EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5490 ISSN 1307-5543 – ejpam.com Published by New York Business Global G-filters and Generalized Complemented Distributive Lattices Jogarao Gunda1, Ramesh Sirisetti2, Ravikumar Bandaru3, Rahul Shukla4,∗ 1 Department of BS & H, Aditya Institute of Technology and Management, Tekkali, Srikakulam, Andhra Pradesh-530021, India 2 Department of Mathematics, GITAM School of Science, GITAM (Deemed to be University), Visakhapatnam, Andhra Pradesh-530045, 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 work, we derive a class of filters (G-filters, normal G-filters, and co-dense filters) in a distributive lattice (with dense elements). We also verify the various algebraic properties of these filters. It is observed that the set of co-dense filters forms an uninduced distributive lattice, and the set of G-filters forms a Boolean algebra. We characterize quasi-complemented distributive lattices using G-filters and normal G-filters. Using normal G-filters, we demonstrate several nec- essary and sufficient requirements for a distributive lattice to become quasi-complemented. Also, we introduce generalized complementation on a distributive lattice and characterize it in terms of quasi-complemented distributive lattices. 2020 Mathematics Subject Classifications: 06D05, 06D15 Key Words and Phrases: Dense elements, Filters, G-filters, Normal G-filters, quasi-complemented distributive lattices and generalized complemented distributive lattices 1. Introduction In the order (lattice) theory, distributive lattices are foundational structures, embody- ing a delicate balance between order and algebraic properties. Within these lattices, filters emerge as essential constructs, offering insights into the dynamics of subsets and their interactions. In this regard, the classification of filters in a distributive lattice was studied extensively by several authors [5–7] and then introduced µ-filters, ω–filters and D- filters, etc. At the same time, within these lattices, complements emerge as fundamental ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5490 Email addresses: jogarao.gunda@gmail.com (J. Gunda), ramesh.sirisetti@gmail.com (R. Sirisetti), 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) J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 2 of 11 constructs, offering profound insights into the nature of duality, negation, and comple- mentation within the lattice framework. Thus, the class of complementations classified by many authors [[1], [8], [3]] is called ortho-complementation, pseudo-complementation, and quasi-complementation, etc. The class of maximal elements is a proper sub-collection of the class of dense elements in lattices. We start working on a distributive lattice with dense elements with this initiation. This paper introduces G-filters, normal G-filters, and co-dense filters in a distribu- tive lattice with dense elements. We derive some algebraic properties from them and obtain the necessary and sufficient conditions for a filter to become a G-filter (normal G-filter). Also, we characterize quasi-complemented distributive lattices using G-filters and normal G-filters. Mainly, we introduce G-complementation on a distributive lattice (which may or may not contain the zero element) and prove several algebraic properties. Every finite distributive lattice is G-complemented, as we have seen. Ultimately, we de- rive certain sufficient and necessary conditions under which a distributive lattice becomes G-complemented. 2. G-Filters In this section, we introduce G-filters and co-dense filters in a distributive lattice and prove several algebraic properties of this class of filters. We prove the set of G-filters forms a Boolean algebra, and the set of co-dense filters forms a distributive lattice. Finally, we observe that every maximal filter is either G-filter or a co-dense filter. In this article, a distributive lattice with dense elements is denoted by A. In [4], Kumar and Rao introduced a class {Sd | S is non-empty subset of A} of filters, where Sd = {v ∈ A | s ∨∗ v is dense, for all s ∈ S}. Definition 1. In a distributive lattice A, filter K is said to be a G-filter, if Kdd = K. Lemma 1. For any filter K of A, (i) Kd and Kdd are G-filters. (ii) Kdd is the smallest G-filter containing K. Lemma 2. Let K be a proper filter containing a non-dense element in A. Then Kd is proper. Proof. Suppose that Kd = A and v is a non-dense element in K. Since 0 ∈ A = Kd, 0 ∨∗ f ∈ D, for all f ∈ K. In particular, for v ∈ K, we have 0 ∨∗ v = v ∈ D. Which is contradiction to v is non-dense. Thus Kd is proper. Let GF (A) represent the set of G-filters of A.. Theorem 1. GF (A) forms a Boolean algebra with the operations K ⊔ G = (K ∨∗ G) dd and K ∧∗ G = Kd ∩Gd, where K,G ∈ GF (A) and the complement of K is Kd. J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 3 of 11 Definition 2. For each v, w ∈ A, v ∧∗ w ∈ K implies v ∈ K or w ∈ K. This indicates that K is a prime G-filter whenever K is a proper G-filter of A. Theorem 2. Let I be an ideal in A such that K ∩ I = ∅, where K be a G-filter. Then, U is a prime G-filter in A such that K ⊆ U and U ∩ I = ∅. Proof. Consider a set Q = {G ∈ GF (A) | G containing K and G ∩ I = ∅}, which is a non-empty set (since K ∈ Q). Suppose that G1 ⊆ G2 ⊆ G3 ⊆ · · · is an increasing chain in Q. Take M = ⊔Gi for i = 1, 2, 3, .... Then M ∈ GF (A) (since GF (A) is a distributive lattice) and I ∩M = I ∩ (⊔Gi) = ⊔(I ∩ (Gi) = ∅ (since I ∩ Gi = ∅ for all i). Therefore M ∈ Q. In this case, the chain’s upper bound in Q is M . Zorn’s lemma states that Q has a maximal element, let’s say U , such that U ∩ I = ∅ and K ⊆ U . Choose v, w ∈ A such that v /∈ U and w /∈ U . Then U ⊆ U ⊔ [v)dd = {Ud ∧∗ [v) d}d = [U ∨∗ [v)] dd and U ⊆ U ⊔ [w)dd = {Ud∧∗ [w) d}d = [U ∨∗ [w)] dd. Since U is maximal, [U ∨∗ [v)] dd∩I ̸= ∅ and [U∨∗[w)] dd∩I ̸= ∅. Therefore {[U∨∗[v)] dd∩I}∩{U∨∗[w)] dd}∩I = {U∨∗[[v) dd∩[w)dd]}∩I = {U ∨∗ [[v) ∩ [w)]}dd ∩ I = {U ∨∗ [v ∨∗ w)}dd ∩ I. If v ∨∗ w ∈ U, then v ∨∗ w ∈ Udd = U (since U is a G-filter) and v ∨∗ w ∈ I. Therefore U ∩ I ̸= ∅. which contradicts itself. So that v ∈ U or w ∈ U . For this reason, U is prime. Corollary 1. Let v /∈ K and K be a G-filter in A. After that, a prime G-filter U exists such that K ⊆ U and v /∈ U . Theorem 3. For every filter K of A, Kdd is the intersection of all prime G-filters con- taining K. Proof. Suppose there is an element v in A, v /∈ Kdd, and K is a filter. Consider U = {G | G is a G-filter of A and v /∈ G and K ⊆ G}. Then U ̸= ϕ (since Kdd ∈ U). Let K1 ⊆ K2 ⊆ · · · be a chain in U. Then (⊔Ki) dd = [(∨∗Ki) dd]dd = (∨∗Ki) dd = ⊔Ki (since GF (A) is a distributive lattice). Therefore ⊔Ki is a G-filter and K ⊆ ⊔Ki and (⊔Ki)∩ (v] = ⊔(Ki ∩ (v]) = ϕ. Consequently, ⊔Ki ∈ U, and an upper bound of the chain in U is ⊔Ki. Zorn’s Lemma states that, the set U has a maximal element, say that the maximal, element is U . In the case where v, w ∈ A, and v /∈ U and w /∈ U , v ∈ {(U ⊔ [v)dd) ∩ (U ⊔ [w)dd)} ⇒ v ∈ {U ⊔ ([v)dd ∩ [w)dd)} ⇒ v ∈ {U ⊔ ([v) ∩ [w))dd} ⇒ v ∈ {U ⊔ ([v ∨∗ w) dd)} ⇒ v ∈ {U ∨∗ [v ∨∗ w) dd}dd. If v∨∗w ∈ U , then [v∨∗w) ⊆ U and [v∨∗w) dd ⊆ Udd = U (since U is G-filter). Therefore v ∈ U . Which is a contradiction. Thus U is prime and v /∈ U . Corollary 2. The intersection of prime G-filters is equal to D. Definition 3. A filter K in A is said to be co-dense, if Kd = D. Lemma 3. Every co-dense filter contains at least one dense element. J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 4 of 11 Theorem 4. With the operations K ⊔G = (K ∨∗G) dd and K ∧∗G = Kd ∩Gd, the set of co-dense filters forms a distributive lattice, where K and G are co-dense filters in A with the largest element A and the least element D. Theorem 5. Every maximal filter is a G-filter or co-dense. Proof. Consider a maximal filter of A to be K. Thus, K ⊆ Kdd is known. Thus, either Kdd = A or K = Kdd. If Kdd = A, then Kd = Kddd = Ad = D. Therefore K is co-dense. Otherwise K = Kdd, and then K is a G-filter. 3. Normal G-Filters In this section, we introduce normal G-filters and obtain several algebraic properties. We prove that several necessary and sufficient conditions to become the set of normal G-filters is a Boolean algebra. Now, let us denote (w)d := {v ∈ A | w∨∗ v is dense}, where w ∈ A. Lemma 4. For any v ∈ A, (i) (v)d is a filter (ii) (0)d = D (iii) [v) ∩ [v)d is a subset of D (iv) (v)ddd = (v)d (v) v ∈ (v)dd (vi) (v)d = A⇔ v ∈ D (vii) D is a subset of (v)d Lemma 5. For any w, x ∈ A, (1) w ≤ x =⇒ (w)d ⊆ (x)d (2) (w)d ⊆ (x)d =⇒ (x)dd ⊆ (w)dd (3) (w)d ∩ (x)d = (w ∧∗ x) d (4) (w ∨∗ x) d = (w)d ⊔ (x)d Remark 1. If v, w ∈ A, then (v)d = (w)d dose not implies v = w. Regarding, have a look at this instance: Example 1. Given a lattice A = {0, v, w, x, 1}, its Hasse-diagram is J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 5 of 11 x v 0 w 1 Then (1)d = A and (x)d = A, but 1 ̸= x. Theorem 6. The class of filters Ad = {(x)d | x ∈ L} with the operations (x)d ⊔ (y)d = (x ∨∗ y) d and (x)d ∩ (y)d = (x ∧∗ y) d forms a distributive lattice. A is said to be quasi-complemented [2], if for each v ∈ A, there exists x ∈ A such that v ∧∗ x = 0 and x ∨∗ v is dense. Theorem 7. If A is quasi-complemented, then (Ad,∩,⊔, D,A) is a Boolean algebra. Proof. Suppose that A is quasi-complemented. Let (v)d ∈ Ad, where v ∈ A. Now, for this v ∈ A, by our assumption, there exists x ∈ A such that v∧∗ x = 0 and v∨∗ x is dense. Therefore (v)d ∩ (x)d = (v ∧∗ x) d = (0)d = D and (v)d ⊔ (x)d = (v ∨∗ x) d = A (since v ∨∗ x is dense). Thus Ad is a Boolean algebra. Theorem 8. If A is a finite distributive lattice, then (Ad,∩,⊔, D,A) is a Boolean algebra if and only if A is quasi-complemented . Proof. If Ad is a Boolean algebra, then A is quasi-complemented, as required by Theorem 7. Assume that (v)d, (w)d ∈ Ad. Since Ad is a Boolean algebra, (x)d, (t)d ∈ Ad exist such that D = (0)d = (v ∧∗ x) d = (v)d ∩ (x)d and A = (v ∨∗ x) d = (v)d ⊔ (x)d. As we have 0 ∈ A = (v ∨∗ x) d, v ∧∗ x = 0 and v ∨∗ x is dense. Thus A is quasi-complemented. Definition 4. If there is a proper filter G such that K ∩G = D and K ∨∗ G = A, then a filter K of A is called a G-factor. Theorem 9. For any v ∈ A, (v)d is a G-factor if and only if (v)d ∨∗ (v) dd = A. Lemma 6. If v, w,∈ A, then we have (i) (v)d = (w)d =⇒ (v ∧∗ x) d = (w ∧∗ x) d, for all x ∈ A. (ii) (v)d = (w)d =⇒ (v ∨∗ x) d = (w ∨∗ x) d, for all x ∈ A. Theorem 10. The following are equivalent for any filter K of A; (i) K is a G-filter J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 6 of 11 (ii) For any v ∈ A, v ∈ K implies (v)dd ⊆ K (iii) Assuming v ∈ K and (v)d = (w)d, for every v, w ∈ A implies w ∈ K (iv) K = ⋃ v∈K [v)dd. Definition 5. A filter K of A is said to be a normal G-filter, if K = (v)d, for some v ∈ A. Let us denote the class of normal G-filters of A asNgF (A). Now, we have the following; Theorem 11. (NgF (A),∩,⊔) is a sublattice of GF (A) in which (0)d is the least and (v)d is the greatest elements in NgF (A), for some v ∈ D. Proof. Assume that K,G ∈ NgF (A). K = (v)d and G = (y)d then exist a pair of v, y ∈ A. It is observe that (x ∨∗ y) d is an upper bound of K,G. Let H ∈ NgF (A) be an upper bound of K,G. Then there exists z ∈ A such that H = (z)d and (v)d, (y)d ⊆ (z)d. So that (z)dd ⊆ (x)dd∩(y)dd = (x∨∗ y) dd. Therefore (x∨∗ y) d ⊆ (z)d. Hence (x∨∗ y) d is the least upper bound of K and G and it is indicated by K ⊔G. Thus NgF (A) is a sub-lattice of GF (A). For any v ∈ A, (v)d∩(0)d = (x∧∗0) d = (0)d. Therefore (0)d is the least element in NgF (A). Let w ∈ D. Then (v)d ⊔ (y)d = (x ∨∗ y) d. Since v ∨∗ y ∈ D, (v)d ⊔ (y)d = A. Therefore (v)d is the greatest element in NgF (A). Thus NgF (A) is bounded. Denote a relation ψ := {(v, w) ∈ A × A | (w)d = (v)d}. Then it is easy to prove that ψ is a congruence relation on A. Lemma 7. For any v ∈ A, (i) v/ψ = {0} if and only if v = 0 (ii) v/ψ = D if and only if v ∈ D. Theorem 12. The quotient lattice A/ψ forms a distributive lattice with the operations v/ψ ∧∗ w/ψ = (v ∧∗ w)/ψ and v/ψ ∨∗ w/ψ = (v ∨∗ w)/ψ.Furthermore, the largest element in A/ψ exists only when A is dense. Theorem 13. The subsequent algebras are equivalent; (1) A is quasi complemented (2) (NgF (A),∩,⊔, D,A) is a Boolean algebra (3) (A/ψ,∧∗,∨∗, 0/ψ, d/ψ) is a Boolean algebra (4) All principal ideals are quasi-complemented. Proof. (1) =⇒ (2): Suppose that A is quasi complemented. Let v ∈ A .Then, v∧∗w = 0 and v∨∗w are dense for some w ∈ A. So that D = (0)d = (v∧∗w) d = (v)d∩(w)d and A = (v ∨∗ w) d = (v)d ⊔ (w)d. Therefore NgF (A) is a Boolean algebra. (2) =⇒ (3): Let v ∈ A. Then (v)d ∈ NgF (A). For this (v)d ∈ NgF (A), there exists J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 7 of 11 w ∈ A such that D = (0)d = (v ∧∗ w) d = (v)d ∩ (w)d and A = (v ∨∗ w) d = (v)d ⊔ (w)d. Therefore v ∨∗ w is dense and v ∧∗ w = 0. Hence {0} = 0/ψ = (v ∧∗ w)/ψ = v/ψ ∧∗ w/ψ and D = (v∨∗w)/ψ = v/ψ∨∗w/ψ(since v∨∗w is dense). Thus A/ψ is a Boolean algebra. (3) =⇒ (4): Let v ∈ A. Then, v/ψ ∧∗ w/ψ = {0} and v/ψ ∨∗ w/ψ = D exist for some w ∈ A. So that v ∨∗ w is dense and v ∧∗ w = 0. So that (v] ∧∗ (w] = (v ∧∗ w] = (0] = {0} and (v] ∨∗ (w] = (v ∨∗ w]. Now, (v ∧∗ w] = {0} and (v ∨∗ w] is a principal ideal generated by a dense element v ∨∗ w. Hence every principal ideal is quasi-complemented. (4) =⇒ (1): Suppose that every principal ideal is quasi-complemented. Assume v ∈ A. Then, (v]∩ (w] = {0} and (v]∨∗ (w] exist for some w ∈ A is the principal ideals produced by v ∨∗ w dense elements. As a result, both v ∨∗ w is dense, and v ∧∗ w = 0 . Thus A is quasi-complemented. 4. Generalized Complemented distributive lattice This section obtains numerous algebraic characteristics and introduces a generalized complementation on a distributive lattice. For a distributive lattice to become a G- complemented one, we give both required and sufficient necessities. Furthermore, we derive sufficient and necessary conditions under which a quasi-complemented distributive lattice can be formed from a generalized complemented distributive lattice. Definition 6. If a unary operation g on A meets these requirements, it is called a gener- alized complementation. (i) for any v ∈ A, v ∨∗ v g ∈ D (ii) for any w ∈ A, v ∨∗ w ∈ D ⇔ vg ≤ w. In this case, vg is called a generalized complement of v and A is called a generalized complemented distributive lattice. Example 2. In Example 1, let us define 0g = x, vg = w,wg = v, xg = 1g = 0. Then g is a generalized complementation on A. Example 3. Every complemented distributive lattice is generalized complemented. Remark 2. The converse of the above statement need not be true. For, in Example 1, A is a generalized complemented distributive lattice, but A is not complemented. Lemma 8. If g is a generalized complementation on A and v, w ∈ A, then (i) 0g ∈ D (ii) d ∈ D =⇒ dg = 0 (iii) v ≤ w =⇒ wg ≤ vg (iv) vgg ≤ v J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 8 of 11 (v) vggg = vg (vi) 0gg = 0 (vii) v ∈ D ⇐⇒ vg = 0 ⇐⇒ vgg ∈ D (viii) vg ≤ 0g (ix) vg ≤ wg ⇐⇒ wgg ≤ vgg (x) v = 0 =⇒ vgg = 0. Lemma 9. If g is a generalized complementation on A, then the following are equivalent; (i) v ∨∗ w ∈ D, for all v, w ∈ A. (ii) vgg ∨∗ w ∈ D, for all v, w ∈ A. (iii) vgg ∨∗ w gg ∈ D, for all v, w ∈ A. (iv) v ∨∗ w gg ∈ D, for all v, w ∈ A. Proof. Let v, w ∈ A. (i) =⇒ (ii): Suppose that v ∨∗ w ∈ D. Then vg ≤ w and vg ∨∗w = w. Now, vgg ∨∗w = vgg ∨∗ (v g ∨∗w) = (vgg ∨∗ v g)∨∗w ∈ D, since vgg ∨∗ v g ∈ D. (ii) =⇒ (iii): Suppose that vgg ∨∗ w ∈ D. Then w ∨∗ v gg ∈ D and hence wgg ∨∗ v gg = vgg ∨∗ w gg ∈ D. (iii) =⇒ (iv): Suppose that vgg ∨∗ w gg ∈ D. By Lemma 8(iv), vgg ≤ v. Therefore v ∨∗ w gg ∈ D. (iv) =⇒ (i): Suppose that v∨∗w gg ∈ D. By Lemma 8(iv), vgg ≤ v. Therefore v∨∗w ∈ D. Lemma 10. If g is a generalized complementation on A and v, w ∈ A, then (i) (v ∧∗ w) g = vg ∨∗ w g (ii) (v ∨∗ w) g ≤ vg ∧∗ w g (iii) (v ∨∗ w) gg = vgg ∨∗ w gg = (vg ∧∗ w g)g (iv) (v ∧∗ w) gg = (vg ∨∗ w g)g = (vgg ∧∗ w gg)gg. Proof. (i): For any v, w ∈ A, v, w ≥ v ∧∗ w. Then (v ∧∗ w) g ≥ vg, wg. Therefore vg ∨∗ w g ≤ (v ∧∗ w) g. Now, (v ∧∗ w) ∨∗ (v g ∨∗ w g) = [v ∨∗ (v g ∨∗ w g)] ∧∗ [w ∨∗ (v g ∨∗ wg)] = [(v ∨∗ v g) ∨∗ w] ∧∗ [v g ∨∗ (w ∨∗ w g)] ∈ D (since v ∨∗ v g, w ∨∗ w g ∈ D). So that (v ∧∗ w) g ≤ vg ∨∗ w g. Therefore (v ∧∗ w) g = vg ∨∗ w g. (ii): We have v, w ≤ v ∨∗ w. Then (v ∨∗ w) g ≤ vg, wg. Therefore (v ∨∗ w) g ≤ vg ∧∗ w g. J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 9 of 11 (iii): By (ii), (vg ∧∗ w g)g ≤ (v ∨∗ w) gg. Then vgg ∨∗ w gg ≤ (v ∨∗ w) gg. On the other hand, (v ∨∗ w) ∨∗ (v ∨∗ w) g ∈ D. Then v ∨∗ [w ∨∗ (v ∨∗ w) g] ∈ D. By Lemma 9., vgg ∨∗ [w ∨∗ (v ∨∗ w) g] ∈ D ⇒ [vgg ∨∗ w] ∨∗ [(v ∨∗ w) g] ∈ D ⇒ w ∨∗ v gg ∨∗ (v ∨∗ w) g ∈ D ⇒ wgg ∨∗ v gg ∨∗ (v ∨∗ w) g ∈ D ⇒ (v ∨∗ w) g ∨∗ v gg ∨∗ w gg ∈ D ⇒ (v ∨∗ w) gg ≤ vgg ∨∗ w gg. Therefore (v ∨∗ w) gg = vgg ∨∗ w gg. (iv): (v∧∗w) gg = [(v∧∗w) g]g = [vg∨∗w g]g = [vg∨∗w g]ggg = [vggg∨∗w ggg]g = (vg∨∗w g)g = (vgg ∧∗ w gg)gg. Theorem 14. Every finite distributive lattice is generalized complemented. Proof. Let A be a finite distributive lattice and D ̸= ∅. For any v ∈ A, define vg = Inf{w ∈ A | v ∨∗ w ∈ D}. Then v ∨∗ v g = v ∨∗ [∧∗{w ∈ A | v ∨∗ w ∈ D}] = ∧∗[{v ∨∗ w | v ∨∗ w ∈ D}] ∈ D. If v ∨∗ x ∈ D, then x ∈ {v ∈ A | v ∨∗ v ∈ D}. Therefore ∧∗{v ∈ A | v ∨∗ v ∈ D} ≤ x and hence vg ≤ x. On other hand, suppose vg ≤ x, then we have v ∨∗ v g ∈ D, v ∨∗ x ∈ D. Hence vg is a generalized complementation of v. Thus A is generalized complemented. Theorem 15. A is generalized complemented if and only if [0, d] is generalized comple- mented, for all dense elements d ∈ A. Proof. Suppose that A is a generalized complemented and g is a generalized comple- mentation on A. Let v ∈ [0, d]. Then there exists vg ∈ A such that v ∨∗ v g ∈ D and for any w ∈ A, v ∨∗ w ∈ D if and only if vg ≤ w. Since v ∨∗ d ∈ D, vg ≤ d. Therefore vg ∈ [0, d]. Hence [0, d] is generalized complemented. Conversely suppose that [0, d] is a generalized complemented distributive lattice with dense element d. Let v ∈ A. Then v ∨∗ d is dense in A. Therefore v ∈ [0, v ∨∗ d]. Hence there exists vg in [0, v ∨∗ d] such that v ∨∗ v g is dense in [0, v ∨∗ d]. Let w ∈ A, such that v ∨∗ w is dense in A. Then (v ∨∗ w)∧∗ d is dense in [0, v ∨∗ d]. Therefore (v ∧∗ d)∨∗ (w ∧∗ d) is dense in [0, v ∨∗ d]. So that (v ∧∗ d) g ≤ w ∧∗ d ≤ w. Also that vg ≤ vg ∨∗ d g ≤ w. If vg ≤ w, then v ∨∗ v g ∈ D. Therefore v ∨∗ w ∈ D. Hence A is generalized complemented distributive lattice with the generalized complementation g. Theorem 16. A is generalized complemented if and only if PI(A) is generalized comple- mented. Proof. Suppose that A is a generalized complemented distributive lattice and g is the generalized complementation on A. Let (v] ∈ PI(A), for some v ∈ A. Then there exists vg ∈ A such that v∨∗ v g ∈ D and, v∨∗w ∈ D if and only if vg ≤ w, for any w ∈ A. Define J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 10 of 11 (v]g = (vg]. For (t] ∈ PI(A), (t] ∈ [(v] ∨∗ (v] g]∗ =⇒ (t] ∈ [(v ∨∗ v g]]∗ =⇒ (t] ∧∗ (v ∨∗ v g] = (0] =⇒ (t ∧∗ (v ∨∗ v g)] = (0] =⇒ t ∧∗ (v ∨∗ v g) = 0 =⇒ t = 0 (since v ∨∗ v g = 0) =⇒ (t] = (0]. Therefore (v] ∨∗ (v] g is dense in PI(A). If (v] ∨∗ (w] is dense in PI(A), for some w ∈ A. Then (v ∨∗ w] is dense in PI(A) if and only if v ∨∗ w is dense in A. Therefore vg ≤ w. So that (vg] ⊆ (w]. Hence (v]g ⊆ (w]. If (v]g ⊆ (w]. Since (v] ∨∗ (vg] is dense in PI(A), (vg] ∨∗ (w] is dense in PI(A). Hence (v] ∨∗ (w] is dense in PI(A). Conversely suppose that PI(A) is generalized complemented. Let v ∈ A. Then (v] ∈ PI(A). Then there exists (v]g in PI(A) such that (v] ∨∗ (v] g is dense in PI(A) and (v] ∨∗ (w] is dense if and only if (v]g ⊆ (w]. Now, we can write (v]g = (v+] for some v+ ∈ A . Then (v]∨∗ (v] g = (v]∨∗ (v +] = (v ∨∗ v +] is dense in PI(A) if and only if v ∨∗ v + is dense in A. If v ∨∗ w ∈ D for some w ∈ A. Then (v] ∨∗ (w] is dense in PI(A). Therefore (v]g ⊆ (w]. Hence (v+] ⊆ (w]. So that (v+]∩(w] = (v+] and also (v+∧∗w] = (v+]. Now v+∧∗w = v+. Hence v+ ≤ w. If v+ ≤ w. Then (v+] ⊆ (w]. Therefore (v+] ⊆ (w]. So that (v]g ⊆ (w] if and only if (v] ∨∗ (w] = (v ∨∗ w] is dense in PI(A). Hence v ∨∗ w is dense in A. Thus + is a generalized complementation on A and A is generalized complemented. Definition 7. A distributive lattice A with a generalized complementation g is said to be p-complemented if for any v ∈ A, there exists vg ∈ A such that v ∧∗ v g = 0. Theorem 17. Every G-complemented distributive lattice is quasi complemented if and only if it is p-complemented. Proof. Let g be a generalized complementation on A. Suppose A is quasi comple- mented. Let v ∈ A. Then there exists w ∈ A such that v ∧∗ w = 0 and v ∨∗ w is dense. Since g is a generalized complementation, vg ≤ w. Now 0 = v ∧∗ w = v ∧∗ (v g ∨∗ w) = (v ∧∗ v g)∨∗ (v ∧∗ w). Therefore v ∧∗ v g = 0. Hence A is p-complemented. Conversely sup- pose that A is p-complemented. Let v ∈ A. Then there exists vg ∈ A such that v∧∗v g = 0. By definition of G-complementation, v ∨∗ v g is dense. Thus A is quasi complemented. 5. Conclusions This paper extensively studied on G-filters, normal G-filters, co-dense filters and gen- eralized complementations in a distributive lattice with dense elements. The class of quasi complemented distributive lattices and the class of generalized complemented distributive lattices are characterized. Further we can co relate the class of relatively complemented distributive lattices and the class of ortho-complemented distributive lattices, using the class of G-filters and the class of generalized complemented distributive lattices. J. Gunda et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5490 11 of 11 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. 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Algebraic Structures and Their Applications, 9:145–159, 2022. [8] P V Venkatanarasimhan. Psuedo-complements in posets. American Mathematical Society, 28:9–17, 1971. Introduction G-Filters Normal G-Filters Generalized Complemented distributive lattice Conclusions