EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5535 ISSN 1307-5543 – ejpam.com Published by New York Business Global A Certain Class of Filters in Generalized Complemented Distributive Lattices Ramesh Sirisetti1, Jogarao Gunda2, Ravikumar Bandaru3, Rahul Shukla4,∗ 1 Department of Mathematics, GITAM School of Science, GITAM (Deemed to be University), Visakhapatnam, Andhra Pradesh-530045, India 2 Department of BS & H, Aditya Institute of Technology and Management, Tekkali, Srikakulam, Andhra Pradesh-530021, India 3 Department of Mathematics, School of Advanced Sciences, VIT-AP University, Andhra Pradesh-522237, India 4 Department of Mathematical Sciences and Computing, Walter Sisulu University, Mthatha 5117, South Africa Abstract. In this paper, we introduce Kg-filters jointly derived from the class of ideals and the class of generalized complementations in a generalized complemented distributive lattice. We obtain some algebraic properties on the obtained class, and we provide some counter-examples. Mainly, we derive some Boolean algebras (distributive lattices) through the class of Kg-filters in a generalized complemented distributive lattice. Finally, we introduce normal Kg-filters in a generalized complemented distributive lattice and then prove that the class of normal Kg-filters is a Boolean algebra. 2020 Mathematics Subject Classifications: 06D05, 06D15 Key Words and Phrases: Ideals, Filters, Distributive Lattices, Generalized Complemented Distributive Lattices, Boolean Algebras 1. Introduction The concept of distributive lattices [2, 3] has been extensively studied by several au- thors by taking a unary operation, like complementation [7, 12], pseudo-complementation [14, 15], quasi-complementation [5, 13] etc., and also by considering the class of ideals (filters). Some of authors recursively studied the class of distributive lattices by taking a binary operations like generalized implementation [11], by taking median graphs [10], by considering canonical extensions [9], by using subordinations [1, 6]. The class of ideals (filters) has a major role in the theory of lattices. It is known that there is a one-to-one ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5535 Email addresses: ramesh.sirisetti@gmail.com (R. Sirisetti), jogarao.gunda@gmail.com (J. Gunda), ravimaths83@gmail.com (R. Bandaru), rshukla@wsu.ac.za (R. Shukla) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 2 of 12 correspondence between the class of prime ideals and the class of prime filters in a dis- tributive lattice. It is easy to observe that we can obtain a class of ideals (filter) from a complementation on a distributive lattice. The authors [8] introduced and studied a generalized complementation (g-complementation) on a distributive lattice with dense el- ements; it is a generalization of complementation and dual of pseudo-complementation in distributive lattices. In this regard, we derive filters from a generalized complementation on a distributive lattice using dense elements. In this paper, we introduce a class of filters (Kg-filters), which are abstract combined from the class of ideals and the class of generalized complementations on a distributive lattice A through dense elements and prove several algebraic properties on them. Also, we obtain some necessary and sufficient conditions for a filter to become Kg-filter in A. We obtain some algebraic results on the class of Kg-filters in a generalized complemented distributive lattice. Mainly, we abstract some structures which are distributive lattices (Boolean algebras) [4] from the class of Kg-filters. Finally we discuss normal Kg-filters in generalized complemented distributive lattices and prove that the class of normalKg-filters forms a Boolean algebra. 2. Some results on generalized complemented distributive lattices In this section, we introduce a filter corresponding to an ideal in a generalized dis- tributive lattice and obtain some properties. Also, we discuss prime filters corresponding to an ideal in a generalized complemented distributive lattice. Definition 2.1. [8] A unary operation g on a distributive lattice A is said to be a g- complementation (generalized complementation) if for any v, w ∈ A, v ∨ vg ∈ D, and v ∨w ∈ D if and only if vg ≤ w. In this case, vg is said to be a g-complement of v, and A is called a g-complemented distributive lattice. Let us consider A means that it is a distributive lattice with dense elements and g is a generalized complementation on A. Proposition 2.1. [8] For any g-complementation g on A, (i) 0g ∈ D (ii) d ∈ D ⇒ dg = 0 (iii) v ≤ w ⇒ wg ≤ vg (iv) vgg ≤ v (v) vggg = vg (vi) 0gg = 0 (vii) v ∈ D ⇐⇒ vg = 0 ⇐⇒ vgg ∈ D R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 3 of 12 (viii) v ≤ 0g, for all v, w ∈ A. Proposition 2.2. [8] For any g-complementation g on A and for any v, w ∈ A, the following are equivalent; (i) v ∨ w ∈ D (ii) vgg ∨ w ∈ D (iii) vgg ∨ wgg ∈ D (iv) v ∨ wgg ∈ D. Proposition 2.3. [8] For any v, w ∈ A, (i) (v ∧ w)g = vg ∨ wg (ii) (v ∨ w)g ≤ vg ∧ wg (iii) (v ∨ w)gg = vgg ∨ wgg = (vg ∧ wg)g (iv) (v ∧ w)gg = (vg ∨ wg)g = (vgg ∧ wgg)gg. Given any ideal K of A, consider a set G(K) = {u ∈ A | ug ∈ K} containing D (because dg = 0 ∈ K, for all d ∈ D). Lemma 2.2. For any ideal K of A, G(K) is a filter. Proof. Let v, w ∈ G(K). Then vg, wg ∈ K. By Proposition 2.3 (i), (v∧w)g = vg∨wg ∈ K (since K is ideal). Therefore v ∧w ∈ G(K). Given u ∈ A, we have v, u ≤ v ∨ u, which implies (v ∨ u)g ≤ vg, ug (By Proposition 2.1(iii)) and then (v ∨ u)g ≤ vg ∧ ug ∈ K (since vg ∈ K, ug ∈ A, and K is an ideal). Therefore v ∨ u ∈ G(K). Thus G(K) is filter. Remark 2.3. D = G(0) = G((0]) and G(A) = A. Lemma 2.4. For any ideals K,H of A, we have; (i) K ⊆ H implies G(K) ⊆ G(H). (ii) G(K) ∩G(H) = G(K ∩H). (iii) G(K) ∨G(H) ⊆ G(K ∨H). Proof. (i) Suppose that K ⊆ H and u ∈ G(K). Then ug ∈ K ⊆ H. Therefore u ∈ G(H) and hence G(K) ⊆ G(H). (ii) and (iii) follows from (i). Lemma 2.5. For any ideal K of A, either G(K) ∩K = ∅ or G(K) = A. R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 4 of 12 Proof. Suppose that G(K) ∩K ̸= ϕ. Let x ∈ G(K) ∩K. Then xg ∈ K, and x ∈ K. Therefore x ∨ xg ∈ D ∩K (since K is an ideal). Since 0g ≤ d for all d ∈ D, 0g ≤ x ∨ xg. Therefore 0 in G(K) and hence G(K) = A (since G(K) is filter). Lemma 2.6. For any ideal K of A and u ∈ A, (i) u ∈ G(K) if and only if ugg ∈ G(K) (ii) u ∈ K implies ug ∈ G(K). Proof. (i) follows from Proposition 2.1(v). (ii) follows from Proposition 2.1(iv). The converse of Lemma 2.6(ii) need not be hold. Check out the following example. Example 2.7. Consider a distributive lattice A = {0, i, j, k, 1}, whose Hasse diagram is showing in below; k i 0 j 1 For the ideal K = {0, i} of A and 0g = k, ig = j, jg = i, kg = 1g = 0, we have G(K) = {j, k, 1}, and kg = 0 = 1g ∈ G(K), but k, 1 /∈ K. Lemma 2.8. For any ideal K of A, K ∩D ̸= ϕ if and only if A = G(K). Proof. Suppose that K ∩ D ̸= ϕ. Let v ∈ K ∩ D. Then v ∈ K and vg = 0 ∈ K. Therefore v ∈ K ∩ G(K). So that K ∩ G(K) is non-empty. By Lemma 2.5., G(K) = A. Conversely suppose that G(K) = A. For 0 ∈ A = G(K), we have 0g ∈ K ∩D. Theorem 2.9. Given an ideal K of A, G(K) is proper if and only if either i /∈ G(K) or ig /∈ G(K). That is., for any i ∈ A, i, ig in G(K) if and only if A = G(K). Proof. Let us consider that G(K) is a proper filter of A. Let i ∈ A. If i ∈ G(K) and ig ∈ G(K), then ig in K and ig in G(K). Therefore G(K) ∩ K ̸= ϕ. By Theorem 2.5., G(K) = A. Which is contradiction. Here i /∈ G(K) or ig /∈ G(K). Conversely suppose that i /∈ G(K) or ig /∈ G(K). If G(K) = A, then i, ig ∈ A = G(K). Which is contradiction. Therefore G(K) ̸= A. Hence G(K) is proper. R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 5 of 12 Theorem 2.10. If A satisfies (v ∨w)g = vg ∧wg, for all v, w ∈ A and K is a prime ideal of A, then G(K) is a prime filter in A. Proof. Let v, w ∈ A such that v ∨w in G(K). Then (v ∨w)g = vg ∧wg in K. Since K is prime, vg ∈ K or wg ∈ K. Therefore v ∈ G(K) or w ∈ G(K) and hence G(K) is prime. The converse of Theorem 2.10 need not be true. Example 2.11. Consider the distributive lattice A = {0, i, j, k, l, 1}, whose Hasse diagram is in below; 0 i j kl 1 For the ideal K = {0, i} and 0g = k, ig = j, jg = i, kg = 1g = 0, G(K) = {j, k, 1} is a prime filter, but K is not prime. Because j ∧ l = 0 ∈ K, j /∈ K and l /∈ K. Remark 2.12. The converse of Theorem 2.10. is true, provided vgg = v, for all v ∈ A. For, Suppose that A does not have a prime ideal K. In such case, v ∧ w ∈ K such that v /∈ K and w /∈ K exist for v, w ∈ A. (v ∧w)gg ≤ v ∧w ∈ K, by Proposition 2.1 (iv). Consequently, (v ∧ w)gg ∈ K. Therefore, (v ∧ w)g ∈ G(K) holds. (v ∧ w)g = vg ∨ wg ∈ G(K), at this point. vg ∈ G(K) or wg ∈ G(K) if G(K) is prime. It follows that either w = wgg ∈ K or v = vgg ∈ K. It is contradictory. Therefore It is not prime, G(K). Theorem 2.13. If vgg = v, for all v ∈ A, then A has a unique dense element. Proof. Suppose that d1 and d2 are two dense elements inA. Now, d1 = dgg1 = (dg1) g = 0g and d2 = dgg2 = (dg2) g = 0g (by Proposition 2.1.(ii)). Therefore d1 = d2 = 0g. Hence A has a unique dense element. In a general lattice, the converse of Theorem 2.13. need not be true. Example 2.14. Consider a lattice A = {0, i, j, k, 1}, whose Hasse diagram is in below; R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 6 of 12 0 i j k 1 and 0g = 1, 1g = 0, ig = k, jg = k, kg = i. Then A has a unique dense element, but jgg = i ̸= j. 3. Kg-filters in g-complemented distributive lattices In this section, we introduce Kg-filters derived from ideals using a generalized comple- mentation on a distributive lattice with dense elements and study some algebraic prop- erties. We obtain some necessary conditions for a filter to become a Kg-filter. Mainly, we prove that the set of Kg-filters is a distributive lattice. Also, we obtain a Boolean al- gebraic structure through Kg-filters in a g-complemented distributive lattice. Finally we introduce normal Kg-filters in a generalized complemented distributive lattice and prove that the set of normal Kg-filters is a Boolean algebra. Definition 3.1. A filter N of A is called a Kg-filter if there exists an ideal K in A such that G(K) = N . It is easy to observe that D is a Kg-filter, because G((0]) = D. Lemma 3.2. For any v ∈ A, the principal filter generated by vg is a Kg-filter of A. Moreover [vg) = G((v]). Proof. Let u ∈ G((v]). Then ug ∈ (v] and ug ≤ v. Since u ∨ ug ∈ D, u ∨ v ∈ D. By Definition 2.1., vg ≤ u. So that u ∈ [vg). Therefore G((v]) ⊆ [vg). For any u ∈ A, u ∈ [vg) ⇒ vg ≤ u ⇒ ug ≤ vgg (Proposition 2.1(iii)) ⇒ ug ≤ v (since vgg ≤ v) ⇒ ug ∈ (v] ⇒ u ∈ G((v]). Therefore [vg) ⊆ G((v]) and hence [vg) = G((v]). Thus [vg) is a Kg-filter. Now, we provide an example of a distributive lattice in which there is a Kg-filter and there is a non-Kg-filter. R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 7 of 12 Example 3.3. Consider a distributive lattice A = {0, i, j, k, 1} whose Hasse diagram is given below; 0 i j k 1 Then Ka = {0},Kb = {0, i},Kc = {0, i, k},Kd = {0, i, j},Ke = A are ideals of A and Na = {1}, Nb = {j, 1}, Nc = {k, 1}, Nd = {i, j, k, 1}, Ne = A are filters of A. Moreover Nd is a Kg-filter but others are not. Theorem 3.4. Let A be a distributive lattice with vgg = v, for all v ∈ A. Then for any prime filter containing D in A is a Kg-filter. Proof. Let N be a prime filter containing D in A. For any v ∈ N , vg ∧ vgg = 0 ∈ A−N ⇒vg ∈ A−N or vgg ∈ A−N (since A−N is prime ideal) ⇒vg ∈ A−N or v ∈ A−N (since vgg = v) ⇒vg ∈ A−N (since v ∈ N) ⇒v ∈ G(A−N). Therefore N ⊆ G(A−N). Let v ∈ G(A−N). Then vg ∈ A−N . Since D ⊆ N , v ∨ vg in N . Then v in N or vg in N . Therefore v in N(since vg /∈ N). So that G(A − N) ⊆ N . Hence G(A−N) = N . Thus N is Kg-filter. Corollary 3.5. Let A be a distributive lattice with vgg = v, for all v ∈ A. Then every maximal filter is a Kg-filter. Let us denote the set of Kg-filters in a distributive lattice A with dense elements by GF(A). Now we have the following theorem. Theorem 3.6. GF(A) is a distributive lattice with the operations G(Ka) ∩ G(Kb) = G(Ka ∧Kb) and G(Ka) ⊔G(Kb) = G(Ka ∨Kb), for all G(Ka), G(Kb) ∈ GF(A). R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 8 of 12 Proof. LetG(Ka), G(Kb) ∈ GF(A), whereKa,Kb in I(A). By Lemma 2.4(i), G(Ka), G(Kb) ⊆ G(Ka ∨ Kb). Therefore G(Ka ∨ Kb) is an upper bound of G(Ka), G(Kb). Let G(Kc) ∈ GF(A) is an upper bound of G(Ka), G(Kb), for some Kc ∈ I(A). For v ∈ G(Ka ∨Kb), vg ∈ Ka ∨Kb ⇒vg = i ∨ j for some i ∈ Ka and j ∈ Kb ⇒igg ∈ Ka and jgg ∈ Kb (since vgg ≤ v forall v ∈ A) ⇒ig ∈ G(Ka) and jg ∈ G(Kb) ⇒ig ∈ G(Kc) and jg ∈ G(Kc) (since G(Ka), G(Kb) ⊆ G(Kc)) ⇒igg ∨ jgg ∈ Kc (since Kc is an ideal) ⇒(i ∨ j)gg ∈ Kc (since Proportion 2.3(iii)) ⇒(vg)gg ∈ Kc (since vg = i ∨ j) ⇒vggg ∈ Kc ⇒vg ∈ Kc (since vggg = vg) ⇒v ∈ G(Kc). Therefore G(Ka∨Kb) ⊆ G(Kc) and hence G(Ka∨Kb) is the least upper bound of G(Ka), G(Kb). Hence it is denoted by G(Ka) ⊔ G(Kb). By Lemma 2.4(ii)., G(Ka) ∧ G(Kb) = G(Ka∧Kb). Since I(A) is distributive, GF(A) is distributive. Hence GF(A) is a distribu- tive lattice with the lease element G({0}) and the greatest element G(A). Corollary 3.7. If I(A) is a Boolean algebra, then GF(A) is a Boolean algebra, but not a sub-Boolean algebra of I(A). Lemma 3.8. For any v ∈ A,G((vgg]) = G((v]). Proof. Let v ∈ A. Then vgg ≤ v. Therefore (vgg] ⊆ (v]. By Lemma 2.4(ii)., G((vgg]) ⊆ G((v]). Let l ∈ G((v]). Then lg ∈ (v] and hence lg ≤ v. By Proposition 2.1., lggg = lg ≤ vgg ≤ v. So that lg ∈ (vgg]. Hence l ∈ G((vgg]). Thus G((v]) = G((vgg]). Lemma 3.9. For any v, w in A, the following (1.), and (2.) are equivalent; (1). (v ∧ w)gg = vgg ∧ wgg (2). (v ∨ w)g = vg ∧ wg. Proof. (1) ⇒ (2) : Assume(1), let v, w ∈ A. Then (v∨w)g = (v∨w)ggg = (vgg∨wgg)g = (vg ∧ wg)gg = vggg ∧ wggg = vg ∧ wg. (2) ⇒ (1): Assume (2), let v, w ∈ A. Then (v ∧ w)gg = (vg ∨ wg)g = vgg ∧ wgg. Lemma 3.10. For any v, w ∈ A, we have (i) G((v]) = G((vgg]) (ii) w ∈ N ⇔ wgg ∈ N if N is Kg-filter in A R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 9 of 12 (iii) If N is a Kg-filter, then N = ⋃ w∈N G((wg]). Proof. (i) Let v ∈ A. By Lemma 3.2., G((v]) = [vg) = [vggg) = G((vgg]). (ii) Let N be a Kg filter of A. Then N = G(K), for some ideal K in A. For any w ∈ A, w in N ⇔ w ∈ G(K) ⇔ wg ∈ K ⇔ wggg ∈ K ⇔ wgg ∈ G(K) ⇔ wgg ∈ N. (iii) Let N be a Kg-filter of A. Then N = G(K) for some ideal K in A. Let v ∈ G((wg]) for some w ∈ N . Then vg ∈ (wg]. Therefore vg ≤ wg ⇒ wgg ≤ vgg. Now, w ∈ N = G(K). Then wg ∈ K. Therefore wggg = wg ∈ K implies wgg ∈ G(K) = N . Since N is filter, vgg ∈ N . So that v ∈ N . Hence ⋃ w∈N G((wg]) ⊆ N . Let w ∈ N . Then wg ∈ K. Therefore wggg = wg ∈ (wg] implies wgg ∈ G((wg]). So that w ∈ G((wg]). Hence N ⊆ ⋃ w∈N G((wg]). Thus N = ⋃ w∈N G((wg]). Let us consider a set D(A) = {v ∈ A | vg ∈ D}. Then it is easy to observe that D(A) is an ideal of A and G(D(A)) = D. Lemma 3.11. For any ideal K of A, and x ∈ A, we have the following equivalent condi- tions: (i) G(K) = D (ii) v ∈ K ⇒ vg ∈ D (iii) K ⊆ D(K) ⊆ D(A) Proof. (i) ⇒ (ii); Assume (i); Let v ∈ K. Then vgg ∈ K. Therefore vg ∈ G(K) = D. Hence vg ∈ D. (ii) ⇒ (iii); Assume (ii); Let v ∈ K. Then vg ∈ D. Therefore v ∈ D(A). Since K ⊆ A, v ∈ D(K). Hence K ⊆ D(K). (iii) ⇒ (i); Assume (iii); Now, K ⊆ D(K) ⊆ D(A) implies G(K) ⊆ G(D(K)) ⊆ G(D(A)). Therefore G(K) ⊆ G(D(K)) ⊆ D. Since D ⊆ G(K), G(K) = D. Theorem 3.12. The set BG(A) = {v ∈ A | vgg = v} is a Boolean algebra with the operations ∨ and v1 ∗ v2 = (vg1 ∨ vg2) g, for all v1, v2 ∈ BG(A). Proof. Let v1, v2 ∈ BG(A). Now, (v1 ∨ v2) gg = vgg1 ∨ vgg2 = v1 ∨ v2 (By proposition 2.3(iii)). Therefore v1∨v2 ∈ BG(A). Now [v1∗v2]gg = [(vg1∨v g 2) g]gg = ((vg1∨v g 2) g = v1∗v2 ∈ BG(A). Therefore BG(A) is closed under ∨ and ∗. Now v1∗v2 = (vg1∨vg2) g = (v1∧v2) gg ≤ v1∧v2 ≤ v1, v2. Therefore v1∗v2 is a lower bound of v1, w2. For any lower bound l of v1, v2 in BG(A), we have l ≤ v1, v2 ⇒ vg1 , v g 2 ≤ lg ⇒ vg1 ∨vg2 ≤ lg ⇒ lgg ≤ (vg1 ∨vg2) g ⇒ l ≤ v1 ∗v2 (since l ∈ BG(A)). Therefore v1 ∗ v2 is the greatest lower bound of v1, v2 in BG(A). Since A is distributive lattice, BG(A) is form a distributive lattice. Let b ∈ BG(A). Then there exists bg ∈ BG(A) such that b∨bg is dense and b∨bg = bgg∨bg = bgg∨(bg)gg = (b∨bg)gg = 0g is the greatest element in BG(A). Now, b ∗ bg = (bg ∨ bgg)g. Since bg ∨ bgg is dense, (bg ∨ bgg)g = 0 = b ∗ bg. Hence BG(A) is a Boolean algebra. R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 10 of 12 Theorem 3.13. There is an onto homomorphism between I(A) to GF(A). Proof. Define a map Φ from I(A) to GF(A) by Φ(K) = G(K) for all K ∈ I(A). Let Ka,Kb ∈ GF(A). Now, Ka = Kb. Then Φ(Ka) = Φ(Kb). Therefore Φ is well defined. Now, Φ(Ka ∩ Kb) = G(Ka ∩ Kb) = G(Ka) ∧ G(Kb) = Φ(Ka) ∧ Φ(Kb), Φ(Ka ∨ Kb) = G(Ka ∨Kb) = G(Ka) ⊔G(Kb) = Φ(Ka) ⊔Φ(Kb) and Φ({0}) = G({0}) = D. Therefore Φ is an onto homomorphism. Remark 3.14. The above homomorphism need not be one-one. In Example 3.3., let Kb = {0, i},Kc = {0, i, k}, then G(Kb) = A = G(Kc). That is Φ(Kb) = Φ(Kc). But Kb ̸= Kc. Hence Φ is not one-one. Lemma 3.15. Let Φ be a homomorphism between I(A) and GF(A). Then kernel of Φ is an ideal of I(A). Proof. Let Ka,Kb ∈ ker(Φ). Then Φ(Ka) = D = Φ(Kb). Therefore G(Ka) = D = G(Kb). Now, Φ(Ka ∨ Kb) = G(Ka ∨ Kb) = G(Ka) ⊔ G(Kb) = D ⊔ D = D. Therefore Ka ∨ Kb ∈ ker(Φ). For any K ∈ I(A),Φ(K ∧ Ka) = G(K ∧ Ka) = G(K) ∧ G(Ka) = G(K) ∧D = D. Therefore K ∧Ka ∈ ker(Φ). Hence ker(Φ) is an ideal of I(A). Theorem 3.16. If Φ is a homomorphism from I(A) to GF(A) defined by Φ(K) = G(K), for all K ∈ I(A), then the following are equivalent; (i) ker(Φ) = {0} (ii) vgg = v, for all v ∈ A. (iii) Φ is one-one. Proof. (ii) ⇒ (iii); Suppose that vgg = v for all v ∈ A. Let Ka,Kb ∈ I(A). Now, Φ(Ka) = Φ(Kb). Then G(Ka) = G(Kb). For any k ∈ A, k ∈ Ka ⇔ kgg ∈ Ka (since kgg = k) ⇔ kg ∈ G(Ka) = G(Kb) ⇔ kg ∈ G(Kb) ⇔ kgg ∈ Kb ⇔ k ∈ Kb. (since kgg = k) Therefore Ka = Kb. Hence Φ is one-one. (iii) ⇒ (i); Suppose that Φ is one-one. Let K ∈ ker(Φ). Then Φ(K) = D. Therefore Φ(K) = G(K) = D = G((0]) = Φ((0]). Since Φ is one-one, K = (0]. Hence ker(Φ) = {0}. (ii) ⇒ (i); Suppose that vgg = v, for all v ∈ A. Let K ∈ ker(Φ). Then Φ(K) = D. Therefore G(K) = D. Let k ∈ K. Then kgg ∈ K. Therefore kg ∈ G(K) = D. For this kg ∈ D, kgg = 0. Since kgg = k = 0. Hence K = {0}. Therefore ker(Φ) = {0}. (iii) ⇒ (ii); Suppose that Φ is one-one. Let v ∈ A. By Lemma 3.3., G((vgg]) = G((v]). Then Φ((vgg]) = Φ((v]). Therefore (vgg] = (v]. Since vgg ∈ (vgg] = (v], (vgg] ≤ v. Similarly v ≤ vgg. Hence vgg = v, for all v ∈ A. R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 11 of 12 For any i ∈ A, G((i]) is a Kg-filter of A and it is called a normal Kg-filter. The set of all normal Kg-filters of A denoted by NG(A) = {G((i]) = [ig)|i ∈ A} and it forms a sublattice of GF(A). Moreover NG(A) is a Boolean algebra. Theorem 3.17. The set NG(A) forms a sublattice of GF(A) with the operations G((i])∧ G((j]) = G((i ∧ j]) and G((i]) ⊔ G((j]) = G((i ∨ j]), for all G((i]), G(j]) ∈ NG(A), for some i, j ∈ A. Moreover NG(A) is a Boolean algebra. Proof. Let G((i]), G((j]) ∈ NG(A). By Lemma 2.4., G((i]) ∧ G((j]) = G((i] ∧ (j]) = G((i ∧ j]) and by Theorem 3.9., G((i]) ⊔ G((j]) = G((i] ∨ (j]) = G((i ∨ j]). Therefore G((i]) ∧ G((j]) ∈ NG(A) and G((i]) ⊔ G((j]) ∈ NG(A). Hence NG(A) is a sublattice of distributive lattice GF(A) with the least element G((0]) and the greatest element G((d]) where d is dense in A. For any G((i]) in NG(A), there exists G((ig]) in NG(A) such that G((i]) ∧ G((ig]) = G((i ∧ ig]). Let x ∈ G((i ∧ ig]) = G((i]) ∧ G((ig]). Then we have xg ∈ (i ∧ ig]. So, xg ≤ i ∧ ig. Hence (i ∧ ig)g = ig ∨ igg ≤ xgg ≤ x. Since ig ∨ igg is dense, x is dense. Then G((i] ∧ G(ig]) ⊆ D. Hence G((i] ∧ G(ig]) = D. Now, G((i]) ⊔G((ig]) = G((i ∨ ig]) = [(i ∨ ig)g). Since (i ∨ ig) is dense, (i ∨ ig)g = 0. Therefore G((i ∨ ig]) = A. Hence G((i]) ⊔G((ig]) = A. Thus NG(A) is a Boolean algebra. Theorem 3.18. For any normal Kg-filter N of A, there exists a prime filter Q of A such that N ⊆ Q. Proof. Let N be a Kg-filter of A. Then there exists an ideal K of A such that G(K) = N . Now, N ∩ K = Φ. Then there exists a prime filter Q of A such that Q ∩K = Φ and N ⊆ Q. 4. Conclusions This paper exponentially enrich algebraic properties of a certain class of filters (Kg- filters) generated by a generalized complementation on a distributive lattice with dense elements. Also, we prove the class of Kg-filters forms a distributive lattice which is not induced. Further, we derive normal Kg-filters in generalized complemented distributive lattices and proven that the class of normal Kg-filters is a Boolean algebra which is not induced. We can classify Kg-filters and normal Kg-filters in different class of lattice. Conflict of Interest The authors declare that there are no conflicts of interest regarding the publication of this paper Acknowledgements The authors wish to thank the anonymous reviewers for their valuable suggestions. This work was supported by Directorate of Research and Innovation, Walter Sisulu Uni- versity, South Africa. R. Sirisetti et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5535 12 of 12 References [1] Sergio A. Subordinations on bounded distributive lattices. Order, 40:1–27, 2023. [2] G. Birkhoff. Lattice theory. Amer. Math. Soc. Collequium Pub, 1967. [3] G. Boole. An investigation of the laws of thought. Reprinted by Open Court Pub- lishing Co., Chelsea, London, 1940. [4] S. Burris and H. P. Sankappanavar. 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