PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 81(95) (2007), 95–102 DOI 102298/PIM0795095R

ON REGULAR ANTI-CONGRUENCE IN ANTI-ORDERED SEMIGROUPS

Daniel Abraham Romano

Abstract. For an anti-congruence q we say that it is regular anti-congruence on semigroup (S, =, ≠, ·, α) ordered under anti-order α if there exists an antiorder θ on S/q such that the natural epimorphism is a reverse isotone homomorphism of semigroups. Anti-congruence q is regular if there exists a quasi-antiorder σ on S under α such that q = σ ∪ σ^{−1}. Besides, for regular anti-congruence q on S, a construction of the maximal quasi-antiorder relation under α with respect to q is shown.

1. Introduction and preliminaries

This short investigation in Bishop’s Constructive Algebra is a continuation of [9] and [10]. Bishop’s Constructive Mathematics is developed on Constructive Logic [11] – logic without the Law of Excluded Middle P ∨ ¬P. Let us note that in the Constructive Logic the ‘Double Negation Law’ P ⇔ ¬¬P does not hold, but the following implication P ⇒ ¬¬P holds even in the Minimal Logic. We have to note that ‘the crazy axiom’ ¬P ⇒ (P ⇒ Q) is included in the Constructive Logic. In the Constructive Logic ‘Weak Law of Excluded Middle’ ¬P ∨ ¬¬P does not hold, too. It is interesting, that in the Constructive Logic the following deduction principle A ∨ B, A ⊢ B holds, but this is impossible to prove without ‘the crazy axiom’. Bishop’s Constructive Mathematics is consistent with the Classical Mathematics.

Relational structure (S, =, ≠), where the relation ≠ is a binary relation on S, which satisfies the following properties:

¬(x ≠ x), x ≠ y ⇒ y ≠ x, x ≠ z ⇒ x ≠ y ∨ y ≠ z, x ≠ y ∧ y = z ⇒ x ≠ z

we call set. Following Heyting, the relation ≠ is called apartness. A relation q on S is a coequality relation on S if and only if it is consistent, symmetric and cotransitive [6]–[8]: q ⊆ ≠, q^{−1} = q, q ⊆ q ∗ q, where ∗ is the filled product between relations [5]. Let β be a consistent relation on S. We put ^{1}β = β and (^{n}β) = β ∗ ⋯ ∗ β (n factors, n ∈ N). Then the relation c(β) = ⋂_{n∈N}(^{n}β), the cotransitive fulfillment of β, is the maximal consistent and cotransitive relation on S under β [5].

2000 Mathematics Subject Classification: Primary 03F65; Secondary 06F05, 20M10.

Key words and phrases: Constructive mathematics, semigroup with apartness, anti-ordered semigroup, anti-congruence, regular anti-congruence, quasi-antiorder.

Supported by the Ministry of sciences and technology of the Republic of Srpska, Bosnia and Herzegovina.

Let (S, =, ≠, ·) be a semigroup with an apartness [6]–[8]. As in [8], a coequality relation q on S is anti-congruence if and only if it is compatible with the semigroup operation in the following sense

(∀x, y, z ∈ S)(((xz, yz) ∈ q ⇒ (x, y) ∈ q) ∧ ((zx, zy) ∈ q ⇒ (x, y) ∈ q)).

A relation α on S is an anti-order [9] on S if and only if α ⊆ ≠, α ⊆ α ∗ α, ≠ ⊆ α ∪ α^{−1} (linearity) and

(∀x, y, z ∈ S)(((xz, yz) ∈ α ⇒ (x, y) ∈ α) ∧ ((zx, zy) ∈ α ⇒ (x, y) ∈ α)).

A relation τ on S is a quasi-antiorder [9] on S if τ ⊆ ≠, τ ⊆ τ ∗ τ and

(∀x, y, z ∈ S)(((xz, yz) ∈ τ ⇒ (x, y) ∈ τ) ∧ ((zx, zy) ∈ τ ⇒ (x, y) ∈ τ)).

Let x be an element of S and A a subset of S. We denote x ▷◁ A if and only if (∀a ∈ A)(x ≠ a), and A^{C} = {x ∈ S : x ▷◁ A}. If τ is a quasi-antiorder on S, then the relation q = τ ∪ τ^{−1} is an anti-congruence on S. Firstly, the relation q^{C} = {(x, y) ∈ S × S : (x, y) ▷◁ q} is a congruence on S compatible with q, in the following sense

(∀a, b, c ∈ S)((a, b) ∈ q^{C} ∧ (b, c) ∈ q ⇒ (a, c) ∈ q).

We can construct the semigroup S/(q^{C}, q) = {aq^{C} : a ∈ S} with

aq^{C} = bq^{C} ⇔ (a, b) ▷◁ q, aq^{C} ≠ bq^{C} ⇔ (a, b) ∈ q, aq^{C} · bq^{C} = (ab)q^{C}.

We can also construct the semigroup S/q = {aq : a ∈ S} with

aq = bq ⇔ (a, b) ▷◁ q, aq ≠ bq ⇔ (a, b) ∈ q, aq · bq = (ab)q.

It is easy to check that S/(q^{C}, q) ≅ S/q. The mapping π : S → S/q, defined by π(a) = aq for any a ∈ S, is a strongly extensional epimorphism. Secondly, note that the relation α^{C} is an order relation on set (S, ¬ ≠, ≠). If the relation ¬α is an order relation on set (S, =, ≠), then, as for example in [1] when the apartness is tight, ¬ ≠ ⊆ = [7], the relation α is called excise relation on S. (The notion of anti-order relation is more general than notion of excise relation.)

For a given anti-ordered semigroup (S, =, ≠, ·, α) it is essential to know if there exists an anti-congruence q on S such that S/q be an anti-ordered semigroup. This plays an important role for studying the structure of anti-ordered semigroups. The following question is natural: If (S, =, ≠, ·, α) is an anti-ordered semigroup and q an anti-congruence on S, is then the set S/q an anti-ordered semigroup? A possible anti-order on S/q could be the relation Θ on S/q defined by the anti-order α on S, Θ = {(xq, yq) ∈ S/q × S/q : (x, y) ∈ α}. But is not an anti-order, in general. The following question arises: Is there any anti-congruence q on S for which S/q is an anti-ordered semigroup? The concept of quasi-antiorder relation was defined in [9]. According to [9], if (S, =, ≠, ·, α) is an anti-ordered semigroup and σ a quasi-antiorder on S, then the relation q on S, defined by q = σ ∪ σ^{−1}, is an anti-congruence on S and the set S/q is an anti-ordered semigroup under anti-order Θ defined by (xq, yq) ∈ Θ ⇔ (x, y) ∈ σ. So, according to the results in [9], each quasi-antiorder σ on an ordered semigroup S under anti-order α induces an anti-congruence q = σ ∪ σ^{−1} on S such that S/q is an ordered semigroup under anti-order Θ. In [10] we proved that the converse of this statement also holds. If (S, =, ≠, ·, α) is an anti-ordered semigroup and q anti-congruence on S and if there exists an anti-order relation Θ_{1} on S/q such that (S/q, =_{1}, ≠_{1}, ∘, Θ_{1}) is an ordered semigroup under anti-order Θ_{1}, then there exists a quasi-antiorder τ on S such that q = τ ∪ τ^{−1} and Θ_{1} = Θ. So, each anti-congruence q on a semigroup (S, =, ≠, ·, α) such that S/q is an anti-ordered semigroup induces a quasi-antiorder on S. This was the motivation for introduction of a new notion. For that we need the following notion: Let f be a strongly extensional mapping of anti-ordered sets from (X, =, ≠, α) into (Y, =, ≠, β). For f we say that it is reverse isotone if

(∀a, b ∈ X)((f(a), f(b)) ∈ β ⇒ (a, b) ∈ α)

holds. An anti-congruence q on S is called regular if there is an anti-order “θ_{1}” on S/q satisfying the following conditions:

(1) (S/q, =_{1}, ≠_{1}, , θ_{1}) is an anti-ordered semigroup;

(2) The mapping π : S ∋ a ↦ aq ∈ S/q is an anti-order reverse isotone epimorphism.

We call the anti-order “θ_{1}” on S/q a regular anti-order with respect to a regular anti-congruence q on S and the anti-order α.

It is obvious that the regular anti-order on S/q with respect to a regular anticongruence q and to the anti-order α on S is in general not unique. The following questions now naturally arise: Does there exist the maximal regular anti-order on S/q with respect to a regular anti-congruence q on S? Are all anti-congruences on anti-ordered semigroups regular? In this note, we give a partial answer on the questions above. In Theorem 1 and Corollary 2 we give necessary and sufficient conditions for anti-congruence on an anti-ordered semigroup to be regular. In Theorem 3 we give a construction of the maximal quasi-antiorder on anti-ordered semigroup S induced by a regular anti-congruence q on S.

For the necessary undefined notions, the reader is referred to books [2]–[4], [11] and to papers [5]–[8].

Lemma 0. Let τ be a quasi-antiorder on set (S, =, ≠). Then xτ (τx) is a strongly extensional subset of S, such that x ▷◁ xτ (x ▷◁ τx), for each x ∈ S. Also, the implication (x, z) ∈ τ ⇒ xτ ∪ τz = S holds for each x, z of S.

Proof. From τ ⊆≠ it follows x ▷◁ xτ. Let y ∈ xτ and let z be an arbitrary element of S. Then, (x, y) ∈ τ and (x, z) ∈ τ ∨(z, y) ∈ τ. So, we have z ∈ xτ ∨y ≠ z. Therefore, xτ is a strongly extensional subset of S such that x ▷◁ xτ.

The proof that τx is a strongly extensional subset of S such that x ▷◁ τx is analogous. Besides, the following implication (x, z) ∈ τ ⇒ xτ ∪ τz = S holds for each x, z of S. Indeed, if (x, z) ∈ τ and y is an arbitrary element of S, then (∀y ∈ S)((x, y) ∈ τ ∨ (y, z) ∈ τ). Thus, S = xτ ∪ τz. □

2. Regular anti-congruences

In order to obtain the relationship between regular anti-congruence and quasiantiorder on S, the following theorem is essential.

Theorem 1. Let (S, =, ≠, ·, α) be an anti-ordered semigroup, q an anti-congru-ence on S. The following statements are equivalent:

(1) q is regular.

(2) There exists a quasi-antiorder σ on S, such that q = σ ∪ σ^{−1}.

Proof. (2) ⇒ (1). By Lemma 1 in [9], since q = σ ∪ σ^{−1}, we have that the quotient semigroup (S/q, =_{1}, ≠_{1}, ·) is an anti-ordered semigroup with respect to the anti-order θ defined as (qx, qy) ∈ θ ⇔ (x, y) ∈ σ. If x, y ∈ S and (qx, qy) ∈ θ, then (x, y) ∈ σ ⊆ α. By definition, q is regular.

(1) ⇒ (2). Let q be a regular anti-congruence. Then there exists an anti-order relation θ on the quotient semigroup (S/q, =_{1}, ≠_{1}, ·) such that (S/q, =_{1}, ≠_{1}, ·, θ) is an anti-ordered semigroup, and π : S → S/q is a strongly extensional reverse isotone homomorphism of anti-ordered semigroups. Let σ = {(x, y) ∈ S ×S : (qx, qy) ∈ θ}. By [10], σ is a quasi-antiorder on S and it is easy to check that q = σ ∪ σ^{−1}. □

Corollary 2. Let (S, =, ≠, ·, α) be an anti-ordered semigroup, q an anticongruence on S. The following statements are equivalent:

(1) q is regular;

(2) There exists an anti-ordered semigroup (T, =, ≠, ·, θ) and a strongly extensional reverse isotone homomorphism φ : S → T such that q = {(a, b) ∈ S × S : φ(a) ≠ φ(b)}.

Proof. (1) ⇒ (2). Let q be regular. Then there exists an anti-order relation θ on the semigroup S/q such that the natural epimorphism π : S → S/q is a reverse isotone mapping. Then, by [10], there exists a quasi-antiorder σ on S such that q = σ ∪ σ^{−1}. So, there exists an anti-ordered semigroup T = (S/q, =_{1}, ≠_{1}, ·) under θ and a strongly extensional reverse isotone homomorphism π : S → T such that σ = {(a, b) ∈ S × S : (π(a), π(b)) ∈ θ}. Further on, we have q = {(a, b) ∈ S × S : π(a) ≠_{1} π(b)}. In fact,

(a, b) ∈ q ⇔ (a, b) ∈ σ ∨ (a, b) ∈ σ^{−1} ⇔ (π(a), π(b)) ∈ θ ∨ (π(a), π(b)) ∈ θ^{−1}

(π(a), π(b)) ∈ θ ∪ θ^{−1} = ≠_{1} ⇔ π(a) ≠_{1} π(b).

(2) ⇒ (1). Let T be an anti-ordered semigroup under an anti-order θ and φ : S → T a strongly extensional reverse isotone homomorphism such that q = {(a, b) ∈ S × S : φ(a) ≠ φ(b)}. Since θ is an anti-order relation on T, then ≠ = θ ∪ θ^{−1} holds. Thus, by Theorem in [10], the relation σ on S, defined by (a, b) ∈ σ if and only if (φ(a), φ(b)) ∈ θ, is a quasi-antiorder relation on S. On the other hand, q = σ ∪ σ^{−1}. In fact, if (a, b) is an arbitrary element of q, then

φ(a) ≠ φ(b) ⇔ (φ(a), φ(b)) ∈ θ ∨ (φ(b), φ(a)) ∈ θ ⇔ (a, b) ∈ σ ∨ (b, a) ∈ σ
⇔ (a, b) ∈ σ ∪ σ^{−1} = q

Let (φ(a), φ(b)) ∈ θ. Then (a, b) ∈ σ ⊆ α. By Theorem 1, q is regular anticongruence on S. □

Recall that, by Lemma 0, any class aq of anti-congruence q, generated by the element a ∈ S, is a strongly extensional subset of S. Besides, we have the following assertion, which is crucial in characterization of regular anti-congruences on an anti-ordered semigroup (S, =, ≠, ·, α): If q is a regular anti-congruence on an anti-ordered semigroup S, then for every q−class aq in S we have

((x, y) ▷◁ α ∧ (y, z) ▷◁ α ∧ x, z ∈ aq) ⇒ y ∈ aq

for any x, y, z, a ∈ S. If q is a regular anti-congruence on a semigroup S, then there exists an anti-order relation θ on S/q such that the natural epimorphism π : S → S/q is a strongly extensive reverse isotone homomorphism. Besides, there exists a quasi-antiorder σ under α, defined by (x, y) ∈ σ ⇔ (xq, yq) ∈ θ such that σ ∪ σ^{−1} = q. Let t be an arbitrary element of aq. Then (a, t) ∈ q = σ ∪ σ^{−1}. Thus (a, t) ∈ σ or (t, a) ∈ σ. Hence, we have

(a, t) ∈ σ ⇒ (a, x) ∈ σ ⊆ q ∨ (x, y) ∈ σ ⊆ α ∨ (y, t) ∈ σ ⊆ q ⊆ ≠ ⇒ t ≠ y;
(t, a) ∈ σ ⇒ (t, y) ∈ σ ⊆ ≠ ∨(y, z) ∈ σ ⊆ α ∨ (z, a) ∈ σ ⊆ q ⇒ t ≠ y.

So, in both cases, we have that t ∈ aq ⇒ t ≠ y. Therefore, y ▷◁ aq. We have

((x, y) ▷◁ α ∧ (y, z) ▷◁ α ∧ y ∈ aq) ⇒ x ∈ aq ∨ z ∈ aq

for any x, y, a ∈ S. Indeed, if x, y, z, a ∈ S such that (x, y) ▷◁ α and (y, z) ▷◁ α and x ∈ aq, then (a, y) ∈ q = σ ∪ σ^{−1} ⇒ ((a, y) ∈ σ ∨ (y, a) ∈ σ). Thus, we have

((a, y) ∈ σ ∨ (y, a) ∈ σ) ⇒

((a, x)∈σ ⊆q ∨ (x, y)∈σ ⊆α) ∨ ((y, z)∈σ ⊆α ∨ (z, a)∈σ ⊆q) ⇒ x ∈ aq ∨ z ∈ aq.

It is not known whether the condition given above on q-classes is sufficient for regularity of an anti-congruence on an anti-ordered semigroup.

Example. We consider the anti-ordered set S = {a, b, c, d, e, f} under the relation α = S×S∖{(a, a), (a, d), (a, e), (b, b), (b, e), (c, c), (c, b), (c, e), (d, d), (d, e), (e, e), (f, f), (f, a), (f, b), (f, c), (f, d), (f, e)}. Define a coequality relation q on S by q = S × S ∖ {(a, a), (b, b), (b, c), (b, d), (c, c), (c, b), (c, d), (d, d), (d, c), (d, b), (e, e), (f, f)}. Then

S/q = {aq = {b, c, d, e, f}, bq = {a, e, f}, cq = {a, e, f},
dq = {a, e, f}, eq = {a, b, c, d, f}, fq = {a, b, c, d, e}}.

The following relation is an anti-order relation on S/q

θ_{1} = ℘(S) × ℘(S) ∖ {({f}, {f}), ({f}, {a}), ({f}, {b, d, c}), ({f}, {e}), ({a}, {a}),

({a}, {b, d, c}), ({a}, {e}), ({b, d, c}, {b, d, c}), ({b, d, c}, {e}), ({e}, {e})}.

Then (S/q, =_{1}, ≠_{1}, θ_{1}) is an anti-ordered set, q is a regular coequality on S. If in S we define the internal operation by the table below, then the set S is an anti-ordered semigroup, q is an anti-congruence on S. It is easy to check that q is a regular anticongruence on the semigroup S. The proof of these facts is straightforward.

· a b c d e f
a d d d d d a
b e e e e e b
c d d d d d c
d d d d d d d
e e e e e e e
f d d d d d f

Remark. If σ is a quasi-antiorder on S, then q = σ∪σ^{−1} is the minimal regular anti-congruence on S which contains σ. In fact, if ρ is a regular anti-congruence on S containing σ, then ρ = ρ ∪ ρ^{−1} ⊇ σ ∪ σ^{−1} = q.

Let q be a regular anti-congruence on an anti-ordered semigroup (S, =, ≠, ·, α). Then there exists anti-order θ on S/q such that the natural homomorphism π : S → S/q is reverse isotone. Hence, by [10], there exists a quasi-antiorder σ under α such that q = σ ∪ σ^{−1} and θ = {(aq, bq) ∈ S/q × S/q : (a, b) ∈ σ}. In the following theorem we show that there exists such maximal quasi-antiorder τ under α and we give a construction of that relation.

Theorem 3. Let q be a regular anti-congruence on an anti-ordered semigroup (S, =, ≠, ·, α). Then there exists the maximal quasi-antiorder relation τ under α such that q = τ ∪ τ^{−1} and θ ⊆ {(aq, bq) ∈ S/q × S/q : (a, b) ∈ τ}. This relation is exactly the following relation c(q ∩ α) = ⋂_{n∈N} (^{n}(q ∩ α)).

Proof. (1) It is clear that c(α ∩ q) ⊆ α ∩ q ⊆ q ⊆ ≠, c(q ∩ α) ⊆ α, and that the relation c(α ∩ q) is cotransitive [4]. In fact, for cotransitivness we need to prove that (a, c) ∈ c(q ∩ α) ⇒ (∀b ∈ S)((a, b) ∈ c(q ∩ α) ∨ (b, c) ∈ c(q ∩ α)), i.e., to prove that

(a, c) ∈ c(q ∩ α) ⇒

(∀b ∈ S)((∀i ∈ N)((a, b) ∈ (^{i}(q ∩ α))) ∨ (∀j ∈ N)((b, c) ∈ (^{j}(q ∩ α)))).

First, we have

(a, c) ∈ c(q ∩ α) ⇒ (a, c) ∈ (^{2}(q ∩ α)) = (q ∩ α) ∗ (q ∩ α)
⇒ (∀b ∈ S)((a, b) ∈ (q ∩ α) ∨ (b, c) ∈ (q ∩ α)).

Second, suppose that the following implication holds for some n ⩾ 2

(a, c) ∈ c(q ∩ α) ⇒ (∀b ∈ S)((a, b) ∈ (^{n}(q ∩ α)) ∨ (b, c) ∈ (^{n}(q ∩ α))).

Thus, since the filled product “∗” is associative, we have

(a, c) ∈ c(q ∩ α) ⇒ (a, c) ∈ (^{2(n+1)}(q ∩ α)) = (^{(n+1)}(q ∩ α)) ∗ (^{(n+1)}(q ∩ α))

⇔ (∀b ∈ S)((a, b) ∈ (^{(n+1)}(q ∩ α)) ∨ (b, c) ∈ (^{(n+1)}(q ∩ α))).

Therefore, for any natural number n, by induction, we have

(a, c) ∈ c(q ∩ α) ⇒

(∀b ∈ S)((∀i ⩽ n)((a, b) ∈ (^{i}(q ∩ α))) ∨ (∀j ⩽ n)((b, c) ∈ (^{j}(q ∩ α)))).

Hence (a, c) ∈ c(q ∩ α) ⇒ (∀b ∈ S)((a, b) ∈ c(q ∩ α) ∨ (b, c) ∈ c(q ∩ α)).

(2) Further on, let a, b, x be arbitrary elements of S such that (ax, bx) ∈ c(α∩q). Then (ax, bx) ∈ q and, by compatibility of α and q in S, we have (a, b) ∈ q. Suppose that the implication (ax, bx) ∈ c(α ∩ q) ⇒ (a, b) ∈ (^{n}(α ∩ q)), n ∈ N, holds for any a, b, x ∈ S. Then, from (ax, bx) ∈ c(α ∩ q) ∈ (^{n+1}(α ∩ q)) = (^{n}(α ∩ q)) ∗ (α ∩ q) it follows

(ax, bx) ∈ c(α ∩ q) ⊆ (^{n+1}(α ∩ q)) = (^{n}(α ∩ q)) ∗ (α ∩ q)
⇒ (∀y ∈ S)((ax, yx) ∈ (^{n}(α ∩ q)) ∨ (yx, bx) ∈ (α ∩ q))
⇒ (∀y ∈ S)((a, y) ∈ (^{n}(α ∩ q)) ∨ (y, b) ∈ (α ∩ q))
⇒ (a, b) ∈ (^{n+1}(α ∩ q)).

So, by induction, we have (ax, bx) ∈ c(α ∩ q) ⇒ (a, b) ∈ c(α ∩ q).

The other implication (xa, xb) ∈ c(α ∩ q) ⇒ (a, b) ∈ c(α ∩ q) can be proved analogously. Therefore, the relation c(α ∩ q) is compatible with the semigroup operation in S.

(3) Let σ be a quasi-antiorder relation under α such that q = σ ∪ σ^{−1} and θ = {(aq, bq) ∈ S/q × S/q : (a, b) ∈ σ}. Then σ ⊆ α ∩ q and σ = c(σ) ⊆ c(α ∩ q) because, by Lemma 0.4.2 in [7], the cotransitive fulfillment satisfies the implication σ ⊆ α ∩ q ⇒ c(σ) ⊆ c(α ∩ q) and, in addition, since σ is a cotransitive relation, σ = c(σ).

So, the relation τ = c(α ∩ q) is the maximal quasi-antiorder under α such that q = τ ∪τ^{−1}. Indeed, q = σ ∪σ^{−1} ⊆ τ ∪τ^{−1} ⊆ q. Besides, it is clear that the relation Θ = {(aq, bq) ∈ S/q × S/q : (a, b) ∈ τ} is an anti-order relation on S/q such that θ ⊆ Θ. □

Corollary 4. Let q be a regular anti-congruence on an anti-ordered semigroup (S, =, ≠, ·, α). Then there exists the maximal anti-order relation on S/q. This relation is exactly the following relation {(aq, bq) ∈ S/q × S/q : (a, b) ∈ c(q ∩ α)}.

Proof. Let θ_{1} be a regular anti-congruence on S/q with respect to α. Then there exists a quasi-antiorder σ on S such that q = σ ∪ σ^{−1} and θ_{1} = {(aq, bq) ∈ S/q×S/q : (a, b) ∈ σ}. Since c(α∩q) is the maximal quasi-antiorder with respect to α, then σ ⊆ τ holds. Thus, we have θ_{1} ⊆ {(aq, bq) ∈ S/q × S/q : (a, b) ∈ c(α ∩ q)}. So, there exists the maximal regular anti-congruence with respect to α. □

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