ON QUASI-LATTICE IMPLICATION ALGEBRAS

Abstract

The notion of quasi-lattice implication algebras is a generalization of lattice implication algebras. In this paper, we give an optimized definition of quasi-lattice implication algebra and show that this algebra is a distributive lattice and that this algebra is a lattice implication algebra. Also, we define a congruence relation Φ_F induced by a filter F and show that every congruence relation on a quasi-lattice implication algebra is a congruence relation Φ_F induced by a filter F.

1. Introduction

The notion of lattice implication algebras was introduced in [8] to research a lattice-valued logic, which is a logical system equipped with a logical implication and an involution unary operation on a lattice. This logical system was studied from the algebraic viewpoint in many literature [5,7,8,9], and some operators on this algebra was studied in [4,11]

A lattice implication algebra is a bounded lattice (L, ∧, ∨, 0, 1) with a binary operation “ → ” and an order-reversing involution “ ′ ” satisfying the following axioms: for all x, y, z ∈ L,

The notion of filters on algebras with implication was introduced and studied in [3,6]. This filter is known as deductive filter and different from the notion of filters on lattices. This filter was proposed and studied as the notion of filters on lattice implication algebras in [1,2,10,12]. On a lattice implication algebra, the filter of lattice is the generalized concept of the filter.

A quasi-lattice implication algebra was introduced in [5, 7] as an algebraic system (L,∧,∨,→, ′, 0, 1) satisfying the axioms (I1)-(I5). This algebra is a generalization of lattice implication algebras and has the binary operation → and the involution ′ in the axioms for definition.

In this paper, the quasi-lattice implication algebra will be more clearly defined, and we show that this algebra is a distributive lattice, and hence this algebra is a lattice implication algebra. Also, an alternative definition of quasilattice implication algebra will be given. In section 3, we define a congruence relation ΦF induced by a filter F and show that every congruence relation on a quasi-lattice implication algebra is a congruence relation ΦF induced by a filter F.

 

2. Quasi-lattice implication algebra

We will define the notion of quasi-lattice implication algebras by the following optimized type, and x → y will be denoted by xy.

Definition 2.1. A quasi-lattice implication algebra is an algebraic system (L, ·, ′, 1) with a binary operation “ · ”, an involution “ ′ ” and an element 1 satisfying the following axioms: for all x, y, z ∈ L,

In the definition of quasi-lattice implication algebra L, the involution ′ is an unary operation on L such that x′′ = x for every x ∈ L.

Lemma 2.2. Let L be a quasi-lattice implication algebra. Then L satisfies the following: for all x ∈ L,

Proof. (1) Let x ∈ L. Then x(1x) = 1(xx) = 11 = 1 by (Q1) and (Q2). Also, we have (1x)x = (x1)1 = (x1)(11) = 1((x1)1) = 1((1x)x) = (1x)(1x) = 1 by (Q3), (Q2) and (Q1). Hence x(1x) = (1x)x = 1. This implies x = 1x by (Q4).

(2) Let x ∈ L. Then x1 = (1x)1 = (1x)(xx) = x((1x)x) = x(x1)1) = (x1)(x1) = 1 by (1) of this lemma, (Q2), (Q1) and (Q3). □

Lemma 2.3. Let L be a quasi-lattice implication algebra. If we define a binary relation " ≤ " by

for any x, y ∈ L, then (L,≤) is a poset with the greatest element 1 and the smallest element 1′.

Proof. For every x ∈ L, x ≤ x by (Q2), and for any x, y ∈ L, x ≤ y and y ≤ x imply x = y by (Q4).

To show the transitivity, let x ≤ y and y ≤ z. Then xy = 1 and yz = 1, and we have xz = x(1z) = x((yz)z) = x((zy)y) = (zy)(xy) = (zy)1 = 1 by 2.2, (Q3) and (Q1). This implies x ≤ z. Hence (L,≤) is a poset.

Also, 1 is the greatest element in L by 2.2(2). Let x ∈ L. Then 1 = x′1 = 1′x′′ = 1′x by 2.2(2), (Q5) and the definition of involution ′. This implies 1′ ≤ x for every x ∈ L. Hence 1′ is the smallest element in L. □

We will denote the smallest element 1′ in L by 0.

Lemma 2.4. Let L be a quasi-lattice implication algebra. Then L satisfies the following: for every x, y, z ∈ L,

Proof. (1) It is clear that 1′ = 0 by 2.3, and 0′ = 1′′ = 1.

(2) For every x ∈ L, x′ = 1x′ = x′′1′ = x0 by 2.2(1), (Q5) and (1) of this lemma.

(3) Let x ≤ yz. Then y(xz) = x(yz) = 1. Hence y ≤ xz.

(4) Let x, y ∈ L. Then we have

Hence xy ≤ (yz)(xz).

(5) Let x, y, z ∈ L. Then zx ≤ (xy)(zy) by (4) of this lemma. Hence xy ≤ (zx)(zy) by (3) of this lemma.

(6) Let x ≤ y. Then 1 = xy ≤ (yz)(xz) by (4) of this lemma. This implies (yz)(xz) = 1. Hence yz ≤ xz. Also, 1 = xy ≤ (zx)(zy) by (5) of this lemma. This implies (zx)(zy) = 1, and hence zx ≤ zy.

(7) Let x, y ∈ L. Then x((xy)y) = (xy)(xy) = 1. Hence x ≤ (xy)y.

(8) Let x, y ∈ L. Then y(xy) = x(yy) = x1 = 1. Hence y ≤ xy.

(9) Let x ≤ y, Then 1 = xy = y′x′. Hence y′ ≤ x′.

(10) Let x, y ∈ L. Then xy ≤ ((xy)y)y by (7) of this lemma. Also, since x ≤ (xy)y, ((xy)y)y ≤ xy by (6) of this lemma. Thus xy = ((xy)y)y. □

Theorem 2.5. A quasi-lattice implication algebra L is a lattice with

for every x, y ∈ L.

Proof. Let x, y ∈ L. Then (xy)y is an upper bound of x and y by (7) and (8) of 2.4. Suppose that u is an upper bound of x and y. Then x ≤ u and y ≤ u, and yu = 1. This implies uy ≤ xy, and

by 2.4(6) and (Q3). Hence (xy)y is the least upper bound of x and y, and x ∨ y = (xy)y.

Also, since x′ ≤ x′ ∨ y′ and y′ ≤ x′ ∨ y′, (x′ ∨ y′)′ ≤ x′′ = x and (x′ ∨ y′)′ ≤ y′′ = y by 2.4(9). Hence (x′ ∨ y′)′ is a lower bound of x and y. Suppose that l ≤ x and l ≤ y. Then x′ ≤ l′ and y′ ≤ l′. This implies x′ ∨ y′ ≤ l′. Hence l = l′′ ≤ (x′ ∨ y′)′. This means (x′ ∨ y′)′ is the greatest lower bound of x and y, and x ∧ y = (x′ ∨ y′)′. □

Lemma 2.6. Let L be a quasi-lattice implication algebra. Then L satisfies the following: for every x, y, z ∈ L,

Proof. (1) Let x, y ∈ L. Then x′ ∧ y′ = (x′′ ∨ y′′)′ = (x ∨ y)′ by 2.5.

(2) Let x, y ∈ L. Then (x ∧ y)′ = (x′ ∨ y′)′′ = x′ ∨ y′ by 2.5.

(3) Let x, y, z ∈ L. Then x ≤ x ∨ y and y ≤ x ∨ y. This implies (x ∨ y)z ≤ xz and (x ∨ y)z ≤ yz by 2.4(6). Hence (x ∨ y)z ≤ (xz) ∧ (yz). Also, since (xz) ∧ (yz) ≤ xz and (xz) ∧ (yz) ≤ yz, we have

by 2.4(3). This implies x ∨ y ≤ ((xz) ∧ (yz))z, and (xz) ∧ (yz) ≤ (x ∨ y)z by 2.4(3). Hence (x ∨ y)z = (xz) ∧ (yz).

(4) Let x, y, z ∈ L. Then we have z(x ∧ y) = (x ∧ y)′z′ = (x′ ∨ y′)z′ = (x′z′) ∧ (y′z′) = (zx) ∧ (zy) by (Q5) and (3) of this lemma. □

Theorem 2.7. Let L be a quasi-lattice implication algebra. Then L is distributive.

Proof. Let x, y, z ∈ L. Then it is clear that (x ∧ y) ∨ (x ∧ z) ≤ x ∧ (y ∨ z) since x ∧ y ≤ x ∧ (y ∨ z) and x ∧ z ≤ x ∧ (y ∨ z). Conversely, we have

since (x′y′) ∧ (x′z′) ≤ x′y′ and (x′y′) ∧ (x′z′) ≤ x′z′. This implies

Hence x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z). □

In a lattice L, it is well known that the following are equivalent:

This mean that a lattice L is a distributive if and only if x∨(y∧z) = (x∨y)∧(x∨z) for every x, y, z ∈ L.

Theorem 2.8. Every quasi-lattice implication algebra is a lattice implication algebra.

Proof. Let (L,→, ′, 1) be a quasi-lattice implication algebra. Then it satisfies the axioms (I1)-(I5) from the definition. Also L is a lattice and satisfies the axiom (L1) by 2.6(3). So we need only to show that it satisfies the axiom (L2).

Let x, y, z ∈ L. Then we have

This implies

since z ≤ (xz) ∨ (yz). Thus L satisfies the axiom (L2) and so it is a lattice implication algebra. □

It is clear that a lattice implication algebra is a quasi-lattice implication algebra. From the above theorem, the notion of quasi-lattice implication algebras is equivalent to that of lattice implication algebras.

Theorem 2.9.A set L is a quasi-lattice implication algebra if and only if there are a binary operation · on L and two elements 0, 1 in L satisfying the following: for every x, y, z ∈ L,

Proof. Let L be a quasi-lattice implication algebra. Then it satisfies the properties (Q1)-(Q4) and (B) by the definition of quasi-lattice implication algebra and 2.3.

Conversely, suppose that L be a set with a binary operation · and two elements 0, 1 satisfying the properties (Q1)-(Q4) and (B). Then we need to show that there is an involution ′ on L satisfying (Q5) : xy = y′x′ for every x, y ∈ L.

It can be proved that 1x = x for every x ∈ L in the same way as the proof of 2.2(1). Let x′ = x0. Then we have x′′ = (x0)0 = (0x)x = 1x = x by (Q3) and (B). So ′ is an involution, and it satisfies the following.

by (Q1). Hence (L, ·,′ , 1) is a quasi-lattice implication algebra. □

 

3. Congruence relations on quasi-lattice implication algebras

A subset F of a quasi-lattice implication algebra L is called a filter of L if it satisfies the following: for any x, y ∈ L,

Lemma 3.1. If F is a filter of a quasi-lattice implication algebra L, then F is a lattice filter of lattice L, i.e., F satisfies the following:

Proof. (1) Let x ∈ F and x ≤ y. Then xy = 1 ∈ F. Hence y ∈ F since x ∈ F.

(2) Let x, y ∈ F. Then y ≤ xy. This implies xy ∈ F by (1) of this lemma, and x(x ∧ y) = (xx) ∧ (xy) = 1 ∧ (xy) = xy ∈ F by 2.6(4). Since x(x ∧ y) ∈ F and x ∈ F, x ∧ y ∈ F. □

The converse of Lemma 3.1 is not true in general, as the following example shows.

Example 3.2. Let Q = {0, a, b, c, d, 1} be a set with a binary operation · defined by the following Cayley table:

If we define x′ = x0 for every x ∈ Q, then (Q, ·, ′, 1) is a quasi-lattice implication algebra. Also, it is a lattice with x∨y = (xy)y and x∧y = (x′∨y′)′ = ((x′y′)y′)′. This lattice is depicted by Hasse diagram of Figure 1. Let F = {a, c, d, 1}. Then

Figure 1.Hasse diagram of a lattice Q

F is a lattice filter of lattice Q, but it is not filter of Q, because a ∈ F and ab = d ∈ F but b ≠ F.

For any filter F of a quasi-lattice implication algebra L, we can define a binary relation ΦF on L by

for any x, y ∈ L.

Lemma 3.3. Let F be a filter of L. Then ΦF is a congruence relation.

Proof. For any x ∈ L, it is clear that xΦFx, since xx = 1 ∈ F, and that xΦFy implies yΦFx.

To show the transitivity of ΦF, let xΦFy and yΦFz. Then xy, yx ∈ F and yz, zy ∈ F. Since xy ≤ (yz)(xz) by 2.4(4), (yz) (xz) ∈ F by 3.1(1). This implies xz ∈ F since yz ∈ F. Also, we can see zx ∈ F in the similar way. Hence xΦFz. Thus ΦF is an equivalence relation in L.

To show that ΦF is a congruence relation on L, let xΦFy and z ∈ L. Then xy, yx ∈ F. Since xy ∈ F and xy ≤ (zx)(zy) by 2.4(5), (zx)(zy) ∈ F. Also, since yx ∈ F and yx ≤ (zy)(zx), (zy)(zx) ∈ F. Thus (zx)ΦF(zy), and ΦF is left compatible. In the similar way, we can show (xz)(yz), (yz)(xz) ∈ F by 2.4(4). That is (xz)ΦF(yz), and ΦF is right compatible. Hence ΦF is a congruence relation on L. □

We will call this relation ΦF as a congruence relation induced by a filter F.

For any equivalence relation Θ on a quasi-lattice implication algebra L, we will write [x]Θ for the equivalence classes.

Lemma 3.4. Let L be a quasi-lattice implication algebra. If Θ is a congruence relation on L, then [1]Θ is a filter of L.

Proof. Let Θ be a congruence relation on L. Then it is trivial that 1 ∈ [1]Θ. If x ∈ [1]Θ and xy ∈ [1]Θ, then xΘ1 and xyΘ1 imply xyΘ1y and xyΘ1, hence yΘ1 since 1y = y. This implies y ∈ [1]Θ. Thus [1]Θ is a filter of L. □

Theorem 3.5. Every congruence relation on a quasi-lattice implication algebra L is a congruence relation induced by a filter.

Proof. Suppose that Θ is a congruence relation on L. Then [1]Θ is a filter by 3.4. Set F = [1]Θ, and we will show that Θ = ΦF.

Let xΘy. Then xyΘyy and yxΘyy, since Θ is a congruence relation. This implies xyΘ1 and yxΘ1, and xy, yx ∈ [1]Θ = F. Thus xΦFy. Also, let xΦFy. Then xy, yx ∈ F = [1]Θ. This implies xyΘ1 and yxΘ1, and (xy)yΘ1y and (yx)xΘ1x, i.e., (xy)yΘy and (yx)xΘx. Hence xΘy since (xy)y = (yx)x.

This mean Θ = ΦF, and Θ is a congruence relation induced by the filter F = [1]Θ. □

Let L be a quasi-lattice implication algebra. Then the family Fil(L) (resp. Con(L)) of all filters of L (resp. all congruence relations on L) is partially ordered by set inclusion, and it is a complete lattice with

for arbitrary subset {Fα | α ∈ Λ} of Fil(L) (resp. {Θα | α ∈ Λ} of Con(L)), where ⟨X⟩ is the filter (resp. the congruence relation ) generated by a subset X of L (resp. of L × L).

Lemma 3.6. Let L be a quasi-lattice implication algebra. Then it satisfies the following:

Proof. (1) Let F ∈ Fil(L). Then we have

Hence [1]ΦF = F.

(2) It was proved in the proof of 3.5.

(3) Let F ⊆ G in Fil(L) and xΦFy. Then xy, yx ∈ F, and xy, yx ∈ G since F ⊆ G. This implies xΦGy. Hence ΦF ⊆ ΦG. Conversely, let ΦF ⊆ ΦG and x ∈ F. Then x1 = 1 ∈ F, since F is a filter, and 1x = x ∈ F. This implies xΦF1, and xΦG1 since ΦF ⊆ ΦG. This implies 1x = x ∈ G. Hence F ⊆ G.

(4) Let Θ,Ψ ∈ Con(L). Then since Θ = Φ[1]Θ and Ψ = Φ[1]Ψ by (2) of this lemma, we have

by (3) of this lemma. □

Theorem 3.7. Let L be a quasi-lattice implication algebra and ϕ : Fil(L) → Con(L) a map defined by ϕ(F) = ΦF for every F ∈ Fil(L). Then it satisfies the following:

Proof. (1) Let ϕ : Fil(L) → Con(L) be a map defined by ϕ(F) = ΦF for every F ∈ Fil(L). Then F ⊆ G if and only if ϕ(F) = ΦF ⊆ ΦG = ϕ(G) by 3.6(3). Hence ϕ is order-embedding. Also let Θ ∈ Con(L). Then there exists F = [1]Θ ∈ Fil(L) such that ϕ(F) = Φ[1]Θ = Θ by 3.6(2). Hence ϕ is onto.

(2) Let {Fα | α ∈ Λ} be an arbitrary subset of Fil(L). Then ϕ(Fβ) ⊆ ϕ(∨α∈Λ Fα) for every β ∈ Λ, since ϕ is order-preserving by (1) of this theorem. Hence ϕ(∨α∈Λ Fα) is an upper bound of the set {ϕ(Fα) | α ∈ Λ}

Suppose that ϕ(Fβ) = ΦFβ ⊆ Θ for every β ∈ Λ. Then by (1) and (4) of 3.6, Fβ = [1]ΦFβ ⊆ [1]Θ for every β ∈ Λ. This implies ∨ α∈Λ Fα ⊆ [1]Θ, and hence

Thus ϕ(∨α∈Λ Fα) is the least upper bound of the set {ϕ(Fα) | α ∈ Λ}. Hence ϕ (∨α∈Λ Fα) = ∨α∈Λ ϕ(Fα). Also, we can show ϕ(∧α∈Λ Fα) = ∧α∈Λ ϕ(Fα) in the similar way. □

From the above theorem, Fil(L) and Con(L) of a quasi-lattice implication algebra L have the same structure as complete lattices.