EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 11, No. 2, 2018, 476-492 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On semilattice congruences on hypersemigroups and on ordered hypersemigroups Niovi Kehayopulu Abstract. We prove that if H is an hypersemigroup (resp. ordered hypersemigroup) and σ is a semilattice congruence (resp. complete semilattice congruence) on H, then there exists a family A of proper prime ideals of H such that σ is the intersection of the semilattice congruences σI , I ∈ A (σI is the known relation defined by aσIb⇔ a, b ∈ I or a, b /∈ I). Furthermore, we study the relation between the semilattices of an ordered semigroup and the ordered hypersemigroup derived by the hyperoperations a ◦ b = {ab} and a ◦ b := {t ∈ S | t ≤ ab}. We introduce the concept of a pseudocomplete semilattice congruence as a semilattice congruence σ for which ≤⊆ σ and we prove, among others, that if (S, ·,≤) is an ordered semigroup, (S, ◦,≤) the hypersemigroup defined by t ∈ a ◦ b if and only if t ≤ ab and σ is a pseudocomplete semilattice congruence on (S, ·,≤), then it is a complete semilattice congruence on (S, ◦,≤). Illustrative examples are given. 2010 Mathematics Subject Classifications: 06F99, 20M99, 06F05 Key Words and Phrases: hypergroupoid, ordered hypersemigroup, semilattice congruence, complete (pseudocomplete) semilattice congruence, filter, prime ideal 1. Introduction Filters play an essential role in studying the structure of semigroups or ordered semi- groups. For a semigroup S –especially for decompositions of a semigroup S– an important role is played by the relation N which is the least semilattice congruence on S and leads to several important results concerning the structure of semigroups (cf. [13]). Using our computer program, we have proved in [11] that for an ordered semigroup S, N is not the least semilattice congruence on S in general, we introduced the concept of the complete semilattice congruence and proved that N is the least complete semilattice congruence on S. We always use the terms “prime”, “weakly prime” instead of “completely prime”, “prime” considered by Petrich in [13]. The present paper is based on our papers in [3, 11] and its aim is to show how we pass from semigroups (ordered semigroups) to hypersemi- groups (ordered hypersemigroups). The main result is that if H is an hypersemigroup (resp. ordered hypersemigroup) and σ a semilattice congruence (resp. complete semilat- tice congruence) on H, then there exists a family A of proper prime ideals of H such that σ = ⋂ I∈A σI, σI is the relation on H defined by aσI ⇔ a, b ∈ I or a, b /∈ I. Then Email address: nkehayop@math.uoa.gr (N. Kehayopulu) http://www.ejpam.com 476 c© 2018 EJPAM All rights reserved. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 477 we prove, among others, that if (S, ·,≤) is an ordered groupoid, and (S, ◦,≤) the ordered hypergroupoid defined by a ◦ b = {ab}, then σ is a semilattice (resp. complete semilat- tice) congruence on (S, ·,≤) if and only if it is a semilattice (resp. complete semilattice) congruence on (S, ◦,≤). As an immediate consequence, in an hypersemigroup H, the complete semilattice congruence N defined by xN y ⇔ N(x) = N(y) (where N(x) is the filter generated by the element x of H) cannot be the least semilattice congruence on H in general. For an ordered groupoid (S, ·,≤) we consider the hypergroupoid on S with the hyperoperation defined by a ◦ b =: {t ∈ S | t ≤ ab} and we prove that if σ is a semilattice congruence on (S, ◦,≤), then it is a semilattice congruence on (S, ·,≤) but the converse statement does not hold in general. In addition, if σ is a complete semilattice congruence on (S, ◦,≤), then it is a complete semilattice congruence on (S, ·,≤). It is natural to ask if there are semilattice congruences on an ordered groupoid (S, ·,≤) that are semilattice congruences on (S, ◦,≤) as well. On this purpose, we introduce the concept of pseudocom- plete semilattice congruences as the semilattice congruences σ such that ≤⊆ σ, and we prove that the pseudocomplete semilattice congruences on an ordered groupoid (S, ·,≤) are complete semilattice congruences on (S, ◦,≤). We could finally mention the following: If (S, ·) is a groupoid and “◦” the hyperoperation on S defined by a ◦ b := {ab}, then F is a filter of (S, ·) if and and only if it is a filter of (S, ◦); for an ordered groupoid (S, ·,≤) with the same hyperoperation, the filters of (S, ·,≤) and the filters of (S, ◦,≤) are also the same. If (S, ·,≤) is an ordered groupoid and “◦” the hyperoperation on S defined by t ∈ a ◦ b ⇔ t ≤ ab, the filters of (S, ◦,≤) are also filters of (S, ·,≤) but the converse statement does not hold in general. An hypergroupoid is a nonempty set H with an hyperoperation ◦ : H ×H → P∗(H) | (a, b)→ a ◦ b on H and an operation ∗ : P∗(H) × P∗(H) → P∗(H) | (A,B) → A ∗ B on P∗(H) (induced by the operation of H) such that A ∗ B = ⋃ (a,b)∈A×B (a ◦ b) for every A,B ∈ P∗(H) (P∗(H) denotes the set of nonempty subsets of H). A nonempty subset A of H is called a subgroupoid of H if A ∗ A ⊆ A, equivalently if, for any a, b ∈ A, we have a ◦ b ⊆ A. The following two properties, though clear, play an essential role in the theory of hypergroupoids: (1) if x ∈ A ∗B, then x ∈ a ◦ b for some a ∈ A, b ∈ B and (2) if a ∈ A and b ∈ B, then a ◦ b ⊆ A ∗B. Moreover, we have {x}∗{y} = x◦y for any x, y ∈ H. An hypergroupoid (H, ◦, ∗) is called hypersemigroup if {x} ∗ (y ◦ z) = (x ◦ y) ∗ {z} for every x, y, z ∈ H. 2. Some results on hypergroupoids If S is a groupoid or an ordered groupoid, an equivalence relation σ on S is called right (resp. left) congruence on S if (a, b) ∈ σ implies (ac, bc) ∈ σ (resp. (ca, cb) ∈ σ) for every c ∈ S. It is called a congruence on S if it is both a right and a left congruence on S. A congruence σ on S is called semilattice congruence if (a2, a) ∈ σ and (ab, ba) ∈ σ for any N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 478 a, b ∈ S [3,13]. These concepts can be naturally transferred to hypergroupoids by replacing the multiplication “·” of the groupoid by the hyperoperation “◦” of the hypergroupoid. But while for a groupoid ac is an element, in case of an hypergroupoid where a ◦ c is a set, we have to declare what the (a ◦ c, b ◦ c) ∈ σ means. We can define it as “for every u ∈ a ◦ c and every v ∈ b ◦ c we have (u, v) ∈ σ” or “for every u ∈ a ◦ c there exists v ∈ b ◦ c such that (u, v) ∈ σ” and get two different definitions of the left congruence, two different definitions for the right congruence; and so two different definitions of a congruence or a semilattice congruence. Although there is one between them that implies the other (it can be easily proved) and so they could be named differently (like congruence– weak congruence; complete congruence–congruence; strong congruence–congruence, for example), we will define as “congruence” both of them and, according to our investigation it will be clear which of them we use. This have been said, we give the Definitions 2.2 and 2.3 below. We first have to introduce the following notation: Notation 2.1. If H is an hypergroupoid, σ an equivalence relation on H and A,B two nonempty subsets of H, then we write (A,B) ∈ σ if for every a ∈ A and every b ∈ B, we have (a, b) ∈ σ. We write (A, b) instead of (A, {b}) and (a,B) instead of ({a}, B). So (A, b) means that, for every a ∈ A, we have (a, b) ∈ σ. If it is convenient we write, for short, A ∗ c instead of A ∗ {c} (A ⊆ H, c ∈ H). Definition 2.2. Let H be an hypergroupoid. An equivalence relation σ on H is called right congruence if (a, b) ∈ σ implies (a ◦ c, b ◦ c) ∈ σ for every c ∈ H. It is called left congruence if (a, b) ∈ σ implies (c ◦ a, c ◦ b) ∈ σ for every c ∈ H. By a congruence on H we mean a relation on H which is both a right and a left congruence on H. Definition 2.3. Let H be an hypergroupoid. A congruence σ on H is called semilattice congruence if, for any a, b ∈ H, we have (a ◦ a, a) ∈ σ and (a ◦ b, b ◦ a) ∈ σ. If (S, ·) is a groupoid, a nonempty subset F of S is called a filter of S [13] if the following assertions are satisfied: (1) if a, b ∈ F , then ab ∈ F and (2) if a, b ∈ S such that ab ∈ F , then a ∈ F and b ∈ F ; in other words, if it is a subgroupoid of S satisfying the property (2). A nonempty subset A of S is called an ideal of S [13] if AS ⊆ A and SA ⊆ A, that is if a ∈ A and s ∈ S implies as ∈ A and sa ∈ A. If (S, ·,≤) is an ordered groupoid, a subset F of S is called a filter of S if it is a filter of (S, ·) and, in addition if a ∈ F and S ∈ b ≥ a implies b ∈ F [1]; it is called an ideal of (S, ·,≤) if it is an ideal of (S, ·) and, in addition if a ∈ A and S 3 b ≤ a implies b ∈ A [2]. A subset T of a groupoid (or ordered groupoid) S is said to be prime [3,13] if a, b ∈ S such that ab ∈ T implies a ∈ T or b ∈ T . It is well known that a nonempty subset F of a groupoid (or an ordered groupoid) S is a filter of S if and only if the complement of F to S is either empty or a prime ideal of S [3,13] and, when we pass from groupoids to hypergroupoids the corresponding result should be satisfied. To manage it, a new condition should be added to the corresponding conditions of the filter and of prime ideals of groupoids we already have. And the concept N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 479 of the filter of groupoids can be naturally transferred to hypergroupoids in the definition below; the prime subsets can be defined in a similar way –adding a new condition. Definition 2.4. (cf. also [7]) Let H be an hypergroupoid. A nonempty subset F of H is called a filter of H if the following assertions are satisfied: (1) if x, y ∈ F , then x ◦ y ⊆ F ; (2) if x, y ∈ H such that x ◦ y ⊆ F , then x ∈ F and y ∈ F ; and (3) for any x, y ∈ H, we have x ◦ y ⊆ F or (x ◦ y) ∩ F = ∅. That is, a filter of H is a subgroupoid of H satisfying the relations (2) and (3). Definition 2.5. Let H be an hypergroupoid. A nonempty subset T of H is called a prime subset of H if the following assertions are satisfied: (1) if a, b ∈ H such that a ◦ b ⊆ T, then a ∈ T or b ∈ T and (2) for every a, b ∈ H, we have a ◦ b ⊆ T or (a ◦ b) ∩ T = ∅. As we see, we keep the definitions of filters and prime subsets of groupoids in which we add condition (3) in case of filters and condition (2) in case of prime subsets. If a subset T of an hypergroupoid satisfies only the condition (2) of Definition 2.5, then we call it half prime subset of H. Remark 2.6. If H is an hypergroupoid, I a half prime subset of H and a, c /∈ I, then a ◦ c * I. So a /∈ I implies a ◦ a * I. As in groupoids, for an element x of H, we denote by N(x) the filter of H generated by x, and by N the equivalence relation on H defined by N := {(x, y) ∈ H ×H | N(x) = N(y)}. Using the Definitions 2.4 and 2.5, with the usual changes we pass from groupoids to hypergroupoids. In an ordered semigroup S, the relation N is a semilattice congruence on S [3], and the same holds for groupoids as well. By a modification of that proof, we have the following proposition; for the sake of completeness we will give its proof. Proposition 2.7. (see also [3; the Proposition]) If H is an hypergroupoid, then the equivalence relation N is a semilattice congruence on H. Proof. Let (x, y) ∈ N and z ∈ H. Then (z ◦ x, z ◦ y) ∈ N . In fact: Let u ∈ z ◦ x and v ∈ z ◦ y. Then (u, v) ∈ N . Indeed: Since u ∈ N(u) and u ∈ z ◦ x, we have (z ◦ x) ∩ N(u) 6= ∅. Since N(u) is a filter of H, we have z ◦ x ⊆ N(u), and z, x ∈ N(u). Since x ∈ N(u), we have N(x) ⊆ N(u), then y ∈ N(u). Since z, y ∈ N(u), we have z ◦ y ⊆ N(u), then v ∈ N(u), and N(v) ⊆ N(u). By symmetry, we get N(u) ⊆ N(v), so we have N(u) = N(v), and (u, v) ∈ N . Thus N is a left congruence on H. In a similar way we prove that N is a right congruence on H, so N is a congruence on H. Let x ∈ H. Then (x ◦ x, x) ∈ N . In fact: Let u ∈ x ◦ x. Then (u, x) ∈ N . Indeed: Since u ∈ N(u), we have (x ◦ x) ∩ N(u) 6= ∅. Since N(u) is a filter of H, we have x ◦ x ⊆ N(u), then x ∈ N(u), and N(x) ⊆ N(u). On the other hand, since x ∈ N(x) and N(x) is a filter of H, we have x ◦ x ⊆ N(x). Then u ∈ N(x), so N(u) ⊆ N(x). Thus we have N(u) = N(x), and (u, x) ∈ N . Let x, y ∈ H. Then (x ◦ y, y ◦x) ∈ N . In fact: Let u ∈ x ◦ y and v ∈ y ◦x. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 480 Then (u, v) ∈ N . Indeed: Since u ∈ N(u), we have (x ◦ y)∩N(u) 6= ∅, then x ◦ y ⊆ N(u), x, y ∈ N(u), and y ◦ x ⊆ N(u). Then v ∈ N(u), and N(v) ⊆ N(u). By symmetry, we get N(u) ⊆ N(v), then N(u) = N(v), and (u, v) ∈ N . � Notation 2.8. For a subset I of H, we denote by σI the equivalence relation on H defined by: σI := {(a, b) ∈ H ×H | a, b ∈ I or a, b /∈ I} (i.e. a, b both belong to I or a, b both do not belong to I). If S is a semigroup or an ordered semigroup and I a prime ideal of S, then the relation σI is a semilattice congruence on S [3,13] (and the same holds if we replace the work “semigroup” by “groupoid”). Recall that for ordered groupoids the semilattice congruences are defined exactly as in groupoids. In an attempt to show the way we pass from semigroups to Γ-semigroups, we transferred this result to Γ-semigroups in [5]. Here we do the same for hypergroupoids using the following proposition. Proposition 2.9. Let H be an hypergroupoid, a, b, c ∈ H and I ⊆ H. Then we have the following: (1) if a ◦ c, b ◦ c ⊆ I, then (a ◦ c, b ◦ c) ∈ σI . Suppose now that, for every a, b ∈ H, we have a ◦ b ⊆ I or (a ◦ b) ∩ I = ∅ (∗) Then the following two conditions are satisfied: (2) if a ◦ c, b ◦ c * I, then (a ◦ c, b ◦ c) ∈ σI . (3) if a /∈ I and a ◦ a * I, then (a, a ◦ a) ∈ σI . Proof. (1) Let a ◦ c, b ◦ c ⊆ I, u ∈ a ◦ c and v ∈ b ◦ c. Then u, v ∈ I, so (u, v) ∈ σI . (2) Let a ◦ c, b ◦ c * I, u ∈ a ◦ c and v ∈ b ◦ c. If u, v ∈ I, then (u, v) ∈ σI . If u /∈ I, then v /∈ I. Indeed, if v ∈ I, then v ∈ (b ◦ c) ∩ I. Since (b ◦ c) ∩ I 6= ∅, by (∗), we have b ◦ c ⊆ I which is impossible. So u, v /∈ I, and (u, v) ∈ σI . If v /∈ I, in a similar way we get u /∈ I, so again (u, v) ∈ σI . (3) Let a /∈ I, a ◦ a * I and u ∈ a ◦ a. If u ∈ I, then (a ◦ a) ∩ I 6= ∅ and, by (∗), a ◦ a ⊆ I which is impossible. Thus we have u 6∈ I. Since a, u /∈ I, we have (a, u) ∈ σI . � If H is an hypergroupoid, a nonempty subset A of H is called an ideal of H if A∗H ⊆ A and H ∗ A ⊆ A, equivalently if a ∈ A and h ∈ H, then a ◦ h ⊆ A and h ◦ a ⊆ A [8]. By a prime ideal of H we clearly mean an ideal of H which is at the same time a prime subset of H. Corollary 2.10. (cf. also [3] and [5; Proposition 2.2]) Let H be an hypergroupoid and I a prime ideal of H. Then the equivalence relation σI is a semilattice congruence on H. Proof. Let (a, b) ∈ σI and c ∈ H. Then (a ◦ c, b ◦ c) ∈ σI . In fact: Since (a, b) ∈ σI , we have a, b ∈ I or a, b /∈ I. Let a, b ∈ I. Since I is an ideal of H, we have a ◦ c, b ◦ c ⊆ I. Then, by Proposition 2.9(1), we have (a ◦ c, b ◦ c) ∈ σI . Let a, b /∈ I. If c ∈ I then, since I is an ideal of H, we have a ◦ c, b ◦ c ⊆ I, then (a ◦ c, b ◦ c) ∈ σI . Let c /∈ I. Since a, b, c /∈ I, by Remark 2.6, we have a ◦ c, b ◦ c * I. Then, by Proposition 2.9(2), we have (a ◦ c, b ◦ c) ∈ σI . Thus σI is a right congruence on H. In a similar way we can prove that σI is a left congruence on H and so it is a congruence on H. Let a ∈ H. Then N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 481 (a ◦ a, a) ∈ σI . In fact: Let u ∈ a ◦ a. If a ∈ I then, since I is an ideal of H, we have a◦a ⊆ I, then u ∈ I; since u, a ∈ I, we have (u, a) ∈ σI . If a /∈ I, then a◦a * I. Then, by Proposition 2.9(3), we have (a, a ◦ a) ∈ σI . Let a, b ∈ H. Then (a ◦ b, b ◦ a) ∈ σI . Indeed: if a ◦ b ⊆ I then, since I is a prime ideal of H, we have a ∈ I or b ∈ I. Since I is an ideal of H, we have b ◦ a ⊆ I. Since a ◦ b, b ◦ a ⊆ I, by Lemma 2.9(1), we have (a ◦ b, b ◦ a) ∈ σI . If a ◦ b * I then b ◦ a * I. This is because if b ◦ a * I then, since I is a prime ideal of H, we have b ∈ I or a ∈ I and, since I is an ideal of H, we have a ◦ b ⊆ I which is impossible. Since a ◦ b, b ◦ a * I, by Proposition 2.9(2), we have (a ◦ b, b ◦ a) ∈ σI . � We have the following: (1) if (x,A) ∈ σ and ∅ 6= B ⊆ A, then (x,B) ∈ σ; (2) if (A,B) ∈ σ, then (B,A) ∈ σ; (3) if (A,B) ∈ σ, (B,C) ∈ σ and B 6= ∅, then (A,C) ∈ σ; indeed, let a ∈ A, c ∈ C. Take an element b ∈ B (B 6= ∅). Since (a, b) ∈ σ and (b, c) ∈ σ, we have (a, c) ∈ σ. Proposition 2.11. Let H be an hypergroupoid, σ a congruence on H and A,B,C,D nonempty subsets of H. If (A,B) ∈ σ and (C,D) ∈ σ, then (A ∗ C,B ∗ D) ∈ σ and (C ∗A,D ∗B) ∈ σ. Proof. Let (A,B) ∈ σ, u ∈ A ∗ C and v ∈ B ∗ D. We have u ∈ a ◦ c for some a ∈ A, c ∈ C and v ∈ b ◦ d for some b ∈ B, d ∈ D. Since a ∈ A, b ∈ B and (A,B) ∈ σ, we have (a, b) ∈ σ and, since σ is a right congruence on H, we have (a ◦ c, b ◦ c) ∈ σ. Since (C,D) ∈ σ, c ∈ C and d ∈ D, we have (c, d) ∈ σ and, since σ is a left congruence on H, we have (b ◦ c, b ◦ d) ∈ σ. By the transitivity relation, we have (a ◦ c, b ◦ d) ∈ σ. Since u ∈ a ◦ c and v ∈ b ◦ d, we have (u, v) ∈ σ. Similarly we get (C ∗A,D ∗B) ∈ σ. � Lemma 2.12. Let H be an hypergroupoid, A,B nonempty subsets of H and c ∈ H. If σ a right congruence on H and (A,B) ∈ σ, then (A ∗ c,B ∗ c) ∈ σ. Proof. Since c ∈ H and σ is a reflexive relation on H, we have ({c}, {c}) ∈ σ. Since (A,B) ∈ σ and ({c}, {c}) ∈ σ, by Proposition 2.11, we have (A ∗ c,B ∗ c) ∈ σ. An independent proof is the following: Let u ∈ A ∗ c and v ∈ B ∗ c. Then u ∈ a ◦ c for some a ∈ A and v ∈ b ◦ c for some b ∈ B. Since a ∈ A, b ∈ B and (A,B) ∈ σ, we have (a, b) ∈ σ. Since σ is a right congruence on H, we have (a ◦ c, b ◦ c) ∈ σ. Since u ∈ a ◦ c and v ∈ b ◦ c, we get (u, v) ∈ σ and so (A ∗ c,B ∗ c) ∈ σ. � In a similar way we have the following lemma. Lemma 2.13. Let H be an hypergroupoid, A,B nonempty subsets of H and c ∈ H. If σ a left congruence on H and (A,B) ∈ σ, then (c ∗A, c ∗B) ∈ σ. As in groupoids, the following proposition holds and one can prove it as a modification of the proof of the corresponding result in [3]. Proposition 2.14. (cf. also [3; the Lemma]) Let H be an hypergroupoid. If H is a filter of H, then the property (∗) is satisfied: N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 482 (∗) either H\F = ∅ or H\F is a prime ideal of H. In particular, any nonempty subset F of H satisfying (∗) is a filter of H. An ideal I of H is called proper if I 6= H. 3. Main result Theorem 3.1. Let H be an hypersemigroup and σ be a semilattice congruence on H. Then there exists a family A of proper prime ideals of H such that σ = ⋂ I∈A σI. Proof. Let x ∈ H. We consider the set Ax := {y ∈ H | (x, x ◦ y) ∈ σ}. The set Ax is a filter of H. Indeed: Since x ∈ H and σ is a semilattice congruence on H, we have (x, x ◦ x) ∈ σ, thus x ∈ Ax and Ax is a nonempty subset of H. Let y, z ∈ Ax. Then y ◦ z ⊆ Ax. In fact: Let u ∈ y ◦ z. Then u ∈ Ax, that is (x, x ◦ u) ∈ σ. Indeed: Let v ∈ x ◦ u. Since y ∈ Ax, we have (x, x ◦ y) ∈ σ. Then, by Lemma 2.12, we have ( x ◦ z, (x ◦ y) ∗ {z} ) ∈ σ. Since z ∈ Ax, we have (x, x ◦ z) ∈ σ and, by the transitivity relation, we have ( x, {x} ∗ (y ◦ z) ) ∈ σ. Since v ∈ x ◦ u ⊆ {x} ∗ (y ◦ z), we have (x, v) ∈ σ. Let y, z ∈ H such that y ◦ z ⊆ Ax. Then y ∈ Ax and z ∈ Ax. In fact: Since y ◦ z ⊆ Ax, we have ( x, {x} ∗ (y ◦ z) ) ∈ σ (1) Indeed: if u ∈ {x} ∗ (y ◦ z), then u ∈ x ◦ t for some t ∈ y ◦ z ⊆ Ax. Since t ∈ Ax, we have (x, x ◦ t) ∈ σ. Then, since u ∈ x ◦ t, we obtain (x, u) ∈ σ, so property (1) is satisfied. By (1) and Lemma 2.12, we have ( x ◦ z, {x} ∗ (y ◦ z) ∗ {z} ) ∈ σ (2) On the other hand, since (z, z ◦ z) ∈ σ, we have ( (x ◦ y) ∗ {z}, (x ◦ y) ∗ (z ◦ z) ) ∈ σ (3) In fact: Since (z, z ◦ z) ∈ σ, by Lemma 2.13, we have ( y ◦ z, {y} ∗ (z ◦ z ) ∈ σ; again by Lemma 2.13, we have ( {x} ∗ (y ◦ z), (x ◦ y) ∗ (z ◦ z) ) ∈ σ and (3) holds. By (1),(2) and (3), we obtain (x, x◦z) ∈ σ, and so z ∈ Ax. It remains to prove that y ∈ Ax. Since z ∈ Ax, we have (x, x ◦ z) ∈ σ. By Lemma 2.12, we have ( x ◦ y, (x ◦ z) ∗ {y} ) ∈ σ. Since (y ◦ z, z ◦ y) ∈ σ, by Lemma 2.13, we have ( {x} ∗ (y ◦ z), {x} ∗ (z ◦ y) ) ∈ σ. Then, by (1), we get (x, x ◦ y) ∈ σ, and so y ∈ Ax. Let y, z ∈ H. Then y ◦ z ⊆ Ax or (y ◦ z) ∩Ax = ∅. In fact: Let y ◦ z * Ax and (y ◦ z)∩Ax 6= ∅. Let u ∈ y ◦ z such that u /∈ Ax, v ∈ y ◦ z and v ∈ Ax. Then we have u ∈ y ◦ z, (x, x ◦ u) /∈ σ, v ∈ y ◦ z, (x, x ◦ v) ∈ σ. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 483 On the other hand, (x, x ◦ v) ∈ σ and v ∈ y ◦ z implies ( x, {x} ∗ (y ◦ z) ) ∈ σ. Indeed: Let a ∈ {x} ∗ (y ◦ z). Then a ∈ x ◦ d for some d ∈ y ◦ z. Since (y ◦ z, y ◦ z) ∈ σ, v ∈ y ◦ z and d ∈ y ◦ z, we have (v, d) ∈ σ, then (x ◦ v, x ◦ d) ∈ σ. Since (x, x ◦ v) ∈ σ and (x ◦ v, x ◦ d) ∈ σ, we have (x, x ◦ d) ∈ σ. Since a ∈ x ◦ d, we have (x, a) ∈ σ. We have x ◦ u ⊆ {x} ∗ (y ◦ z) and ( x, {x} ∗ (y ◦ z) ) ∈ σ, thus we have (x, x ◦ u) ∈ σ which is impossible. Since Ax is a filter of H, by Proposition 2.14, we have H\Ax = ∅ or H\Ax is a prime ideal of H. Then H\Ax = ∅ or H\Ax is a proper prime ideal of H (indeed, if H\Ax = H then, since Ax ⊆ H, we have Ax = ∅ which is not possible). We consider the set {H\Az | z ∈ H, H\Az proper prime ideal of H}. We have σ = ⋂ z∈H σH\Az . In fact: Let (x, y) ∈ σ and z ∈ H. Then (x, y) ∈ σH\Az . Indeed: Since x ∈ H, we have x ∈ H\Az or x /∈ H\Az. (a) If x /∈ H\Az, then x ∈ Az, so (z, z◦x) ∈ σ. Since (x, y) ∈ σ, we have (z◦x, z◦y) ∈ σ. Then (z, z ◦ y) ∈ σ, and y ∈ Az, so y /∈ H\Az. (b) Let x ∈ H\Az. If y /∈ H\Az, then in a similar way as in (a), we prove that x /∈ H\Az which is impossible. Thus we have y ∈ H\Az. Since both x and y belong toH\Az or both do not belong toH\Az, we have (x, y) ∈ σH\Az . Let now (x, y) ∈ σH\Az for every z ∈ H. Then (x, y) ∈ σ. Indeed: Since x ∈ Ax, we have x /∈ H\Ax and, since (x, y) ∈ σH\Ax , we have y /∈ H\Ax, so y ∈ Ax, that is (x, x ◦ y) ∈ σ. Since y ∈ Ay, we have y /∈ H\Ay, then x /∈ H\Ay, so x ∈ Ay, thus (y, y ◦ x) ∈ σ. Since σ is a semilattice congruence on H, we have (x ◦ y, y ◦ x) ∈ σ. Since (x, x ◦ y) ∈ σ, (x ◦ y, y ◦ x) ∈ σ and (y ◦ x, y) ∈ σ, we have (x, y) ∈ σ. � The following proposition holds for hypergroupoids and its proof is exactly the same as the proof of the corresponding result in [3] (no change is needed). Proposition 3.2. (cf. also [3; the Proposition]) Let H be an hypergroupoid and I(H) the set of prime ideals of H. Then we have N = ⋂ I∈I(H) σI . Corollary 3.3. If H is an hypersemigroup, then the relation N is the least semilattice congruence on H. Proof. Let σ be a semilattice congruence on H. Then N ⊆ σ. In fact: By Theorem 3.1, there exists a family A of proper prime ideals of H such that σ = ⋂ I∈A σI. By Proposition 3.2, N = ⋂ I∈I(H) σI , where I(H) is the set of prime ideals of H. On the other hand, N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 484⋂ I∈A σI ⊇ ⋂ I∈I(H) σI. Indeed, if (x, y) ∈ σI for every prime ideal of H, then clearly (x, y) ∈ σI for every proper prime ideal of H, and so for the elements of A as well. Hence we obtain N ⊆ σ. � Proposition 3.4. Let (S, ·) be a groupoid and “◦” the hypergroupoid with the hyperoper- ation “◦” defined by ◦ : S × S → P∗(S) | (a, b)→ a ◦ b := {ab}. Then F is a filter of (S, ·) if and only if it is a filter of (S, ◦). Proof. =⇒. Let F be a filter of (S, ·). If a, b ∈ F and x ∈ a ◦ b, then x = ab ∈ F and so a ◦ b ⊆ F . If a, b ∈ S such that a ◦ b ⊆ F , then ab ∈ a ◦ b ⊆ F , ab ∈ F and so a ∈ F and b ∈ F . Let (a ◦ b) ∩ F 6= ∅, u ∈ a ◦ b and u ∈ F . Then u = ab and u ∈ F , then ab ∈ F , then a ◦ b = {ab} ⊆ F and so a ◦ b ⊆ F . ⇐=. Let F be a filter of (S, ◦). If a, b ∈ F , then {ab} = a ◦ b ⊆ F , thus ab ∈ F . If a, b ∈ S such that ab ∈ F , then a ◦ b = {ab} ⊆ F , so a ◦ b ⊆ F , then a ∈ F and b ∈ F and so F is a filter of (S, ·). � 4. Complete semilattice congruences on ordered hypersemigroups Let us consider now the case of ordered hypergroupoids. For the necessary definitions and notations on ordered hypergroupoids we refer to [6] and [9]. For an ordered hyper- groupoid, the semilattice congruence is defined exactly as in hypergroupoids. The concept of complete semilattice congruences of ordered groupoids introduced by Kehayopulu and Tsingelis in [11] can be naturally transferred to ordered hypergroupoids by the following definition. Definition 4.1. If (S, ◦,≤) is an ordered hypergroupoid, a semilattice congruence σ on S is called complete if, for every a, b ∈ S, the relation a ≤ b implies (a, a ◦ b) ∈ σ. Proposition 4.2. Let (S, ·,≤) be an ordered groupoid and “◦” the hyperoperation on S defined by a ◦ b := {ab}. Then (S, ◦,≤) is an ordered hypergroupoid. The relation σ is a semilattice (resp. complete semilattice) congruence on (S, ·,≤) if and only if it is semilattice (resp. complete semilattice) congruence on (S, ◦,≤). If (S, ·,≤) is an ordered semigroup, then (S, ◦,≤) is an ordered hypersemigroup as well. Proof. If (S, ·,≤) is an ordered groupoid, a ≤ b, c ∈ S and u ∈ a ◦ c, then u = ac ≤ bc, so for the element v := bc ∈ b ◦ c we have u ≤ v; similarly c ◦ a � c ◦ b and so (S, ◦,≤) is an ordered hypergroupoid. Let σ be a semilattice congruence on (S, ·,≤). If (a, b) ∈ σ and c ∈ S, then (a ◦ c, b ◦ c) ∈ σ. Indeed, if u ∈ a ◦ c and v ∈ b ◦ c, then u = ac, v = bc and (ac, bc) ∈ σ, so (u, v) ∈ σ. Similarly σ is a left congruence on (S, ◦,≤). Let a, b ∈ S. Then (a ◦ a, a) ∈ σ. Indeed, if u ∈ a ◦ a, then u = a2 and (a2, a) ∈ σ, thus we get (u, a) ∈ σ. We have (a ◦ b, b ◦ a) ∈ σ. Indeed, if u ∈ a ◦ b and v ∈ b ◦ a, then u = ab, v = ba and (ab, ba) ∈ σ, thus we get (u, v) ∈ σ and so σ is a semilattice congruence on (S, ◦,≤). Let σ be a complete semilattice congruence on (S, ·,≤), a ≤ b, and u ∈ a◦ b. Since u = ab and N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 485 a ≤ b, we have (a, ab) ∈ σ, then (a, u) ∈ σ and so σ is a complete semilattice congruence on (S, ◦,≤). Let σ be a semilattice congruence on (S, ◦,≤). If (a, b) ∈ σ and c ∈ S then, since (a ◦ c, b ◦ c) ∈ σ, ac ∈ a ◦ c and bc ∈ b ◦ c, we have (ac, bc) ∈ σ. Similarly σ is a left congruence on (S, ·,≤). If a ∈ S, then (a ◦ a, a) ∈ σ and, since a2 ∈ a ◦ a, we have (a2, a) ∈ σ. If a, b ∈ S, then (a ◦ b, b ◦ a) ∈ σ and, since ab ∈ a ◦ b and ba ∈ b ◦ a, we have (ab, ba) ∈ σ. Hence σ is a semilattice congruence on (S, ·,≤). Let now σ be a complete semilattice congruence on (S, ◦,≤) and a ≤ b. Since (a, a ◦ b) ∈ σ and ab ∈ a ◦ b, we have (a, ab) ∈ σ, so σ is a complete semilattice congruence on (S, ·,≤). Let now (S, ·,≤) be an ordered semigroup, a, b, c ∈ S and x ∈ {a} ∗ (b ◦ c). Then x ∈ a ◦ u for some u ∈ b ◦ c, then x = au and u = bc. Then we have x = a(bc) ∈ {a(bc)} = {(ab)c} = (ab) ◦ c ⊆ {ab} ∗ {c} = (a ◦ b) ∗ {c}, then {a} ∗ (b ◦ c) ⊆ (a ◦ b) ∗ {c}. Similarly (a ◦ b) ∗ {c} ⊆ {a} ∗ (b ◦ c) and so (S, ◦,≤) is an ordered hypersemigroup. � Definition 4.3. Let (S, ◦,≤) be an ordered hypergroupoid. A subset F of S is called a filter of (S, ◦,≤) if it is a filter of the hypergroupoid (S, ◦) and, in addition, if a ∈ F and S 3 b ≥ a implies b ∈ F. By Proposition 3.4, we have the following Proposition 4.4. Let (S, ·,≤) be an ordered groupoid and “◦” the hyperoperation on S defined by a◦b := {ab}. Then F is a filter of (S, ·,≤) if and only if it is a filter of (S, ◦,≤). Remark 4.5. According to Proposition 4.4, if the hyperoperation “◦” is defined by a ◦ b = {ab}, then the filters of the groupoid (S, ·,≤) and the filters of the hypergroupoid (S, ◦,≤) are the same. Also, by Proposition 4.2, the semilattice congruences on (S, ·,≤) and the semilattice congruences on (S, ◦,≤) are the same. As we have seen in [11], in an ordered hypersemigroup S, the relation N is not the least semilattice congruence on S in general, so according to Proposition 4.2, in an ordered hypersemigroup, the relation N cannot be the least semilattice congruence as well, in general. Let us see it in the following example. Example 4.6. We get the ordered semigroup defined in [11] with the following table and figure. · a b c d f g a b b a d a a b b b b d b b c a b c d c c d d d d d d d f a b c d c c g a b c d f g Table 1. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 486 d gf c ba Figure 1. For this semigroup, N(a) = N(b) = {a, b, c, f, g}, N(c) = N(f) = N(g) = {c, f, g} and N(d) = S. We consider the semilattice congruences on S. They are eight and they are the following: σ1 = {(a, a), (a, b), (b, a), (b, b), (c, c), (c, f), (d, d), (f, c), (f, f), (g, g)}. σ2 = {(a, a), (a, b), (b, a), (b, b), (c, c), (c, f), (c, g), (d, d), (f, c), (f, f), (f, g), (g, c), (g, f), (g, g)} = N . σ3 = {(a, a), (a, b), (a, d), (b, a), (b, b), (b, d), (c, c), (c, f), (d, a), (d, b), (d, d), (f, c), (f, f), (g, g)}. σ4 = {(a, a), (a, b), (a, c), (a, f), (b, a), (b, b), (b, c), (b, f), (c, a), (c, b), (c, c), (c, f), (d, d), (f, a), (f, b), (f, c), (f, f), (g, g)}. σ5 = {(a, a), (a, b), (a, d), (b, a), (b, b), (b, d), (c, c), (c, f), (c, g), (d, a), (d, b), (d, d), (f, c), (f, f), (f, g), (g, c), (g, f), (g, g)}. σ6 = {(a, a), (a, b), (a, c), (a, d), (a, f), (b, a), (b, b), (b, c), (b, d), (b, f), (c, a), (c, b), (c, c), (c, d), (c, f), (d, a), (d, b), (d, c), (d, d), (d, f), (f, a), (f, b), (f, c), (f, d), (f, f), (g, g)}. σ7 = {(a, a), (a, b), (a, c), (a, f), (a, g), (b, a), (b, b), (b, c), (b, f), (b, g), (c, a), (c, b), (c, c), (c, f), (c, g), (d, d), (f, a), (f, b), (f, c), (f, f), (f, g), (g, a), (g, b), (g, c), (g, f), (g, g)}. σ8 = S × S. σ1 is the least semilattice congruence on S, the relations σ2, σ5, σ7, σ8 are complete semilattice congruences on S, N (= σ2) ⊆ σ5, σ7, σ8, that is N is the least complete semilattice congruence on S and σ1 6= N . According to Proposition 4.2, the ordered hypersemigroup (S, ◦,≤) defined by the table below and the same Figure 1 is an ordered hypersemigroup and the semilattice congruences N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 487 on (S, ·,≤) and on (S, ◦,≤) are the same, so σ1 is the least semilattice congruence on (S, ◦,≤), N is the least complete semilattice congruence on (S, ◦,≤) and N is different than σ1. ◦ {a} {b} {c} {d} {f} {g} a {b} {b} {a} {d} {a} {a} b {b} {b} {b} {d} {b} {b} c {a} {b} {c} {d} {c} {c} d {d} {d} {d} {d} {d} {d} f {a} {b} {c} {d} {c} {c} g {a} {b} {c} {d} {f} {g} Table 2. In the above example the ordered semigroup has been found using our computer pro- gram. Let us give another example which is easier to check by hand. Example 4.7. (cf. [4; Example 1]) The ordered semigroup given by the multiplication “·” and the figure below is an example of an ordered semigroup for which the relation N is not the least semilattice congruence on S. · a b c d e a b a a a a b a b b b b c a b b b b d a b b d d e a b c d e Table 3. c e a b d Figure 2. We have N(a) = N(b) = N(c) = S and N(d) = N(f) = {d, e}. We give all the semilattice congruences on S. They are four and they are the following: σ1 = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c), N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 488 (d, d), (e, e)}. σ2 = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c), (d, d) (d, e), (e, d), (e, e)} = N . σ3 = {(a, a), (a, b), (a, c), (a, d), (b, a), (b, b), (b, c), (b, d), (c, a), (c, b), (c, c), (c, d), (d, a), (d, b), (d, c), (d, d), (e, e)}. σ4 = S × S. The relation σ1 is the least semilattice congruence on S, the relationN is the least complete semilattice congruence on S, and σ1 6= N . According to Proposition 4.2, the ordered semigroup (S, ·,≤) with the operation “◦” on S defined by a◦b := {ab} is an ordered hypersemigroup and the semilattice congruences on (S, ·,≤) and on (S, ◦,≤) coincide. In other words, the set S with the multiplication given by the table ◦ a b c d e a {b} {a} {a} {a} {a} b {a} {b} {b} {b} {b} c {a} {b} {b} {b} {b} d {a} {b} {b} {d} {d} e {a} {b} {c} {d} {e} Table 4. and the same order as in (S, ·,≤) (: Figure 2) is an ordered hypersemigroup, the relation σ1 is the least semilattice congruence on S and it is different than N . In [4] there are also examples of ordered semigroups S for which the complete semilat- tice congruence N is at the same time the least semilattice congruence on S. Let us get one of them and pass from the ordered semigroup to ordered hypersemigroup. Example 4.8. If we take the ordered hypersemigroup (S, ·,≤) given in the Example 2 in [4] and define the hyperoperation as a◦b := {ab}, then we get the ordered hypersemigroup defined by the table and the figure below. ◦ a b c d f a {b} {b} {d} {d} {d} b {b} {b} {d} {d} {d} c {d} {d} {c} {d} {c} d {d} {d} {d} {d} {d} f {d} {d} {c} {d} {c} Table 5. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 489 a d f b c Figure 3. We have N(a) = N(b) = {a, b}, N(c) = N(f) = {c, f} and N(d) = S. There are four semilattice congruences on (S, ◦ ≤) and they are the following σ1 = {(a, a), (a, b), (b, a), (b, b), (c, c), (c, f), (d, d), (f, c), (f, f)} = N . σ2 = {(a, a), (a, b), (a, d), (b, a), (b, b), (b, d), (c, c), (c, f), (d, a), (d, b), (d, d), (f, c), (f, f)}. σ3 = {(a, a), (a, b), (b, a), (b, b), (c, c), (c, d), (c, f), (d, c), (d, d), (d, f), (f, c), (f, d), (f, f)}. σ4 = S × S. The relation σ1 is the least semilattice congruence on (S, ◦,≤) and at the same time the least complete semilattice congruence on (S, ◦,≤). In the Example 3 in [4] and in its “dual” given immediately after the Example 3, the relation σ1 mentioned in them is at the same time the least semilattice congruence and the least complete semilattice congruence; as a consequence it is so in the corresponding ordered hypersemigroup defined by the hyperoperation a ◦ b = {ab}. Proposition 4.9. (cf. also [12; Remark 1]) If (H, ◦,≤) is an ordered hypergroupoid, then the semilattice congruence N is a complete semilattice congruence on H. Proof. Let a ≤ b. Then (a, a ◦ b) ∈ N . In fact: Let u ∈ a ◦ b. Then (a, u) ∈ N , that is N(a) = N(u). Indeed: Since N(a) 3 a ≤ b, we have b ∈ N(a). Since a, b ∈ N(a), we have a ◦ b ⊆ N(a), then u ∈ N(a), and so N(u) ⊆ N(a). On the other hand, since u ∈ a ◦ b and u ∈ N(u), we have (a ◦ b) ∩ N(u) 6= ∅, then a ◦ b ⊆ N(u), then a ∈ N(u), and N(a) ⊆ N(u). Hence we obtain N(u) = N(a) and the proof is complete. � Theorem 4.10. (cf. also [11]) Let H be an ordered hypersemigroup and σ a complete semilattice congruence on H. Then there exists a family A of proper prime ideals of H such that σ = ⋂ I∈A σI. Proof. Following Theorem 3.1, it remains to prove that for the set Ax := {y ∈ H | (x, x ◦ y) ∈ σ} N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 490 the following property is satisfied if y ∈ Ax and H 3 z ≥ y, then z ∈ Ax. Indeed: Since y ∈ Ax, we have (x, x◦y) ∈ σ then, by Lemma 2.12, ( x◦z, (x◦y)∗{z} ) ∈ σ, that is ( (x ◦ y) ∗ {z}, x ◦ z ) ∈ σ. Since y ≤ z and σ is a complete semilattice congruence on H, we have (y, y ◦ z) ∈ σ and, by Lemma 2.13, ( x ◦ y, {x} ∗ (y ◦ z) ) ∈ σ. Hence we obtain (x, x ◦ z) ∈ σ that is, z ∈ Ax. � By Proposition 3.2, Proposition 4.9 and Theorem 4.10, we have the following Corollary 4.11. (cf. also [11; the Proposition]) If H is an ordered hypersemigroup, then the relation N is the least complete semilattice congruence on H. If (S, ·,≤) is an ordered groupoid and “◦” the hyperoperation on S defined by: ◦ : S×S → P∗(S) | (a, b)→ a◦ b, where a◦ b := {t ∈ S | t ≤ ab}, then (S, ◦,≤) is an ordered hypergroupoid [10; Lemma 1]. Proposition 4.12. Let (S, ·,≤) be an ordered groupoid and “◦” the hyperoperation on S defined by ◦ : S × S → P∗(S) | (a, b)→ a ◦ b := {t ∈ S | t ≤ ab}. If F is a filter of (S, ◦,≤), then it is a filter of (S, ·,≤) as well. The converse statement does not hold in general. Proof. Let a, b ∈ F . Since F is a filter of (S, ◦,≤), we have a ◦ b ⊆ F . Since ab ∈ a ◦ b, we have ab ∈ F . Let now a, b ∈ S such that ab ∈ F . Since ab ∈ a ◦ b and ab ∈ F , we have (a ◦ b) ∩ F 6= ∅, then a ◦ b ⊆ F , and then a, b ∈ F . For the converse statement, consider the ordered semigroup (S, ·,≤) of the Example 4.6 given by Table 1 and Figure 1 and the ordered hypersemigroup defined by the same order and the hyperoperation x ◦ y := {t ∈ S | t ≤ xy} in the following table. ◦ a b c d f g a {b, d} {b, d} {a, d} {d} {a, d} {a, d} b {b, d} {b, d} {b, d} {d} {b, d} {b, d} c {a, d} {b, d} {c, d, f, g} {d} {c, d, f, g} {c, d, f, g} d {d} {d} {d} {d} {d} {d} f {a, d} {b, d} {c, d, f, g} {d} {c, d, f, g} {c, d, f, g} g {a, d} {b, d} {c, d, f, g} {d} {d, f} {d, g} Table 6. The set {b, c, f} is a filter of (S, ·,≤), but it is not a filter of (S, ◦,≤). Indeed, for example, b ◦ c = {d, b} * {b, c, f}. � Proposition 4.13. Let (S, ·,≤) be an ordered groupoid and (S, ◦,≤) the ordered hyper- groupoid defined by the hyperoperation a ◦ b := {t ∈ S | t ≤ ab}. If σ is a semilattice N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (2) (2018), 476-492 491 (resp. complete semilattice) congruence on (S, ◦,≤), then it is a semilattice (resp. com- plete semilattice) congruence on (S, ·,≤). If σ is a semilattice congruence on (S, ·,≤), then it is not a semilattice congruence on (S, ◦,≤) in general. Proof. Let σ be a semilattice congruence on (S, ◦,≤). Let (a, b) ∈ σ and c ∈ S. Since (a◦c, b◦c) ∈ σ, ac ∈ a◦c, and bc ∈ b◦c, we have (ac, bc) ∈ σ. Similarly we get (ca, cb) ∈ σ, and σ is a congruence on (S, ·,≤). Let now a, b ∈ S. Since (a ◦ a, a) ∈ σ and a2 ∈ a ◦ a, we have (a2, a) ∈ σ. Since (a ◦ b, b ◦ a) ∈ σ, ab ∈ a ◦ b and ba ∈ b ◦ a, we have (ab, ba) ∈ σ, so σ is a semilattice congruence on (S, ·,≤). Let σ be a complete semilattice congruence on (S, ◦,≤) and a ≤ b. Since (a, a ◦ b) ∈ σ and ab ∈ a ◦ b, we have (a, ab) ∈ σ, thus σ is a complete semilattice congruence on (S, ·,≤). For the converse statement, consider the ordered semigroup of the Example 4.6 given by Table 1 and Figure 1 and the ordered hypersemigroup defined by the same order and the hyperoperation x ◦ y := {t ∈ S | t ≤ xy} in Table 6. As we have seen, the relation σ1 = {(a, a), (a, b), (b, a), (b, b), (c, c), (c, f), (d, d), (f, c), (f, f), (g, g)} is a semilattice congruence on (S, ·,≤). On the other site, σ1 is not a semilattice congruence on (S, ◦,≤). It is enough to observe that (a, b) ∈ σ1 but (a ◦ c, b ◦ c) /∈ σ1, since a ∈ a ◦ c, d ∈ b ◦ c but (a, d) /∈ σ1. � The following question is natural. Under what restrictions a semilattice congruence on (S, ·,≤) is a semilattice congruence on (S, ◦,≤)? For this purpose, we introduce the concept of pseudocomplete semilattice congruences as follows: Definition 4.14. Let (S, ·,≤) be an ordered groupoid. A semilattice congruence σ on S is called pseudocomplete if ≤⊆ σ. Example 4.15. The relation σ2 (= N ) in the example 4.7 is an example of a pseudocom- plete semilattice congruence on (S, ·,≤). There is no proper pseudocomplete semilattice congruence on (S, ·,≤) in the Example 4.6, in fact the only pseudocomplete semilattice congruence on (S, ·,≤) is the set S × S. Proposition 4.16. Let (S, ·,≤) be an ordered groupoid and σ a semilattice congruence on S. If σ is pseudocomplete, then it is complete. Proof. Let a ≤ b. Then (a, ab) ∈ σ. Indeed: Since a ≤ b and σ is pseudocomplete, we have (a, b) ∈ σ. Since σ is a semilattice congruence, we have (a2, ab) ∈ σ and (a, a2) ∈ σ, thus we get (a, ab) ∈ σ. � Proposition 4.17. Let (S, ·,≤) be an ordered groupoid and “◦” the hyperoperation on S defined by ◦ : S × S → S | (a, b)→ a ◦ b := {t ∈ S | t ≤ ab}. If σ is a pseudocomplete semilattice congruence on (S, ·,≤), then it is a complete semilattice congruence on (S, ◦,≤). Proof. Let (a, b) ∈ σ and c ∈ S. Then (a ◦ c, b ◦ c) ∈ σ and (c ◦ a, c ◦ b) ∈ σ. Indeed: Let u ∈ a ◦ c and v ∈ b ◦ c. Then u ≤ ac and v ≤ bc. Since σ is pseudocomplete, we have REFERENCES 492 (u, ac) ∈ σ and (v, bc) ∈ σ. Moreover (ac, bc) ∈ σ, thus we get (u, v) ∈ σ, and σ is a right congruence on (S, ◦,≤). Similarly σ is a left congruence on (S, ◦,≤). Let a ∈ S. Then (a ◦ a, a) ∈ σ. In fact: Let u ∈ a ◦ a. Then u ≤ a2, thus (u, a2) ∈ σ. Moreover (a2, a) ∈ σ, and then (u, a) ∈ σ. Let a, b ∈ S. Then (a ◦ b, b ◦ a) ∈ σ. In fact: Let u ∈ a ◦ b and v ∈ b ◦ a. Then u ≤ ab, v ≤ ba, from which (u, ab) ∈ σ, (v, ba) ∈ σ. Moreover (ab, ba) ∈ σ, thus we get (u, v) ∈ σ. Let a ≤ b. Then (a, a ◦ b) ∈ σ. Indeed: Let u ∈ a ◦ b. Then u ≤ ab, so (u, ab) ∈ σ. Since a ≤ b, we have (a, b) ∈ σ, then (a2, ab) ∈ σ. Moreover (a, a2) ∈ σ, and then (a, u) ∈ σ. Hence σ is a complete semilattice congruence on (S, ◦,≤). � With my best thanks to the two anonymous referees for their time to read the paper carefully, their interest on my work and their prompt reply. References [1] N. Kehayopulu. On weakly commutative poe-semigroups. Semigroup Forum 34(3):367–370, 1987. [2] N. Kehayopulu. On weakly prime ideals of ordered semigroups. Math. Japon. 35(6):1051–1056, 1990. [3] N. Kehayopulu. Remark on ordered semigroups. Math. Japon. 35(6):1061–1063, 1990. [4] N. Kehayopulu. On intra-regular ordered semigroups. Semigroup Forum 46(3):271–278, 1993. [5] N. Kehayopulu. Green’s relations and the relation N in Γ-semigroups. Quasigroups Related Systems 22(1):89–96, 2014. [6] N. Kehayopulu. Left regular and intra-regular ordered nypersemigroups in terms of semiprime and fuzzy semiprime subsets. Sci. Math. Jpn. 80(3):295–305, 2017. [7] N. Kehayopulu. Fuzzy sets in ≤-hypergroupoids. Sci. Math. Jpn. 80(3):307–314, 2017. [8] N. Kehayopulu. How we pass from semigroups to hypersemigroups. Lobachevskii J. Math. 39(1):121–128, 2018. [9] N. Kehayopulu. On ordered hypersemigroups with idempotent ideals, prime or weakly prime ideals. Eur. J. Pure Appl. Math. 11(1):10–22, 2018. [10] N. Kehayopulu. On ordered hypersemigroups given by a table of multiplication and a figure. Turkish J. Math., submitted. [11] N. Kehayopulu, M. Tsingelis. Remark on ordered semigroups. In: Partitions and holomorphic mappings of semigroups (Russian) 50–55, “Obrazovanie”, St. Petersburg 1992. [12] N. Kehayopulu, M. Tsingelis. On the decomposition of prime ideals of ordered semigroups into their N -classes. Semigroup Forum 47(3):393–395, 1993. [13] M. Petrich. Introduction to Semigroups, Charles E. Merrill Publishing Company, A Bell & Howell Company. Columbus, Ohio 1973. viii+198 pp.