EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 1, 2023, 577-586 ISSN 1307-5543 – ejpam.com Published by New York Business Global First and Third Isomorphism Theorems for the Dual B-Algebra Jethro Elijah Bolima1,∗, Katrina Belleza Fuentes1 1 Department of Computer, Information Sciences, and Mathematics, School of Arts and Sciences, University of San Carlos, 6000 Cebu City, Philippines Abstract. In this paper, some properties of the dual B-homomorphism are provided, along with the natural dual B-homomorphism and the fundamental theorem of dual B-homomorphisms for dual B-algebras. The first and third isomorphism theorems for the dual B-algebra are also presented in the paper. 2020 Mathematics Subject Classifications: 47L45, 08A35 Key Words and Phrases: Dual B-algebra, quotient dual B-algebra, fundamental theorem of dual B-homomorphism, dual B-isomorphism 1. Introduction K. Belleza and J.P. Vilela in their paper in 2019 [2] introduced the dual B-Algebra, its relationship with other algebras, and its characteristics. More studies were then conducted on the said topic. One of the recent papers published by K. Belleza and J.R. Albaracin in 2022 [1] discussed about dual B-filters and dual B-subalgebras in a topological dual B-algebra, wherein the researchers first constructed a congruence relation on a dual B- algebra which is necessary in creating a natural homomorphism from one dual B-algebra onto another; an important first step in this study. While many other algebraic structures prepared different approaches in constructing isomorphism to their respective algebras (see [6], [3], [5]), J. Neggers and H.S. Kim in particular, presented a fundamental theorem of B-homomorphism for B-algebras and using the said theorem created the 1st and 3rd isomorphism theorems for the B-algebra in 2002 [7]. Later in 2015, J.C. Endam and J.P. Vilela also provided more insights on the properties of normal subsets of B-algebra and B-homomorphism, and presented proof for the 2nd isomorphism theorem for the B-algebras [4]. This, in turn warrants a need for investigation of the dual B-algebra as to whether the isomorphism theorems can be constructed within the dual B-algebra since there exists a close relationship between the B-algebra and the dual B-algebra [2]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4675 Email addresses: 15101364@usc.edu.ph (J.E. Bolima), kebelleza@usc.edu.ph (K.B. Fuentes) https://www.ejpam.com 577 © 2023 EJPAM All rights reserved. J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 578 2. Preliminaries Definition 1. [2] A Dual B-Algebra, (or dB-algebra), X is a triple (X, ·, 1) where X is a non-empty set with a binary operation “ · ” and a constant 1 satisfying the following axioms for all x, y, z in X: (DB1) x · x = 1; (DB2) 1 · x = x; (DB3) x · (y · z) = ((y · 1) · x) · z. Example 1. [1] Let X = {1, a, b, c} with the binary operation · as defined in the table: · 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 Then (X, ·, 1) is a dB-algebra. Lemma 1. [2] Let (X, ·, 1) be a dB-algebra, then for any x, y ∈ X, x · y = 1 implies x = y Definition 2. [1] Let X be a dB-algebra and S a nonempty subset of X. Then S is called a dual B-subalgebra, (or dB-subalgebra), of X if S itself is a dB-algebra with binary operation of X on S. Remark 1. [1] If S is a dB-subalgebra of X, then 1 ∈ S. Theorem 1. [1] S is a dB-subalgebra if and only if for any x, y ∈ S, x · y ∈ S. Example 2. Consider the dB-algebra X = {1, a, b, c} with the binary operation · as defined in the table: · 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 By Remark 2.6, A = (1, a) is a dB-subalgebra of X but B = (1, a, c) is not a dB-subalgebra of X since a · c = b ̸∈ B. Definition 3. [1] Let X be a dB-algebra. A subset F of X is called a dual B-filter, (or dB-filter), if it satisfies the following: (i.) 1 ∈ F ; (ii.) for each x, y ∈ X, x · y ∈ F and x ∈ F imply y ∈ F . J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 579 Proposition 1. [1] If F is a dB-filter of a dB-algebra X, then F is a dB-subalgebra of X. Definition 4. [1] Let X be a dB-algebra and N a nonempty subset of X. Then N is a normal subset of X if for any a · b, x · y ∈ N, (a · x) · (b · y) ∈ N . A dB-filter F of a dB-algebra X is called a normal dB-filter if F is a normal subset of X. A dB-subalgebra S of a dB-algebra X is called a normal dB-subalgebra if S is a normal subset of X. Example 3. [1] Let X = {1, a, b, c, d, e} with the binary operation · as defined in the table: · 1 a b c d e 1 1 a b c d e a b 1 a d e c b a b 1 e c d c c d e 1 a b d d e c b 1 a e e c d a b 1 Then (X, ·, 1) is a dB-algebra. So, (a) The set A={1, a, e} is not a dB-filter since ∃e · c = a but c ̸∈ A (b) The set B={1, c} is a dB-filter but is not normal since ∃c · 1 = a · e = c ∈ B but (c · a) · (1 · e) = d · e = a ̸∈ B (c) The set C={1, a, b} is a normal dB-filter. Theorem 2. [1] Let (X, ·, 1) be a dB-algebra and S a normal dB-subalgebra of X. The relation defined by x ∼ y if and only if x · y, y · x ∈ S is a congruence relation on X for any x, y ∈ X. Definition 5. [1] Let (X, ·, 1) be a dB-algebra and S a normal dB-subalgebra of X. Define a congruence class [x]S by [x]S = {y ∈ X|y ∼ x} and define X/S to be the set of all congruence classes of X, that is X/S = { [x]S |x ∈ X } . 3. Results Lemma 2. Let S be a normal dB-subalgebra of a dB-algebra (X, ·, 1) and x, y ∈ X. Then [x]S = [y]S if and only if x ∼ y. Proof. Suppose [x]S = [y]S . Then z ∈ [x]S implies that z ∈ [y]S . We have that z ∼ x, z ∼ y and since ∼ is symmetric and transitive by Theorem 2, then z ∼ x, z ∼ y implies x ∼ z, z ∼ y which implies that x ∼ y. Now, suppose x ∼ y. Then x · y, y · x ∈ S. Let z ∈ [x]S , then z ∼ x. We have that z ∼ x, x ∼ y implies z ∼ y which implies that z ∈ [y]S . Hence, [x]S ⊆ [y]S . Similarly, let a ∈ [y]S , then a ∼ y. We have that a ∼ y, x ∼ y implies a ∼ y, y ∼ x which implies that a ∼ x and so a ∈ [x]S . Thus [y]S ⊆ [x]S and it follows that [x]S = [y]S . J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 580 Theorem 3. Let S be a normal dB-subalgebra of a dB-algebra (X, ·, 1X). Then (X/S, ∗, [1]S) with the binary operation ∗ on X/S defined by [x]S ∗ [y]S = [x · y]S for all x, y ∈ X is a dB-algebra. X/S is called the quotient dB-algebra of X by S. Proof. Let x1, x2, y1, y2 ∈ X such that [x1]S = [x2]S and [y1]S = [y2]S . Then x1 ∼ x2 and y1 ∼ y2. Since ∼ is a congruence relation, we have that x1 · y1 ∼ x2 · y2 and by Lemma 2, [x1 · y1]S = [x2 · y2]S which implies that [x1]S ∗ [y1]S = [x2]S ∗ [y2]S . Hence ∗ is well-defined. Now, for all x, y, z ∈ X, [x]S ∗ [x]S = [x · x]S = [1]S (DB1) [1]S ∗ [x]S = [1 · x]S = [x]S (DB2) [x]S ∗ ( [y]S ∗ [z]S ) = [x]S ∗ ( [y · z]S ) = [x · (y · z)]S = [((y · 1) · x) · z]S = [(y · 1) · x]S ∗ [z]S = ( [y · 1]S ∗ [x]S ) ∗ [z]S = ( ([y]S ∗ [1]S) ∗ [x]S ) ∗ [z]S (DB3) Hence, (X/S, ∗, [1]S) is a dB-algebra. Proposition 2. Let (X, ·, 1) be a dB-algebra and S be a subset of X. Then S is a normal dB-subalgebra of X if and only if S is a normal dB-filter of X. Proof. Suppose S is a normal dB-filter of X. It follows from Proposition 1 and S as a normal subset of X that S is a normal dB-subalgebra of X. Now, suppose S is a normal dB-subalgebra of X. Let x, y ∈ X such that x · y ∈ S and x ∈ S. Since S is a dB-subalgebra, then 1 ∈ S. Since 1, x ∈ S and S is closed by Theorem 1, we have that x · 1 ∈ S and since x · y ∈ S, it follows that (x · x) · (1 · y) ∈ S since S is normal. Then, y = 1 ·y = 1 · (1 ·y) = (x ·x) · (1 ·y) ∈ S. Hence, S is a normal dB-filter. Definition 6. Let (X, ·, 1X) and (Y, ∗, 1Y ) be dB-algebras. A mapping Φ : X → Y is called a dual B-homomorphism (or dB-homomorphism), from X into Y if Φ(x · y) = Φ(x) ∗ Φ(y) for any x, y ∈ X. A dB-homomorphism Φ is called dB-monomorphism, dB-epimorphism, or dB-isomorphism (denoted by X ∼= Y ), if Φ is one-to-one, onto, or a bijection, respectively. An isomorphism Φ : X → X is called dB-automorphism. The kernel of the dB-homomorphism Φ, denoted by kerΦ, is the set whose elements of X are mapped to 1Y . J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 581 Example 4. Let (R+, ·, 1) be a dB-algebra with the binary operator · be defined as x·y = y x for all x, y in R+. Define Φ : R+ → R+ by Φ(x) = x2 for all x ∈ R+. For all x, y ∈ R+, x = y implies x2 = y2 which implies that Φ(x) = Φ(y). Hence, Φ is well-defined. Now, Φ(x · y) = Φ ( y x ) = y2 x2 = x2 · y2 = Φ(x) · Φ(y) for all x, y ∈ R+. Hence, Φ is a dB-homomorphism. Suppose that Φ(x) = Φ(y) for all x, y ∈ R+, then x2 = y2 implies that x = y which means Φ is one-to-one. Now, for all y ∈ R+, ∃x ∈ R+ such that x = √ y implies x2 = y which implies that Φ(x) = y and so Φ is onto. Consequently, Φ is a dB-automorphism. The kernel of this dB-automorphism is kerΦ = {x ∈ R+|Φ(x) = 1} = {x ∈ R+|x2 = 1} = {x ∈ R+|x = 1} = {1} The next corollary, which is needed for the following results, is immediate from Lemma 1 and DB1. Corollary 1. Let (X, ·, 1) be a dB-algebra, then for any x, y ∈ X, x = y implies that x · y = 1. Theorem 4. Let Φ : X → Y be a dB-homomorphism, (X, ·, 1X), (Y, ∗, 1Y ) be dB- algebras, and S ⊆ X, then (i) Φ(1X) = 1Y (ii) Φ is a dB-monomorphism, if and only if kerΦ = {1X} (iii) Im(Φ) is a dB-subalgebra of Y . (iv) kerΦ is a dB-filter of X and consequently a dB-subalgebra of X. (v) If S is a dB-filter of X, then Φ(S) is a dB-filter of Y and consequently a dB- subalgebra of Y . Proof. Suppose Φ : X → Y be a dB-homomorphism and S ⊆ X, (i) Since Φ is a dB-homomorphism and by DB1, Φ(1X) = Φ(1X · 1X) = Φ(1X) ∗ Φ(1X) = 1Y . J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 582 (ii) Suppose Φ is a dB-monomorphism. It follows from (i.) that 1X ∈ kerΦ. Let x ∈ kerΦ. then Φ(x) = 1Y = Φ(1X). Since Φ is one-to-one, Φ(x) = Φ(1X) implies x = 1X . Hence, kerΦ = {1X}. Conversely, suppose kerΦ = {1X} and x, y ∈ X such that Φ(x) = Φ(y). By Corollary 1, Φ(x) ∗ Φ(y) = 1Y = Φ(x · y). Then x · y ∈ kerΦ. Since kerΦ = {1X}, x · y = 1X and it follows that x = y by Lemma 1. Hence, Φ is one-to-one i.e. Φ is a dB-monomorphism. (iii) Let x, y ∈ Im(Φ). Then there exists a, b ∈ X such that x = Φ(a), y = Φ(b). This implies that x ∗ y = Φ(a) ∗ Φ(b) = Φ(a · b) ∈ Im(Φ) since a · b ∈ X. Thus, Im(Φ) is a dB-subalgebra of Y . (iv) By Definition 6, kerΦ ⊆ X and by (i.), 1X ∈ kerΦ which also implies that kerΦ ̸= ∅. Let x · y ∈ kerΦ and x ∈ kerΦ. Then for all y ∈ X, Φ(y) = 1Y ∗ Φ(y) = Φ(x) ∗ Φ(y) = Φ(x · y) = 1Y . Hence, y ∈ kerΦ and it follows that kerΦ is a dB-filter of X. Consequently, by Theorem 1, kerΦ is a dB-subalgebra of X. (v) Let S be a dB-filter of X, then 1X ∈ S and by (i.), Φ(1X) = 1Y ∈ Φ(S). Now, for all x, y ∈ X such that x ∈ S and x · y ∈ S implies that Φ(x) ∈ Φ(S) and Φ(x) ∗Φ(y) = Φ(x · y) ∈ S. Since S is a dB-filter of X, then y ∈ S also implies that Φ(y) ∈ Φ(S). Hence, Φ(S) is a dB-filter of Y . Consequently, by Theorem 1, Φ(S) is a dB-subalgebra of Y . Theorem 5. Let S be a normal dB-subalgebra (normal dB-filter) of a dB-algebra (X, ·, 1). Then the mapping Φ : (X, ·, 1) → (X/S, ∗, [1]S) given by Φ(x) = [x]S for all x ∈ X is a dB-epimorphism and kerΦ = S. The mapping Φ in this case is called the natural dB- homomorphism of X onto X/S. Proof. Let x, y ∈ X such that x = y which by Corollary 1, x · y = 1 ∈ S and y · x = 1 ∈ S. Then x ∼ y implies [x]S = [y]S which implies that Φ(x) = Φ(y). Hence, Φ is well-defined. Now, let a, b ∈ X. Then Φ(a · b) = [a · b]S = [a]S ∗ [b]S = Φ(a) ∗Φ(b). This shows that Φ is a dB-homomorphism. Since Φ(X) = { Φ(a) : a ∈ X } = { [a]S : a ∈ X } = X/S, it shows that Φ is onto and so Φ is a dB-epimorphism. To show that kerΦ = S, let x ∈ kerΦ. Then [x]S = Φ(x) = [1]S and so x ∼ 1. It follows that x · 1 ∈ S and 1 · x ∈ S. Since 1 ∈ S and S is also a dB-filter by Proposition 2, then x ∈ S and so kerΦ ⊆ S. Now, let x ∈ S. By Remark 1, 1 ∈ S, and since S is closed by Theorem 1, 1 · x ∈ S and x · 1 ∈ S. Then x ∼ 1, and so [x]S = [1]S . Since Φ(x) = [x]S = [1]S , then x ∈ kerΦ. This implies that S ⊆ kerΦ and it follows that kerΦ = S. Lemma 3. Let f : (X, ·, 1X) → (Y, ∗, 1Y ) and g : (Y, ∗, 1Y ) → (Z, ∗′, 1Z) be dB- homomorphisms, then g ◦ f : (X, ·, 1X) → (Z, ∗′, 1Z) is also a dB-homomorphism (◦ is the usual composition of functions). J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 583 Proof. Let x, y ∈ X. Since f and g are dB-homomorphisms, then (g ◦ f)(x · y) = g ( f(x · y) ) = g ( f(x) ∗ f(y) ) = g ( f(x) ) ∗′ g ( f(y) ) = (g ◦ f)(x) ∗′ (g ◦ f)(y). Hence, g ◦ f is a dB-homomorphism. Theorem 6. Fundamental Theorem of dB-homomorphism for dB-Algebras Let Φ be a dB-homomorphism of a dB-algebra (X, ·, 1X) onto a dB-algebra (Y, ∗, 1Y ), S ⊆ kerΦ be a normal dB-subalgebra (normal dB-filter) of X, and g be the natural dB- homomorphism of X onto (X/S, θ, [1]S). Then there exists a unique dB-homomorphism h of X/S onto Y such that Φ = h ◦ g. Furthermore, h is one-to-one if and only if S = kerΦ. Proof. Define the map h : X/S → Y by h ( [x]S ) = Φ(x) for all [x]S ∈ X/S. Let [x]S , [y]S ∈ X/S such that [x]S = [y]S . Then x ∼ y, so x · y ∈ S and y · x ∈ S. Since S ⊆ kerΦ, x · y ∈ kerΦ and y · x ∈ kerΦ. Thus Φ(x) ∗ Φ(y) = Φ(x · y) = 1Y and Φ(y) ∗Φ(x) = Φ(y ·x) = 1Y . By Lemma 1, Φ(x) = Φ(y) and so h ( [x]S ) = h ( [y]S ) . Hence, h is well-defined. Let [x]S , [y]S ∈ X/S. Then h ( [x]Sθ[y]S ) = h ( [x · y]S ) = Φ(x · y) = Φ(x) ∗ Φ(y) = h ( [x]S ) ∗ h ( [y]S ) . Thus, h is a dB-homomorphism. Since Φ is onto, for all y ∈ Y there exists x ∈ X such that Φ(x) = y. As h ( [x]S ) = Φ(x) for all [x]S ∈ X/S, it follows that there exists [x]S ∈ X/S such that h ( [x]S ) = y for all y ∈ Y . Hence, h is onto. Suppose h′ : X/S → Y is another function such that Φ = h′ ◦ g. Let [x]S ∈ X/S, then h′ ( [x]S ) = h′ ( g(x) ) = (h′ ◦ g)(x) = Φ(x) = h ( [x]S ) . Thus, h′ ( [x]S ) = h ( [x]S ) for all [a]S ∈ X/S, i.e. h is unique. Now, to show that h is one-to-one if and only if S = kerΦ, suppose h is one-to-one and x ∈ kerΦ. Then h ( [x]S ) = Φ(x) = 1Y = h ( [1]S ) and since h is one-to-one, [x]S = [1]S . It follows that x ∼ 1X , and so x · 1X ∈ S and 1X · x ∈ S. Since 1X ∈ S and S is a dB-filter, x ∈ S. Thus, kerΦ ⊆ S and since S ⊆ kerΦ by hypothesis, kerΦ = S. Suppose that kerΦ = S and [x]S , [y]S ∈ X/S such that h ( [x]S ) = h ( [y]S ) . Then Φ(x) = Φ(y). By Corollary 1, 1Y = Φ(x) ∗ Φ(y) = Φ(x · y) which implies that x · y ∈ kerΦ = S. Similarly, 1Y = Φ(y) ∗Φ(x) = Φ(y ·x) implies that y ·x ∈ S. Hence, x ∼ y and it follows that [x]S = [y]S , showing that h is one-to-one. Theorem 7. First Isomorphism Theorem for the dB-Algebra Let Φ be a dB-homomorphism of a dB-algebra (X, ·, 1X) into a dB-algebra (Y, ∗, 1Y ), then (X/ kerΦ, θ, [1]kerΦ) ∼= Φ(X). Proof. Let S = kerΦ, g be the natural dB-homomorphism from X onto X/S, the mapping f : X/S → Φ(X) be defined by f ( [x]S ) = Φ(x) for all [x]S ∈ X/S, and recall that Φ(X) is a dB-subalgebra of Y by Theorem 4 (iii) which implies that Φ(X) has the same binary operator as Y . J.E. Bolima, K.B. Fuentes / Eur. J. Pure Appl. Math, 16 (1) (2023), 577-586 584 Let [x]S , [y]S ∈ X/S such that [x]S = [y]S . Then x ∼ y and it follows that x · y ∈ S and y · x ∈ S. Since S = kerΦ, Φ(x · y) = Φ(x) ∗ Φ(y) = 1Y = Φ(y) ∗ Φ(x) = Φ(y · x) and by Lemma 1, Φ(x) = Φ(y) which is f ( [x]S ) = f ( [y]S ) . Hence, f is well-defined. Let [x]S , [y]S ∈ X/S. Then f ( [x]Sθ[y]S ) = f ( [x · y]S ) = Φ(x · y) = Φ(x) ∗ Φ(y) = f ( [x]S ) ∗ f ( [y]S ) Thus, f is a dB-homomorphism. Let [x]S , [y]S ∈ X/S such that f ( [x]S ) = f ( [y]S ) . Then Φ(x) = Φ(y). It follows that 1Y = Φ(x) ∗ Φ(y) = Φ(x · y) which implies that x · y ∈ kerΦ = S. Similarly, 1Y = Φ(y) ∗ Φ(x) = Φ(y · x) implies that y · x ∈ S. Thus, x ∼ y which implies that [x]S = [y]S , so f is one-to-one. Let y ∈ Φ(X), then there exists x ∈ X such that y = Φ(x) and [x]S ∈ X/S. Then f ( [x]S ) = Φ(x) = y. Hence, f is onto and consequently, f is a dB-isomorphism. . Proposition 3. Suppose f : (G, ·, 1G) → (G/H1, ∗, [1]H1 ) is a dB-epimorphism of dB- algebras. If H2 is a normal dB-subalgebra of G, then f(H2) is a normal dB-subalgebra of G/H1. Proof. It follows from Theorem 4 (iii) that f(H2) is a dB-subalgebra of G/H1. Now to show that f(H2) is normal, let [x]H1 ∗ [y]H1 , [a]H1 ∗ [b]H1 ∈ f(H2) for any [x]H1 , [y]H1 , [a]H1 , and [b]H1 ∈ G/H1. Since f is onto, then there exists j, k, l,m ∈ G such that f(j) = [x]H1 , f(k) = [y]H1 , f(l) = [a]H1 , f(m) = [b]H1 . Suppose j ·k, l·m ∈ H2. Then (j ·l)·(k ·m) ∈ H2 since H2 is normal, which then implies that f ( (j · l) · (k ·m) ) ∈ f(H2). It follows that f ( (j · l) · (k ·m) ) = f(j · l) ∗ f(k ·m) = ( f(j) ∗ f(l) ) ∗ ( f(k) ∗ f(m) ) = ( [x]H1 ∗ [a]H1 ) ∗ ( [y]H1 ∗ [b]H1 ) Thus, f(H2) is normal and consequently, f(H2) is a normal dB-subalgebra of G/H1. Theorem 8. Third Isomorphism Theorem for the dB-Algebra Let f be a natural dB-homomorphism of a dB-algebra (G, ·, 1G) onto a dB-algebra (G/H1, ∗, [1]H1 ), H2 be a normal dB-subalgebra of G such that ker f = H1 ⊆ H2, and g, g′ be the natural dB-homomorphisms of G onto (G/H2, ·′, [1]H2 ) and G/H1 onto ((G/H1)/(H2/H1), ∗′, [1]H2/H1 ), respectively. Then there exists a unique dB-isomorphism h of G/H2 onto (G/H1)/(H2/H1), that is G/H2 ∼= (G/H1)/(H2/H1), where g ′ ◦ f = h◦ g. Proof. Since f(H2) is a normal dB-subalgebra of G/H1 by Proposition 3, we have that f(H2) = ker g′ by Theorem 5. Suppose a ∈ H2, then f(a) ∈ f(H2) which implies that f(a) ∈ ker g′. Then (g′ ◦ f)(a) = g′ ( f(a) ) = g′ ( [1]H1 ) = [ [1]H1 ] H2/H1 by Theorem 4 (i). This implies that a ∈ ker(g′ ◦ f) and so, H2 ⊆ ker(g′ ◦ f). REFERENCES 585 Conversely, suppose a ∈ ker(g′ ◦ f), then (g′ ◦ f)(a) = g′ ( f(a) ) = g′ ( [a]H1 ) = [ [a]H1 ] H2/H1 = [1]H2/H1 . By Theorem 4 (i), we have that g′ ( [1]H1 ) = [ [1]H1 ] H2/H1 = [1]H2/H1 . Then [1]H1 ∼ [a]H1 . This implies that [a]H1 ∗′ [1]H1 ∈ H2/H1 and [1]H1 ∗′ [a]H1 ∈ H2/H1 which by DB2 implies that [a]H1 ∈ H2/H1. It follows that f(a) ∈ H2/H1 = ker g′ by Theorem 5 and so, f(a) ∈ f(H2). Since f is onto, there exists x ∈ H2 such that f(x) = f(a) implies [x]H1 = [a]H1 which implies that x ∼ a. So, x · a ∈ H1 and a · x ∈ H1. Since H1 ⊆ H2, x · a ∈ H2 and a · x ∈ H2 and it follows that since H2 is also a normal dB-filter by Proposition 2, a ∈ H2. This implies that ker(g′ ◦ f) ⊆ H2, and consequently ker(g′ ◦ f) = H2. By Theorem 6, there exists a unique dB-isomorphism h of G/H2 onto (G/H1)/(H2/H1) such that g′ ◦ f = h ◦ g. 4. Conclusion In this paper, it is shown that the necessary and sufficient condition for a db-filter to be a db-subalgebra and vice versa is normality. Using the quotient dB-algebra, along with some properties (such as normality) of the dB-filter, dB-subalgebra, and dB-homomorphism presented in the paper, the natural dB-homomorphism is determined; this then led to the creation of the fundamental theorem of dB-homomorphisms for dB-algebras. Following the aforementioned theorem, the first and third isomorphism theorems for the dB-algebra are constructed. Acknowledgements The authors would like to thank the Department of Science and Technology - Acceler- ated Science and Technology Human Resource Development Program (DOST-ASTHRDP) and the University of San Carlos for funding this research. References [1] Katrina Belleza and Jimboy Albaracin. On dual b-filters and dual b-subalgebras in a topological dual b-algebra. Journal of Mathematics and Computer Science, 28:1–10, 04 2022. [2] Katrina Belleza and Jocelyn P Vilela. The dual b-algebra. European Journal of Pure and Applied Mathematics, 12(4):1497–1507, 2019. REFERENCES 586 [3] Muhammad Chaudhry, Muhammad Qureshi, Asfand Fahad, and Muhammad Bashir. Isomorphism theorems in generalized d-algebras. Journal of Prime Research in Math- ematics, 17:149–158, 06 2022. [4] Joemar C Endam and Jocelyn P Vilela. The second isomorphism theorem for b- algebras. Applied Mathematical Sciences, 8(38):1865–1872, 2014. [5] Aiyared Iampan. The up-isomorphism theorems for up-algebras. arXiv preprint arXiv:1808.07370, 2018. [6] YB Jun, SM Hong, XL Xin, and EH Roh. Chinese remainder theorems in bci-algebras. Soochow Journal of Mathematics, 24(3):219–230, 1998. [7] Joseph Neggers and Hee Sik Kim. A fundamental theorem of b-homomorphism for b-algebras. Int. Math. J., 2, 3:207–214, 2002.