideal, we get that b ∈ (s,E). Therefore b ∈ OE(M) and hence OE(M) is an ideal of R. Clearly, we have that E ⊆ OE(M). Thus OE(M) is an E-ideal of R. Let a ∈ OE(M). Then there exists s /∈ M such that a ∈ (s,E). That implies a ∧ s ∈ E ⊆ M. Since M is prime, we get that a ∈ M. Hence OE(M) ⊆ M. � 14 Int. J. Anal. Appl. (2023), 21:85 Corollary 3.8. For any prime E-ideal M of R, M is minimal if and only if OE(M) = M. Theorem 3.13. Every minimal prime E-ideal of R belonging to OE(M) is contained in M. Proof. Let N be any minimal prime E-ideal belonging to OE(M). We prove that N ⊆ M. Suppose N * M. Choose a ∈ N\M. Then there exists b /∈ N such that a∧b ∈OE(M). Hence a∧b ∈ (s,E), for some s /∈ M. That implies b ∧ (a ∧ s) ∈ E ⊆ M. Since a /∈ M,s /∈ M, and M is prime, we get a∧ s /∈ M. Therefore b ∈OE(M) ⊆ N, which is a contradiction. Hence N ⊆ M. � Theorem 3.14. For any prime E-ideal M of an ADL R, OE(M) is the intersection of all minimal prime E-ideals contained in M. Proof. Let M be a prime E-ideal of R. By Zorn’s lemma, M contains a minimal prime E-ideal. Let {Sα}α∈M be the set of all minimal prime E-ideals contained in M. Let x ∈OE(M). Then x ∈ (a,E), for some a /∈ M. Since each Sα ⊆ M, we have that a /∈ Sα, for all α ∈M . Since x ∧a ∈ E ⊆ Sα and a /∈ Sα, for all α ∈M, we get x ∈ Sα for all α ∈M. Hence x ∈ ⋂ α∈M Sα. Therefore OE(M) ⊆ ⋂ α∈M Sα. Let x /∈ OE(M). Consider S = (R \M) ∨ [x). Suppose E ∩S 6= ∅. Choose a ∈ E ∩S. Since a ∈ S, we get a = t ∧x, for some t ∈ R\M. Since a ∈ E, we get that t ∧x ∈ E. Hence x ∈ (t,E), where t /∈ M. Thus x ∈OE(M), which is a contradiction. Therefore S ∩E = ∅. Let M be a maximal filter such that S ⊆ M and M ∩E = ∅. Then R\M is a minimal prime E-ideal such that R\M ⊆ M and x /∈ R\M, since x ∈ S ⊆ M. Hence x /∈ ⋂ α∈M Sα. Therefore ⋂ α∈M Sα ⊆OE(M). � Proposition 3.3. Let M1 and M2 be two prime E-ideals in an ADL R with M1 ⊆ M2. Then OE(M2) ⊆ OE(M1). Proof. Let x ∈ OE(M2). Then there exists an element a /∈ M2 such that x ∈ (a,E). That implies x ∈ (a,E) and a /∈ M1. So that x ∈OE(M1). Therefore OE(M2) ⊆OE(M1). � Proposition 3.4. For any non zero element a ∈ R with a /∈ E, there is a minimal prime E-ideal not containing a. Proof. Let a be any non zero element of R with a /∈ E. By Corollary 3.2, there exists a prime E-ideal P of R such that a /∈ P. Consider F = {Q | Q is a prime E − ideal of R,a /∈ Q and Q ⊆ P}. It satisfies the hypothesis of Zorn’s Lemma. So that F has a minimal element say M. i.e. M is minimal and a /∈ M. � Theorem 3.15. For any prime E-ideal M of an ADL R, the following are equivalent: (1) M is minimal prime E-ideal (2) M = OE(M) (3) for any x ∈ R,M contains precisely one of x or (x,E). Int. J. Anal. Appl. (2023), 21:85 15 Proof. (1) ⇒ (2) Assume (1). Let x ∈ M. Then there exists y /∈ M such that x∧y ∈ E. This implies that x ∈OE(M). So that M ⊆OE(M). Since OE(M) ⊆ M, we get that M = OE(M). (2) ⇒ (3) Assume (2). Let x ∈ R. Suppose x /∈ M. Let a ∈ (x,E). Then a ∧ x ∈ E. That implies a∧x ∈ M. So that a ∈ M. Since x /∈ M. Therefore (x,E) ⊆ M. (3) ⇒ (1) Let Q be any prime E-ideal of R with Q ( M. Then choose x ∈ M such that x /∈ Q. That implies (x,E) ⊆ Q ( M. So that (x,E) ( M which is a contradiction. � Corollary 3.9. Let P be a minimal prime E-ideal of an ADL R and a ∈ R. Then a ∈ P if and only if ((a,E),E) ⊆ P. Proof. Assume that a ∈ P. Then (a,E) * P. Let t ∈ ((a,E),E). Then (a,E) ⊆ (t,E). Suppose t /∈ P. Then (a,E) ⊆ (t,E) ⊆ P, which is a contradiction. That implies t ∈ P, which gives ((a,E),E) ⊆ P. The converse follows from the fact that a ∈ ((a,E),E). � Definition 3.6. An ADL R with maximal elements is called an E- semi complemented if for each non maximal element x ∈ R, there exists a non zero element y /∈ E such that x ∧y ∈ E. Example 3.4. From the Example 3.2, clearly we have that R is an E-semi complemented ADL. Theorem 3.16. Let R be an ADL with maximal elements. Then R is E-semi complemented if and only if the intersection of all maximal filters disjoint with E is M. Proof. Assume that R is E-semi complemented. Consider K = ⋂{ M | M is a maximal filter of R and M ∩E = ∅}. We have to prove that K = M. Let x ∈ K with x is not a maximal element. Then x ∈ M, for all maximal filter M disjoint with E. Then x /∈ E. Since x is non maximal and R is E- semi complemented, there exists a non zero element y /∈ E such that x ∧ y ∈ E. Then x ∧ y /∈ M. That implies M∨[x∧y) = R. Since y /∈ E, there exists a minimal prime E-ideal N of R such that y /∈ N. That implies y ∈ R\N and (R\N)∩E = ∅, where R\N is maximal filter of R. So that x,y ∈ R\N. We have x∧y ∈ R\N. Therefore (R\N)∩E 6= ∅, which is a contradiction. Therefore x is a maximal element. Hence K = M. Conversely, assume that ⋂ {M | M is a maximal filter of R and M ∩E = ∅} = M. Let x be any non maximal element of R. Then there exists a maximal filter M such that x /∈ M and M ∩ E = ∅. That implies M ∨ [x) = R. So that a ∧ x = 0, for some a ∈ M. Since a ∈ M and M ∩E = ∅, we get a /∈ E. Clearly, a ∧ x ∈ E. That is, for any non maximal element x of R, there exists a non zero element a /∈ E such that a∧x ∈ E. Hence R is E-semi complemented. � Definition 3.7. An ADL R is said to be E-normal if for any a,b ∈ R such that a∧b ∈ E, there exists x ∈ (a,E) and y ∈ (b,E) such that x ∨y is maximal. From the Example 3.2, clearly we have that R is a D−normal ADL. The following result is a direct consequence of the above definition. 16 Int. J. Anal. Appl. (2023), 21:85 Theorem 3.17. R is E-normal if and only if (a,E) ∨ (b,E) = R, for any a,b ∈ R, with a∧b ∈ E. Definition 3.8. Two E-ideals G1 and G2 of R are said to be co-maximal if G1 ∨G2 = R. Example 3.5. From the Example 3.2, we have that I2, I3, I4, I5 are E-ideals of R. Clearly, I4∨I5 = R. Therefore I4 and I5 are co-maximal. Also, we have I2 ∨ I3 6= R. Therefore I2, I3 are not co-maximal. Theorem 3.18. In an ADL R, the following are equivalent: (1) for any a,b ∈ R with a∧b ∈ E, (a,E) ∨ (b,E) = R (2) for any a,b ∈ R, (a,E) ∨ (b,E) = (a∧b,E) (3) any two distinct minimal prime E-ideals are co-maximal (4) every prime E-ideal contains a unique minimal prime E-ideal (5) for any prime E-ideal P, OE(P ) is prime. Proof. (1) ⇒ (2) Assume (1). Let a,b ∈ R with x ∈ (a ∧ b,E). Then x ∧ (a ∧ b) ∈ E and hence (x ∧ a) ∧ (x ∧ b) ∈ E. By (1), we have that (x ∧ a,E) ∨ (x ∧ b,E) = R. That implies x ∈ (x ∧a,E) ∨ (x ∧b,E). Then there exists r ∈ (x ∧a,E) and s ∈ (x ∧b,E) such that x = r ∨ s. Since r ∈ (x ∧ a,E), s ∈ (x ∧ b,E) we get that r ∧ x ∈ (a,E) and s ∧ x ∈ (b,E). That implies (x ∧ r) ∨ (x ∧ s) ∈ (a,E) ∨ (b,E) and hence x ∧ (r ∨ s) ∈ (a,E) ∨ (b,E). Since x = r ∨ s, we get that x ∈ (a,E)∨(b,E). Therefore (a∧b,E) ⊆ (a,E)∨(b,E). Since (a,E)∨(b,E) ⊆ (a∧b,E), we get that (a,E) ∨ (b,E) = (a∧b,E), for all a,b ∈ R. (2) ⇒ (3) Assume (2). Let M and N be two distinct minimal prime E-ideals of R. Choose elements x,y ∈ R such that x ∈ M \N and y ∈ N \M. Since M and N are minimal, x ∧a ∈ E, y ∧b ∈ E, for some a /∈ M, b /∈ N. That implies x ∧a ∧ y ∧b ∈ E and hence R = (x ∧a ∧ y ∧b,E). By (2), we get that (x ∧b,E) ∨ (a∧ y,E) = R. Since a /∈ M and y /∈ M, we get that a∧ y /∈ M. That implies (a∧y,E) ⊆ M. Similarly, we have that (x ∧b,E) ⊆ N. That implies ((x ∧b) ∧ (a∧y),E) ⊆ M ∨N and hence R = M ∨N. Therefore M and N are co-maximal. (3) ⇒ (4) Assume (3). Let M be a prime E-ideal of R. Suppose M contains two distinct minimal prime E-ideals, say N1 and N2. By (3), we get that R = N1 ∨ N2 ⊆ M, we get a contradiction. Therefore every prime E-ideal contains a unique minimal prime E-filter. (4) ⇒ (5) Assume that every prime E-ideal of R contains a unique minimal prime E-ideal. Then by Corollary 3.8, we get that OE(P ) is a prime E-ideal. (5) ⇒ (1) Assume (5). Let a,b ∈ R be such that a∧b ∈ E. Suppose (a,E)∨(b,E) 6= R. Then there exists a maximal E-ideal M such that (a,E) ∨ (b,E) ⊆ M. That implies (a,E) ⊆ M and (b,E) ⊆ M. That implies a /∈ OE(M) and b /∈ OE(M). Since OE(M) is prime, we get a ∧b /∈ OE(M). So that E *OE(M), which is a contradiction. Therefore (a,E) ∨ (b,E) = R. � Theorem 3.19. In an ADL R with maximal elements, the following conditions are equivalent: (1) R is E-normal Int. J. Anal. Appl. (2023), 21:85 17 (2) for any two distinct maximal filters G1 and G2 of R with G1 ∩E = ∅, G2 ∩E = ∅ there exist a /∈ G1 and b /∈ G2 such that a∨b is maximal (3) for any maximal filter G with G∩E = ∅, G is the unique maximal filter containing R\OE(P ). Proof. (1) ⇒ (2) Assume that R is E-normal. Let G1 and G2 be two distinct maximal filters of R with G1 ∩E = ∅,G2 ∩E = ∅. Then R \G1 and R \G2 are distinct minimal prime E-ideals of R. By our assumption, we get R \G1 and R \G2 are co-maximal. That is, (R\G1)∨(R\G2) = R. Then, there exist a ∈ R\G1 and b ∈ R\G2 such that a∨b is maximal. (2) ⇒ (3) Assume (2). Let G be any maximal filter of R with G∩E = ∅ and R\OE(P ) ⊆ G. Let G1 be any maximal filter of R with G1 ∩E = ∅ and R \OE(P ) ⊆ G1. We prove that G = G1. Suppose G 6= G1. By our assumption, there exists a /∈ G and b /∈ G1 such that a∨b is maximal. That implies a,b /∈ R \OE(P ). So that a,b ∈ OE(P ). This implies that a∨b ∈ OE(P ). Therefore OE(P ) = R, which is a contradiction. We conclude that G = G1. (3) ⇒ (1) For any maximal filter G with G ∩ E = ∅, G is the unique maximal filter containing R \OE(P ). Let P be a prime E-ideal of R. Suppose P contains two minimal prime E-ideals say Q1 and Q2. That is, Q1 ⊆ P and Q2 ⊆ P. That implies OE(P ) ⊆OE(Q1) and OE(P ) ⊆OE(Q2). We get P ⊆OE(Q1) and P ⊆OE(Q2). So that Q2 ⊆ Q1 and Q1 ⊆ Q2. This concludes that Q1 = Q2. � Let F be a filter of R. For any x,y ∈ R, define a binary relation φF on R as φF = {(x,y) ∈ R×R | x ∧a = y ∧a, for some a ∈ F}. Proposition 3.5. For any filter F of an associative ADL R, φF is a congruence relation on R. For any ADL R, it can be easily verified that the quotient R/φF is also an ADL with respect to the following operations: [a]φF ∧ [b]φF = [a ∧ b]φF and [a]φF ∨ [b]φF = [a ∨ b]φF where [a]φF is the congruence class of a modulo φF . It can be routinely verified that the mapping Φ : R → R/φF defined by Φ(a) = [a]φF is a homomorphism. Theorem 3.20. In an ADL R, we have the following: (1) if x is a dual dense element of R, then [x]φF is a dual dense element of R/φF (2) if G is a E-ideal of R/φF , then Φ−1(G) is a E-ideal of R (3) if G is a prime E-ideal of R/φF , then Φ−1(G) is a prime E-ideal of R. Definition 3.9. Let F be a filter of an ADL R. For any ideal G of R, define G̃ = {[a]φF | a ∈ G}. The following result can be proved easily. Lemma 3.6. Let G be an E-ideal of R. Then G̃ is an E-ideal of R/φF . Proposition 3.6. Let G be a prime E-ideal and F a filter of an ADL R such that G∩F = ∅. We have the following: 18 Int. J. Anal. Appl. (2023), 21:85 (1) x ∈ G if and only if [x]φF ∈ G̃ (2) G̃ ∩ F̃ = ∅ (3) if G is a prime E-ideal of R, then G̃ is a prime E-ideal of R/φF . Proof. (1) Assume that x ∈ G. Then we have [x]φF ∈ G̃. Conversely assume that [x]φF ∈ G̃. Then there exists y ∈ G such that [x]φF = [y]φF . That implies (x,y) ∈ φF . So there exists a ∈ F such that x ∧a = y ∧a ∈ G. Since G ∩F = ∅, we get a /∈ G. Since x ∧a ∈ G and a /∈ G, we get that x ∈ G. (2) Suppose G̃ ∩ F̃ 6= ∅. Then choose an element x ∈ R such that [x]φF ∈ G̃ ∩ F̃ . Then [x]φF ∈ G̃ and [x]φF ∈ F̃ . Since [x]φF ∈ G̃ and by (1), we get x ∈ G. Since [x]φF ∈ F̃ , there exists y ∈ F such that [x]φF = [y]φF . Then (x,y) ∈ φF . So there exist a ∈ F such that x ∧a = y ∧a. Since y ∧a ∈ F, we get that x ∧a ∈ F. Since x ∈ G, we have that x ∧a ∈ G ∩F. That implies G ∩F 6= ∅, we get a contradiction. Hence G̃ ∩ F̃ = ∅. (3) Clearly, we have that G̃ is a proper ideal of R/φF . Let [x]φF ∈ Ẽ. Then x ∈ E ⊆ G. That implies [x]φF ∈ G and hence G̃ is an E-ideal of R/φF . Let [x]φF , [y]φF ∈ R/φF such that [x]φF ∧ [y]φF ∈ G̃. Then [x ∧ y]φF ∈ G̃. By (1) we have that x ∧ y ∈ G. Since G is prime, we get that x ∈ G or y ∈ G Again by(1) we get that [x]φF ∈ G̃ or [y]φF ∈ G̃. Hence G̃ is a prime E-ideal in R/φF . � Proposition 3.7. Let F be a filter of an ADL R. Then there is an order isomorphism of the set of all prime E-ideals of R disjoint from F onto the set of all prime E-ideals of R/φF . Proof. Let G and H be two prime E-ideals of R such that G ∩ F = ∅ and H ∩ F = ∅. Then by Proposition 3.6(1), we get that G ⊆ H if and only if G̃ ⊆ H̃. Let G be a prime E-ideal of R with G ∩ F = ∅. Then by Proposition 3.6(3), we get that G̃ is a prime E-ideal of R/φF . Let Q be a prime E-ideal of R/φF . Consider G = {a ∈ R|[a]φF ∈ Q}. Since Q is a E-ideal of R/φF , we get that G is a E-ideal of R. Let a,b ∈ R with a ∧ b ∈ G. Then [a]φF ∧ [b]φF = [a ∧ b]φF ∈ Q. Since Q is prime, we get [a]φF ∈ Q or [b]φF ∈ Q. Therefore a ∈ G or b ∈ G. Hence G is a prime E-ideal of R. Clearly G̃ = Q. Suppose G ∩ F 6= ∅. Then choose an element s ∈ G ∩ F. That implies [s]φF ∈ Q and s ∈ F. Let [b]φF ∈ R/φF . Since s ∈ F and b ∧ s = b ∧ s ∧ s, we get that (b,b ∧ s) ∈ F. That implies [b]φF = [b∧ s]φF = [b]φF ∧ [s]φF ∈ Q. Therefore [b]φF ∈ Q. and hence R/φF = Q, which is a contradiction. Thus G ∩F = ∅. � Corollary 3.10. Let R be an ADL. Then the above map induces a one-to-one correspondence between the set of all minimal prime E-ideals of R which are disjoint from F and the set of all minimal prime E-ideals of R/φF . Theorem 3.21. For any filter F of an ADL R, the following are equivalent: (1) any two distinct minimal prime E-ideals of R are co-maximal (2) any two distinct minimal prime E-ideals of R/φF are co-maximal. Int. J. Anal. Appl. (2023), 21:85 19 Proof. (1) ⇒ (2) Assume (1). Let G1,G2 be two distinct minimal prime E-ideals of R/φF . Then by the corollary 3.10, there exist two minimal prime E-ideals H1 and H2 of R such that H1∩F = ∅ and H2 ∩F = ∅. Also H̃1 = G1 and H̃2 = G2. Since G1 and G2 are distinct, we get that H1 and H2 are distinct. By the assumption, we have H1∨H2 = R. Let a ∈ R. There exist a1 ∈ H1 and a2 ∈ H2 such that a = a1 ∨a2. Since a1 ∈ H1 and a2 ∈ H2 we get [a1]φF ∈ H̃1 = G1 and [a2]φF ∈ H̃2 = G2. Now, [a]φF = [a1∨a2]φF = [a1]φF ∨ [a2]φF ∈ G1∨G2. That implies [a]φF ∈ G1∨G2, for all a ∈ R.Therefore G1 ∨G2 = R/φF . (2) ⇒ (1) Assume (2). Let P be a prime E-ideal of R. Suppose P contains two distinct minimal prime E-ideals, say G1 and G2. Consider K = R \P. Clearly K is a filter of R and G1 ∩K = ∅ = G2 ∩K. By Corollary 3.10, we get that G̃1 and G̃2 are distinct minimal prime E-ideals of R/φF such that G̃1, G̃2 ⊆ P̃. That implies P̃ is containing two distinct minimal prime E-ideals of R/φF , which is a contradiction. Hence P contains a unique minimal prime E-ideal. By Theorem 3.18, any two distinct minimal prime E-ideals of R are co-maximal. � 4. On the space prime E-ideals In this section, some topological properties of the space of all prime E-ideals and the space of all minimal prime E-ideals of an ADL are studied. Let us denote the set of all prime E-ideals of an ADL R by SpecEI (R). For any A ⊆ R, define α(A) = {P ∈ SpecEI (R)|A * P} and for any a ∈ R, α(a) = {P ∈ Spec E I (R)|a /∈ P}. Then we have the following result whose proof is straightforward. Lemma 4.1. Let R be an ADL and a,b ∈ R. Then the following conditions hold: (1) ⋃ a∈R α(a) = SpecEI (R) (2) α(a) ∩α(b) = α(a∧b) (3) α(a) ∪α(b) = α(a∨b) (4) α(a) = ∅ if and only if a ∈ E (5) α(a) = SpecEI (R) if and only if a ∈M. From the above result, it can be easily observed that the collection {α(a)|a ∈ R} forms a base for a topology on SpecEI (R). The topology generated by this base is precisely {α(A | A ⊆ R} and is called the hull-kernel topology on SpecEI (R). Under this topology, we have the following result. Theorem 4.1. In an ADL R, we have the following: (1) for any a ∈ R,α(a) is compact in SpecEI (R) (2) if C is a compact open subset of SpecEI (R), then C = α(a) for some a ∈ R (3) SpecEI (R) is a T0-space (4) the map a 7→ α(a) is an epimorphism from R onto the lattice of all compact open subsets of SpecEI (R). 20 Int. J. Anal. Appl. (2023), 21:85 Proof. (1) Let a ∈ R. Let X ⊆ R be such that α(a) ⊆ ⋃ x∈X α(x). Let J be a E-ideal generated by the set X. Suppose a /∈ J. Then there exists a prime E-ideal P such that J ⊆ P and a /∈ P. Since X ⊆ J ⊆ P, we get P /∈ α(x) for all x ∈ X. Since a /∈ P, we get P ∈ α(a), which is a contradiction. Hence a ∈ J. So we can write a = ( n∨ i=1 xi ) ∧a for some x1,x2, . . . ,xn ∈ X and n ∈ N. Then, we get α(a) = α(( n∨ i=1 xi ) ∧ a) ⊆ α( n∨ i=1 xi ) = n⋃ i=1 α(xi ) which is finite subcover for α(a). Therefore α(a) is compact. (2) Let C be a compact open subset of SpecEI (R). Since C is open, we get C = ⋃ x∈X α(x) for some X ⊆ R. Since C is compact, there exist x1,x2, . . . ,xn ∈ X such that C = n⋃ i=1 α(xi ) = α( n∨ i=1 ) Therefore C = α(x) for some x ∈ R. (3) Let P and Q be two distinct prime E-ideals of R. Without loss of generality, assume that P * Q. Choose x ∈ R such that x ∈ P and x /∈ Q. Hence P /∈ α(x) and Q ∈ α(x). Therefore SpecEI (R) is a T0-space. (4) It can be obtained from (1), (2) and by the above lemma. � Proposition 4.1. In an ADL R, the following are equivalent: (1) SpecEI (R) is a Hausdorff space (2) for each P ∈ SpecEI (R),P is the unique member of Spec E I (R) such that O E(P ) ⊆ P (3) every prime E-ideal is minimal (4) every prime E-ideal is maximal. Proof. (1) ⇒ (2) Assume (1). Let P ∈ SpecEI (R). Clearly O E(P ) ⊆ P. Suppose Q ∈ SpecEI (R) such that Q 6= P and OE(P ) ⊆ Q. Since SpecIE(R) is Hausdorff, there exists a,b ∈ R such that P ∈ α(a),Q ∈ α(b) and α(a∧b) = α(a) ∩α(b) = ∅. Hence a /∈ P,b /∈ Q and a∧b ∈ E. Therefore b ∈ OE(P ) ⊆ Q, which is a contradiction to that b /∈ Q. Hence P = Q. Therefore P is the unique member of SpecEI (R) such that O E(P ) ⊆ P. (2) ⇒ (3) Assume (2). Let P be a prime E-ideal of R. Let Q be a prime E-ideal in R such that Q ⊆ P. Hence OE(Q) ⊆ Q ⊆ P. Therefore P is a minimal prime E-ideal of R. (3) ⇒ (4) It is clear. (4) ⇒ (1) Assume (4). Let P and Q be two distinct elements of SpecEI (R). Hence O E(Q) * P. Choose a ∈OE(Q) such that a /∈ P. Since a ∈OE(Q), there exists b /∈ Q such that a ∈ (b,E). Hence a∧b ∈ E. Thus it yields, P ∈ α(a),Q ∈ α(b). Since a∧b ∈ E, we get that α(a)∩α(b) = α(a∧b) = ∅. Therefore SpecEI (R) is Hausdorff. � Theorem 4.2. For any E-ideal G of an ADL R, (G,E) = ⋂ {P ∈ SpecEI (R) | G * P}. Proof. Let G be an E-ideal of L. Consider K = ⋂ {P ∈ SpecEI (R) | G * P}. Let P ∈ α(G). Then G * P. Since G ∩ (G,E) = E ⊆ P and P is prime, we get (G,E) ⊆ P. Hence every prime E-ideal Int. J. Anal. Appl. (2023), 21:85 21 P of R such that G * P contains (G,E). Therefore (G,E) ⊆ K. Let x /∈ (G,E). Then there exists y ∈ G such that x ∧ y /∈ E. Let K = {G | G is an E − ideal of L and x ∧ y /∈ G}. Clearly, E ∈ K and so P = ∅. Clearly, (K,⊆) is a partially ordered set and it satisfies the hypothesis of the Zorn’s lemma, K has a maximal element, say N. Then N is an E-ideal of R and x ∧y /∈ N. Therefore x /∈ N and y /∈ N. Since y ∈ G, we get G * N. We now show that N is prime. Let a,b ∈ R with a /∈ N and b /∈ N. Then N ( N ∨ (a)E and N ( N ∨ (b)E. By the maximality of N, we get x ∧y ∈ N ∨ (a)E and x ∧y ∈ N ∨ (b)E. Hence, x ∧y ∈{N ∨ (a)E}∩{N ∨ (b)E} = N ∨{(a)E ∩ (b)E} = N ∨ (a∧b)E. If a∧b ∈ N, then x ∧y ∈ N which is a contradiction. Thus N is a prime E-ideal of R such that G * N and x /∈ N. Therefore x /∈ K. Hence K ⊆ (G,E). � Corollary 4.1. For any ADL R and a ∈ R, (a,E) = ⋂ {P ∈ SpecEI (R) | a /∈ P}. Let MinEI (R) denote the set of all minimal prime E-ideals of ADL R. For any a ∈ R, write αm(x) = α(x) ∩MinEI (R). Theorem 4.3. For any ADL R, the following conditions hold in R : (1) Every prime E-ideals contains a minimal prime E-ideal (2) ⋂ P∈MinE I (R) P = E (3) for any subset A with E ⊆ A, (A,E) = ⋂ P∈αm(A) (P ). Proof. (1) Let P be a prime E-ideal of R. Consider X = {N ∈ SpecEI (R) | N ⊆ P}. Clearly X is a partially ordered set under set inclusion and hence it satisfies the hypothesis of the Zorn’s lemma, X has a minimal element say M. Clearly M will be the required minimal prime E-ideal of R. (2) Since E is contained in every minimal prime E-ideal of R and so contained in the intersection of all minimal prime E-ideals. Let x /∈ E. Then there exists a prime E-ideal P of L such that x /∈ P. By (1), there exists a minimal prime E-ideal of R such that M ⊆ P. Since x /∈ P, we get x /∈ M. That implies M is a minimal prime E-ideal of R such that x /∈ M. Hence x is not in the intersection of all minimal prime. Thus intersection of all minimal prime E-ideals of R is equal to E. (3) Let P ∈ MinEI (R) such that A * P. Choose x ∈ A such that x /∈ P. Then (A,E) ⊆ (x,E) ⊆ P. That implies (A,E) is contained in every minimal prime E-ideal of R such that A * P. Hence (A,E) ⊆ ⋂ P∈αm(A) (P ). Let x /∈ (A,E). Then x ∧ y /∈ E, for some y ∈ A. By the condition (2), there exists a minimal prime E-ideal P of R such that x ∧y /∈ P. That implies x /∈ P and y /∈ P. Therefore x /∈ ⋂ P∈αm(A) P and hence (A,E) = ⋂ P∈αm(A) P. � Lemma 4.2. For any a,b ∈ R, we have following: (1) (a,E) ⊆ (b,E) if and only if αm(b) ⊆ αm(a) (2) αm(a) = ∅ if and only if a ∈ E (3) αm(a) = MinEI (R) if and only if (a,E) = E. 22 Int. J. Anal. Appl. (2023), 21:85 Proof. (1) Let a,b ∈ R. Assume that (a,E) ⊆ (b,E). Let P ∈ αm(b) Then b /∈ P. That implies (a,E) ⊆ (b,E) ⊆ P. Therefore a /∈ P and hence P ∈ αm(a). Thus αm(b) ⊆ αm(a). Conversely, assume that αm(b) ⊆ αm(a). Now, (a,E) = ⋂ P∈αm(a) P ⊆ ⋂ P∈αm(b) P = (b,E). Hence (a,E) ⊆ (b,E). (2) Suppose MinEI (R) = ∅. Then a ∈ P for all P ∈ Min E I (R). That implies a ∈ ⋂ P∈MinE I (R) P. Since a ∈ ⋂ P∈MinE I (R) P = E, we get a ∈ E. The converse is clear. (3) Assume αm(a) = MinEI (R). Then (a,E) = ⋂ P∈αm(a) P = ⋂ P∈MinE I (R) P = E. Therefore(a,E) = E. Conversely, assume (a,E) = E. Then (a,E) = E ⊆ P. That implies a /∈ P, for all P ∈ MinEI (R). Therefore αm(a) = MinEI (R). � For any E-ideal G of an ADL R, define βm(G) = {P ∈ MinEI (R) | G ⊆ P}. Lemma 4.3. Let G be an E-ideal of an ADL R. If βm(G) = ∅, then (G,E) = E. Proof. Let βm(G) = ∅. Then βm(G) = MinEI (R). That implies (G,E) = ⋂ P∈αm(F) P ⊆ ⋂ P∈MinE I (R) P = E. Therefore (G,E) = E. � For any ADL R, define K = {x ∈ R | (x,E) = E}. Lemma 4.4. For any ADL R,K is a filter of R. Proof. Clearly, we have that for any m ∈M, m ∈ K. Let x,y ∈ K. Then ((x∧y,E),E) = ((x,E),E)∩ ((y,E),E) = (E,E) ∩ (E,E) = R ∩ R = R. That implies ((x ∧ y),E) = (R,E) = E. Therefore x∧y ∈ K. Let x ∈ K. Then (x,E) = E. Let y ∈ R. Now, (x∨y,E) = (x,E)∩(y,E) = E∩(y,E) = E. Therefore x ∨y ∈ K. Hence K is a filter of R. � Theorem 4.4. Let G be an E-ideal of an ADL R. Then MinEI (R) is compact if and only if βm(G) = ∅ implies G ∩K 6= ∅. Proof. Assume that MinEI (R) is compact. Let G be an E-ideal R such that βm(G) = ∅. Then αm(G) = Min E I (R). Since Min E I (R) is compact, there exists a ∈ G such that αm(a) = Min E I (R). That implies (a,E) = E. Therefore a ∈ K and hence G ∩ K 6= ∅. Conversely, assume that for any E-ideal G of R,βm(G) = ∅ implies G ∩ K 6= ∅. Let A ⊆ R be such that MinEI (R) = ⋃ a∈A αm(A) = αm(A) = αm(G) where G = AE. Since MinEI (R) = αm(G), we get βm(G) = ∅. By the assumption, we get G ∩E 6= ∅. Choose d ∈ G ∩K. Since d ∈ G and G = AE, there exists a1,a2, . . . ,an ∈ A such that d = (a1∨a2∨ . . .∨an)∧d. Since d ∈ E, MinEI (R) = αm(d) ⊆ αm ( n∨ i=1 ai ) = n⋃ i=1 αm(ai ). Hence MinEI (R) is compact. � Theorem 4.5. Let R be an ADL. For any Y ⊆ MinEI (R), the closure of Y in Min E I (R) is βm( ⋂ P∈Y P ) and, in particular, αm(F ) = βm((G,E)), for any E ⊆ G ⊆ R. Int. J. Anal. Appl. (2023), 21:85 23 Proof. Let Y ⊆ MinEI (R). Then Y in Min E I (R) = {Y in Spec E I (R)} ∩ Min E I (R) = H( ⋂ P∈Y P ) ∩ MinEI (R) = βm( ⋂ P∈Y P ). In particular, for any E ⊆ G ⊆ R, we have αm(G) = βm( ⋂ P∈αm(G) P ) = βm( ⋂ I*P, P∈MinE I (R) P ) = βm((F,E)). � Proposition 4.2. Let F,G be two E-ideals of an ADL R. Then the following are equivalent: (1) G ⊆ (F,E) (2) G ∩F = E (3) αm(G) ∩αm(F ) = ∅. Proof. (1) ⇒ (2) Assume that G ⊆ (F,E). Then G ∩F ⊆ (F,E) ∩F = E. Therefore G ∩F = E. (2) ⇒ (3) Assume that G ∩F = E. Let P ∈ αm(G) ∩αm(F ) = αm(G ∩F ). Then E = G ∩F * P, which is a contradiction. Therefore αm(G) ∩αm(F ) = ∅. (3) ⇒ (1) Assume that αm(G) ∩ αm(F ) = ∅. Let x ∈ G. Suppose x /∈ (F,E). Then there exists y ∈ F such that x ∧ y /∈ E. Then there exists P ∈ MinEI (R) such that x ∧ y /∈ P. That implies x /∈ P and y /∈ P. Hence G * P and F * P. Therefore P ∈ αm(G) and P ∈ αm(F ). Therefore P ∈ αm(G) ∩αm(F ), which is a contradiction. So x ∈ (F,E). Therefore G ⊆ (F,E). � Corollary 4.2. Let G be an E-ideal of an ADL R and x ∈ R. Then x ∈ (G,E) if and only if αm(x) ∩αm(G) = ∅. Proof. By taking G = {x}, in the above proposition. � Theorem 4.6. Every open subset of MinEI (R) is closed if and only if for any E-ideal of R, (G,E) = E implies βm(G) = ∅. Proof. Assume that every open set of MinEI (R) is closed. Let G be an E-ideal of R. Then βm(G) is an open set in MinEI (R). Now, βm(G) 6= ∅. Then there exists x ∈ R \E such that αm(x) ⊆ βm(G). That implies αm(x) ∩ αm(G) = ∅. Therefore x ∈ (G,E) and x /∈ E. Hence (G,E) 6= E. Thus (G,E) = E, which gives βm(G) = ∅. Conversely, assume that the condition holds. Let H be an open subset of MinEI (R). Then H = αm(G), for some E-ideal G of L. By Theorem 4.5, we have αm(G) = βm((G,E)). It is enough to show that βm((G,E)) = αm(G). Since ((G ∨ (G,E)),E) = E, by the assumption, we get βm(G ∨ (G,E)) = ∅. Now, for any P ∈ MinEI (R), we have P ∈ αm(G) ⇔ G * P ⇔ (G,E) ⊆ P ⇔ P ∈ βm(G). Hence αm(G) = βm(G). Therefore H is closed in MinEI (R). � Theorem 4.7. In an ADL R, MinEI (R) is a Hausdorff space. Proof. Let P and Q be distinct elements of MinEI (R). Then there exists a ∈ P such that a /∈ Q. Since P is minimal, we get (a,E) * P. Then there exists b ∈ (a,E) such that b /∈ P. That implies a∧b ∈ E and hence αm(a)∩αm(b) = ∅. Since a /∈ Q and b /∈ P, we get Q ∈ αm(a) and P ∈ αm(b). Therefore MinEI (R) is a Hausdorff space. � 24 Int. J. Anal. Appl. (2023), 21:85 Acknowledgment: This research project was supported by the Thailand Science Research and Inno- vation Fund and the University of Phayao (Grant No. FF66-UoE017). Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi- cation of this paper. References [1] G. Birkhoff, Lattice theory, Colloq. Publ. XXV, Amer. Math. Soc. Providence, 1967. [2] W.H. Cornish, Normal lattices, J. Aust. Math. Soc. 14 (1972), 200-215. https://doi.org/10.1017/ s1446788700010041. [3] G. 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Soc. 31 (1981), 77-91. https://doi.org/10. 1017/s1446788700018498. https://doi.org/10.1017/s1446788700010041 https://doi.org/10.1017/s1446788700010041 https://doi.org/10.1007/978-3-0348-7633-9 https://doi.org/10.5817/AM2021-3-157 https://doi.org/10.1017/s1446788700018498 https://doi.org/10.1017/s1446788700018498 1. Introduction 2. Preliminaries 3. E-ideals of ADLs 4. On the space prime E-ideals References