Balanced factor congruences of double MS-algebras

Abstract

In this paper, we introduce the concepts of Stone elements, central elements and Birkhoof central elements of a double MS-algebra and study their related properties. We observe that the center C(L) of a double MS -algebra L is precisely the Birkhoof center BC(L) of L. A complete description of factor congruences on a double MS-algebra is given by means of the central elements. A characterization of balanced factor congruences of double MS-algebra is obtained. A one-to-one correspondence between the class of all balanced factor congruences of a double MS-algebra L and the central elements of L is obtained.

Introduction

Blyth and Varlet [1] introduced MS-algebras as a generalization of both de Morgan algebras and Stone algebras. Blyth and Varlet [2] characterized the subvarieties of the class MS of all MS-algebras. Badawy, Guffova, and Haviar [3] introduced and characterized the class of principal MS-algebras and the class of decomposable MS-algebras by means of triples. Badawy [4] introduced and studied many properties of d_L-filters of principal MS-algebras. Also, Badawy and El-Fawal [5] considered homomorphisms and subalgebras of decomposable MS-algebras.

Blyth and Varlet [6] introduced the class of double MS-algebras and showed that every de Morgan algebra M can be represented non-trivially as the skeleton of the double MS-algebra M^[2]={(a,b)∈M×M:a≤b}. The class of double MS-algebras satisfying the complement property has been introduced by Congwen [7]. Haviar [8] studied affine complete of double MS-algebras from the class K₂, of all double K-algebras. Wang [9] introduced the notion of congruence pairs of double K₂-algebras. Recently, Badawy [10] introduced and constructed the class of double MS-algebras satisfying the generalized complement property that is containing the class of double MS -algebras satisfying the complement property.

In this paper, we introduce the notion of Stone elements in double MS -algebras. Then, we prove that the set of Stone elements of a double MS-algebra L forms the greatest Stone subalgebra of L. We introduce the concept of central elements of a double MS-algebra L and we show that the set C(L) of all central elements forms the greatest Boolean subalgebra of L. For a principal ideal (a] (filter [a)) of a double MS-algebra (L;^∘,⁺), it is observed that a relativized algebra $\phantom {\dot {i}\!}(a]_{L}=((a];\vee,\wedge,^{\circ _{a}},^{+_{a}},0,a)$ ($\phantom {\dot {i}\!}[a)_{L}=([a);\vee,\wedge,^{\circ _{a}},^{+_{a}},a,1)$) is a double MS-algebra if and only if a is a central element of L, where $\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a$ and $\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\wedge a$ ($\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\vee a$ and $\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\vee a$). Also, we introduce the Birkhoof center of a double MS-algebra, then we showed that the Birokhoof center of a double MS-algebra L can be identified with the center of L. Factor congruences of a double MS-algebra are investigated by means of central elements. Finally, we study and characterize balanced factor congruences of a double MS-algebra. There is one-to-one correspondence between the class of balanced factor congruences of a double MS-algebra L and the center C(L) of L.

Preliminaries

In this section, some definitions and results were introduced in [1, 2, 6, 11, 12].

A de Morgan algebra is an algebra (L;∨,∧,⁻,0,1) of type (2,2,1,0,0) where (L;∨,∧,0,1) is a bounded distributive lattice and ⁻ the unary operation of involution satisfies:

$\overline {\overline {x}}=x,\overline {(x\vee y)}=\overline {x}\wedge \overline { y},\overline {(x\wedge y)}=\overline {x}\vee \overline {y}.$

An MS-algebra is an algebra (L;∨,∧,^∘,0,1) of type (2,2,1,0,0) where (L;∨,∧,0,1) is a bounded distributive lattice and a unary operation ^∘ satisfies:

x≤x^∘∘,(x∧y)^∘=x^∘∨y^∘,1^∘=0.

The basic properties of MS-algebras are given in the following theorem.

Theorem 1

(Blyth and Varlet [6]) For any two elements a,b of an MS-algebra L, we have

(1) 0^∘∘=0 and 1^∘∘=1,

(2) a≤b⇒b^∘≤a^∘,

(3) a^∘∘∘=a^∘,

(4) a^∘∘∘∘=a^∘∘,

(5) (a∨b)^∘=a^∘∧b^∘,

(6) (a∨b)^∘∘=a^∘∘∨b^∘∘,

(7) (a∧b)^∘∘=a^∘∘∧b^∘∘.

A dual MS-algebra is an algebra (L;∨,∧,⁺,0,1) of type (2,2,1,0,0) where (L;∨,∧,0,1) is a bounded distributive lattice and the unary operation ⁺ satisfies:

x≥x⁺⁺,(x∧y)⁺=x⁺∨y⁺,0⁺=1.

Proposition 1

For any two elements a,b of a dual MS-algebra (L;⁺), we have

(1) 0⁺⁺=0 and 1⁺⁺=1,

(2) a≤b⇒b⁺≤a⁺,

(3) a⁺⁺⁺=a⁺,

(4) a⁺⁺⁺⁺=a⁺⁺,

(5) (a∨b)⁺=a⁺∧b⁺,

(6) (a∨b)⁺⁺=a⁺⁺∨b⁺⁺,

(7) (a∧b)⁺⁺=a⁺⁺∧b⁺⁺.

A double MS-algebra is an algebra (L;^∘,⁺) such that (L;^∘) is an MS-algebra, (L;⁺) is a dual MS-algebra, and the unary operations ^∘,⁺ are linked by the identities x^+∘=x⁺⁺ and x^∘+=x^∘∘, for all x∈L.

For any element x of a double MS-algebra L, it is clear that x⁺⁺≤x^∘∘ and consequently x^∘≤x≤x⁺.

Some subsets of a double MS-algebra play a significant role in the investigation, by the skeleton L^∘∘ of a double MS-algebra L we mean a de Morgan algebra

L^∘∘={x∈L:x=x^∘∘}=L⁺⁺={x∈L:x=x⁺⁺}={x∈L:x^∘=x⁺}.

An equivalence relation θ on a lattice L is called a lattice congruence on L if (a,b)∈θ and (c,d)∈θ implies (a∨c,b∨d)∈θ and (a∧c,b∧d)∈θ.

Theorem 2

(Blyth [12]) An equivalence relation on a lattice L is a lattice congruence on L if and only if (a,b)∈θ implies (a∨z,b∨z)∈θ and (a∧z,b∧z)∈θ for all z∈L.

A lattice congruence θ on a double MS-algebra (L;^∘,⁺) is called a congruence on L if (a,b)∈θ implies (a^∘,b^∘)∈θ and (a⁺,b⁺)∈θ.

We use ∇=L×L for the universal congruence on a lattice L and Δ={(a,a):a∈L} for the equality congruence on L.

We say the congruences θ,ψ on a lattice L are permutable if θ∘ψ=ψ∘θ, that is, x≡y(θ) and y≡z(ψ) imply x≡r(ψ) and r≡z(θ) for some y,r∈L.

Center and Birkhoof center of a double MS-algebra

We introduce the concept of Stone elements of a double MS-algebra L. Then, we show that the set L_S of all Stone elements of L is the greatest Stone subalgebra of L.

Definition 1

An element x of a double MS-algebra L is called a Stone element of L if x^∘∨x^∘∘=1 and x⁺∧x⁺⁺=0. Let L_S denote the set of all Stone elements of L, that is, L_S={x∈L:x^∘∨x^∘∘=1,x⁺∧x⁺⁺=0}.

Definition 2

Let L₁ be a bounded sublattice of a double MS-algebra L. Then, L₁ is called a subalgebra of L if x^∘,x⁺∈L₁ for every x∈L₁.

Definition 3

A subalgebra L₁ of a double MS-algebra L is called a Stone subalgebra if x^∘∨x^∘∘=1 and x⁺∧x⁺⁺=0, for all x∈L₁.

Proposition 2

L_S is the greatest Stone subalgebra of a double MS-algebra L.

Proof

It is clear that 0,1∈L_S. Let x,y∈L_S. Then, x^∘∨x^∘∘=1, x⁺∧x⁺⁺=0, y^∘∨y^∘∘=1, and y⁺∧y⁺⁺=0. Thus, we get

$$\begin{array}{@{}rcl@{}} (x\vee y)^{\circ}\vee(x\vee y)^{\circ\circ}&=&(x^{\circ}\wedge y^{\circ})\vee(x^{\circ\circ}\vee y^{\circ\circ})\text{ by Theorem 1(5),(6)}\\ &=&(x^{\circ}\vee x^{\circ\circ}\vee y^{\circ\circ})\wedge(y^{\circ}\vee x^{\circ\circ}\vee y^{\circ\circ})\text{ by distributivity of }L\\ &=&1\wedge 1=1\text{ as }x^{\circ}\vee x^{\circ\circ}=1,y^{\circ}\vee y^{\circ\circ}=1,\\ (x\vee y)^{+}\wedge(x\vee y)^{++}&=&(x^{+}\wedge y^{+})\wedge(x^{++}\vee y^{++})\text{ by Proposition 1(5),(6)}\\ &=&(x^{+}\wedge y^{+}\wedge x^{++})\vee(x^{+}\wedge y^{+}\wedge y^{++})\text{ by distributivity of }L\\ &=&0\vee 0=0\text{ as }x^{+}\wedge x^{++}=0,y^{+}\wedge y^{++}=0. \end{array} $$

Then, x∨y∈L_S. Using a similar way, we get x∧y∈L_S. Therefore, (L_S,∨,∧,0,1) is a bounded distributive sublattice of L. Now, we prove that x⁺∈L_S for all x∈L_S.

$$\begin{array}{@{}rcl@{}} x^{+\circ}\vee x^{+\circ\circ}&=&x^{++}\vee x^{+++}\text{as }x^{+\circ}=x^{++}\\ &=&(x^{+}\wedge x^{++})^{+}\text{ by Proposition 1(5)}\\ &=&0^{+}=1\text{ as }x^{+}\wedge x^{++}=0,\\ x^{+\circ}\wedge x^{+\circ\circ}&=&x^{++}\wedge x^{+++}\text{ as }x^{+\circ}=x^{++}\\ &=&x^{++}\wedge x^{+}=0\text{ by Proposition 1(3)}. \end{array} $$

Hence, x⁺∈L_S. Similarly, we can prove that x^∘∈L_S for all x∈L_S. Therefore, L_S is a subalgebra of a double MS-algebra L. Since x^∘∨x^∘∘=1 and x⁺∧x⁺⁺=0 for every x∈L_S, then L_S is a Stone subalgebra of L. To prove that L_S is the greatest Stone subalgebra of L, let S be any Stone subalgebra of L. Let x∈S. Then, x is a Stone element of L, and hence, x∈L_S. So S⊆L_S as claimed. □

On the following, we introduce the notion of central elements of a double MS-algebra L and prove that the set C(L) of all central elements of L is the greatest Boolean subalgebra of L. Also, the relationship among L_S, C(L), and L^∘∘ is investigated.

Definition 4

An element a of double MS-algebra L is called a central element if x∨x^∘=1 and x∧x⁺=0. The set of all central elements of L is called the center of L and is denoted by C(L), that is, C(L)={x∈L:x∨x^∘=1,x∧x⁺=0}.

Example 1

Consider the bounded distributive lattice L in Fig. 1. Define unary operations ^∘,⁺ on L by

$$ b^{\circ}=x^{\circ}=a, d^{\circ}=y^{\circ}=c, 1^{\circ}=z^{\circ}=0,0^{\circ}=1,c^{\circ}=d,a^{\circ}=b $$

(1)

Fig. 1

L

and

$$ a^{+}=z^{+}=b, c^{+}=y^{+}=d, 0^{+}=x^{+}=1,b^{+}=a,d^{+}=c,1^{+}=0. $$

(2)

It is clear that (L;^∘,⁺)is a double MS-algebra. Then, L^∘∘,L_S, and C(L) are given in Figs. 2, 3, and 4, respectively.

Fig. 2

L ^∘∘

Fig. 3

L _s

Fig. 4

C(L)

Theorem 3

Let L be a double MS-algebra. Then

(1) C(L)=L^∘∘∩L_S,

(2) C(L) is the greatest Boolean subalgebra of L, L_S, and L^∘∘,

(3) C(L)=C(L_S)=C(L^∘∘).

Proof

(1). Let x∈C(L). Then, x∨x^∘=1 and x∧x⁺=0. Then

$$\begin{array}{@{}rcl@{}} x^{++}&=&x^{++}\vee 0\\ &=&x^{++}\vee(x\wedge x^{+})\\&=&(x^{++}\vee x)\wedge(x^{++}\vee x^{+})\text{ by distributivity of }L\\&=&x\wedge(x^{+}\wedge x)^{+}\text{ as }x\geq x^{++}\\&=&x\wedge0^{+}=x\wedge 1=x. \end{array} $$

Thus, x∈L^∘∘. Also,

$$\begin{array}{@{}rcl@{}} x^{++}\wedge x^{+}&=&x^{++}\wedge x^{+++}\text{ by Proposition 1(3)}\\ &=&(x\wedge x^{+})^{++}\text{ by Proposition 1(7)}\\&=&0^{++}=0\text{ by Proposition 1(1)},\\ x^{\circ\circ}\vee x^{\circ}&\geq& x\vee x^{\circ}\hbox { as }x^{\circ\circ}\geq x\\&=&1. \end{array} $$

Then, x⁺⁺∧x⁺=0 and x^∘∘∨x^∘=1 imply x∈L_S. Therefore, C(L)⊆L^∘∘∩L_S. Conversely, let x∈L^∘∘∩L_S. Then, x=x^∘∘=x⁺⁺, x^∘∨x^∘∘=1, and x⁺∧x⁺⁺=0. Now,

$$\begin{array}{@{}rcl@{}} x^{\circ}\vee x&=&x^{\circ}\vee x^{\circ\circ} =1,\\ x\wedge x^{+} &=&x^{++}\wedge x^{+}=0. \end{array} $$

Thus, x∈C(L), and hence, L^∘∘∩L_S⊆C(L).

(2) Clearly 0,1∈C(L). Let a,b∈C(L). Then, we have

$$\begin{array}{@{}rcl@{}} (a\vee b)\vee(a\vee b)^{\circ}&=&a\vee b\vee(a^{\circ}\wedge b^{\circ})\text{ by Proposition 1(5)}\\ &=&(a\vee b\vee a^{\circ})\wedge(a\vee b\vee b^{\circ})\text{ by distributivity of }L\\ &=&1\wedge 1=1,\\ (a\vee b)\wedge(a\vee b)^{+}&=&(a\vee b)\wedge(a^{+}\wedge b^{+})\text{ by Theorem 1(5)}\\ &=&(a\wedge a^{+}\wedge a^{+})\vee(b\wedge a^{+}\wedge b^{+})\text{ by distributivity of }L\\ &=&0\vee 0=0. \end{array} $$

Then, a∨b∈C(L). Similarly a∧b∈C(L). Therefore, (C(L);∨,∧,0,1) is a bounded sublattice of L. Now, we observe that a^∘∈C(L) for all a∈C(L),

$$\begin{array}{@{}rcl@{}} a^{\circ\circ}\vee a^{\circ\circ\circ}&=&a\vee a^{\circ}\text{ as }a^{\circ\circ}=a,\forall a\in C(L)\text{ and }a^{\circ\circ\circ}=a^{\circ}\\ &=&1,\\ a^{\circ+}\wedge a^{\circ++}&=&a^{\circ\circ}\wedge a^{\circ\circ\circ}\text{ as }a^{\circ+}=a^{\circ\circ}\\ &=&a^{\circ\circ}\wedge a^{\circ}\text{ as }a^{\circ\circ\circ}=a^{\circ}\\ &=&(a^{\circ}\vee a)^{\circ}=1^{\circ}=0\text{ by Theorem 1(5)}. \end{array} $$

Since a^∘=a⁺ for all a∈C(L), then ^∘ coincide with ⁺ on C(L). Therefore, (C(L),∨,∧,^∘,0,1) is a subalgebra of L. Since a∨a^∘=1 and a∧a^∘=a^∘∘∧a^∘=(a^∘∨a)^∘=1^∘=0 for all a∈C(L), then (C(L),∨,∧,^∘,0,1) is a Boolean subalgebra of L. Suppose that B is any Boolean subalgebra of L and x∈B. Then, a∨a^∘=1 and a∧a⁺=a∧a^∘=0. Hence, a is a central element of L and a∈C(L). So, B⊆C(L) and C(L) is the greatest Boolean subalgebra of L. Using similar agrement, we can show that C(L) is also the greatest Boolean subalgebra of both L_S and L^∘∘.

(3) It follows (1) and (2). □

The following theorem shows that the centers of isomorphic double MS-algebras are isomorphic Boolean algebras.

Theorem 4

If L and M are isomorphic double MS-algebras, then their centers are isomorphic.

Proof

Let h:L→M be an isomorphism and a∈C(L). Then, a∨a^∘=1 and a∧a⁺=0. Hence, h(a∨a^∘)=h(a)∨h(a^∘)=h(a)∨(h(a))^∘=h(1)=1 and h(a)∧(h(a))⁺=h(0)=0. Therefore, h_C(L)(a)=h(a)∈C(M). It is clear that h_C(L) is an injective (0,1) lattice homomorphism. Let b∈C(M). Then, there exists b∈L such that b=h(a)=h_C(L)(a) as h is onto. It follows that b^∘∘=(h(a))^∘∘=h(a^∘∘)=h(a)=h_C(L)(a). Thus, h_C(L) is onto. Obviously, h_C(L) preserves ^∘ and ⁺. Then, h_C(L) is an isomorphism, and hence, C(L)≅C(M). □

For an MS-algebra (L,^∘), it is proved in [3] that $\phantom {\dot {i}\!}(a]_{L}=((a],^{\circ _{a}})$ is an MS-algebra if and only if a^∘∈C(L), where (a]={x∈L:x≤a}=[0,a] is a principal ideal of L generated by the element a of L, a unary operation $\phantom {\dot {i}\!}^{\circ _{a}}$ is defined on (a] by $\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a$ for all x∈(a] and C(L)={x∈L:x∨x^∘=1} is the center of L.

For a double MS-algebra (L;^∘,⁺), the answer of the following question is given: Under what conditions a principal ideal (a],a∈L constructs a double MS-algebra?

Theorem 5

Let L be a double MS-algebra. Suppose that a∈C(L), then the relativized algebra $\phantom {\dot {i}\!}(a]_{L}=((a],\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)$ is a double MS-algebra, where $\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a$ and $\phantom {\dot {i}\!}x^{+_{a}}=x^{+} \wedge a$. Conversely, if $\phantom {\dot {i}\!}(a]_{L}=((a],\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)$ is a double MS-algebra, then a∈L_S.

Proof

Assume that a∈C(L). Hence, a^∘∈C(L). Then, by ([13], Theorem 3.5), $\phantom {\dot {i}\!}((a],\vee,\wedge,^{\circ _{a}},0,a)$ is an MS-algebra, whenever $\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a$. Now, we prove that $\phantom {\dot {i}\!}((a],\vee,\wedge,^{+_{a}},0,a)$ is a dual MS-algebra, where $\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\wedge a$ for any x∈(a]. Let x∈(a], we have

$$\begin{array}{@{}rcl@{}} x^{+_{a}\circ+_{a}}\vee x&=&(x^{+}\wedge a)^{+_{a}}\vee x\\ &=&((x^{+}\wedge a)^{+}\wedge a)\vee x\\ &=&((x^{++}\vee a^{+})\wedge a)\vee x \\ &=&(x^{++}\wedge a)\vee(a^{+}\wedge a)\vee x\text{ by distributivity of }L\\ &=&(x^{++}\wedge a)\vee x\text{ as }a^{+}\wedge a=0\\ &=&x\text{ as }x\geq x^{++}\geq x^{++}\wedge a. \end{array} $$

Then, $\phantom {\dot {i}\!}x\geq x^{+_{a}+_{a}}$. Let x,y∈(a]

$$\begin{array}{@{}rcl@{}} (x\wedge y)^{+_{a}}&=&(x\wedge y)^{\circ}\wedge a\\ &=&(x^{+}\vee y^{\circ+})\wedge a\\ &=&(x^{+}\wedge a)\vee (y^{+}\wedge a)\text{ by distributivity of }L\\ &=&x^{+_{a}}\vee y^{+_{a}}, \end{array} $$

Also, $\phantom {\dot {i}\!}0^{+_{a}}=a$. Now, for every x∈(a], we have

$$\begin{array}{@{}rcl@{}} x^{\circ_{a}+_{a}}&=&(x^{\circ}\wedge a)^{+_{a}}\\ &=&(x^{\circ}\wedge a)^{+}\wedge a\\ &=&(x^{\circ+}\vee a^{+})\wedge a\\ &=&(x^{\circ\circ}\vee a^{+})\wedge a\\ &=&(x^{\circ\circ}\wedge a)\vee (a^{+}\wedge a)\\ &=&x^{\circ\circ}\wedge a\text{ as }a^{+}\wedge a=0,\\ x^{\circ_{a}\circ_{a}}&=&(x^{\circ}\wedge a)^{\circ}\wedge a\\ &=&(x^{\circ\circ}\vee a^{\circ})\wedge a\\ &=&(x^{\circ\circ}\wedge a)\vee(a^{\circ}\wedge a)\text{ by distributivity}\\ &=&x^{\circ\circ}\wedge a\text{ as }a^{+}\wedge a=0. \end{array} $$

This deduce that $\phantom {\dot {i}\!}x^{\circ _{a}+_{a}}=x^{\circ \circ }$. Also, we can get $\phantom {\dot {i}\!}x^{+_{a}\circ _{a}}=x^{+_{a}+_{a}}$. Therefore, $\phantom {\dot {i}\!}(a]_{L}=((a],\vee,\wedge,^{\circ _{a}},^{+_{a}},0,a)$ is a double MS-algebra.

Conversely, suppose that a∈L, $\phantom {\dot {i}\!}(a]_{L}=((a],\vee,\wedge,^{\circ _{a}},^{+_{a}},0,a)$ is a double MS-algebra with $\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a$ and $\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\wedge a$. Since a is the greatest element of (a]_L, then $\phantom {\dot {i}\!}a^{+_{a}}=0$ and $\phantom {\dot {i}\!}a^{\circ _{a}}=0$. This gives a⁺∧a=0 and a^∘∧a=0, respectively. Consequently, a⁺∧x⁺⁺=(a⁺∧a)⁺⁺=0⁺⁺=0 and a^∘∘∨a^∘=(a^∘∧a)^∘=0^∘=1. Therefore, a is a Stone element of L. □

Similarly for the principal filter [a) of a double MS-algebra, we establish the following result, where [a)={x∈L:x≥a}=[a,1].

Theorem 6

Let L be a double MS-algebra. If a∈C(L), then the relativized algebra $\phantom {\dot {i}\!}[a)_{L}=([a),\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)$ is a double MS-algebra, where $\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\vee a$ and $\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\vee a$. Conversely, if $\phantom {\dot {i}\!}[a)_{L}=([a),\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)$ is a double MS-algebra, then a∈L_S.

Let L₁,L₂ are double MS-algebras. Then, L₁×L₂ is a double MS-algebra, where ^∘ and ⁺ are defined by (x,y)^∘=(x^∘,y^∘) and (x,y)⁺=(x⁺,y⁺). Moreover, $(L_{1}\times L_{1})^{\circ \circ }=L_{1}^{\circ \circ }\times L_{2}^{\circ \circ }$ and C(L₁×L₂)=C(L₁)×C(L₂).

As a consequence of Theorem 5 and Theorem 6, we have

Theorem 7

Let L be a double MS-algebra. If a∈C(L), then ((a]_L×[a)_L,^∘,⁺) is a double MS-algebra, where

(a]_L×[a)_L={(x,y):x∈(a]_L,y∈[a)_L},

and

(x,y)^∘=(x^∘∧a,y^∘∨a) and (x,y)⁺=(x⁺∧a,y⁺∨a) for all (x,y)∈(a]_L×[a)_L.

Now, we introduce the concept of Birkhoff center for a double MS-algebra.

Definition 5

An element a of a double MS-algebra L is called a Birkhoff central element if there exist double MS-algebras L₁ and L₂ and an isomorphism from L to L₁×L₂ such that a is mapped to (1,0). The set BC(L) of all Birkhoff central elements of L is called the Birkhoff center.

Theorem 8

Let L be a double MS-algebra. Then, BC(L)=C(L).

Proof

Let a∈BC(L). Then, there exist double MS-algebras L₁ and L₂ and an isomorphism h from L to L₁×L₂ such that h(a)=(1,0). By Theorem 4, C(L) is isomorphic to C(L₁×L₁)=C(L₁)×C(L₂). Thus, (1,0)∈C(L₁)×C(L₂). Therefore, a=h⁻¹(1,0)∈C(L) and BC(L)⊆C(L).Conversely, let a∈C(L). Then, by Theorem 5 and Theorem 6, L₁=(a]_L and L₂=[a)_L are double MS-algebras, respectively. The direct product L₁×L₂=(a]_L×[a)_L is a double MS-algebra, by Theorem 7. Notice that $1_{L_{1}}=a$ is the greatest element of L₁ and $0_{L_{2}}=a$ is the smallest element of L₂. Define h:L→L₁×L₂ by h(x)=(a∧x,a∨x). It is already seen that h is an isomorphism of L onto L₁×L₂. Then, $h(a)=(a,a)=(1_{L_{1}},0_{L_{2}})$ implies a∈BC(L). Therefore, C(L)⊆BC(L). □

Balanced factor congruences of a double MS-algebra

In [14], Badawy investigated the relationship between congruences and de Morgan filters of decomposable MS-algebras. In this section, we study the connection between congruences and central elements of a double MS-algebra.

Let a be an element of a double MS-algebra L. Define a binary relation θ_a on L by

(x,y)∈θ_a iff x∨a=y∨a.

Proposition 3

For any two elements a and b of a double MS-algebra L, we have

(1) θ_a is a lattice congruence on L with Ker θ_a=(a],

(2) a≤b iff θ_a⊆θ_b,

(3) a=b iff θ_a=θ_b,

(4) θ₀=Δ and θ₁=∇,

(5) θ_a is the smallest lattice congruence containing (0,a).

Proof

(1). Obviously θ_a is an equivalence relation on L. Let (x,y)∈θ_a. Then, x∨a=y∨a. For all z∈L, by associativity and commutativity of ∨, we have

$$\begin{array}{@{}rcl@{}} (x\vee z)\vee a&=&x\vee (z\vee a)\\&=&x\vee (a\vee z)\\&=&(x\vee a)\vee z\\ &=&y\vee (a\vee z\\ &=&y\vee(z\vee a)\\ &=&(y\vee z)\vee a, \end{array} $$

and

$$\begin{array}{@{}rcl@{}} (x\wedge z)\vee a&=&(x\vee a)\wedge(z\vee a)\text{ by distributivity of }L\\ &=&(y\vee a)\wedge(z\vee a)\\ &=&(y\wedge z)\vee a\text{ by distributivity of }L. \end{array} $$

Then, by Theorem 2, θ_a is a lattice congruence on L. Now

$$\begin{array}{@{}rcl@{}} Ker~\theta_{a}&=&\{x\in L:(0,x)\in \theta_{a}\}\\ &=&\{x\in L:a=0\vee a=x\vee a\}\\ &=&\{x\in L:x\leq a\}=[a). \end{array} $$

(2) Let a≤b and (x,y)∈θ_a. Hence, x∨a=y∨a. Then, x∨a∨b=y∨a∨b implies x∨b=y∨b. This gives (x,y)∈θ_a and θ_a⊆θ_b. Conversely, let θ_a⊆θ_b. Since (a∧b)∨a=a=a∨a, then (a∧b,a)∈θ_a. By hypotheses, (a∧b,a)∈θ_b. Then, (a∧b)∨b=a∨b implies b=a∨b. Therefore, a≤b.

(3) It is obvious.

(4) Since for any (x,y)∈θ₀, we have x=y. Then, θ₀=Δ. For all x,y∈L, we have x∨1=1=y∨1 and hence (x,y)∈θ₁. Hence, θ₁=∇.

(5) Let θ be a lattice congruence containing (0,a). Suppose that (x,y)∈θ_a. Then, x∨a=y∨a. Since (x,x),(0,a)∈θ, then (x,x∨a)∈θ. Also, (y,y),(0,a)∈θ give (y,y∨a)∈θ. Then, (x,x∨a),(x∨a,y)∈θ imply (x,y)∈θ. So, θ_a⊆θ. □

Proposition 4

For any two elements a and b of a double MS-algebra L, we have

(1) θ_a∧b=θ_a∩θ_b,

(2) θ_a∨b=θ_a∨θ_b,

(3) θ_a∘θ_b=θ_b∘θ_a,

(4) θ_a∘θ_b=θ_a∨θ_b,

Proof

(1). Since a∧b≤a,b, then by Proposition 3(2), θ_a∧b⊆θ_a,θ_b. Thus, θ_a∧b⊆θ_a∩θ_b. Conversely, let (x,y)∈θ_a∩θ_b. Then

$$\begin{array}{@{}rcl@{}} (x,y)\in\theta_{a}\cap\theta_{b}&\Rightarrow&(x,y)\in\theta_{a} \hbox { and } (x,y)\in\theta_{b}\\ &\Rightarrow&x\vee a=y\vee a\text{ and }x\vee b=y\vee b\\ &\Rightarrow&(x\vee a)\wedge(x\vee b)=(y\vee a)\wedge(y\vee b)\\ &\Rightarrow&x\vee(a\wedge b)=y\vee(a\wedge b)\text{ by distributivity of }L\\ &\Rightarrow&(x,y)\in\theta_{a\wedge b}. \end{array} $$

Therefore, θ_a∩θ_b⊆θ_a∧b and θ_a∧b=θ_a∩θ_b.

(2) Since a,b≤a∨b, then θ_a,θ_b⊆θ_a∨b. Hence, θ_a∨b is an upper bound of θ_a and θ_b. Assume that θ_c is an upper bound of θ_a and θ_b. Then, by Proposition 3(2), θ_a,θ_b⊆θ_c imply that a,b≤c. We prove that θ_a∨b⊆θ_c. Let (x,y)∈θ_a∨b. Then, x∨a∨b=y∨a∨b. Hence, x∨a∨b∨c=y∨a∨b∨c implies x∨c=y∨c and (x,y)∈θ_c. This shows that θ_a∨b is the least upper bound of θ_a and θ_b, that is, θ_a∨b=θ_a∨θ_b.

(3) Let (x,y)∈θ_a∘θ_b. Then, there exists q∈L such that (x,q)∈θ_a and (q,y)∈θ_b. Thus, x∨a=q∨a and q∨b=y∨b. Put s=(a∨y)∧(b∨x). Now

$$\begin{array}{@{}rcl@{}} a\vee s&=&a\vee\{(a\vee y)\wedge(b\vee x)\}\\ &=&(a\vee a\vee y)\wedge(a\vee b\vee x)\text{ by distributivity of }L\\ &=&(a\vee y)\wedge(a\vee b\vee q)\text{ as }a\vee q=a\vee x\\ &=&(a\vee y)\wedge(a\vee b\vee y)\text{ as }b\vee q=b\vee y\\ &=&a\vee\{y\wedge(b\vee y)\}\text{ by distributivity of }L\\ &=&a\vee y\text{ by the absorbtion identity}. \end{array} $$

Then, (s,y)∈θ_a. Also

$$\begin{array}{@{}rcl@{}} b\vee s&=&b\vee\{(a\vee y)\wedge(b\vee x)\}\\ &=&(a\vee b\vee y)\wedge(b\vee x)\text{ by distributivity of }L\\ &=&(b\vee a\vee q)\wedge(b\vee x)\text{ as }b\vee q=b\vee y\\ &=&(b\vee a\vee x)\wedge(b\vee x)\text{ as }x\vee a=q\vee a\\ &=&b\vee\{(a\vee x)\wedge x\}\text{ by distributivity of }L\\ &=&b\vee x\text{ by the absorbtion identity}. \end{array} $$

Then, (x,s)∈θ_b. Therefore, (x,y)∈θ_b∘θ_a and θ_a∘θ_b⊆θ_b∘θ_a. Conversely, let (x,y)∈θ_b∘θ_a. Then, there exists s∈L such that (x,s)∈θ_b and (s,y)∈θ_a. Set t=(b∨y)∧(a∨x). Then, we can get a∨t=a∨x and b∨t=b∨y which means (x,t)∈θ_a and (t,y)∈θ_b. Therefore, (x,y)∈θ_a∘θ_b. So, θ_b∘θ_a⊆θ_a∘θ_b. (4) Let (x,y)∈θ_a∘θ_b. Then, there exists q∈L such that (x,q)∈θ_a and (q,y)∈θ_b. Then, x∨a=q∨a and q∨b=y∨b. Using associativity and commutativity of ∨, we get

$$\begin{array}{@{}rcl@{}} (a\vee b)\vee x=(a\vee x)\vee b =(a\vee q)\vee b =a\vee(q\vee b) =a\vee(y\vee b) =(a\vee b)\vee y. \end{array} $$

Then, (x,y)∈θ_a∨b. Conversely, let (x,y)∈θ_a∨b. Then, a∨b∨x=a∨b∨y. Set q=(a∨x)∧(b∨y). We have

$$\begin{array}{@{}rcl@{}} a\vee q&=&a\vee \{(a\vee x)\wedge(b\vee y)\} \\ &=&(a\vee x)\wedge (a\vee b\vee y)\text{ by distributivity of }L\\ &=&(a\vee x)\wedge (a\vee b\vee x)\\ &=&a\vee x\text{ as }a\vee x\leq a\vee b\vee x. \end{array} $$

Then, (x,q)∈θ_a. Also, we can get (q,y)∈θ_b. Therefore, (x,y)∈θ_a∘θ_b and θ_a∨b⊆θ_a∘θ_b. □

Theorem 9

For any two elements a and b of a double MS-algebra L, we have

(1) θ_a is compatible with ^∘ if and only if a∨a^∘=1,

(2) θ_a is compatible with ⁺ if and only if a⁺∧a⁺⁺=0,

(3) θ_a is a congruence on L if and only if a∈C(L).

Proof

(1). Let (x,y)∈θ_a and a^∘∨a=1. Then, x∨a=y∨a. We prove that (x,y)∈θ_a implies (x^∘,y^∘)∈θ_a.

$$\begin{array}{@{}rcl@{}} (x,y)\in\theta_{a}&\Rightarrow&x\vee a=y\vee a\\ &\Rightarrow&x^{\circ}\wedge a^{\circ}=(x\vee a)^{\circ}=(y\vee a)^{\circ}=y^{\circ}\wedge a^{\circ}\text{ by Theorem 1(5)}\\ &\Rightarrow&(x^{\circ}\wedge a^{\circ})\vee a=(y^{\circ}\wedge a^{\circ})\vee a\text{ by joining two sides with }a\\ &\Rightarrow&(x^{\circ}\vee a)\wedge(a^{\circ}\vee a)=(x^{\circ}\vee a)\wedge(a^{\circ}\vee a)\text{ by the distributivity of }L\\ &\Rightarrow&x^{\circ}\vee a=x^{\circ}\vee a\text{ as }a^{\circ}\vee a=1\\ &\Rightarrow&(x^{\circ},y^{\circ})\in\theta_{a} \end{array} $$

Then, (x^∘,y^∘)∈θ_a. Conversely, let θ_a is compatible with ^∘. Since (0,a)∈θ_a by Proposition 3(5), then (1,a^∘))∈θ_a. Hence, (a,a),(1,a^∘))∈θ_a implies (1,a∨a^∘)∈θ_a. Therefore, 1=1∨a=a∨(a∨a^∘)=a∨b.

(2) Let a⁺∧a⁺⁺=0. Using the properties of dual MS-algebra (L;⁺) and Proposition 1, we get a⁺∨a≥a⁺∨a⁺⁺=(a⁺∧a⁺⁺)⁺=0⁺=1 and hence a⁺∨a=1. Now, let (x,y)∈θ_a. We have

$$\begin{array}{@{}rcl@{}} (x,y)\in\theta_{a}&\Rightarrow&x\vee a=y\vee a\\ &\Rightarrow&x^{+}\wedge a^{+}=(x\vee a)^{+}=(y\vee a)^{+}=y^{+}\wedge a^{+}\text{ by Proposition 1(5)}\\ &\Rightarrow&(x^{+}\wedge a^{+})\vee a=(y^{+}\wedge a^{+})\vee a\text{ by joining two sides with }a\\ &\Rightarrow&(x^{+}\vee a)\wedge(a^{+}\vee a)=(x^{+}\vee a)\wedge(a^{+}\vee a)\text{ by the distributivity of }L\\ &\Rightarrow&x^{+}\vee a=x^{+}\vee a\text{ as }a^{+}\vee a=1. \end{array} $$

Then, (x⁺,y⁺)∈θ_a. Conversely, let θ_a is compatible with ⁺. Then, (0,a)∈θ_a implies (1,a⁺))∈θ_a. Since (a,a),(1,a⁺))∈θ_a, then (1,a∨a⁺)∈θ_a. Hence, 1=1∨a=a∨a⁺. It follows that a⁺∧a⁺⁺=(a∨a⁺)⁺=1⁺=0.

(3) As a∈C(L), then a∨a^∘=1, a∧a⁺=0, and a=a^∘∘, the proof follows (1) and (2). □

Now, we introduce the concept of factor congruences for double MS-algebras.

Definition 6

A congruence θ on a double MS-algebra L is called a factor congruence if there is a congruence ψ on L such that θ∧ψ=Δ, θ∨ψ=∇ and θ permutes with ψ.

Theorem 10

Let L be a double MS-algebra and θ a congruence on L. Then, θ is a factor congruence on L if and only if θ=θ_a for some a∈C(L).

Proof

Let a∈C(L). Hence, a^∘∈C(L). Using Theorem 9(3), we deduce that θ_a and $\phantom {\dot {i}\!}\theta _{a^{\circ }}$ are congruences on L. Hence, we get

$$\begin{array}{@{}rcl@{}} \theta_{a}\vee\theta_{a^{\circ}}&=&\theta_{a\vee a^{\circ}}\text{ by Proposition 4(2)}\\ &=&\theta_{1}\text{ as }a\vee a^{\circ}=1\\ &=&\nabla\text{ by Proposition 3(4)},\\ \theta_{a}\cap\theta_{a^{\circ}}&=&\theta_{a\wedge a^{\circ}}\text{ by Proposition 4(1)}\\ &=&\theta_{0}\text{ as }a\wedge a^{\circ}=0\\ &=&\Delta\text{ by Proposition 3(4)},\\ \theta_{a}\circ\theta_{a^{\circ}}&=&\theta_{a^{\circ}}\circ\theta_{a}\text{ by Proposition 4(3)}. \end{array} $$

Therefore, θ_a is a factor congruence on L, whenever a∈C(L). Conversely, let θ be a factor congruence on L. Then, there exists a congruence ψ on L such that θ∨ψ=∇ and θ∩ψ=Δ. Since (0,1)∈∇=θ∨ψ=θ∘ψ, then there exists x∈L such that (0,x)∈θ and (x,1)∈ψ. Thus, (0,x^∘∘)∈θ and (x^∘∘,1)∈ψ. We prove that $\phantom {\dot {i}\!}\theta =\theta _{x^{\circ \circ }}$ such that x^∘∘∈C(L). Since (0,x^∘∘)∈θ, then by Proposition 3(5), $\theta _{x^{\circ \circ }}\subseteq \theta$. Now, let (p,q)∈θ. Then, (p,q),(x^∘∘,x^∘∘)∈θ implies (p∨x^∘∘,q∨x^∘∘)∈θ. Since (x^∘∘,1),(p,p),(q,q)∈ψ, then (x^∘∘∨p,1),(x^∘∘∨q,1)∈ψ. Hence, (x^∘∘∨p,x^∘∘∨q)∈ψ. Therefore, (x^∘∘∨p,x^∘∘∨q)∈θ∩ψ=Δ. It follows that x^∘∘∨p=x^∘∘∨q and hence $\phantom {\dot {i}\!}(p,q)\in \theta _{x^{\circ \circ }}$. So, θ⊆θ_∘∘ and $\phantom {\dot {i}\!}\theta =\theta _{x^{\circ \circ }}$. This deduce that $\phantom {\dot {i}\!}\theta _{x^{\circ \circ }}$ is a congruence on L. So, by Theorem 9(3), x^∘∘∈C(L). □

Now, we introduce the concept of balanced factor congruences of a double MS-algebra.

Definition 7

A congruence θ on a double MS-algebra L is called balanced if $(\theta \vee \alpha)\cap (\theta \vee \acute {\alpha })=\theta$ for all factor congruence α and its complement $\acute {\alpha }$. The set B(L) of all balanced factor congruences which admit a balanced complement is called the Boolean center of L.

Example 2

Consider the double MS-algebra L as in Example 1. Factor congruences on L are given as follows:

$$\begin{array}{@{}rcl@{}} \theta_{0}=\Delta, \theta_{1}=\nabla, \theta_{a}=\{\{0,c,a\},\{x,y,z\},\{b,d,1\}\}, \theta_{b}=\{\{0,x,b\},\{c,y,d\},\{a,z,1\}\}. \end{array} $$

It is observed that the Boolean lattice B(L), of all balanced factor congruences is B(L)={θ₀,θ_a,θ_b,θ₁} which is represented in Fig. 5. Clearly C(L) and B(L) are isomorphic Boolean lattices.

Fig. 5

B(L)

Lemma 1

Let L be a double MS-algebra and x∈C(L). Then, θ_x is balanced.

Proof

Let α be a factor congruence on L and $\acute {\alpha }$ be its complement. Using Theorem 10, there exist a,b∈C(L) such that α=θ_a and $\acute {\alpha }=\theta _{b}$. Hence, $\alpha \cap \acute {\alpha }=\Delta$ and $\alpha \vee \acute {\alpha }=\nabla$. We have

$$\begin{array}{@{}rcl@{}} (\theta_{x}\vee\alpha)\cap(\theta_{x}\vee\acute{\alpha})&=&(\theta_{x}\vee\theta_{a})\cap(\theta_{x}\vee\theta_{b})\\ &=&\theta_{x\vee a}\cap\theta_{x\vee b}\text{ by Proposition 4(2)}\\ &=&\theta_{(x\vee a)\wedge(x\vee b)}\text{ by Proposition 4(1)}\\ &=&\theta_{x\vee(a\wedge b)}\text{ by distributivity of }L\\ &=&\theta_{x}\vee(\theta_{a}\cap\theta_{b})\text{ by Proposition 4(2) and (1), respectively}\\ &=&\theta_{x}\vee(\alpha\cap\acute{\alpha})\\ &=&\theta_{x}\vee\Delta\hbox { as }\alpha\cap\acute{\alpha}=\Delta\\ &=&\theta_{x}\hbox { as }\Delta\subseteq\theta_{x}\text{for all }x\in L. \end{array} $$

Then, θ_x is balanced. □

We close this section with the following two important results.

Theorem 11

Let L be a double MS-algebra. Then, the Boolean center B(L) of L is precisely the set {θ_a:a∈C(L)}.

Theorem 12

Let L be a double MS-algebra. Then, the Boolean center B(L) is a Boolean algebra and the mapping a↦θ_a is an isomorphism of C(L) onto B(L).

Proof

The set of all balanced factor congruences of L is B(L)={θ_a:a∈C(L)} by Theorem 11. It is clear that θ₁=∇ is the greatest element of B(L) and θ₀=Δ is the smallest element of B(L) by Proposition 3(4). Also, by Proposition 4(1),(2), respectively, we have θ_a∩θ_b=θ_a∧b and θ_a∨θ_b=θ_a∨b for all θ_a,θ_b∈B(L). Then, (B(L);∩,∨,θ₀,θ₁) is a bounded lattice. For all θ_a,θ_b,θ_c∈B(L), by distributivity of C(L), we get θ_a∩(θ_b∨θ_c)=θ_a∩θ_b∨c=θ_a∧(a∨b)=θ_{(a∨b)∧(a∨c)}=θ_a∨b∩θ_a∨c=(θ_a∨θ_b)∩(θ_a∨θ_c). Thus, B(L) is a distributive lattice. The complement of θ_a is $\phantom {\dot {i}\!}\theta _{a^{\circ }}$. Then, B(L) is a Boolean algebra. The proof of the rest part of this theorem is straightforward. □