EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 4180-4194 ISSN 1307-5543 – ejpam.com Published by New York Business Global Implicative Negatively Partially Ordered Ternary Semigroups Kansada Nakwan1, Panuwat Luangchaisri1, Thawhat Changphas1,∗ 1 Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand Abstract. In this paper, we introduce and examine the notion of implicative negatively partially ordered ternary semigroups, for short implicative n.p.o. ternary semigroup, which include an ele- ment that serves as both the greatest element and the multiplicative identity. We study the notion of implicative homomorphisms between these ternary semigroups, and have that any implicative homomorphism is a homomorphism. Let φ : T1 −→ T2 be an implicative homomorphism from a commutative implicative n.p.o. ternary semigroup T1 onto T2. We construct a quotient commuta- tive implicative n.p.o. ternary semigroup T1/ρKerφ, where ρKerφ is a congruence relation defined by Kerφ. We prove that there exists an implicative homomorphism ψ such that ψ ◦ η = φ, where η is a canonical homomorphism from T1 onto T1/ρKerφ. 2020 Mathematics Subject Classifications: 20M12, 06F99, 06A06, 06A12 Key Words and Phrases: Implicative semilattice, Implicative n.p.o. (negatively partially or- dered) ternary semigroup, Implicative homomorphism, Filter 1. Introduction An implicative semilattice (L,≤,∧, ∗) consists of a non-empty set L, a partial order ≤, a greatest lower bound (with respect to ≤) ∧, and a binary multiplication ∗ such that z ≤ x ∗ y ⇔ z ∧ x ≤ y for any x, y, z ∈ L. The notion have been explored in the work of W. C. Nemitz in [13], the author investigated relationships between homomorphisms of implicative semilattices and their kernels. T. S. Blyth in [1] generalized some results of Nemitz by introducing the notion of Brouwerian semigroups. The results of Blyth [1] have been generalized further by M. F. Janowitz and C. S. Johnson Jr in [9]. In [10] Y. B. Jun introduced a special set in an implicative semigroup, from which the author derived an equivalent condition ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5511 Email addresses: kansada.n@kkumail.com (K. Nakwan), panulu@kku.ac.th (P. Luangchaisri), thacha@kku.ac.th (T. Changphas) https://www.ejpam.com 4180 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4181 for a filter and proved that a filter can be represented by the union of such sets. In [14] D. A. Romano introduced the concept of an anti-filter within implicative semigroups and provided several equivalent conditions for when the special inhabited proper subset of an implicative semigroup qualifies as an ordered anti-filter. In [6], a partially ordered semigroup (S, ·,≤) consists of a semigroup (S, ·) together with a partial order ≤ on S that is compatible with the semigroup operation, that is for any x, y, z ∈ S, if x ≤ y then xz ≤ yz and zx ≤ zy. A partially ordered semigroup (S, ·,≤) is called a negatively partially ordered semigroup, for short n.p.o. semigroup, if for any x, y, z ∈ S, xy ≤ x and xy ≤ y. An n.p.o. semigroup (S, ·,≤) with an additional binary multiplication ∗ such that z ≤ x ∗ y ⇔ zx ≤ y for any x, y, z ∈ S is called an implicative n.p.o. semigroup. Inspired by the works of Nemitz [13] and Blyth [1], In [2], M. W. Chan and K. P. Shum introduced the concept of implicative negatively partially ordered semigroups and explored the homomorphisms between these structures. Their work generalizes and expands upon some results by Nemitz regarding implicative semilattices. A non-empty set T together with a ternary multiplication [ ] defined on the set T is called a ternary semigroup if the ternary multiplication [ ] satisfies the following associative law: [[xyz]uv] = [x[yzu]v] = [xy[zuv]] for all x, y, z, u, v ∈ T . This notion has been introduced and studied by S. Banach (cf. J. Los [12]) who is credited with an example of a ternary semigroup which does not reduce to a semigroup. D. H. Lehmer in [11] studied a system called triplexes which turns out to be a commutative ternary group. J. Los in [12] proved that any ternary semigroup can be embedded in a semigroup. F. M. Sioson in [17] considered ideals and radicals of ternary semigroups; various concepts such as primality, semiprimality, and regularity were intro- duced. A. Chronowski in [3] investigated ternary semigroups of mappings of sets; these algebraic structures are used for constructing the natural examples of ternary algebras. V. N. Dixit and S. Dewan in [5] studied quasi-ideals and bi-ideals of ternary semigroups; the authors proved that every quasi-ideals is a bi-ideal and gave several examples in different contexts to prove that the converse is not true in general. M. L. Santiago and S. Sri Bala in [15] investigated regular ternary semigroups and studied several properties. A ternary semigroup (T, [ ]) is called an ordered ternary semigroup if there is a partial order ≤ such that for any a, b, x, y ∈ T , if a ≤ b then [axy] ≤ [bxy], [xay] ≤ [xby], [xya] ≤ [xyb]. A. Iampan in ([7], [8]) discussed ordered ternary semigroups and characterized the minimality and maximality of ordered lateral ideals in ordered ternary semigroups; the author also considered ideal extensions. V. R. Daddi and Y. S. Pawar in [4] introduced and studied quasi-ideals and bi-ideals in ordered ternary semigroups. The purpose of this paper is to introduce and study the notion of implicative n.p.o. ternary semigroups. Main idea of this work is inspired by [2]. We apply concept of im- plicative n.p.o. semigroups to establish implicative n.p.o. ternary semigroups. In addition, we introduce and study implicative homomorphisms related to homomorphisms between implicative n.p.o. ternary semigroups. Moreover, we construct quotient commutative implicative n.p.o. ternary semigroups, and prove the homomorphism theorem. K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4182 2. Implicative ternary semigroups We start this section with the definition of n.p.o. ternary semigroups. Definition 1. An n.p.o. ternary semigroup (T, [ ],≤) consists of a non-empty set T together with a partial order ≤ and a ternary multiplication [ ] on T such that the following conditions are satisfied: for any x, y, z, u, v ∈ T , (1) [[xyz]uv] = [x[yzu]v] = [xy[zuv]]; (2) if x ≤ y, then [xuv] ≤ [yuv], [uxv] ≤ [uyv] and [uvx] ≤ [uvy]; (3) [xyz] ≤ x, [xyz] ≤ y and [xyz] ≤ z. Definition 2. An n.p.o. ternary semigroup (T, [ ],≤) with an additional ternary multi- plication [ ]∗ on T such that u ≤ [xyz]∗ ⇔ [uxy] ≤ z for any x, y, z, u ∈ T is called an implicative n.p.o. ternary semigroup. The ternary multiplication [ ]∗ is called a ternary implication. An element 1 of a ternary semigroup (T, [ ]) is a multiplicative identity of T if [11x] = [1x1] = [x11] = x for any x ∈ T . The following example shows that the greatest element of an implicative n.p.o. ternary semigroup need not be identity. Example 1. Let T = {1, a, 0}. Let us consider the implicative n.p.o. ternary semigroup (T, [ ],≤, [ ]∗) with a ternary multiplication, a ternary implication, and an order relation defined on T as follows: [ ] 1 a 0 11 1 0 0 1a 0 0 0 10 0 0 0 [ ] 1 a 0 aa 0 0 0 a1 0 0 0 a0 0 0 0 [ ] 1 a 0 00 0 0 0 01 0 0 0 0a 0 0 0 [ ]∗ 1 a 0 11 1 1 1 1a 1 1 1 10 1 1 1 [ ]∗ 1 a 0 aa 1 1 1 a1 1 1 1 a0 1 1 1 [ ]∗ 1 a 0 00 1 1 1 01 1 1 1 0a 1 1 1 and ≤= {(0, 0), (1, 1), (a, a), (a, 1), (0, a), (0, 1)}. To express the calculation [x1x2x3] using a multiplication table, we place x1x2 in the first column and x3 in the first row. It is observed that 1 is the greatest element. However, 1 is not the identity since [11a] = 0 ̸= a. K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4183 The next example shows that not every n.p.o. ternary semigroup with identity admits the implicative structure. Example 2. Let T = {0, 1, a, b}. Let us consider the n.p.o. ternary semigroup (T, [ ],≤) with a ternary multiplication and an order relation defined on T as follows: [ ] 1 a b 0 11 1 a b 0 1a a 0 0 0 1b b 0 b 0 10 0 0 0 0 [ ] 1 a b 0 aa 0 0 0 0 a1 a 0 0 0 ab 0 0 0 0 a0 0 0 0 0 [ ] 1 a b 0 bb b 0 b 0 b1 b 0 b 0 ba 0 0 0 0 b0 0 0 0 0 [ ] 1 a b 0 00 0 0 0 0 01 0 0 0 0 0a 0 0 0 0 0b 0 0 0 0 and ≤= {(0, 0), (1, 1), (a, a), (b, b), (0, a), (0, b), (0, 1), (a, 1), (b, 1)}. Observed that 1 is the greatest element. Notice that T is not an implicative n.p.o. ternary semigroup. Indeed, if T is an implicative n.p.o. ternary semigroup with a ternary implica- tion [ ]∗, then a ≤ [1ab]∗ and b ≤ [1ab]∗ since [a1a] = 0 ≤ b and [b1a] = 0 ≤ b, respectively. It follows that [1ab]∗ = 1. As 1 ≤ [1ab]∗, we have a = [11a] ≤ b. This is a contradiction. An implicative n.p.o. ternary semigroup (T, [ ],≤, [ ]∗) is said to be commutative [16] if [xyz] = [yzx] = [zxy] = [yxz] = [zyx] = [xzy] for all elements x, y, z in T . The following example shows an infinite commutative im- plicative n.p.o. ternary semigroup with 1 as its greatest element. Example 3. Let (Z+, [ ]) be the ternary semigroup of positive integers with the ternary multiplication induced by usual multiplication. For a, b ∈ Z+, an order relation ≤ on Z+ is defined by a ≤ b⇔ b | a. Here, b|a means b divides a. We have that (Z+, [ ],≤) is a commutative n.p.o. ternary semigroup with 1 as its greatest element. Indeed: it is easy to see that (Z+, [ ]) is a commutative ternary semigroup. Let x, y, u, v ∈ Z+ with x ≤ y. Since y |x, there exists q ∈ Z+ such that x = qy. From xuv = q(yuv), it follows that yuv |xuv. Hence, [xuv] ≤ [yuv]. Similarly, we get [uxv] ≤ [uyv] and [uvx] ≤ [uvy]. Let x, y, z ∈ Z+. Since xyz = xyz, x |xyz, and so [xyz] ≤ x. Similarly, we get [xyz] ≤ y and [xyz] ≤ z. Since 1 |x for all x ∈ Z+, x ≤ 1 for all x ∈ Z+. Thus, 1 is the greatest element. Moreover, we have the ternary implication on Z+ defined by [xyz]∗ = z gcd(xy,z) for all x, y, z ∈ Z+. To see this, let x, y, z, u ∈ Z+ with gcd(xy, z) = d. Assume that [uxy] ≤ z; K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4184 then z d |u xy d . Since gcd(xyd , z d) = 1, z d |u. Therefore, u ≤ z d . Conversely, assume that u ≤ z d . Then there exists p ∈ Z+ such that u = p zd , and then uxy = (pxyd )z. This implies that z |uxy, that is, [uxy] ≤ z. The following theorem shows that an implicative n.p.o. ternary semigroup always contains the greatest element. Theorem 1. Let (T, [ ],≤, [ ]∗) be an implicative n.p.o. ternary semigroup. Then the following properties hold: (1) x ≤ [xxx]∗; (2) [xxx]∗ = [yyy]∗; (3) T contains the greatest element, namely [xxx]∗, for any x, y ∈ T . Proof. As [xxx] ≤ x, x ≤ [xxx]∗, so (1) is hold. Since [[xxx]∗yy] ≤ y, then [xxx]∗ ≤ [yyy]∗. Similarly, [yyy]∗ ≤ [xxx]∗. Thus, (2) holds. By y ≤ [yyy]∗ ≤ [xxx]∗, it follows that T contains greatest element, namely [xxx]∗. Hence, (3) holds. Let 1 be the greatest element of an n.p.o. ternary semigroup (T, [ ],≤, [ ]∗) if exists. It is observed that if 1 is the multiplicative identity then it can be verified that [xyz] = 1 if and only if x = y = z = 1 for any x, y, z ∈ T . Indeed, if x, y, z ∈ T such that [xyz] = 1 then 1 = [xyz] ≤ x ≤ 1; so x = 1. In a similar argument we can deduce that y = 1 and z = 1. Clearly, if x = y = z = 1 then [xyz] = 1. Throughout the rest of the paper, we deal with an implicative n.p.o. ternary semigroup with 1 which is both the greatest element and the multiplicative identity. The following theorem collects several properties of elements of implicative n.p.o. ternary semigroups. Theorem 2. Let (T, [ ],≤, [ ]∗) be an implicative n.p.o. ternary semigroup. Then for any x, y, z, u, v ∈ T , the following conditions hold: (1) x ≤ 1, [xxx]∗ = 1, x = [11x]∗; (2) x ≤ [yz[xyz]]∗; (3) x ≤ [xx[xxx]]∗; (4) x ≤ [yzx]∗; (5) if x ≤ y, then [yuv]∗ ≤ [xuv]∗ and [uvx]∗ ≤ [uvy]∗; (6) x ≤ y ⇔ [x1y]∗ = 1 ⇔ [1xy]∗ = 1; (7) [xy[zuv]∗]∗ = [[xyz]uv]∗ = [x[yzu]v]∗. K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4185 Proof. (1) It is clear that x ≤ 1 and [xxx]∗ = 1. As [x11] = x ≤ x, we get that x ≤ [11x]∗. Since [11x]∗ ≤ [11x]∗, we have [11x]∗ = [[11x]∗11] ≤ x. (2) From [xyz] ≤ [xyz], we get x ≤ [yz[xyz]]∗. (3) The assertion follows by (2). (4) This is clear because [xyz] ≤ x. (5) Assume that x ≤ y. Since [yuv]∗ ≤ [yuv]∗, [[yuv]∗yu] ≤ v. By assumption, it follows that [[yuv]∗xu] ≤ [[yuv∗]yu] ≤ v. Then [yuv]∗ ≤ [xuv]∗. Similarly, if x ≤ y, then [uvx]∗ ≤ [uvy]∗. (6) If x ≤ y, then [1x1] ≤ [11y]. Hence, 1 ≤ [x1[11y]]∗ = [x1y]∗ ≤ 1. Thus, [x1y]∗ = 1. Similarly, if [x1y]∗ = 1, then [1xy]∗ = 1. Conversely, if [1xy]∗ = 1, then 1 ≤ [1xy]∗, and so x = [11x] ≤ y. (7) Let s = [xy[zuv]∗]∗ and t = [[xyz]uv]∗. We have [sxy] ≤ [zuv]∗], and thus, [s[xyz]u] = [[sxy]zu] ≤ v. Hence, s ≤ [[xyz]uv]∗ = t. By [[txy]zu] = [t[xyz]u] ≤ v, it follows that [txy] ≤ [zuv]∗. Then t ≤ [xy[zuv]∗]∗ = s. Hence, s = t. Now, let s = [xy[zuv]∗]∗ and w = [x[yzu]v]∗. Then [sxy] ≤ [zuv]∗], and thus, [sx[yzu]] = [[sxy]zu] ≤ v, so s ≤ [x[yzu]v]∗ = w. As [[wxy]zu] = [wx[yzu]] ≤ v, we have [wxy] ≤ [zuv]∗. Then w ≤ [xy[zuv]∗]∗ = s. Thus, s = w. 3. Implicative homomorphisms We begin this section with the definition of implicative homomorphisms between im- plicative n.p.o. ternary semigroups. Definition 3. Let (T1, [ ]1,≤1, [ ]∗1) and (T2, [ ]2,≤2, [ ]∗2) be implicative n.p.o. ternary semigroups. A mapping φ : T1 −→ T2 from T1 onto T2 such that φ([xyz]∗1) = [φ(x)φ(y)φ(z)]∗2 for all x, y, z ∈ T1 is called an implicative homomorphism from T1 onto T2. To study the notion of quotient structures of implicative n.p.o. ternary semigroups, we need the concept of filters. Definition 4. Let (T, [ ],≤) be an n.p.o. ternary semigroup. A non-empty subset F of T is called a filter of T if the following coditions hold: (1) [xyz] ∈ F for any x, y, z ∈ F , that is F is a ternary subsemigroup of T ; (2) for any x, y ∈ T , if x ≤ y and x ∈ F , then y ∈ F . Example 4. Let T = {1, a, b, c, d}. Let us consider the n.p.o. ternary semigroup (T, [ ],≤) with a ternary multiplication [ ] and an order relation ≤ defined on T as follows: K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4186 [ ] 1 a b c d 11 1 a b c d 1a a a d c d 1b b d b d d 1c c c d c d 1d d d d d d [ ] 1 a b c d a1 a a d c d aa a a d c d ab d d d d d ac c c d c d ad d d d d d [ ] 1 a b c d b1 b d b d d ba d d d d d bb b d b d d bc d d d d d bd d d d d d [ ] 1 a b c d c1 c c d c d ca c c d c d cb d d d d d cc c c d c d cd d d d d d [ ] 1 a b c d d1 d d d d d da d d d d d db d d d d d dc d d d d d dd d d d d d and ≤ = {(1, 1), (a, a), (b, b), (c, c), (d, d), (b, 1), (a, 1), (c, a), (c, 1), (d, c), (d, a), (d, b), (d, 1)}. Observe that 1 is the greatest element. It is easy to verify that F1 = {1}, F2 = {1, a}, F3 = {1, b}, F4 = {1, a, c}, F5 = T are filters, but {1, a, b} is not a filter. Now, we investigate some properties of implicative homomorphisms. Theorem 3. Let (T1, [ ]1,≤1, [ ]∗1) and (T2, [ ]2,≤2, [ ]∗2) be implicative n.p.o. ternary semigroups. Let φ : T1 −→ T2 be an implicative homomorphism from T1 onto T2. Then the following conditions hold: (1) φ(1) = 1′, where 1 and 1′ are the identities as well as the greatest elements of T1 and of T2, respectively; (2) φ is isotonic, that is for any x, y ∈ T1, if x ≤1 y then φ(x) ≤2 φ(y); (3) φ is a (ternary semigroup) homomorphism (i.e., for any x, y, z ∈ T1, φ[xyz]1 = [φ(x)φ(y)φ(z)]2); (4) φ−1(1′) is a filter of T1, when φ −1(1′) = {x ∈ T1 : φ(x) = 1′}; (5) φ is an (ternary semigroup) isomorphism if and only if φ−1(1′) = {1}. K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4187 Proof. (1) By Theorem 2 (1), we have 1 = [111]∗1. Consequently, φ(1) = φ([111]∗1) = [φ(1)φ(1)φ(1)]∗2 = 1′. (2) If x, y ∈ T1 such that x ≤1 y, then by Theorem 2 (6) we get that [x1y]∗1 = 1. Hence, [φ(x)1′φ(y)]∗2 = [φ(x)φ(1)φ(y)]∗2 = φ([x1y]∗1) = φ(1) = 1′. By Theorem 2 (6), we have φ(x) ≤2 φ(y). (3) Let x, y, z ∈ T1; then φ(u) = [φ(x)φ(y)φ(z)]2 for some u ∈ T1. Since φ is an implicative homomorphism, we have [φ([xyz]1)φ(1)φ(u)]∗2 = φ([[xyz]11u]∗1) = φ([xy[z1u]1] ∗ 1) = [φ(x)φ(y)φ([z1u]1)] ∗ 2 = [φ(x)φ(y)[φ(z)φ(1)φ(u)]2] ∗ 2 = [[φ(x)φ(y)φ(z)]2φ(1)φ(u)]∗2 = [φ(u)φ(1)φ(u)]∗2 = 1′. Then, by Theorem 2 (6), φ([xyz]1) ≤2 [φ(x)φ(y)φ(z)]2. From [xyz]1 ≤1 [xyz]1, it follows that x ≤1 [yz[xyz]1] ∗ 1. By (2), φ(x) ≤2 φ([yz[xyz]1] ∗ 1) = [φ(y)φ(z)φ([xyz]1)] ∗ 2. That is, [φ(x)φ(y)φ(z)]2 ≤2 φ([xyz]1). Hence, φ([xyz]1) = [φ(x)φ(y)φ(z)]2. (4) Let x, y, z ∈ φ−1(1′), that is, φ(x) = φ(y) = φ(z) = 1′. By (3), φ([xyz]1) = [φ(x)φ(y)φ(z)]2 = [1′1′1′]2 = 1′. Thus, [xyz] ∈ φ−1(1′). Assume that x, y ∈ T1 such that x ≤1 y and x ∈ φ−1(1′). Then 1′ = φ(1) = φ([x1y]∗1) = [φ(x)φ(1)φ(y)]∗2 = [1′1′φ(y)]∗2 = φ(y), so y ∈ φ−1(1′). Hence, φ−1(1′) is a filter. (5) Assume that φ−1(1′) = {1}. Let x, y ∈ T1 be such that φ(x) = φ(y). Thus, φ([x1y]∗1) = [φ(x)φ(1)φ(y)]∗2 = [φ(x)1′φ(x)]∗2 = 1′. This means that [x1y]∗1 ∈ φ−1(1′), that is, [x1y]∗1 = 1. Then by Theorem 2 (6), we get that x ≤1 y. Similarly, we have [y1x]∗1 = 1, then y ≤1 x. Hence, x = y. On the other hand, assume that φ is an isomorphism. If x ∈ φ−1(1′), then φ(x) = 1′ = φ(1), so by assumption we have x = 1. Hence, φ−1(1′) = {1}. K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4188 4. Commutative implicative n.p.o. ternary semigroups Hereafter, we deal with commutative implicative n.p.o. ternary semigroups. Proposition 1. If F is a filter of a commutative implicative n.p.o. ternary semigroup (T, [ ],≤, [ ]∗), then 1 ∈ F . Proof. The assertion follows by 1 is the greatest element of T . Let (T, [ ]) be a ternary semigroup. An equivalence relation α on T is called a congru- ence if for any x, y, s, t ∈ T , if (x, y) ∈ α, then ([stx], [sty]), ([xst], [yst]), ([sxt], [syt]) ∈ α. Definition 5. Let (T, [ ],≤, [ ]∗) be a commutative implicative n.p.o. ternary semigroup, and let F be a filter of T . For any x, y ∈ T , the relation ρF defined on T as follows: xρF y ⇔ there exist a, b ∈ F such that [abx] ≤ y and [aby] ≤ x. Lemma 1. Let (T, [ ],≤, [ ]∗) be a commutative implicative n.p.o. ternary semigroup, and let F be a filter of T . Then the relation ρF is a congruence defined on T . Proof. Let x, y, z ∈ T . As 1 ∈ F and [11x] ≤ x, we have xρFx. If xρF y, then there exist a, b ∈ F such that [abx] ≤ y and [aby] ≤ x, hence, yρFx. Assume that xρF y and yρF z. Then there exist a, b, c, d ∈ F such that [abx] ≤ y and [aby] ≤ x, [cdy] ≤ z and [cdz] ≤ y. We have [[abc]dx] = [[cab]dx] = [c[abd]x] = [c[dab]x] = [cd[abx]] ≤ [cdy] ≤ z and [[abc]dz] = [ab[cdz]] ≤ [aby] ≤ x. Thus, xρF z. Therefore, ρF is an equivalence relation on T . Suppose that xρF y. Then there exist a, b ∈ F such that [abx] ≤ y and [aby] ≤ x. Let s, t ∈ T . We have [ab[xst]] = [[abx]st] ≤ [yst] and [ab[yst]] = [[aby]st] ≤ [xst]; so [xst]ρF [yst]. Similarly, [stx]ρF [sty] and [sxt]ρF [syt]. Hence, ρF is a congruence relation on T . Let (T, [ ],≤, [ ]∗) be a commutative implicative n.p.o. ternary semigroup, and let F be a filter of T . As usual, for each x ∈ T the corresponding element in T/ρF , denoted by [x]ρF , is the equivalence class [x]ρF = {y ∈ T : yρFx}, that is T/ρF = {[x]ρF : x ∈ T}. Define the ternary multiplication [[ ]] : T/ρF × T/ρF × T/ρF −→ T/ρF by [[[x]ρF [y]ρF [z]ρF ]] = [[xyz]]ρF for all [x]ρF , [y]ρF , [z]ρF ∈ T/ρF . Then (T/ρF , [[ ]]) is a commutative ternary semigroup. Indeed, let [x]ρF [y]ρF [z]ρF ∈ T/ρF . We have [[[x]ρF [y]ρF [z]ρF ]] = [[xyz]]ρF = [[yzx]]ρF = [[[y]ρF [z]ρF [x]ρF ]]. Similarly, we get [[[x]ρF [y]ρF [z]ρF ]] = [[[z]ρF [x]ρF [y]ρF ]], [[[x]ρF [y]ρF [z]ρF ]] = [[[y]ρF [x]ρF [z]ρF ]], [[[x]ρF [y]ρF [z]ρF ]] = [[[z]ρF [y]ρF [x]ρF ]], [[[x]ρF [y]ρF [z]ρF ]] = [[[x]ρF [z]ρF [y]ρF ]]. The order relation ⪯ on T/ρF is induced by the relation ≤ as follows: for any [x]ρF , [y]ρF ∈ T/ρF , defined ⪯ by [x]ρF ⪯ [y]ρF if for any a ∈ [x]ρF , b ∈ [y]ρF , there exist c, d ∈ F such that [cda] ≤ b. K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4189 Lemma 2. Let (T, [ ],≤, [ ]∗) be a commutative implicative n.p.o. ternary semigroup, and let F be a filter of T . The ordered relation ⪯ is a partial order on T/ρF . Proof. If x′ ∈ [x]ρF , then by 1 ∈ F we have [11x′] ≤ x′; so [x]ρF ⪯ [x]ρF . Assume [x]ρF ⪯ [y]ρF and [y]ρF ⪯ [x]ρF . Since x ∈ [x]ρF and y ∈ [x]ρF , there exist c1, d1, c2, d2 ∈ F such that [c1d1x] ≤ y and [c2d2y] ≤ x. We have [[c1d1c2]d2x] = [c1d1[c2d2x]] = [c1d1[xc2d2]] = [[c1d1x]c2d2] ≤ [yc2d2] ≤ y. Also, [[c1d1c2]d2y] = [c1d1[c2d2y]] ≤ [c1d1x] ≤ x. Thus, xρF y, that is, [x]ρF = [y]ρF . Assume that [x]ρF ⪯ [y]ρF and [y]ρF ⪯ [z]ρF . Let x′ ∈ [x]ρF , y′ ∈ [y]ρF and z′ ∈ [z]ρF . Then there exist c3, d3, c4, d4 ∈ F such that [c3d3x ′] ≤ y′ and [c4d4y ′] ≤ z′. From [[c3d3c4]d4x ′] = [c3d3[c4d4x ′]] = [c3d3[x ′c4d4]] = [[c3d3x ′]c4d4] ≤ [y′c4d4] = [c4d4y ′] ≤ z′ it follows that [x]ρF ⪯ [z]ρF . These show that ⪯ is a partial order on T/ρF . Lemma 3. Let (T, [ ],≤) be a commutative n.p.o. ternary semigroup, and let F be a filter of T . Then (T/ρF , [[ ]],⪯) is a commutative n.p.o. ternary semigroup. Proof. We have seen that (T/ρF , [[ ]]) is a commutative ternary semigroup. By Lemma 2, ⪯ is a partial order on T/ρF . Let [x]ρF , [y]ρF , [u]ρF , [v]ρF ∈ T/ρF be such that [x]ρF ⪯ [y]ρF . Let a ∈ [[xuv]]ρF and b ∈ [[yuv]]ρF . Since x ∈ [x]ρF , y ∈ [y]ρF , and [x]ρF ⪯ [y]ρF , there exist c, d ∈ F such that [cdx] ≤ y. From a ∈ [[xuv]]ρF , there exist c1, d1 ∈ F such that [c1d1[xuv]] ≤ a and [c1d1a] ≤ [xuv]. Similarly, by b ∈ [[yuv]]ρF , there exist c2, d2 ∈ F such that [c2d2[yuv]] ≤ b and [c2d2b] ≤ [yuv]. Consider: [[c2d2[cdc1]]d1a] = [c2d2[[cdc1]d1a]] = [c2d2[cd[c1d1a]]] ≤ [c2d2[cd[xuv]]] = [c2d2[[cdx]uv]] ≤ [c2d2[yuv]] ≤ b. Then [[c2d2[cdc1]]d1a] ≤ b, so [[xuv]]ρF ⪯ [[yuv]]ρF . Hence, [[[x]ρF [u]ρF [v]ρF ]] = [[xuv]]ρF ⪯ [[yuv]]ρF = [[[y]ρF [u]ρF [v]ρF ]]. Similarly, [[[u]ρF [x]ρF [v]ρF ]] ⪯ [[[u]ρF [y]ρF [v]ρF ]], and [[[u]ρF [v]ρF [x]ρF ]] ⪯ [[[u]ρF [v]ρF [y]ρF ]]. Let [x]ρF , [y]ρF , [z]ρF ∈ T/ρF . To show that [[[x]ρF [y]ρF [z]ρF ]] ⪯ [x]ρF , let a′ ∈ [[xyz]]ρF , b′ ∈ [x]ρF . As [xyz]ρFa ′, there exist c1, d1 ∈ F such that [c1d1[xyz]] ≤ a′ and [c1d1a ′] ≤ [xyz]. By xρF b ′, there exist c2, d2 ∈ F such that [c2d2x] ≤ b′ and [c2d2b ′] ≤ x. We have K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4190 [[c1d1c2]d2a ′] = [c1d1[c2d2a ′]] = [c1d1[a ′c2d2]] = [[c1d1a ′]c2d2] ≤ [[xyz]c2d2] = [c2d2[xyz]] = [[c2d2x]yz] ≤ [b′yz] ≤ b′. Then [[c1d1c2]d2a ′] ≤ b′; hence, [[xyz]]ρF ⪯ [x]ρF . That is [[[x]ρF [y]ρF [z]ρF ]] ⪯ [x]ρF . In the same manner, we can prove that [[[x]ρF [y]ρF [z]ρF ]] ⪯ [y]ρF and [[[x]ρF [y]ρF [z]ρF ]] ⪯ [z]ρF . Lemma 4. Let (T, [ ],≤, [ ]∗) be a commutative implicative n.p.o. ternary semigroup, and let F be a filter of T . Then (T/ρF , [[ ]],⪯, [[ ]]∗) is a commutative implicative n.p.o. ternary semigroup with the ternary implication [[[x]ρF [y]ρF [z]ρF ]]∗ = [[xyz]∗]ρF for all [x]ρF , [y]ρF , [z]ρF ∈ T/ρF . Proof. We first show that [[ ]]∗ is well-defined on T/ρF . Let [x]ρF , [y]ρF , [z]ρF , [x′]ρF , [y′]ρF , [z′]ρF ∈ T/ρF be such that [x]ρF = [x′]ρF , [y]ρF = [y′]ρF and [z]ρF = [z′]ρF . Then there exist c1, d1, c2, d2, c3, d3 ∈ F such that [c1d1x] ≤ x′, [c1d1x ′] ≤ x, [c2d2y] ≤ y′, [c2d2y ′] ≤ y, [c3d3z] ≤ z′ and [c3d3z ′] ≤ z. We denote [xyz]∗ by u and [x′y′z′]∗ by t. Since u ≤ [xyz]∗, [uxy] ≤ z. Using the associativity and commutativity of T , we have [[[[c1d1c2]d2c3]d3u]x′y′] ≤ z′. This shows that [[[[c1d1c2]d2c3]d3[xyz] ∗]] ≤ [x′y′z′]∗. In the same manner, we have [[[[c1d1c2]d2c3]d3[x ′y′z′]∗]] ≤ [xyz]∗. Thus we have shown that [xyz]∗ρF [x′y′z′]∗ and hence [[xyz]∗]ρF = [[x′y′z′]∗]ρF . Let [x]ρF , [y]ρF , [z]ρF , [u]ρF ∈ T/ρF be such that [[[u]ρF [x]ρF [y]ρF ]] ⪯ [z]ρF ; then [[uxy]]ρF ⪯ [z]ρF . To show that [u]ρF ⪯ [[xyz]∗]ρF , that is, [u]ρF ⪯ [[[x]ρF [y]ρF [z]ρF ]]∗. Let a ∈ [u]ρF and b ∈ [[xyz]∗]ρF . Since [[uxy]]ρF ⪯ [z]ρF , there exist c, d ∈ F such that [cd[uxy]] ≤ z. By a ∈ [u]ρF , there exist c1, d1 ∈ F such that [c1d1u] ≤ a and [c1d1a] ≤ u. Similarly, by b ∈ [[xyz]∗]ρF , there exist c2, d2 ∈ F such that [c2d2[xyz] ∗] ≤ b and [c2d2b] ≤ [xyz]∗. As [cd[uxy]] ≤ z, we have [cdu] ≤ [xyz]∗. Consider: [[c2d2[cdc1]]d1a] = [c2d2[[cdc1]d1a]] = [c2d2[cd[c1d1a]]] = [c2d2[cdu]] = [c2d2[xyz] ∗] ≤ b. Then [u]ρF ⪯ [[[x]ρF [y]ρF [z]ρF ]]∗. Conversely, suppose that [u]ρF ⪯ [[[x]ρF [y]ρF [z]ρF ]]∗; then [u]ρF ⪯ [[xyz]∗]ρF . To show that [[[u]ρF [x]ρF [y]ρF ]] ⪯ [z]ρF , that is, [uxy]ρF ⪯ [z]ρF . Let s ∈ [[uxy]]ρF and t ∈ [z]ρF . Since [xyz]∗ ∈ [[xyz]∗]ρF , u ∈ [u]ρF , and [u]ρF ⪯ [[xyz]∗]ρF , there exist c′, d′ ∈ F such that [c′d′u] ≤ [xyz]∗. By s ∈ [[uxy]]ρF , there exist c3, d3 ∈ F such that [c3d3[uxy]] ≤ s and [c3d3s] ≤ [uxy]. Similarly, by t ∈ [z]ρF , there exist c4, d4 ∈ F such that [c4d4z] ≤ t and [c4d4t] ≤ z. Consider: [[c5d5[c ′d′c3]]d3s] = [c5d5[[c ′d′c3]d3s]] K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4191 = [c5d5[c ′d′[c3d3s]]] ≤ [c5d5[c ′d′[uxy]]] = [c5d5[[c ′d′u]xy]] ≤ [c5d5z] ≤ t. Hence, [[[u]ρF [x]ρF [y]ρF ]] ⪯ [z]ρF . In view of Lemma 4, the mapping η : (T, [ ],≤, [ ]∗) −→ (T/ρF , [[ ]],⪯, [[ ]]∗) defined by x 7→ [x]ρF is a surjective mapping. Definition 6. Let φ be an implicative homomorphism from a commutative implicative n.p.o. ternary semigroup (T1, [ ]1,≤1, [ ]∗1) onto a commutative implicative n.p.o. ternary semigroup (T2, [ ]2,≤2, [ ]∗2). The kernel of φ, denoted by Kerφ, is defined to be the set Kerφ = {x ∈ T1 : φ(x) = 1′}, where 1 and 1′ are the identities and the greatest elements of T1 and T2, respectively. Remark 1. By Theorem 3, we get that Kerφ is a filter of T1. Lemma 5. Let (T, [ ],≤, [ ]∗) be a commutative implicative n.p.o. ternary semigroup and let F be a filter of T and ρF a congruence on T . Then the canonical homomorphism η : T −→ T/ρF is an implicative homomorphism from (T, [ ],≤, [ ]∗) onto (T/ρF , [[ ]],⪯, [[ ]]∗). Proof. If x, y, z ∈ T , then η([xyz]∗) = [[xyz]∗]ρF = [[[x]ρF [y]ρF [z]ρF ]]]∗ = [[η([x]ρF )η([y]ρF )η([z]ρF )]]∗. Hence, the assertion holds. Theorem 4. Let (T1, [ ]1,≤1, [ ]∗1) and (T2, [ ]2,≤2, [ ]∗2) be any two commutative implica- tive n.p.o. ternary semigroups, with 1 and 1′ are the identities and the greatest elements of T1 and T2, respectively. Let φ : T1 −→ T2 be an implicative homomorphism from T1 onto T2, with F = Kerφ, and let η : T1 −→ T1/ρF be a canonical homomorphism from T1 onto T1/ρF . Then there exists an implicative homomorphism ψ : T1/ρF −→ T2 from T1/ρF onto T2 such that the following diagram is commutative: (T1, [ ]1,≤1, [ ]∗1) (T2, [ ]2,≤2, [ ]∗2) (T1/ρF , [[ ]],⪯, [[ ]]∗) φ η ψ K. Nakwan, P. Luangchaisri, T. Changphas / Eur. J. Pure Appl. Math, 17 (4) (2024), 4180-4194 4192 Moreover, if Ker η = φ−1(1′) , then ψ is an implicative isomorphism, that is (T1/ρF , [[ ]],⪯, [[ ]]∗) ∼= (T2, [ ]2,≤2, [ ]∗2). Proof. Define ψ : T1/ρF −→ T2 by ψ([x]ρF ) = φ(x) for any x ∈ T1. We have ψ is well-defined. Indeed, let [x]ρF , [y]ρF ∈ T1/ρF be such that [x]ρF = [y]ρF . Since xρF y, there exist c, d ∈ F such that [cdx]1 ≤1 y and [cdy]1 ≤1 x. By Theorem 3, φ([cdx]1) ≤2 φ(y) and φ([cdy]1) ≤2 φ(x). Since c, d ∈ F , we have φ(c) = 1′ and φ(d) = 1′. From φ([cdx]1) = [φ(c)φ(d)φ(x)]2 = [1′1′φ(x)]2 = φ(x) and φ([cdy]1) = [φ(c)φ(d)φ(y)]2 = [1′1′φ(y)]2 = φ(y) it follows that φ(x) ≤2 φ(y) and φ(y) ≤2 φ(x). Hence, φ(x) = φ(y). Next, we have ψ is an implicative homomorphism. In fact, for [x]ρF , [y]ρF , [z]ρF ∈ T1/ρF , we have ψ([[[x]ρF [y]ρF [z]ρF ]]∗) = ψ([[xyz]∗1]ρF ) = φ([xyz]∗1) = [φ(x)φ(y)φ(z)]∗2 = [ψ([x]ρF )ψ([y]ρF )ψ([z]ρF )]∗2. If t ∈ T2, since φ is onto, then φ(s) = t for some s ∈ T1. Consequently, ψ(η(s)) = ψ([s]ρF ) = φ(s) = t. For any s ∈ T1, we have ψ ◦ η(s) = ψ(η(s)) = ψ([s]ρF ) = φ(s). This shows that the diagram is commutative. Finally, we assume that Ker η = φ−1(1′). To show that ψ is an implicative iso- morphism, we can only show ψ is one-to-one. Let [x]ρF , [y]ρF ∈ T1/ρF be such that ψ([x]ρF ) = ψ([y]ρF ); then φ(x) = φ(y). By Theorem 2 (6), we have φ([1xy]∗1) = [φ(1)φ(x)φ(y)]∗2 = [1′φ(x)φ(x)]∗2 = 1′. Hence, [1xy]∗1 ∈ φ−1(1′) = Ker η. Similarly, [1yx]∗1 ∈ φ−1(1′) = Ker η. 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