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Balanced factor congruences of double MS-algebras
Journal of the Egyptian Mathematical Society volume 27, Article number: 6 (2019)
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 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: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 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 x∘∨x∘∘=1 and x+∧x++=0. Let LS denote the set of all Stone elements of L, that is, LS={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
LS is the greatest Stone subalgebra of a double MS-algebra L.
Proof
It is clear that 0,1∈LS. Let x,y∈LS. Then, x∘∨x∘∘=1, x+∧x++=0, y∘∨y∘∘=1, and y+∧y++=0. Thus, we get
Then, x∨y∈LS. Using a similar way, we get x∧y∈LS. Therefore, (LS,∨,∧,0,1) is a bounded distributive sublattice of L. Now, we prove that x+∈LS for all x∈LS.
Hence, x+∈LS. Similarly, we can prove that x∘∈LS for all x∈LS. Therefore, LS is a subalgebra of a double MS-algebra L. Since x∘∨x∘∘=1 and x+∧x++=0 for every x∈LS, 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 x∈S. Then, x is a Stone element of L, and hence, x∈LS. So S⊆LS 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 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
and
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.
Theorem 3
Let L be a double MS-algebra. Then
(1) C(L)=L∘∘∩LS,
(2) C(L) is the greatest Boolean subalgebra of L, LS, and L∘∘,
(3) C(L)=C(LS)=C(L∘∘).
Proof
(1). Let x∈C(L). Then, x∨x∘=1 and x∧x+=0. Then
Thus, x∈L∘∘. Also,
Then, x++∧x+=0 and x∘∘∨x∘=1 imply x∈LS. Therefore, C(L)⊆L∘∘∩LS. Conversely, let x∈L∘∘∩LS. Then, x=x∘∘=x++, x∘∨x∘∘=1, and x+∧x++=0. Now,
Thus, x∈C(L), and hence, L∘∘∩LS⊆C(L).
(2) Clearly 0,1∈C(L). Let a,b∈C(L). Then, we have
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),
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 LS and L∘∘.
(3) It follows (1) and (2). â–¡
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, hC(L)(a)=h(a)∈C(M). It is clear that hC(L) is an injective (0,1) lattice homomorphism. Let b∈C(M). Then, there exists b∈L 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 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∈LS.
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
Then, \(\phantom {\dot {i}\!}x\geq x^{+_{a}+_{a}}\). Let x,y∈(a]
Also, \(\phantom {\dot {i}\!}0^{+_{a}}=a\). Now, for every x∈(a], we have
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∈LS.
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 θa on L by
(x,y)∈θa iff x∨a=y∨a.
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) a≤b iff θa⊆θb,
(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, x∨a=y∨a. For all z∈L, by associativity and commutativity of ∨, we have
and
Then, by Theorem 2, θa is a lattice congruence on L. Now
(2) Let a≤b and (x,y)∈θa. Hence, x∨a=y∨a. Then, x∨a∨b=y∨a∨b implies x∨b=y∨b. This gives (x,y)∈θa and θa⊆θb. Conversely, let θa⊆θb. Since (a∧b)∨a=a=a∨a, then (a∧b,a)∈θa. By hypotheses, (a∧b,a)∈θb. 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)∈θa. 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, θa⊆θ. □
Proposition 4
For any two elements a and b of a double MS-algebra L, we have
(1) θa∧b=θa∩θb,
(2) θa∨b=θa∨θb,
(3) θa∘θb=θb∘θa,
(4) θa∘θb=θa∨θb,
Proof
(1). Since a∧b≤a,b, then by Proposition 3(2), θa∧b⊆θa,θb. Thus, θa∧b⊆θa∩θb. Conversely, let (x,y)∈θa∩θb. Then
Therefore, θa∩θb⊆θa∧b and θa∧b=θa∩θb.
(2) Since a,b≤a∨b, then θa,θb⊆θa∨b. Hence, θa∨b 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,θb⊆θc imply that a,b≤c. We prove that θa∨b⊆θc. Let (x,y)∈θa∨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)∈θc. This shows that θa∨b is the least upper bound of θa and θb, that is, θa∨b=θa∨θb.
(3) Let (x,y)∈θa∘θb. Then, there exists q∈L such that (x,q)∈θa and (q,y)∈θb. Thus, x∨a=q∨a and q∨b=y∨b. Put s=(a∨y)∧(b∨x). Now
Then, (s,y)∈θa. Also
Then, (x,s)∈θb. Therefore, (x,y)∈θb∘θa and θa∘θb⊆θb∘θa. Conversely, let (x,y)∈θb∘θa. Then, there exists s∈L such that (x,s)∈θb and (s,y)∈θa. Set t=(b∨y)∧(a∨x). Then, we can get a∨t=a∨x and b∨t=b∨y which means (x,t)∈θa and (t,y)∈θb. Therefore, (x,y)∈θa∘θb. So, θb∘θa⊆θa∘θb. (4) Let (x,y)∈θa∘θb. Then, there exists q∈L such that (x,q)∈θa and (q,y)∈θb. Then, x∨a=q∨a and q∨b=y∨b. Using associativity and commutativity of ∨, we get
Then, (x,y)∈θa∨b. Conversely, let (x,y)∈θa∨b. Then, a∨b∨x=a∨b∨y. Set q=(a∨x)∧(b∨y). We have
Then, (x,q)∈θa. Also, we can get (q,y)∈θb. Therefore, (x,y)∈θa∘θb and θa∨b⊆θa∘θb. □
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 a∨a∘=1,
(2) θa is compatible with + if and only if a+∧a++=0,
(3) θa is a congruence on L if and only if a∈C(L).
Proof
(1). Let (x,y)∈θa and a∘∨a=1. Then, x∨a=y∨a. We prove that (x,y)∈θa implies (x∘,y∘)∈θa.
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,a∨a∘)∈θa. 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)∈θa. We have
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,a∨a+)∈θa. 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=θa for some a∈C(L).
Proof
Let a∈C(L). Hence, a∘∈C(L). Using Theorem 9(3), we deduce that θa and \(\phantom {\dot {i}\!}\theta _{a^{\circ }}\) are congruences on L. Hence, we get
Therefore, θa 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:
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.
Lemma 1
Let L be a double MS-algebra and x∈C(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,b∈C(L) such that α=θa and \(\acute {\alpha }=\theta _{b}\). Hence, \(\alpha \cap \acute {\alpha }=\Delta \) and \(\alpha \vee \acute {\alpha }=\nabla \). We have
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: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↦θa is an isomorphism of C(L) onto B(L).
Proof
The set of all balanced factor congruences of L is B(L)={θa: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 θa∩θb=θa∧b and θa∨θb=θa∨b for all θa,θb∈B(L). Then, (B(L);∩,∨,θ0,θ1) is a bounded lattice. For all θa,θb,θc∈B(L), by distributivity of C(L), we get θa∩(θb∨θc)=θa∩θb∨c=θa∧(a∨b)=θ(a∨b)∧(a∨c)=θa∨b∩θa∨c=(θa∨θb)∩(θa∨θc). 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. □
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Badawy, A.EM. Balanced factor congruences of double MS-algebras. J Egypt Math Soc 27, 6 (2019). https://doi.org/10.1186/s42787-019-0008-y
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DOI: https://doi.org/10.1186/s42787-019-0008-y
Keywords
- De Morgan algebra
- MS-algebra
- Double MS-algebra
- Factor congruences
- Balanced factor congruences
Mathematics Subject Classification
- Primary 06D30
- Secondary 06D15