Congruency, Homomorphism  and Isomorphism on Autometrized Algebras

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We demonstrate that in a normal autometrized algebra, which satisfies certain conditions, the set of all congruence relations forms a complete sublattice within the set of all equivalence relations. Furthermore, we investigate the property that a congruence-permutable autometrized algebra is also congruence-modular. We also explore several fundamental properties related to congruence relations. Additionally, we introduce the kernel of a homomorphism and establish that it is a congruence relation. Lastly, we examine the homomorphism, isomorphism, and correspondence theorems of autometrized algebra using the concept of congruence." } { "@context": "http://schema.org", "@type": "BreadcrumbList", "itemListElement": [ { "@type": "ListItem", "position": "1", "item": { "@id": "https://f1000research.com/", "name": "Home" } }, { "@type": "ListItem", "position": "2", "item": { "@id": "https://f1000research.com/browse/articles", "name": "Browse" } }, { "@type": "ListItem", "position": "3", "item": { "@id": "https://f1000research.com/articles/14-22/v2", "name": "Congruency, Homomorphism and Isomorphism on Autometrized Algebras" } } ] } Home Browse Congruency, Homomorphism and Isomorphism on Autometrized Algebras ALL Metrics - Views Downloads Get PDF Get XML Cite How to cite this article Tilahun GY. Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.12688/f1000research.159591.2 ) NOTE: If applicable, it is important to ensure the information in square brackets after the title is included in all citations of this article. Close Copy Citation Details Export Export Citation Sciwheel EndNote Ref. Manager Bibtex ProCite Sente EXPORT Select a format first Track Share ▬ ✚ Research Article Revised Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] Gebrie Yeshiwas Tilahun https://orcid.org/0000-0002-7010-4847 Gebrie Yeshiwas Tilahun https://orcid.org/0000-0002-7010-4847 PUBLISHED 22 Aug 2025 Author details Author details Department of Mathematics, Assosa University, Asosa, Benishangul-Gumuz, Ethiopia Gebrie Yeshiwas Tilahun Roles: Writing – Review & Editing OPEN PEER REVIEW DETAILS REVIEWER STATUS Abstract This paper presents a study of congruence relations on autometrized algebras. We demonstrate that in a normal autometrized algebra, which satisfies certain conditions, the set of all congruence relations forms a complete sublattice within the set of all equivalence relations. Furthermore, we investigate the property that a congruence-permutable autometrized algebra is also congruence-modular. We also explore several fundamental properties related to congruence relations. Additionally, we introduce the kernel of a homomorphism and establish that it is a congruence relation. Lastly, we examine the homomorphism, isomorphism, and correspondence theorems of autometrized algebra using the concept of congruence. READ ALL READ LESS Keywords autometrized algebra, normal autometrized algebra, equivalent relation, congruence, homomorphism, isomorphism Corresponding Author(s) Gebrie Yeshiwas Tilahun ( [email protected] ) Close Corresponding author: Gebrie Yeshiwas Tilahun Competing interests: No competing interests were disclosed. Grant information: The author(s) declared that no grants were involved in supporting this work. Copyright: © 2025 Tilahun GY. This is an open access article distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. How to cite: Tilahun GY. Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.12688/f1000research.159591.2 ) First published: 03 Jan 2025, 14 :22 ( https://doi.org/10.12688/f1000research.159591.1 ) Latest published: 22 Aug 2025, 14 :22 ( https://doi.org/10.12688/f1000research.159591.2 ) Revised Amendments from Version 1 This version incorporates: 1. the definition of equivalent class and quotient set and explain by using examples. 2. prove the set of all equivalent relation forms a complete lattice. This version incorporates: 1. the definition of equivalent class and quotient set and explain by using examples. 2. prove the set of all equivalent relation forms a complete lattice. See the author's detailed response to the review by Sileshe Gone Korma See the author's detailed response to the review by Abdelmohsen Badawy READ REVIEWER RESPONSES 1. Introduction The concept of autometrized algebras was first introduced by K. N. Swamy (1964) to develop a unified theory containing previously established autometrized algebras, including Boolean algebras ( Blumenthal 1952 ; Ellis 1951 ), Brouwerian algebras ( Nordhaus & Lapidus 1954 ), Newman algebras ( Roy 1960 ), autometrized lattices ( Nordhaus & Lapidus 1954 ), and commutative lattice-ordered groups, or l-groups ( Narasimha Swamy 1964 ). The study of congruences within autometrized algebras was introduced by K. Swamy & Rao (1977) . Subsequent developments in the theory of autometrized algebras were contributed by K. Swamy & Rao (1977) , Rachŭnek (1987 , 1989 , 1990 , 1998 ), Hansen (1994) , Kovář (2000) , and Chajda & Rachunek (2001) . Furthermore, the notion of representable autometrized algebras was explored by Subba Rao & Yedlapalli (2018) and later by Rao et al. (2019 , 2021 , 2022) . In their research, Tilahun et al. (2023a , b , c , d) established a theory concerning subalgebras, ideals, and homomorphisms. They analyzed the relationship between normal autometrized l-algebras and representable autometrized algebras. Additionally, they introduced the concepts of direct and subdirect products of autometrized algebras, convex subalgebras, and congruences within autometrized algebras. However, previous studies have not explored the properties and applications of congruence relations in autometrized algebras. Therefore, our motivation is to address these gaps. This paper presents the concept of congruency in autometrized algebras, along with the introduction and proof of various theorems and facts related to congruence relations. The study will also cover homomorphisms, isomorphisms, and correspondence theorems for autometrized algebras through the concept of congruency. The structure of this paper is as follows: Section 2 will provide definitions and terminology. Section 3 will introduce the concept of congruency in autometrized algebras. Section 4 will present theorems related to homomorphisms and isomorphisms in autometrized algebras. Finally, Section 5 will conclude the paper. 2. Preliminaries In this section, we’ll look at some essential concepts, definitions, and terms; that are important in other sections. Definition 2.1 ( K. N. Swamy 1964 ) A system A = ( A , + , 0 , ≤ , ∗ ) is called an autometrized algebra if (i) ( A , + , 0 ) is a commutative monoid. (ii) ( A , ≤ ) is a partial ordered set, and ≤ is translation invariant, that is, ∀ a , b , c ∈ A ; a ≤ b ⇒ a + c ≤ b + c . (iii) ∗ : A × A → A is autometric on A , that is, ∗ satisfies metric operation axioms: ( M 1 ) ∀ a , b ∈ A ; a ∗ b ≥ 0 and, a ∗ b = 0 ⇔ a = b , ( M 2 ) ∀ a , b ∈ A ; a ∗ b = b ∗ a , ( M 3 ) ∀ a , b , c ∈ A ; a ∗ c ≤ a ∗ b + b ∗ c . Definition 2.2 ( K. Swamy & Rao 1977 ) An autometrized algebra A = ( A , + , 0 , ≤ , ∗ ) is called normal if and only if (i) a ≤ a ∗ 0 ∀ a ∈ A . (ii) ( a + c ) ∗ ( b + d ) ≤ ( a ∗ b ) + ( c ∗ d ) ∀ a , b , c , d ∈ A . (iii) ( a ∗ c ) ∗ ( b ∗ d ) ≤ ( a ∗ b ) + ( c ∗ d ) ∀ a , b , c , d ∈ A . (iv) For any a and b in A, a ≤ b ⇒ ∃ x ≥ 0 such that a + x = b . Definition 2.3 ( K. Swamy & Rao 1977 ) Let A = ( A , + , 0 , ≤ , ∗ ) be a system. Then A is said to be a lattice ordered autometrized algebra (or) autometrized l-algebra if (i) ( A , + , 0 ) is a commutative semigroup with 0. (ii) ( A , ≤ ) is a lattice, and ≤ is translation invariant, that is, ∀ a , b , c ∈ A ; a + ( b ∨ c ) = ( a + b ) ∨ ( a + c ) a + ( b ∧ c ) = ( a + b ) ∧ ( a + c ) (iii) ∗ : A × A → A is autometric on A , that is, ∗ satisfies metric operation axioms: M 1 , M 2 and M 3 . Definition 2.4 ( Tilahun et al. 2023d ) Let A = ( A , + , 0 , ≤ , ∗ ) be autometrized algebra. Let B ⊆ A . Then B is said to be a subalgebra of A if; (i) ( B , + , 0 ) is a commutative monoid. (ii) ( B , ≤ ) is a subposet of A, and ≤ is translation invariant, that is, a ≤ b ⇒ a + c ≤ b + c for any a , b , c ∈ B . (iii) ∗ | B : B × B → B is metric. Definition 2.5 ( Tilahun et al. 2023d ) A nonempty subset I of an autometrized algebra A = ( A , + , 0 , ≤ , ∗ ) is called an ideal if and only if (i) a , b ∈ I imply a + b ∈ I . (ii) a ∈ I , b ∈ A and b ∗ 0 ≤ a ∗ 0 imply b ∈ I . Definition 2.6 ( K. Swamy & Rao 1977 ) Let A be an autometrized algebra. Then A is said to be semiregular if for any a ∈ A , a ≥ 0 ⇒ a ∗ 0 = a . Definition 2.7 ( K. Swamy & Rao 1977 ) Let A be a normal autometrized algebra. An equivalence relation Θ on A is called a congruence relation if (i) ( a , b ) , ( c , d ) ∈ Θ ⇒ ( a + c , b + d ) ∈ Θ , ∀ a , b , c , d ∈ A , (ii) ( a , b ) , ( c , d ) ∈ Θ ⇒ ( a ∗ c , b ∗ d ) ∈ Θ , ∀ a , b , c , d ∈ A , (iii) ( a , b ) ∈ Θ and x ∗ y ≤ a ∗ b ⇒ ( x , y ) ∈ Θ , ∀ a , b , x , y ∈ A . Definition 2.8 ( K. Swamy & Rao 1977 ) Let A be a normal autometrized algebra and Θ be a congruence on A . Then, A / Θ = the set of all equivalence classes of Θ in A = { x ¯ = Θ ( x ) | x ∈ A } . Theorem 2.9 ( K. Swamy & Rao 1977 ) Let A be a normal autometrized algebra. Let Θ be a congruence relation on A . For any a ¯ , b ¯ ∈ A / Θ , Define: a ¯ + b ¯ = a + b ¯ . a ¯ ∗ b ¯ = a ∗ b ¯ . a ¯ ≤ b ¯ ⇔ ∃ x ≥ 0 such that ( a + x , b ) ∈ Θ . Then ( A / Θ , + , ≤ , ∗ ) is a normal autometrized algebra if and only if Θ has the following property: For any a , b ∈ A and z 1 ≥ 0 , z 2 ≥ 0 ; ( a + z 1 , b ) ∈ Θ and ( b + z 2 , a ) ∈ Θ ⇒ ( a , b ) ∈ Θ . Theorem 2.10 ( K. Swamy & Rao 1977 ) Let A be normal autometrized algebra. Let ℐ ( A ) = set of all ideals of A . Let Con ( A ) = set of all congruences on A . Then ℐ ( A ) and Con ( A ) are in one-to-one correspondence. Definition 2.11 ( Tilahun et al. 2023d ) Let A = ( A , + , 0 , ≤ , ∗ ) and B = ( B , + , 0 , ≤ , ∗ ) be autometrized algebras. Let f : A → B be a map. Then f is said to be a homomorphism from A to B if and only if (i) f ( a + b ) = f ( a ) + f ( b ) ∀ a , b ∈ A , (ii) f ( a ∗ b ) = f ( a ) ∗ f ( b ) ∀ a , b ∈ A and (iii) a ≤ b ⇒ f ( a ) ≤ f ( b ) ∀ a , b ∈ A . A homomorphism f : A → B is called (i) an epimorphism if and only if f is onto. (ii) a monomorphism(embedding) if and only if f is one-to-one. (iii) an isomorphism if and only if f is a bijection. Definition 2.12 ( Tilahun et al. 2023d ) Let A , B be normal autometrized algebras. Let f : A → B be a homomorphism. Then ker f = { x ∈ A | f ( x ) = 0 ¯ } where 0 ¯ is the zero element of B . Clearly, f is one-to-one if and only if ker f = { 0 } . Definition 2.13 Let A and B be autometrized algebras. Let f : A → B be a map. If a ≤ b ⇔ f ( a ) ≤ f ( b ) , ∀ a , b ∈ A , then f is said to be an order-embedding of A in to B . That is, f is both order-preserving and order-reversing. Remark 2.14 If f is an order-embedding of A in to B , then f is necessary injective. Since f ( a ) = f ( b ) implies a ≤ b and b ≤ a and in turn a = b according to antisymmetry of ≤ . Theorem 2.15 ( K. Swamy & Rao 1977 ) Let A be a normal autometrized algebra. Then every ideal of A is a kernel of an epimorphism of A . That is; (i) Kernel of any epimorphism of A is an ideal of A . (ii) If I is an ideal of A , then there exists an epimorphism f on A such that ker f = I . 3. Congruency on Autometrized Algebra In this section, we demonstrate that in a normal autometrized algebra, which satisfies certain conditions, the set of all congruence relations forms a complete sublattice within the set of all equivalence relations. We prove that a congruence-permutable autometrized algebra is also congruence-modular. Definition 3.1 Let A be an autometrized algebra. Then, any subset of A×A is called a relation on A . Definition 3.2 Let A be an autometrized algebra. Then Δ = { ( a , b ) ∈ A × A | a = b } = { ( a , a ) | a ∈ A } is called diagonal relation on A . And also ∇ = A × A is called all relation on A . Definition 3.3 Let A be an autometrized algebra. The set of all equivalence relations on A is denoted by Eq ( A ) . Definition 3.4 Let Θ 1 , Θ 2 be relations on an autometrized algebra A . Then their product is denoted by Θ 1 , Θ 2 , defined as: ( a , b ) ∈ Θ 1 ◦ Θ 2 ⇔ ∃ c ∈ A such that ( a , c ) ∈ Θ 1 and ( c , b ) ∈ Θ 2 . That is; Θ 1 ◦ Θ 2 = {( a , b )| ∃ c ∈ A such that ( a , c ) ∈ Θ 1 and ( c , b ) ∈ Θ 2 }. Example 3.5 Let A = { 0 , a , b , c } with 0 ≤ a ≤ b ≤ c . Define ∗ and + by the following tables. ∗ 0 a b c 0 0 a b c a a 0 b c b b b 0 c c c c c 0 + 0 a b c 0 0 a b c a a a b c b b b b c c c c c c It is clear to show that A is an autometrized algebra. Let us consider the equivalent relations Θ 1 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, a ), ( a ,0)} and Θ 2 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, b ), ( b , 0)} . Therefore, Θ 1 ◦ Θ 2 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, b ), (0, a ), ( a , 0), ( a , b )}and Θ 2 ◦ Θ 1 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, a ), ( a , 0), ( a , b ), (0, b ), ( b , a )}. Here, Θ 1 ◦ Θ 2 ≠ Θ 2 ◦ Θ 1 . In general, Θ 1 ◦ Θ 2 ≠ Θ 2 ◦ Θ 1 . Remark 3.6 If Θ 1 , Θ 2 ∈ Eq(A) , then Θ 1 ◦ Θ 2 need not be an equivalent relation on A. Example 3.7 It is clear that Example (3.5) , A is an autometrized algebra. And let’s consider the equivalent relations. Θ 1 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, a ), ( a ,0)} and Θ 2 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, b ), ( b , 0)} . Then, Θ 1 ◦ Θ 2 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, b ), (0, a ), ( a , 0), ( a , b )} is not an equivalent reletion on A. Remark 3.8 Let A be an autometrized algebra. Let Θ 1 , Θ 2 ∈ Eq( A ). Then i. Θ 1 , Θ 2 ⊆ Θ 1 ◦ Θ 2 . Let ( a , b ) ∈ Θ 1 . We know that ( b , b ) ∈ Θ 2 . So, ( a , b ) ∈ Θ 1 ◦ Θ 2 . Therefore, Θ 1 ⊆ Θ 1 ◦ Θ 2 . Similarly, Θ 2 ⊆ Θ 1 ◦ Θ 2 . ii. Θ 1 = Θ 1 ◦ Θ 1 . By (i); Θ 1 ⊆ Θ 1 ◦ Θ 1 . Let ( a , b ) ∈ Θ 1 ◦ Θ 1 . Then, ∃ c ∈ A such that ( a , c ) ∈ Θ 1 and ( c , b ) ∈ Θ 1 . By transitivity property, ( a , b ) ∈ Θ 1 . So, Θ 1 ◦ Θ 1 ⊆ Θ 1 . Hence, Θ 1 = Θ 1 ◦ Θ 1 . Theorem 3.9 Let A be an autometrized algebra. Then, i. (Eq( A ), ⊆ ) is a complete lattice. ii. For any Θ , Φ ∈ Eq( A ); Θ ∧ Φ = Θ ∩ Φ Θ ∨ Φ = Θ ∪ ( Θ ◦ Φ ) ∪ ( Θ ◦ Φ ◦ Θ ) ∪ ( Θ ◦ Φ ◦ Θ ◦ Φ ) ∪ ⋯ . That is, ( a , b ) ∈ Θ ∨ Φ ⇔ ∃ a = c 1 , c 2 ,…, c n = b ∈ A such that ( c i , c i + 1 ) ∈ Θ or Φ for 1 ≤ i ≤ n − 1 . iii. For { Θ i } i ∈ I ⊆ Eq( A ), ∧ i ∈ I Θ i = ∩ i ∈ I Θ i ∨ i ∈ I Θ i = ∪ { Θ 0 ◦ Θ 1 ◦ ⋯ ◦ Θ k | 0 , 1 , ⋯ , k ∈ I } . Proof. i. It is clear that ∇ ∈ Eq ( A ) is the greatest element of the poset (Eq( A ), ⊆ ). For any { Θ i } i ∈ I ⊆ Eq( A ), Inf Eq ( A ) { Θ i } i ∈ I = ∩ i ∈ I Θ i . Hence, (Eq( A ), ⊆ ) is a complete lattice. ii. Let Θ , Φ ∈ Eq( A ). Clearly, Θ ∧ Φ = Θ ∩ Φ . To show that Ψ = Θ ∪ ( Θ ◦ Φ ) ∪ ( Θ ◦ Φ ◦ Θ ) ∪ ( Θ ◦ Φ ◦ Θ ◦ Φ ) ∪ ⋯ = {( a , b ) | ∃ a = c 1 , c 2 ,…, c n = b ∈ A such that ( c i , c i + 1 ) ∈ Θ or Φ for 1 ≤ i ≤ n − 1 } is the least upper bound of {· Θ , Φ }. Here Ψ satisfies the three properties of equivalent relation and it is an equivalent relation. By Remark (3.8) ; Θ ⊆ Ψ and Φ ⊆ Θ ◦ Φ ⊆ Ψ . Therefore, Ψ is the least upper bound of {· Θ , Φ }. Let ϒ be equivalent relation on A and any upper bound of {· Θ , Φ }. To show that Ψ ⊆ ϒ . Let ( a , b ) ∈ Θ ◦ Φ . Then ∃ c ∈ A such that ( a , c ) ∈ Θ and ( c , b ) ∈ Φ . So, ( a , c ) ∈ ϒ and ( c , b ) ∈ ϒ . Therefore, ( a , b ) ∈ ϒ . Thus, Θ ◦ Φ ⊆ ϒ . Similarly, Θ ◦ Φ ◦ Θ ⊆ ϒ . As a result, Θ , Θ ◦ Φ , Θ ◦ Φ ◦ Θ , … ⊆ ϒ . So, ∪ Θ ◦ Φ ∪ Θ ◦ Φ ◦ Θ ∪ ⋯ ⊆ ϒ . Therefore, Ψ ⊆ ϒ . Consequently, Ψ is the least upper bound of {· Θ , Φ }. Hence, Θ ∨ Φ = Θ ∪ ( Θ ◦ Φ ) ∪ ( Θ ◦ Φ ◦ Θ ) ∪ ( Θ ◦ Φ ◦ Θ ◦ Φ ) ∪ ⋯ . iii. Similarly, we can prove as (ii). Definition 3.10 Let A be an autometrized algebra. and Θ ∈ Eq( A ). Then, for any x ∈ A ; define a Θ = Θ ( a ) = a ¯ = {b ∈ A | ( b , a ) ∈ Θ } is called the equivalence class of a with respect to Θ . Further, the set of all equivalence class of Θ is denoted by A Θ and A Θ = { a Θ | a ∈ A }. Example 3.11 It is clear that Example (3.5) , A is an autometrized algebra. Let’s consider the following equivalent relation. Θ 1 = {(0,0), ( a , a ), ( b , b ), ( c , c ), (0, a ), ( a ,0)}. Then, 0 Θ 1 = Θ 1 ( 0 ) = 0 ¯ = { y ∈ A | ( y , 0) ∈ Θ } = {0, a }. a Θ 1 = Θ 1 ( a ) = a ¯ = { y ∈ A | ( y , a ) ∈ Θ } = {0, a }. b Θ 1 = Θ 1 ( b ) = b ¯ = { y ∈ A | ( y , b ) ∈ Θ } = { b }. c Θ 1 = Θ 1 ( c ) = c ¯ = { y ∈ A | ( y , c ) ∈ Θ } = { c }. Therefore, A Θ 1 = 0 Θ 1 ∪ a Θ 1 ∪ b Θ 1 ∪ c Θ 1 = {0, a , b , c } = A . Definition 3.12 Let A be an autometrized algebra. Let Θ be a relation on A. Then Θ is said to be satisfied compatibility property, if • (i) ( a , b ), ( c , d )∈Θ⇒( a + c , b + d )∈Θ∀ a , b , c , d ∈ A , • (ii) ( a , b ), ( c , d )∈Θ⇒( a ∗c, b ∗ d )∈Θ∀ a , b , c , d ∈ A , • (iii) ( a , b )∈Θ and x ∗ y ≤ a ∗ b ⇒( x , y )∈Θ∀ a , b , x , y ∈ A . Definition 3.13 Let A be an autometrized algebra. Let Θ∈Eq( A ) . Then Θ is said to be a congruence on A if it satisfies compatibility property. Definition 3.14 Let A be an autometrized algebra. The set of all congruence relations on A is denoted by Con ( A ). Example 3.15 Let A = {0, a , b , c } with 0 ≤ a ≤ b ≤ c . Define ∗ and + by the following tables. ∗ 0 a b c 0 0 a b c a a 0 b c b b b 0 c c c c c 0 + 0 a b c 0 0 a b c a a a b c b b b b c c c c c c It is clear to show that A is an autometrized algebra. Let us consider the equivalent relations Δ={(0, 0), ( a , a ), ( b , b ), ( c , c )} , Θ 1 ={(0,0), ( a , a ), ( b , b ), ( c , c ), (0, a ), ( a , 0)} , Θ 2 ={(0,0), ( a , a ), ( b , b ), ( c , c ), ( a , 0), (0, a ), (0, b ), ( b , 0), ( a , b ), ( b , a )} and ∇=A×A on A . Then Δ , Θ 1 , Θ 2 and ∇ are all the congruence relations on A. Therefore, Con ( A )={Δ, Θ 1 , Θ 2 ,∇}. Definition 3.16 Let A be an autometrized algebra and X ⊆ A . Then the smallest congruence containing X × X is called the congruence generated by X × X or X and denoted by Θ ( X × X ) or Θ ( X ) . Further, if X = { a , b } , then Θ ( X ) = Θ ( a , b ) is called principal congruence. Example 3.17 In Example (3.15) , A is an autometrized algebra. Take X = { a , b } . Then, X × X = { ( a , a ) , ( b , b ) , ( a , b ) , ( b , a ) } . Therefore, Θ ( X × X ) = Θ ( X ) = { ( 0 , 0 ) , ( a , a ) , ( b , b ) , ( c , c ) , ( a , 0 ) , ( 0 , a ) , ( 0 , b ) , ( b , 0 ) , ( a , b ) , ( b , a ) } . Hence, Θ ( X ) is a principal congruence. Theorem 3.18 If Θ 1 , Θ 2 are two congruence relations on a normal autometrized algebra A satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a , b , c ∈ A and b ≠ c , then Θ 1 ◦ Θ 2 satisfies compatibility property. Proof. Now to show that Θ 1 ◦ Θ 2 satisfies compatibility property. (i) Suppose ( a , b ) , ( c , d ) ∈ Θ 1 ◦ Θ 2 . There exist x , y ∈ A such that ( a , x ) ∈ Θ 1 , ( x , b ) ∈ Θ 2 and ( c , y ) ∈ Θ 1 , ( y , d ) ∈ Θ 2 . Since Θ 1 and Θ 2 are congruence relations; ( a + c , x + y ) ∈ Θ 1 and ( x + y , b + d ) ∈ Θ 2 . Thus, ( a + c , c + d ) ∈ Θ 1 ◦ Θ 2 . (ii) Suppose ( a , b ) , ( c , d ) ∈ Θ 1 ◦ Θ 2 . There exist x , y ∈ A such that ( a , x ) ∈ Θ 1 , ( x , b ) ∈ Θ 2 and ( c , y ) ∈ Θ 1 , ( y , d ) ∈ Θ 2 . Since Θ 1 and Θ 2 are congruence relations; ( a ∗ c , x ∗ y ) ∈ Θ 1 and ( x ∗ y , b ∗ d ) ∈ Θ 2 . Thus, ( a ∗ c , c ∗ d ) ∈ Θ 1 ◦ Θ 2 . (iii) Suppose ( a , b ) ∈ Θ 1 ◦ Θ 2 and x ∗ y ≤ a ∗ b . To show that ( x , y ) ∈ Θ 1 ◦ Θ 2 . There exists z ∈ A such that ( a , z ) ∈ Θ 1 and ( z , b ) ∈ Θ 2 . It is clear that (1) x ∗ y ≤ a ∗ b ≤ a ∗ z + b ∗ z . Since Θ 1 is a congruence relation; ( z , z ) , ( b , b ) ∈ Θ 1 . As a result, ( a ∗ z , 0 ) , ( a ∗ b , b ∗ z ) ∈ Θ 1 . This implies that ( a ∗ z + b ∗ z , a ∗ b ) ∈ Θ 1 . We know that 0 ≤ a ∗ b . By the given condition ( a ∗ z + b ∗ z ) ∗ 0 ≤ ( a ∗ z + b ∗ z ) ∗ ( a ∗ b ) . Since A is normal; equation (1) becomes x ∗ y ≤ a ∗ b ≤ a ∗ z + b ∗ z ≤ ( a ∗ z + b ∗ z ) ∗ 0 ≤ ( a ∗ z + b ∗ z ) ∗ ( a ∗ b ) . Since ( a ∗ z + b ∗ z , a ∗ b ) ∈ Θ 1 and x ∗ y ≤ ( a ∗ z + b ∗ z ) ∗ ( a ∗ b ) ; by definition of congruence ( x , y ) ∈ Θ 1 . It is clear that ( y , y ) ∈ Θ 2 . Therefore, ( x , y ) ∈ Θ 1 ◦ Θ 2 . Hence, Θ 1 ◦ Θ 2 satisfies compatibility property. Theorem 3.19 Let Θ 1 , Θ 2 be two congruence relations on a normal autometrized algebra A satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a, b, c ∈ A and b ≠ c. Suppose that Θ 1 ◦ Θ 2 = Θ 2 ◦ Θ 1 . Then Θ 1 ◦ Θ 2 is a congruence relation on A. Proof. To show that the product Θ 1 ◦ Θ 2 is an equivalence relation. (i) Reflexive : Let x ∈ A. We know that ( x , x ) ∈ Θ 1 and ( x , x ) ∈ Θ 2 . Therefore ( x , x ) ∈ Θ 1 ◦ Θ 2 ∀ x ∈ A. (ii) Symmetric : Suppose ( x , y ) ∈ Θ 1 ◦ Θ 2 . There exists z ∈ A such that ( x , z ) ∈ Θ 1 and ( z , y ) ∈ Θ 2 . Since Θ 1 and Θ 2 are equivalence relations; ( z , x ) ∈ Θ 1 and ( y , z ) ∈ Θ 2 . As a result, ( y , x ) ∈ Θ 2 ◦ Θ 1 . Since Θ 1 ◦ Θ 2 = Θ 2 ◦ Θ 1 ; and hence ( y , x ) ∈ Θ 1 ◦ Θ 2 . (iii) Transitivity : Suppose ( x , y ), ( y , z ) ∈ Θ 1 ◦ Θ 2 . There exist r , q ∈ A such that ( x , r ) ∈ Θ 1 , ( r , y ) ∈ Θ 2 and ( y , q ) ∈ Θ 1 , ( q , z ) ∈ Θ 2 . Consequently, ( r , q ) ∈ Θ 2 ◦ Θ 1 , ( r , z ) ∈ (Θ 2 ◦ Θ 1 ) ◦ Θ 2 and ( x , z ) ∈ Θ 2 ◦ (Θ 2 ◦ Θ 1 ) ◦ Θ 2 . Since Θ 1 ◦ Θ 2 = Θ 2 ◦ Θ 1 ; ( x , z ) ∈ Θ 2 ◦ (Θ 1 ◦ Θ 2 ) ◦ Θ 2 . This implies that there exist f , g ∈ A such that ( x , f ) ∈ Θ 1 , ( f , g ) ∈ Θ 1 ◦ Θ 2 and ( g , z ) ∈ Θ 2 . Since ( f , g ) ∈ Θ 1 ◦ Θ 2 ; there exists l ∈ A such that ( f , l ) ∈ Θ 1 and ( l , g ) ∈ Θ 2 . Since Θ 1 and Θ 2 are equivalence relations; ( x , l ) ∈ Θ 1 and ( l , z ) ∈ Θ 2 . Hence ( x , z ) ∈ Θ 1 ◦ Θ 2 . Hence, Θ 1 ◦ Θ 2 is an equivalence relation on A. By Theorem (3.18) ; we know that Θ 1 ◦ Θ 2 satisfies compatibility property. Hence, Θ 1 ◦ Θ 2 is a congruence relation on A. Example 3.20 It is clear that in Example (3.15), A is a normal autometrized algebra satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a , b , c ∈ A and b ≠ c. And we know that Θ 1 = {(0, 0), ( a , a ), ( b , b ), ( c , c ), (0, a ), ( a , 0)}, Θ 2 = {(0, 0), ( a , a ), ( b , b ), ( c , c ), ( a , 0), (0, a ), (0, b ), ( b , 0), ( a , b ), ( b , a )}. Therefore, Θ 1 ◦ Θ 2 = Θ 2 ◦ Θ 1 = Θ 2 . Hence, Θ 1 ◦ Θ 2 is a congruence relation on A. Theorem 3.21 Let A be a normal autometrized algebra A satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a , b , c ∈ A and b ≠ c. Then ( Con ( A ), ⊆) is a complete sublattice of ( Eq ( A ), ⊆). Proof. Let {Θ i } i ∈ I ⊆ Con ( A ). Therefore, {Θ i } i ∈ I ⊆ Eq ( A ). By Theorem (3.9), Inf Eq ( A ) { Θ i } i ∈ I = ∧ i ∈ I Θ i = ∩ i ∈ I Θ i . Sup Eq ( A ) { Θ i } i ∈ I = ∨ i ∈ I Θ i = ∪ { Θ 0 ◦ Θ 1 ◦ ⋯ ◦ Θ k | 0 , 1 , ⋯ k ∈ I } . Clearly, ∩ i ∈ I Θ i ∈ Con ( A ) . So, Inf Eq ( A ) { Θ i } i ∈ I = ∧ i ∈ I Θ i ∈ Con ( A ) . To show that Sup Eq ( A ) { Θ i } i ∈ I = ∨ i ∈ I Θ i ∈ Con ( A ) . Let Ψ = ∪ { Θ 0 ◦ Θ 1 ◦ ⋯ ◦ Θ k | 0 , 1 , ⋯ k ∈ I } . To show that Ψ ∈ Con ( A ) . Clearly, Ψ ∈ Eq ( A ) . To show that Ψ satisfies compatibility property. Let ( a , b ) ∈ Ψ and x ∗ y ≤ a ∗ b . Therefore, ( a , b ) ∈ Θ 0 ◦ Θ 1 ◦ ⋯ ◦ Θ i for i = 0 , ⋯ , k . By theorem (3.18) ; ( x , y ) ∈ Θ 0 ◦ Θ 1 ◦ ⋯ ◦ Θ i . It is clear that ( x , y ) ∈ Ψ . Therefore, Sup Eq ( A ) { Θ i } i ∈ I = ∨ i ∈ I Θ i ∈ Con ( A ) . Hence, Con ( A ) is a complete sublattice of Eq ( A ) . Definition 3.22 Let A be an autometrized algebra and θ, φ ∈ Con ( A ) . Then θ, φ are said to be permutable congruences if θ ◦ φ = φ ◦ θ. Definition 3.23 It is clear that in Example (3.15) ; Θ 1 , Θ 2 are permutable congruences. Since Θ 1 ◦ Θ 2 = Θ 2 ◦ Θ 1 = Θ 2 . Definition 3.24 Let A be an autometrized algebra. If every pair of elements in Con ( A ) are permutable, then A is said to be a congruence-permutable autometrized algebra. Example 3.25 It is clear that in Example (3.15) ; Con ( A ) = { Δ , Θ 1 , Θ 2 , ∇ } . Further, Δ ◦ Θ 1 = Θ 1 ◦ Δ , Δ ◦ Θ 2 = Θ 2 ◦ Δ , Δ ◦ ∇ = ∇ ◦ Δ , Θ 1 ◦ Θ 2 = Θ 2 ◦ Θ 1 , Θ 1 ◦ Δ = Δ ◦ Θ 1 , and Θ 2 ◦ Δ = Δ ◦ Θ 2 . Hence, A is a congruence-permutable autometrized algebra. Definition 3.26 Let A be an autometrized algebra. Then A is said to be a congruence-modular algebra if Con ( A ) is a modular lattice. Theorem 3.27 ( Burris 1981 ) Let A be an autometrized algebra and θ , ϕ ∈ Eq ( A ) . Then the following are equivalent. (i) θ ◦ ϕ = ϕ ◦ θ . (ii) θ ∨ ϕ = θ ◦ ϕ . (iii) θ ◦ ϕ ≤ ϕ ◦ θ . Theorem 3.28 Let A be a congruence-permutable autometrized algebra. Then A is congruence-modular. Proof. Suppose A is a congruence-permutable. That is every pair of congruences on A are permutable. Now to show that A is congruence-modular. To show that Con ( A ) is a modular. Let Θ 1 , Θ 2 , Θ 3 ∈ Con ( A ) and Θ 1 ⊆ Θ 3 . Clearly, Θ 1 ∨ ( Θ 2 ∧ Θ 3 ) ≤ ( Θ 1 ∨ Θ 2 ) ∧ Θ 3 . To show that ( Θ 1 ∨ Θ 2 ) ∧ Θ 3 ≤ Θ 1 ∨ ( Θ 2 ∧ Θ 3 ) . Let ( a , b ) ∈ ( Θ 1 ∨ Θ 2 ) ∧ Θ 3 . Therefore, ( a , b ) ∈ Θ 1 ∨ Θ 2 and ( a , b ) ∈ Θ 3 . Since Θ 1 , Θ 2 are permutable; Θ 1 ∨ Θ 2 = Θ 1 ◦ Θ 2 . As a result, ( a , b ) ∈ Θ 1 ◦ Θ 2 . There exists c ∈ A such that ( a , c ) ∈ Θ 1 and ( c , b ) ∈ Θ 2 . Since Θ 1 ⊆ Θ 3 ; ( a , c ) ∈ Θ 3 . Since Θ 3 is a congruence relation; ( c , a ) ∈ Θ 3 . Now we have ( c , a ) , ( a , b ) ∈ Θ 3 . Clearly, ( c , b ) ∈ Θ 3 . As a result, ( c , b ) ∈ Θ 2 ∧ Θ 3 . Now we have ( a , c ) ∈ Θ 1 and ( c , b ) ∈ Θ 2 ∧ Θ 3 . Therefore, ( a , b ) ∈ Θ 1 ◦ ( Θ 2 ∧ Θ 3 ) . Since A is a congruence-permutable; ( a , b ) ∈ Θ 1 ∨ ( Θ 2 ∧ Θ 3 ) . Therefore, ( Θ 1 ∨ Θ 2 ) ∧ Θ 3 ≤ Θ 1 ∨ ( Θ 2 ∧ Θ 3 ) . Consequently, Θ 1 ∨ ( Θ 2 ∧ Θ 3 ) = ( Θ 1 ∨ Θ 2 ) ∧ Θ 3 . Hence, Con ( A ) is modular. 4. Homomorphism and Isomorphism on Autometrized Algebra In this section, we discuss the homomorphism, isomorphism, and correspondence theorems of autometrized algebra using the concept of congruence. Theorem 4.1 Let A , B be normal autometrized algebras. Let f : A → B be a homomorphism. Then the set Ker f = { ( a , b ) ∈ A × A | f ( a ) = f ( b ) } , is called kernel of the homomorphism f . Theorem 4.2 Let A , B be normal autometrized algebras. Let f : A → B be a homomorphism. Then, ker f is a congruence on A . That is ker f ∈ Con ( A ) . Proof. It is clear that ker f is an equivalence relation. (i) Suppose ( a , b ) , ( c , d ) ∈ ker f . So, f ( a ) = f ( b ) and f ( c ) = f ( d ) . Since f is a homomorphism; f ( a + c ) = f ( a ) + f ( c ) = f ( b ) + f ( d ) = f ( b + d ) . Therefore, ( a + c , b + d ) ∈ ker f . (ii) Suppose ( a , b ) , ( c , d ) ∈ ker f . So, f ( a ) = f ( b ) and f ( c ) = f ( d ) . Since f is a homomorphism; f ( a ∗ c ) = f ( a ) ∗ f ( c ) = f ( b ) ∗ f ( d ) = f ( b ∗ d ) . Therefore, ( a ∗ c , b ∗ d ) ∈ ker f . (iii) Suppose ( a , b ) ∈ ker f and x ∗ y ≤ a ∗ b . To show that ( x , y ) ∈ ker f . Clearly, f ( a ) = f ( b ) . Since f is a homomorphism; f ( x ∗ y ) ≤ f ( a ∗ b ) . As result, f ( x ) ∗ f ( y ) ≤ f ( a ) ∗ f ( b ) . Therefore, f ( x ) ∗ f ( y ) ≤ 0 . So, f ( x ) = f ( y ) . Hence, ( x , y ) ∈ ker f . Thus, ker f ∈ Con ( A ) . Theorem 4.3 Let A be a normal autometrized algebra and Θ ∈ Con ( A ) . Define a map f : A → A / Θ by f ( a ) = a / Θ = a ¯ . Then f is an epimorphism and ker f = Θ . Proof. Clearly Θ satisfies the condition. Hence A / Θ is a normal automertrized algebra. Clearly f is onto. To show that f is homomorphism. Let a , b ∈ A . f ( a + b ) = a + b ¯ = a ¯ + b ¯ = f ( a ) + f ( b ) . f ( a ∗ b ) = a ∗ b ¯ = a ¯ ∗ b ¯ = f ( a ) ∗ f ( b ) . Let a ≤ b . Then ∃ x ≥ 0 such that a + x = b . Therefore ( a + x , b ) ∈ Θ . Which implies that a ¯ ≤ b ¯ . That is; f ( a ) ≤ f ( b ) . Hence f is a homomorphism. So that f is an epimorphism from A to A / Θ . Consider; ker f = { ( a , b ) ∈ A × A | f ( a ) = f ( b ) } . = { ( a , b ) ∈ A × A | a ¯ ≤ b ¯ } . = { ( a , b ) ∈ A × A | a / Θ = b / Θ } . = { a ∈ A | ( a , b ) ∈ Θ } . = ( A × A ) ∩ Θ = Θ . Theorem 4.4 (Homomorphism theorem) Let A, B be normal autometrized algebras. Let f: A → B be a homomorphism. Then A/ ker f ≅ Imf. In particular if f is onto, then A/ ker f ≅ B. Proof. Let ker f = Θ. By the above theorem (4.3) , the map g : A → A / Θ is an epimorphism and ker g = Θ . Define a map h : A / Θ → Imf by h ( a ¯ ) = f ( a ) . Clearly, h is onto map. To show that h is well-defined and one to one. Suppose a ¯ = b ¯ , then a ¯ = b ¯ ⇔ ( a , b ) ∈ Θ . ⇔ ( a , b ) ∈ ker f . ⇔ f ( a ) = f ( b ) . ⇔ h ( a ¯ ) = h ( b ¯ ) . Therefore, h is well-defined and one to one. Now to show that h is homomorphism. Let a ¯ , b ¯ ∈ A / Θ . Consider; h ( a ¯ + b ¯ ) = h ( a + b ¯ ) . = f ( a + b ) . = f ( a ) + f ( b ) . = h ( a ¯ ) + h ( b ¯ ) . Again consider; h ( a ¯ ∗ b ¯ ) = h ( a ∗ b ¯ ) . = f ( a ∗ b ) . = f ( a ) ∗ f ( b ) . = h ( a ¯ ) ∗ h ( b ¯ ) . Suppose let a ¯ ≤ b ¯ . Therefore ∃ x ≥ 0 such that ( a + x , b ) ∈ Θ . Hence, ( a + x , b ) ∈ ker f . ⇒ f ( a + x ) = f ( b ) . Hence f ( a ) + f ( x ) = f ( b ) . Since x ≥ 0 ⇒ f ( x ) ≥ 0 ¯ . Then f ( a ) ≤ f ( b ) . Therefore h ( a ¯ ) ≤ h ( b ¯ ) . And thus, h is a homomorphism. Therefore, h is an isomorphism from A / Θ to Imf . That is A / Θ ≅ Imf . If f is onto map, then Imf = B . Hence A / Θ ≅ B . Definition 4.5 Let A and B be autometrized algebras and θ , ϕ ∈ Con ( A ) . Suppose θ ⊆ ϕ . Define, ϕ / θ = { ( a / θ , b / θ ) ∈ A / θ × A / θ | ( a , b ) ∈ ϕ } , and for any ( a / θ , b / θ ) , ( c / θ , d / θ ) ∈ ϕ / θ ; a ∗ b / θ ≤ c ∗ d / θ ⇔ a ∗ b ≤ c ∗ d . Theorem 4.6 Let A be a normal autometrized algebra and θ , ϕ ∈ Con ( A ) . Suppose θ ⊆ ϕ . Then, ϕ / θ ∈ Con ( A / θ ) . Proof. First let us show that ϕ / θ is an equivalence relation on A. 1. Reflexive : Let a / θ ∈ A / θ . Therefore, a ∈ A . Since ϕ is a congruence on A ; ( a , a ) ∈ ϕ . Hence, ( a / θ , a / θ ) ∈ ϕ / θ ∀ a ∈ A . 2. Symmetric : Suppose ( a / θ , b / θ ) ∈ ϕ / θ . Then, ( a , b ) ∈ ϕ . Since ϕ is a congruence on A ; ( b , a ) ∈ ϕ . Hence, ( b / θ , a / θ ) ∈ ϕ / θ . 3. Transitivity : Suppose ( a / θ , b / θ ) , ( b / θ , c / θ ) ∈ ϕ / θ . Which implies that: ( a , b ) , ( b , c ) ∈ ϕ . Thus, ( a , c ) ∈ ϕ . Therefore, ( a / θ , c / θ ) ∈ ϕ / θ . Hence, ϕ / θ is an equivalence relation. To show that ϕ / θ is congruence on A. Let ( a / θ , b / θ ) , ( c / θ , d / θ ) ∈ ϕ / θ . Then, ( a , b ) , ( c , d ) ∈ ϕ . (i) Since ϕ is a congruence; ( a + c , b + d ) ∈ ϕ . Since A / θ is a normal autometrized algebra; ( a + c / θ , b + d / θ ) ∈ ϕ / θ . (ii) Since ϕ is a congruence; ( a ∗ c , b ∗ d ) ∈ ϕ . Since A / θ is a normal autometrized algebra; ( a ∗ c / θ , b ∗ d / θ ) ∈ ϕ / θ . (iii) Let ( a / θ , b / θ ) ∈ ϕ / θ . Then ( a , b ) ∈ ϕ . Suppose x / θ ∗ y / θ ≤ a / θ ∗ b / θ . Since A / θ is normal autometrized algebra; x ∗ y / θ ≤ a ∗ b / θ . This implies that, ∃ z ≥ 0 such that ( x ∗ y + z , a ∗ b ) ∈ θ . Since θ ⊆ ϕ ; we obtain; ( x ∗ y + z , a ∗ b ) ∈ ϕ and implies x ∗ y / ϕ ≤ a ∗ b / ϕ . Therefore, x ∗ y ≤ a ∗ b . Since ϕ is a congruence on A ; ( x , y ) ∈ ϕ . Hence, ( x / θ , y / θ ) ∈ ϕ / θ . Therefore, ϕ / θ ∈ Con ( A ) . Theorem 4.7 (Second Isomorphism Theorem) Let A be a normal autometrized algebras. Let θ,ϕ ∈ Con ( A ) and θ ⊆ ϕ. Then A / θ ϕ / θ ≅ A / ϕ . Proof. Define a map g : A / θ → A / ϕ by g ( a / θ ) = a / ϕ . To show that g is well-defined. Suppose; a ¯ = b ¯ . That is; a / θ = b / θ ⇔ ( a , b ) ∈ θ . ⇔ ( a , b ) ∈ ϕ . [ Since θ ⊆ ϕ ] . ⇔ a / ϕ = b / ϕ . Therefore, g ( a / θ ) = g ( b / θ ) . Hence, g is well-defined and one to one. Clearly g is onto map. To show that g is a homomorphism. Let a / θ , b / θ ∈ A / θ . g ( a / θ + b / θ ) = g ( a + b / θ ) . = ( a + b ) / ϕ . = a / ϕ + b / ϕ . = g ( a / θ ) + g ( b / θ ) . Again consider; g ( a / θ ∗ b / θ ) = g ( a ∗ b / θ ) . = ( a ∗ b ) / ϕ . = a / ϕ ∗ b / ϕ . = g ( a / θ ) ∗ g ( b / θ ) . Suppose a ¯ ≤ b ¯ . That is; a / θ ≤ b / θ . Then ∃ x ≥ 0 such that ( a + x , b ) ∈ θ . Since θ ⊆ ϕ ; we obtain ( a + x , b ) ∈ ϕ . Therefore, a / ϕ ≤ b / ϕ . Hence g ( a / θ ) ≤ g ( b / θ ) . Now let us consider the kernel of g . ker g = { ( a / θ , b / θ ) | g ( a / θ ) = g ( b / θ ) } . = { ( a / θ , b / θ ) | a / ϕ = b / ϕ } . = { ( a / θ , b / θ ) | ( a , b ) ∈ ϕ } . = ϕ / θ . Therefore, by first isomorphism theorem; A / θ ker g ≅ A / ϕ . Hence; A / θ ϕ / θ ≅ A / ϕ . Theorem 4.8 Let A be a normal autometrized algebras. Let B ⊆ A and θ ∈ Con ( A ) . Define B θ = { a ∈ A | B ∩ ( a / θ ) is non‐empty } . Then, B ⊆ B θ . Proof. Let b ∈ B . Since θ ∈ Con ( A ) implies that ( b , b ) ∈ θ . Then b ∈ b / θ . Therefore, b ∈ B ∩ ( b / θ ) . And thus, B ∩ ( b / θ ) is non-empty. So, b ∈ B θ . Hence, B ⊆ B θ . Definition 4.9 Let A be a normal autometrized algebras. Let B be a normal subalgebra of A. Let θ ∈ Con ( A ) . Then, define, θ | B = θ ∩ ( B × B ). Theorem 4.10 Let A be a normal autometrized algebras. Let B be a normal subalgebra of A. Let θ ∈ Con ( A ) . Then, B θ is a normal subalgebra of A. Proof. Since 0 ∈ B and 0 ∈ 0/ θ ; B ∩ (0/ θ ) is non-empty. Hence 0 ∈ B θ . Let a , b ∈ B θ . Then, B ∩ ( a / θ ) and B ∩ ( b / θ ) are non-empty. Choose x ∈ B ∩ ( a / θ ) and y ∈ B ∩ ( b / θ ). Therefore; x , y ∈ B and x ∈ a / θ , y ∈ b / θ. Therefore, ( x , a ), ( y , b ) ∈ θ. Since B is a normal subalgebra of A ; (2) x ∗ y , x + y ∈ B . Since θ ∈ Con ( A ) ; ( x + y , a + b ) ∈ θ . Therefore, (3) x + y ∈ a + b / θ . By equations (2) and (3); we get: x + y ∈ B ∩ ( a + b / θ ) . Hence, B ∩ ( a + b / θ ) is non-empty. Therefore, a + b ∈ B θ . Similarly, Since θ ∈ Con ( A ) ; ( x ∗ y , a ∗ b ) ∈ θ . Therefore, (4) x ∗ y ∈ a ∗ b / θ . By equations (2) and (4); we get: x ∗ y ∈ B ∩ ( a ∗ b / θ ) . Hence, B ∩ ( a ∗ b / θ ) is non-empty. Therefore, a ∗ b ∈ B θ . Thus, B θ is a subalgebra. Now to show that B θ is a normal subalgebra. Let a , b ∈ B θ . Then, B ∩ ( a / θ ) and B ∩ ( b / θ ) are non-empty. Suppose a ≤ b . Since B is normal; ∃ x ≥ 0 such that b = a + x . Now to show that x ∈ B θ . Clearly; x ∈ x / θ . Therefore, B ∩ ( x / θ ) is non-empty. This implies that x ∈ B θ . Hence, B θ is a normal subalgebra of A . Theorem 4.11 Let A be a normal autometrized algebras. Let B be a normal subalgebra of A. Let θ ∈ Con ( A ) . Then, θ | B is a congruence on B. Proof. Clearly, θ ∩ B × B is an equivalence relation on B. Let ( a , b ), ( c , d ) ∈ θ / B = θ ∩ ( B × B ). Then, ( a , b ), ( c , d ) ∈ B × B and θ. Therefore a , b , c , d ∈ B. (a) Since B is a normal subalgebra; a + c , b + d ∈ B. Therefore, ( a + c , b + d ) ∈ B × B. And also, θ is a congruence; ( a + c , b + d ) ∈ θ. Hence, ( a + c , b + d ) ∈ θ | B. (b) Since B is a normal subalgebra; a ∗ c , b ∗ d ∈ B. Therefore, ( a ∗ c , b ∗ d ) ∈ B × B. And also, θ is a congruence; ( a ∗ c , b ∗ d ) ∈ θ. Hence, ( a ∗ c , b ∗ d ) ∈ θ | B. (c) Suppose ( a , b ) ∈ θ | B = θ ∩ B × B and x ∗ y ≤ a ∗ b ∀ x , y , a , b ∈ B. Therefore, ( a , b ) ∈ θ and ( a , b ) ∈ B × B. So, a , b ∈ B and a ∗ b ∈ B. Since θ is congruence; implies ( x , y ) ∈ θ. And clearly, ( x , y ) ∈ B × B. Hence; ( x , y ) ∈ θ ∩ B × B. And thus, θ | B is a congruence on B. Theorem 4.12 (The Third Isomorphism Theorem) Let A be a normal autometrized algebras. Let B be a normal subalgebra of A. Let θ ∈ Con ( A ) . Then, B θ | B ≅ B θ θ | B θ . Proof. Clearly B θ θ | B θ is a normal autometrized algebra. Define a map f : B → B θ θ | B θ by f ( b ) = b θ | B θ . To show that f is well-defined. Let b 1 , b 2 ∈ B . Suppose; b 1 = b 2 . Therefore, b 1 = b 2 ⇒ b 1 θ | B θ = b 2 θ | B θ . ⇒ f ( b 1 ) = f ( b 2 ) . Therefore, f is well-defined. To show that f is onto. Let c θ | B θ ∈ B θ θ | B θ . Therefore, c ∈ B θ . Hence B ∩ ( c / θ ) is non-empty. Then ∃ b ∈ B ∩ ( c / θ ) . Then b ∈ B and b ∈ c / θ . Since B ⊆ B θ ; b ∈ B θ . Therefore, ( b , c ) ∈ B θ × B θ and ( b , c ) ∈ θ . Clearly ( b , c ) ∈ θ ∩ ( B θ × B θ ) = θ | B θ . Whence, b θ | B θ = c θ | B θ . Therefore, f ( b ) = c θ | B θ . Hence, f is on to map. To show that f is a homomorphism. Let a , b ∈ B . Consider; f ( a + b ) = a + b θ | B θ . = a θ | B θ + b θ | B θ . = f ( a ) + f ( b ) . Again consider; f ( a ∗ b ) = a ∗ b θ | B θ . = a θ | B θ ∗ b θ | B θ . = f ( a ) ∗ f ( b ) . Suppose a ≤ b . Then ∃ x ≥ 0 such that a + x = b . Therefore, a + x θ | B θ = b θ | B θ . Which implies that ( a + x , b ) ∈ θ | B θ . Since B θ θ | B θ is normal implies that a θ | B θ ≤ b θ | B θ . Thus, f ( a ) ≤ f ( b ) . Hence, f is a homomorphism. Now consider the kernel of f . Therefore, ker f = { ( a , b ) ∈ B × B | f ( a ) = f ( b ) } . = { ( a , b ) ∈ B × B | a θ | B θ = b θ | B θ } . = { ( a , b ) ∈ B × B | ( a , b ) ∈ θ | B θ } . = { ( a , b ) ∈ B × B | ( a , b ) ∈ θ ∩ B θ × B θ } . = ( B × B ) ∩ θ ∩ ( B θ × B θ ) . = ( B × B ) ∩ θ . [ Since B ⊆ B θ ] = θ | B . Therefore, by first isomorphism theorem; B ker f ≅ B θ θ | B θ . Hence; B θ | B ≅ B θ θ | B θ . Theorem 4.13 (Correspondence Theorem) Let A be a normal autometrized algebra. Let θ ∈ Con ( A ). Then [ θ , ∇ A ] is lattice isomorphic to Con ( A / θ ). Proof. We know that I ( A ) and Con ( A ) are lattices. We know that by theorem (2.10) , I ( A ) and Con ( A ) are in one to one correspondence. Therefore, [ θ , ∇ A ] is an interval in Con ( A ). Therefore, [ θ , ∇ A ] is a sublattice of Con ( A ). We know that Con ( A / θ ) is a lattice. Define a map f : [ θ , ∇ A ] → Con ( A / θ ) by f ( ϕ ) = ϕ / θ. Since ϕ ∈ [ θ , ∇ A ]; implies that θ ⊆ ϕ ⊆ ∇ A . Therefore, by theorem (4.6) ; ϕ / θ ∈ Con ( A / θ ). To show that f is well-defined and one to one. Let ϕ , ψ ∈ [ θ , ∇ A ]. Therefore, θ ⊆ ϕ, ψ ⊆ ∇ A . Suppose ϕ = ψ. Therefore, ϕ / θ = ψ / θ. Which implies that f ( ϕ ) = f ( ψ ). Hence, f is well-defined. Suppose f ( ϕ ) = f ( ψ ). Therefore, ϕ / θ = ψ / θ. Let ( a , b ) ∈ ϕ. Therefore, ( a / θ , b / θ ) ∈ ϕ / θ . ( a / θ , b / θ ) ∈ ψ / θ . ( a , b ) ∈ ψ . ϕ ⊆ ψ . And similarly, let ( a , b ) ∈ ψ . Therefore, ( a / θ , b / θ ) ∈ ψ / θ . ( a / θ , b / θ ) ∈ ϕ / θ . ( a , b ) ∈ ϕ . ψ ⊆ ϕ . Whence, ϕ = ψ . Hence, f is one to one. To show that f is on to. Let Θ 1 ∈ Con ( A / θ ) . Define Θ 2 = { ( a , b ) | a / θ Θ 1 = b / θ Θ 1 } . Θ 2 = { ( a , b ) | ( a / θ , b / θ ) ∈ Θ 1 } Now, to show that Θ 2 ∈ Con ( A ) . Let ( a , b ) , ( c , d ) ∈ Θ 2 . So ( a / θ , b / θ ) , ( c / θ , d / θ ) ∈ Θ 1 . Since Θ 1 ∈ Con ( A / θ ) ; ( a / θ + c / θ , b / θ + d / θ ) ∈ Θ 1 and ( a / θ ∗ c / θ , b / θ ∗ d / θ ) ∈ Θ 1 . Which implies that ( a + c , b + d ) ∈ Θ 2 and ( a ∗ c , b ∗ d ) ∈ Θ 2 . Let ( a , b ) ∈ Θ 2 and x ∗ y ≤ a ∗ b . So ( a / θ , b / θ ) ∈ Θ 1 . By definition (4.5) ; x ∗ y / θ ≤ a ∗ b / θ ⇔ x ∗ y ≤ a ∗ b . Since Θ 1 is congruence; ( x / θ , y / θ ) ∈ Θ 1 . Whence, ( x , y ) ∈ Θ 2 . Hence, Θ 2 ∈ Con ( A ) . Let ( a , b ) ∈ θ . Therefore, a / θ = b / θ . a / θ Θ 1 = b / θ Θ 1 . ( a , b ) ∈ Θ 2 . Therefore, θ ⊆ Θ 2 ⊆ ∇ A . Hence Θ 2 ∈ [ θ , ∇ A ] . Now to show that f ( Θ 2 ) = Θ 1 . To show that Θ 2 / θ = Θ 1 . Let ( a / θ , b / θ ) ∈ Θ 2 / θ . ( a , b ) ∈ Θ 2 . a / θ Θ 1 = b / θ Θ 1 . ( a / θ , b / θ ) ∈ Θ 1 . Therefore, Θ 2 / θ ⊆ Θ 1 . Let ( a / θ , b / θ ) ∈ Θ 1 . Therefore, a / θ Θ 1 = b / θ Θ 1 . ( a , b ) ∈ Θ 2 . ( a / θ , b / θ ) ∈ Θ 2 / θ . Therefore, Θ 1 ⊆ Θ 2 / θ . Hence, Θ 2 / θ = Θ 1 . Thus, f ( Θ 2 ) = Θ 1 . Therefore, f is onto. To show that Θ 1 ⊆ Θ 2 ⇔ f ( Θ 1 ) ⊆ f ( Θ 2 ) . Let Θ 1 , Θ 2 ∈ [ θ , ∇ A ] . Suppose Θ 1 ⊆ Θ 2 . Let ( a / θ , b / θ ) ∈ Θ 1 / θ . ( a , b ) ∈ Θ 1 . ( a , b ) ∈ Θ 2 . ( a / θ , b / θ ) ∈ Θ 2 / θ . Therefore, Θ 1 / θ ⊆ Θ 2 / θ . Hence, f ( Θ 1 ) ⊆ f ( Θ 2 ) . Conversely, suppose f ( Θ 1 ) ⊆ f ( Θ 2 ) . Therefore, Θ 1 / θ ⊆ Θ 2 / θ . Let ( a , b ) ∈ Θ 1 . ( a / θ , b / θ ) ∈ Θ 1 / θ . ( a / θ , b / θ ) ∈ Θ 2 / θ . ( a , b ) ∈ Θ 2 . Therefore, Θ 1 ⊆ Θ 2 . Whence, Θ 1 ⊆ Θ 2 ⇔ f ( Θ 1 ) ⊆ f ( Θ 2 ) . Therefore, f is an ordered isomorphism of [ θ , ∇ A ] onto Con ( A / θ ) . Hence, [ θ , ∇ A ] ≅ Con ( A / θ ) . 5. Conclusion This paper presented the study of congruence relation on autometrized algebras. We introduced in normal autometrized algebra satisfying some conditions the set of all congruence relations form a complete sublattice of the set of all equivalence relations. We also examined that a congruence-permutable autometrized algebra is congruence-modular. We explored several fundamental properties related to congruence relations. 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Tilahun GY, Parimi RK, Melesse MH: Equivalent conditions of normal autometrized l-algebra. Res. Math. 2023c; 10 (1): 2215037. Publisher Full Text Tilahun GY, Parimi RK, Melesse MH: Structure of an autometrized algebra. Res. Math. 2023d; 10 (1): 2192856. Publisher Full Text Comments on this article Comments (0) Version 2 VERSION 2 PUBLISHED 03 Jan 2025 ADD YOUR COMMENT Comment Author details Author details Department of Mathematics, Assosa University, Asosa, Benishangul-Gumuz, Ethiopia Gebrie Yeshiwas Tilahun Roles: Writing – Review & Editing Competing interests No competing interests were disclosed. Grant information The author(s) declared that no grants were involved in supporting this work. Article Versions (2) version 2 Revised Published: 22 Aug 2025, 14:22 https://doi.org/10.12688/f1000research.159591.2 version 1 Published: 03 Jan 2025, 14:22 https://doi.org/10.12688/f1000research.159591.1 Copyright © 2025 Tilahun GY. 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F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.186585.r408526 ) The direct URL for this report is: https://f1000research.com/articles/14-22/v2#referee-response-408526 NOTE: it is important to ensure the information in square brackets after the title is included in this citation. Close Copy Citation Details Reviewer Report 24 Sep 2025 Alachew Amaneh Mechderso , University of KabriDahar, Kabri Dahar, Ethiopia Approved VIEWS 0 https://doi.org/10.5256/f1000research.186585.r408526 The paper titled "Congruency, Homomorphism and Isomorphism on Autometrized Algebra" delivers its content in a clear, organized, and coherent way. The authors effectively break down and explain complex algebraic concepts, using precise language that enhances understanding. Their structured presentation makes ... Continue reading READ ALL The paper titled "Congruency, Homomorphism and Isomorphism on Autometrized Algebra" delivers its content in a clear, organized, and coherent way. The authors effectively break down and explain complex algebraic concepts, using precise language that enhances understanding. Their structured presentation makes the material both informative and approachable for readers with prior knowledge in the field. Is the work clearly and accurately presented and does it cite the current literature? Yes Is the study design appropriate and is the work technically sound? Yes Are sufficient details of methods and analysis provided to allow replication by others? Yes If applicable, is the statistical analysis and its interpretation appropriate? Yes Are all the source data underlying the results available to ensure full reproducibility? Yes Are the conclusions drawn adequately supported by the results? Yes Competing Interests: No competing interests were disclosed. Reviewer Expertise: fuzzy Algebra, Fuzzy Uniform Toplogy I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard. Close READ LESS CITE CITE HOW TO CITE THIS REPORT Mechderso AA. Reviewer Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.186585.r408526 ) The direct URL for this report is: https://f1000research.com/articles/14-22/v2#referee-response-408526 NOTE: it is important to ensure the information in square brackets after the title is included in all citations of this article. COPY CITATION DETAILS Report a concern Respond or Comment COMMENT ON THIS REPORT Version 1 VERSION 1 PUBLISHED 03 Jan 2025 Views 0 Cite How to cite this report: Korma SG. Reviewer Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.175346.r358302 ) The direct URL for this report is: https://f1000research.com/articles/14-22/v1#referee-response-358302 NOTE: it is important to ensure the information in square brackets after the title is included in this citation. Close Copy Citation Details Reviewer Report 20 Jun 2025 Sileshe Gone Korma , Hawassa University, Hawassa, Ethiopia Approved VIEWS 0 https://doi.org/10.5256/f1000research.175346.r358302 This paper investigates the structure and properties of congruence relations in autometrized algebras. It establishes that, under specific conditions, the congruence relations in a normal autometrized algebra form a complete sublattice within the equivalence relations. The study also shows that ... Continue reading READ ALL This paper investigates the structure and properties of congruence relations in autometrized algebras. It establishes that, under specific conditions, the congruence relations in a normal autometrized algebra form a complete sublattice within the equivalence relations. The study also shows that congruence-permutable autometrized algebras exhibit congruence-modularity. Key fundamental properties of congruence relations are explored, including the introduction of the kernel of a homomorphism as a congruence relation. Additionally, the paper examines homomorphism, isomorphism, and correspondence theorems in the context of autometrized algebra, utilizing the framework of congruence. Authors should include the following corrections Comments Section 1 Defnition 2.4 (ii); ( B, ≤) is subposet of A , and ≤ is translation invariant, that is, ∀a, b, c ∈ A; a ≤ b ⇒ a+c ≤ b+c Section 4 Theorem 4.1; line 2; f(a) = f(b) . Theorem 4.2; item (iii); line 1 and 2; make Ker f and make (x, y) ∈ ker f . Theorem 4.3; line 12; make ker f= a, b ∈A×A| a, b ∈θ . Theorem 4.7; line 1 and 2; substitute φ by ϕ . Theorem 4.10; equation 2; make x*y . Theorem 4.13; line 10; ϕ θ = ψ θ . Section 5 Conclusion; line 4; make kernel. Is the work clearly and accurately presented and does it cite the current literature? Yes Is the study design appropriate and is the work technically sound? Yes Are sufficient details of methods and analysis provided to allow replication by others? Yes If applicable, is the statistical analysis and its interpretation appropriate? Not applicable Are all the source data underlying the results available to ensure full reproducibility? No source data required Are the conclusions drawn adequately supported by the results? Yes Competing Interests: No competing interests were disclosed. Reviewer Expertise: Algebra specifically in lattices and fuzzy algebra I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard. Close READ LESS CITE CITE HOW TO CITE THIS REPORT Korma SG. Reviewer Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.175346.r358302 ) The direct URL for this report is: https://f1000research.com/articles/14-22/v1#referee-response-358302 NOTE: it is important to ensure the information in square brackets after the title is included in all citations of this article. COPY CITATION DETAILS Report a concern Author Response 05 Sep 2025 Gebrie Yeshiwas Tilahun , Department of Mathematics, Assosa University, Asosa, Ethiopia 05 Sep 2025 Author Response We incorporated all suggested comments and submitted the corrected version. Competing Interests: No competing interests were disclosed. We incorporated all suggested comments and submitted the corrected version. We incorporated all suggested comments and submitted the corrected version. Competing Interests: No competing interests were disclosed. Close Report a concern Respond or Comment COMMENTS ON THIS REPORT Author Response 05 Sep 2025 Gebrie Yeshiwas Tilahun , Department of Mathematics, Assosa University, Asosa, Ethiopia 05 Sep 2025 Author Response We incorporated all suggested comments and submitted the corrected version. Competing Interests: No competing interests were disclosed. We incorporated all suggested comments and submitted the corrected version. We incorporated all suggested comments and submitted the corrected version. Competing Interests: No competing interests were disclosed. Close Report a concern COMMENT ON THIS REPORT Views 0 Cite How to cite this report: Badawy A. Reviewer Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.175346.r362466 ) The direct URL for this report is: https://f1000research.com/articles/14-22/v1#referee-response-362466 NOTE: it is important to ensure the information in square brackets after the title is included in this citation. Close Copy Citation Details Reviewer Report 25 Feb 2025 Abdelmohsen Badawy , Mathematics, Faculty of science, Tanta University, Tanta, Gharbia Governorate, Egypt Approved with Reservations VIEWS 0 https://doi.org/10.5256/f1000research.175346.r362466 Dear Professor Thank you. I go through the paper, it has small new results, it needs a major improve English, improve the presentation of the paper, and the author must add the basic concepts which are needed as equivalent ... Continue reading READ ALL Dear Professor Thank you. I go through the paper, it has small new results, it needs a major improve English, improve the presentation of the paper, and the author must add the basic concepts which are needed as equivalent class, quotient set and explain how the set Eq of all equivalent relation forms a lattice. Also Must construct example to explore the results. I listed some remarks in the attached report. Is the work clearly and accurately presented and does it cite the current literature? Partly Is the study design appropriate and is the work technically sound? Partly Are sufficient details of methods and analysis provided to allow replication by others? Partly If applicable, is the statistical analysis and its interpretation appropriate? Partly Are all the source data underlying the results available to ensure full reproducibility? Partly Are the conclusions drawn adequately supported by the results? Partly Competing Interests: No competing interests were disclosed. Reviewer Expertise: Algebra- Lattice Theory - MS-algebras I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard, however I have significant reservations, as outlined above. Close READ LESS CITE CITE HOW TO CITE THIS REPORT Badawy A. Reviewer Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.175346.r362466 ) The direct URL for this report is: https://f1000research.com/articles/14-22/v1#referee-response-362466 NOTE: it is important to ensure the information in square brackets after the title is included in all citations of this article. COPY CITATION DETAILS Report a concern Author Response 05 Sep 2025 Gebrie Yeshiwas Tilahun , Department of Mathematics, Assosa University, Asosa, Ethiopia 05 Sep 2025 Author Response We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the ... Continue reading We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the proof of the set of all equivalent relation forms a complete lattice is incorporated. 3. English and the presentation of the paper are improved. We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the proof of the set of all equivalent relation forms a complete lattice is incorporated. 3. English and the presentation of the paper are improved. Competing Interests: No competing interests were disclosed. Close Report a concern Respond or Comment COMMENTS ON THIS REPORT Author Response 05 Sep 2025 Gebrie Yeshiwas Tilahun , Department of Mathematics, Assosa University, Asosa, Ethiopia 05 Sep 2025 Author Response We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the ... Continue reading We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the proof of the set of all equivalent relation forms a complete lattice is incorporated. 3. English and the presentation of the paper are improved. We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the proof of the set of all equivalent relation forms a complete lattice is incorporated. 3. English and the presentation of the paper are improved. Competing Interests: No competing interests were disclosed. Close Report a concern COMMENT ON THIS REPORT Comments on this article Comments (0) Version 2 VERSION 2 PUBLISHED 03 Jan 2025 ADD YOUR COMMENT Comment keyboard_arrow_left keyboard_arrow_right Open Peer Review Reviewer Status info_outline Alongside their report, reviewers assign a status to the article: Approved The paper is scientifically sound in its current form and only minor, if any, improvements are suggested Approved with reservations A number of small changes, sometimes more significant revisions are required to address specific details and improve the papers academic merit. Not approved Fundamental flaws in the paper seriously undermine the findings and conclusions Reviewer Reports Invited Reviewers 1 2 3 Version 2 (revision) 22 Aug 25 read Version 1 03 Jan 25 read read Abdelmohsen Badawy , Tanta University, Tanta, Egypt Sileshe Gone Korma , Hawassa University, Hawassa, Ethiopia Alachew Amaneh Mechderso , University of KabriDahar, Kabri Dahar, Ethiopia Comments on this article All Comments (0) Add a comment Sign up for content alerts Sign Up You are now signed up to receive this alert Browse by related subjects keyboard_arrow_left Back to all reports Reviewer Report 0 Views copyright © 2025 Mechderso A. This is an open access peer review report distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 24 Sep 2025 | for Version 2 Alachew Amaneh Mechderso , University of KabriDahar, Kabri Dahar, Ethiopia 0 Views copyright © 2025 Mechderso A. This is an open access peer review report distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. format_quote Cite this report speaker_notes Responses (0) Approved info_outline Alongside their report, reviewers assign a status to the article: Approved The paper is scientifically sound in its current form and only minor, if any, improvements are suggested Approved with reservations A number of small changes, sometimes more significant revisions are required to address specific details and improve the papers academic merit. Not approved Fundamental flaws in the paper seriously undermine the findings and conclusions The paper titled "Congruency, Homomorphism and Isomorphism on Autometrized Algebra" delivers its content in a clear, organized, and coherent way. The authors effectively break down and explain complex algebraic concepts, using precise language that enhances understanding. Their structured presentation makes the material both informative and approachable for readers with prior knowledge in the field. Is the work clearly and accurately presented and does it cite the current literature? Yes Is the study design appropriate and is the work technically sound? Yes Are sufficient details of methods and analysis provided to allow replication by others? Yes If applicable, is the statistical analysis and its interpretation appropriate? Yes Are all the source data underlying the results available to ensure full reproducibility? Yes Are the conclusions drawn adequately supported by the results? Yes Competing Interests No competing interests were disclosed. Reviewer Expertise fuzzy Algebra, Fuzzy Uniform Toplogy I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard. reply Respond to this report Responses (0) Mechderso AA. Peer Review Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.186585.r408526) NOTE: it is important to ensure the information in square brackets after the title is included in this citation. The direct URL for this report is: https://f1000research.com/articles/14-22/v2#referee-response-408526 keyboard_arrow_left Back to all reports Reviewer Report 0 Views copyright © 2025 Korma S. This is an open access peer review report distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 20 Jun 2025 | for Version 1 Sileshe Gone Korma , Hawassa University, Hawassa, Ethiopia 0 Views copyright © 2025 Korma S. This is an open access peer review report distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. format_quote Cite this report speaker_notes Responses (1) Approved info_outline Alongside their report, reviewers assign a status to the article: Approved The paper is scientifically sound in its current form and only minor, if any, improvements are suggested Approved with reservations A number of small changes, sometimes more significant revisions are required to address specific details and improve the papers academic merit. Not approved Fundamental flaws in the paper seriously undermine the findings and conclusions This paper investigates the structure and properties of congruence relations in autometrized algebras. It establishes that, under specific conditions, the congruence relations in a normal autometrized algebra form a complete sublattice within the equivalence relations. The study also shows that congruence-permutable autometrized algebras exhibit congruence-modularity. Key fundamental properties of congruence relations are explored, including the introduction of the kernel of a homomorphism as a congruence relation. Additionally, the paper examines homomorphism, isomorphism, and correspondence theorems in the context of autometrized algebra, utilizing the framework of congruence. Authors should include the following corrections Comments Section 1 Defnition 2.4 (ii); ( B, ≤) is subposet of A , and ≤ is translation invariant, that is, ∀a, b, c ∈ A; a ≤ b ⇒ a+c ≤ b+c Section 4 Theorem 4.1; line 2; f(a) = f(b) . Theorem 4.2; item (iii); line 1 and 2; make Ker f and make (x, y) ∈ ker f . Theorem 4.3; line 12; make ker f= a, b ∈A×A| a, b ∈θ . Theorem 4.7; line 1 and 2; substitute φ by ϕ . Theorem 4.10; equation 2; make x*y . Theorem 4.13; line 10; ϕ θ = ψ θ . Section 5 Conclusion; line 4; make kernel. Is the work clearly and accurately presented and does it cite the current literature? Yes Is the study design appropriate and is the work technically sound? Yes Are sufficient details of methods and analysis provided to allow replication by others? Yes If applicable, is the statistical analysis and its interpretation appropriate? Not applicable Are all the source data underlying the results available to ensure full reproducibility? No source data required Are the conclusions drawn adequately supported by the results? Yes Competing Interests No competing interests were disclosed. Reviewer Expertise Algebra specifically in lattices and fuzzy algebra I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard. reply Respond to this report Responses (1) Author Response 05 Sep 2025 Gebrie Yeshiwas Tilahun, Department of Mathematics, Assosa University, Asosa, Ethiopia We incorporated all suggested comments and submitted the corrected version. View more View less Competing Interests No competing interests were disclosed. reply Respond Report a concern Korma SG. Peer Review Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.175346.r358302) NOTE: it is important to ensure the information in square brackets after the title is included in this citation. The direct URL for this report is: https://f1000research.com/articles/14-22/v1#referee-response-358302 keyboard_arrow_left Back to all reports Reviewer Report 0 Views copyright © 2025 Badawy A. This is an open access peer review report distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 25 Feb 2025 | for Version 1 Abdelmohsen Badawy , Mathematics, Faculty of science, Tanta University, Tanta, Gharbia Governorate, Egypt 0 Views copyright © 2025 Badawy A. This is an open access peer review report distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. format_quote Cite this report speaker_notes Responses (1) Approved With Reservations info_outline Alongside their report, reviewers assign a status to the article: Approved The paper is scientifically sound in its current form and only minor, if any, improvements are suggested Approved with reservations A number of small changes, sometimes more significant revisions are required to address specific details and improve the papers academic merit. Not approved Fundamental flaws in the paper seriously undermine the findings and conclusions Dear Professor Thank you. I go through the paper, it has small new results, it needs a major improve English, improve the presentation of the paper, and the author must add the basic concepts which are needed as equivalent class, quotient set and explain how the set Eq of all equivalent relation forms a lattice. Also Must construct example to explore the results. I listed some remarks in the attached report. Is the work clearly and accurately presented and does it cite the current literature? Partly Is the study design appropriate and is the work technically sound? Partly Are sufficient details of methods and analysis provided to allow replication by others? Partly If applicable, is the statistical analysis and its interpretation appropriate? Partly Are all the source data underlying the results available to ensure full reproducibility? Partly Are the conclusions drawn adequately supported by the results? Partly Competing Interests No competing interests were disclosed. Reviewer Expertise Algebra- Lattice Theory - MS-algebras I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard, however I have significant reservations, as outlined above. reply Respond to this report Responses (1) Author Response 05 Sep 2025 Gebrie Yeshiwas Tilahun, Department of Mathematics, Assosa University, Asosa, Ethiopia We incorporated all comments as the reviewer's suggestions. Such as: 1. the definition of as equivalent class and quotient set are given. Also, they are illustrated by examples. 2. the proof of the set of all equivalent relation forms a complete lattice is incorporated. 3. English and the presentation of the paper are improved. View more View less Competing Interests No competing interests were disclosed. reply Respond Report a concern Badawy A. Peer Review Report For: Congruency, Homomorphism and Isomorphism on Autometrized Algebras [version 2; peer review: 2 approved, 1 approved with reservations] . F1000Research 2025, 14 :22 ( https://doi.org/10.5256/f1000research.175346.r362466) NOTE: it is important to ensure the information in square brackets after the title is included in this citation. The direct URL for this report is: https://f1000research.com/articles/14-22/v1#referee-response-362466 Alongside their report, reviewers assign a status to the article: Approved - the paper is scientifically sound in its current form and only minor, if any, improvements are suggested Approved with reservations - A number of small changes, sometimes more significant revisions are required to address specific details and improve the papers academic merit. 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