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  • Open Access

Balanced factor congruences of double MS-algebras

Journal of the Egyptian Mathematical Society201927:6

https://doi.org/10.1186/s42787-019-0008-y

  • Received: 26 May 2018
  • Accepted: 1 September 2018
  • Published:

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.

Keywords

  • De Morgan algebra
  • MS-algebra
  • Double MS-algebra
  • Factor congruences
  • Balanced factor congruences

Mathematics Subject Classification

  • Primary 06D30
  • Secondary 06D15

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 dL-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:ab}. 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:

xx,(xy)=xy,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) abba,

(3) a=a,

(4) a=a,

(5) (ab)=ab,

(6) (ab)=ab,

(7) (ab)=ab.

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:

xx++,(xy)+=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) abb+a+,

(3) a+++=a+,

(4) a++++=a++,

(5) (ab)+=a+b+,

(6) (ab)++=a++b++,

(7) (ab)++=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 xL.

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

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={xL:x=x}=L++={xL:x=x++}={xL: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 (ac,bd)θ and (ac,bd)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 (az,bz)θ and (az,bz)θ for all zL.

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):aL} for the equality congruence on L.

We say the congruences θ,ψ on a lattice L are permutable if θψ=ψθ, that is, xy(θ) and yz(ψ) imply xr(ψ) and rz(θ) for some y,rL.

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 LS 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 xx=1 and x+x++=0. Let LS denote the set of all Stone elements of L, that is, LS={xL:xx=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 xL1.

Definition 3

A subalgebra L1 of a double MS-algebra L is called a Stone subalgebra if xx=1 and x+x++=0, for all xL1.

Proposition 2

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

Proof

It is clear that 0,1LS. Let x,yLS. Then, xx=1, x+x++=0, yy=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, xyLS. Using a similar way, we get xyLS. Therefore, (LS,,,0,1) is a bounded distributive sublattice of L. Now, we prove that x+LS for all xLS.
$$\begin{array}{@{}rcl@{}} x^{+\circ}\vee x^{+\circ\circ}&=&x^{++}\vee x^{+++}\text{as }x^{+\circ}=x^{++}\\ &=&(x^{+}\wedge x^{++})^{+}\text{ by Proposition 1(5)}\\ &=&0^{+}=1\text{ as }x^{+}\wedge x^{++}=0,\\ x^{+\circ}\wedge x^{+\circ\circ}&=&x^{++}\wedge x^{+++}\text{ as }x^{+\circ}=x^{++}\\ &=&x^{++}\wedge x^{+}=0\text{ by Proposition 1(3)}. \end{array} $$

Hence, x+LS. Similarly, we can prove that xLS for all xLS. Therefore, LS is a subalgebra of a double MS-algebra L. Since xx=1 and x+x++=0 for every xLS, then LS is a Stone subalgebra of L. To prove that LS is the greatest Stone subalgebra of L, let S be any Stone subalgebra of L. Let xS. Then, x is a Stone element of L, and hence, xLS. So SLS 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 LS, C(L), and L is investigated.

Definition 4

An element a of double MS-algebra L is called a central element if xx=1 and xx+=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)={xL:xx=1,xx+=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
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,LS, and C(L) are given in Figs. 2, 3, and 4, respectively.
Fig. 2
Fig. 2

L

Fig. 3
Fig. 3

L s

Fig. 4
Fig. 4

C(L)

Theorem 3

Let L be a double MS-algebra. Then

(1) C(L)=LLS,

(2) C(L) is the greatest Boolean subalgebra of L, LS, and L,

(3) C(L)=C(LS)=C(L).

Proof

(1). Let xC(L). Then, xx=1 and xx+=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, xL. 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 xx=1 imply xLS. Therefore, C(L)LLS. Conversely, let xLLS. Then, x=x=x++, xx=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, xC(L), and hence, LLSC(L).

(2) Clearly 0,1C(L). Let a,bC(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, abC(L). Similarly abC(L). Therefore, (C(L);,,0,1) is a bounded sublattice of L. Now, we observe that aC(L) for all aC(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 aC(L), then coincide with + on C(L). Therefore, (C(L),,,,0,1) is a subalgebra of L. Since aa=1 and aa=aa=(aa)=1=0 for all aC(L), then (C(L),,,,0,1) is a Boolean subalgebra of L. Suppose that B is any Boolean subalgebra of L and xB. Then, aa=1 and aa+=aa=0. Hence, a is a central element of L and aC(L). So, BC(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 LS 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:LM be an isomorphism and aC(L). Then, aa=1 and aa+=0. Hence, h(aa)=h(a)h(a)=h(a)(h(a))=h(1)=1 and h(a)(h(a))+=h(0)=0. Therefore, hC(L)(a)=h(a)C(M). It is clear that hC(L) is an injective (0,1) lattice homomorphism. Let bC(M). Then, there exists bL such that b=h(a)=hC(L)(a) as h is onto. It follows that b=(h(a))=h(a)=h(a)=hC(L)(a). Thus, hC(L) is onto. Obviously, hC(L) preserves and +. Then, hC(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 aC(L), where (a]={xL:xa}=[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)={xL:xx=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],aL constructs a double MS-algebra?

Theorem 5

Let L be a double MS-algebra. Suppose that aC(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 aLS.

Proof

Assume that aC(L). Hence, aC(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 aL, \(\phantom {\dot {i}\!}(a]_{L}=((a],\vee,\wedge,^{\circ _{a}},^{+_{a}},0,a)\) 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 aa=0, respectively. Consequently, a+x++=(a+a)++=0++=0 and aa=(aa)=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)={xL:xa}=[a,1].

Theorem 6

Let L be a double MS-algebra. If aC(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 aLS.

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(L1C(L2).

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

Theorem 7

Let L be a double MS-algebra. If aC(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)=(xa,ya) 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 aBC(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(L1C(L2). Thus, (1,0)C(L1C(L2). Therefore, a=h−1(1,0)C(L) and BC(L)C(L).Conversely, let aC(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:LL1×L2 by h(x)=(ax,ax). 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 aBC(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 θa on L by

(x,y)θa iff xa=ya.

Proposition 3

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

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

(2) ab iff θaθb,

(3) a=b iff θa=θb,

(4) θ0=Δ and θ1=,

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

Proof

(1). Obviously θa is an equivalence relation on L. Let (x,y)θa. Then, xa=ya. For all zL, 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, θa 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 ab and (x,y)θa. Hence, xa=ya. Then, xab=yab implies xb=yb. This gives (x,y)θa and θaθb. Conversely, let θaθb. Since (ab)a=a=aa, then (ab,a)θa. By hypotheses, (ab,a)θb. Then, (ab)b=ab implies b=ab. Therefore, ab.

(3) It is obvious.

(4) Since for any (x,y)θ0, we have x=y. Then, θ0=Δ. For all x,yL, we have x1=1=y1 and hence (x,y)θ1. Hence, θ1=.

(5) Let θ be a lattice congruence containing (0,a). Suppose that (x,y)θa. Then, xa=ya. Since (x,x),(0,a)θ, then (x,xa)θ. Also, (y,y),(0,a)θ give (y,ya)θ. Then, (x,xa),(xa,y)θ imply (x,y)θ. So, θaθ. □

Proposition 4

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

(1) θab=θaθb,

(2) θab=θaθb,

(3) θaθb=θbθa,

(4) θaθb=θaθb,

Proof

(1). Since aba,b, then by Proposition 3(2), θabθa,θb. Thus, θabθaθb. Conversely, let (x,y)θaθb. 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, θaθbθab and θab=θaθb.

(2) Since a,bab, then θa,θbθab. Hence, θab is an upper bound of θa and θb. Assume that θc is an upper bound of θa and θb. Then, by Proposition 3(2), θa,θbθc imply that a,bc. We prove that θabθc. Let (x,y)θab. Then, xab=yab. Hence, xabc=yabc implies xc=yc and (x,y)θc. This shows that θab is the least upper bound of θa and θb, that is, θab=θaθb.

(3) Let (x,y)θaθb. Then, there exists qL such that (x,q)θa and (q,y)θb. Thus, xa=qa and qb=yb. Put s=(ay)(bx). 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)θa. 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)θb. Therefore, (x,y)θbθa and θaθbθbθa. Conversely, let (x,y)θbθa. Then, there exists sL such that (x,s)θb and (s,y)θa. Set t=(by)(ax). Then, we can get at=ax and bt=by which means (x,t)θa and (t,y)θb. Therefore, (x,y)θaθb. So, θbθaθaθb. (4) Let (x,y)θaθb. Then, there exists qL such that (x,q)θa and (q,y)θb. Then, xa=qa and qb=yb. 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)θab. Conversely, let (x,y)θab. Then, abx=aby. Set q=(ax)(by). 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)θa. Also, we can get (q,y)θb. Therefore, (x,y)θaθb and θabθaθb. □

Theorem 9

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

(1) θa is compatible with if and only if aa=1,

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

(3) θa is a congruence on L if and only if aC(L).

Proof

(1). Let (x,y)θa and aa=1. Then, xa=ya. We prove that (x,y)θa implies (x,y)θa.
$$\begin{array}{@{}rcl@{}} (x,y)\in\theta_{a}&\Rightarrow&x\vee a=y\vee a\\ &\Rightarrow&x^{\circ}\wedge a^{\circ}=(x\vee a)^{\circ}=(y\vee a)^{\circ}=y^{\circ}\wedge a^{\circ}\text{ by Theorem 1(5)}\\ &\Rightarrow&(x^{\circ}\wedge a^{\circ})\vee a=(y^{\circ}\wedge a^{\circ})\vee a\text{ by joining two sides with }a\\ &\Rightarrow&(x^{\circ}\vee a)\wedge(a^{\circ}\vee a)=(x^{\circ}\vee a)\wedge(a^{\circ}\vee a)\text{ by the distributivity of }L\\ &\Rightarrow&x^{\circ}\vee a=x^{\circ}\vee a\text{ as }a^{\circ}\vee a=1\\ &\Rightarrow&(x^{\circ},y^{\circ})\in\theta_{a} \end{array} $$

Then, (x,y)θa. Conversely, let θa is compatible with . Since (0,a)θa by Proposition 3(5), then (1,a))θa. Hence, (a,a),(1,a))θa implies (1,aa)θa. Therefore, 1=1a=a(aa)=ab.

(2) Let a+a++=0. Using the properties of dual MS-algebra (L;+) and Proposition 1, we get a+aa+a++=(a+a++)+=0+=1 and hence a+a=1. Now, let (x,y)θa. 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+)θa. Conversely, let θa is compatible with +. Then, (0,a)θa implies (1,a+))θa. Since (a,a),(1,a+))θa, then (1,aa+)θa. Hence, 1=1a=aa+. It follows that a+a++=(aa+)+=1+=0.

(3) As aC(L), then aa=1, aa+=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 θψ=Δ, θψ= 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=θa for some aC(L).

Proof

Let aC(L). Hence, aC(L). Using Theorem 9(3), we deduce that θa 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, θa is a factor congruence on L, whenever aC(L). Conversely, let θ be a factor congruence on L. Then, there exists a congruence ψ on L such that θψ= and θψ=Δ. Since (0,1)=θψ=θψ, then there exists xL 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 xC(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 (px,qx)θ. Since (x,1),(p,p),(q,q)ψ, then (xp,1),(xq,1)ψ. Hence, (xp,xq)ψ. Therefore, (xp,xq)θψ=Δ. It follows that xp=xq and hence \(\phantom {\dot {i}\!}(p,q)\in \theta _{x^{\circ \circ }}\). So, θθ 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), xC(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,θa,θb,θ1} which is represented in Fig. 5. Clearly C(L) and B(L) are isomorphic Boolean lattices.
Fig. 5
Fig. 5

B(L)

Lemma 1

Let L be a double MS-algebra and xC(L). Then, θx is balanced.

Proof

Let α be a factor congruence on L and \(\acute {\alpha }\) be its complement. Using Theorem 10, there exist a,bC(L) such that α=θa 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, θx 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 {θa:aC(L)}.

Theorem 12

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

Proof

The set of all balanced factor congruences of L is B(L)={θa:aC(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 θaθb=θab and θaθb=θab for all θa,θbB(L). Then, (B(L);∩,,θ0,θ1) is a bounded lattice. For all θa,θb,θcB(L), by distributivity of C(L), we get θa∩(θbθc)=θaθbc=θa(ab)=θ(ab)(ac)=θabθac=(θaθb)∩(θaθc). Thus, B(L) is a distributive lattice. The complement of θa 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. □

Declarations

Acknowledgements

We thank the referees for valuable comments and suggestions for improving the paper.

Authors’ contributions

The author read and approved the final manuscript.

Competing interests

The author declares that he has no competing interests.

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Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Tanta University, Tanta, Egypt

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