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ON QUASI-LATTICE IMPLICATION ALGEBRAS
ON QUASI-LATTICE IMPLICATION ALGEBRAS
Journal of Applied Mathematics & Informatics. 2015. Sep, 33(5_6): 739-748
Copyright © 2015, Korean Society of Computational and Applied Mathematics
  • Received : April 21, 2015
  • Accepted : June 29, 2015
  • Published : September 30, 2015
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YONG HO YON

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 . AMS Mathematics Subject Classification : 03G10, 06B10.
Keywords
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 ,
  • (I1)x→ (y→z) =y→ (x→z),
  • (I2)x→x= 1,
  • (I3) (x→y) →y= (y→x) →x,
  • (I4)x→y=y→x= 1 ⇒x=y,
  • (I5)x→y=y'→x',
  • (L1) (x∨y) →z= (x→z) ∧ (y→z),
  • (L2) (x∧y) →z= (x→z) ∨ (y→z).
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 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 .
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 ,
  • (Q1)x(yz) =y(xz),
  • (Q2)xx= 1,
  • (Q3) (xy)y= (yx)x,
  • (Q4)xy= 1 andyx= 1 implyx=y,
  • (Q5)xy=y′x′.
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 ,
  • (1) 1x=x,
  • (2)x1 = 1.
Proof . (1) Let x L . Then x (1 x ) = 1( xx ) = 11 = 1 by (Q1) and (Q2). Also, we have (1 x ) x = ( x 1)1 = ( x 1)(11) = 1(( x 1)1) = 1((1 x ) x ) = (1 x )(1 x ) = 1 by (Q3), (Q2) and (Q1). Hence x (1 x ) = (1 x ) x = 1. This implies x = 1 x by (Q4).
(2) Let x L . Then x 1 = (1 x )1 = (1 x )( xx ) = x ((1 x ) x ) = x ( x 1)1) = ( x 1)( x 1) = 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
PPT Slide
Lager Image
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 (1 z ) = 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 ,
  • (1) 1′ = 0 and 0′ = 1,
  • (2)x′=x0,
  • (3)x≤yz implies y≤xz,
  • (4) (xy) ≤ (yz)(xz),
  • (5) (xy) ≤ (zx)(zy),
  • (6)x≤y implies yz≤xz and zx≤zy,
  • (7)x≤ (xy)y,
  • (8)y≤xy,
  • (9)x≤y implies y′≤x′,
  • (10) ((xy)y)y=xy.
Proof . (1) It is clear that 1′ = 0 by 2.3, and 0′ = 1′′ = 1.
(2) For every x L, x′ = 1 x′ = x′′ 1′ = x 0 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
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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 ) = x 1 = 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
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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
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Lager Image
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 ,
  • (1) (x∨y)′ =x′∧y′,
  • (2) (x∧y)′ =x′∨y′,
  • (3) (x∨y)z= (xz) ∧ (yz),
  • (4)z(x∧y) = (zx) ∧ (zy).
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
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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
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since ( x′y′ ) ∧ ( x′z′ ) ≤ x′y′ and ( x′y′ ) ∧ ( x′z′ ) ≤ x′z′ . This implies
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Hence x ∧ ( y z ) = ( x y ) ∨ ( x z ). □
In a lattice L , it is well known that the following are equivalent:
  • (1)x∧ (y∨z) = (x∧y) ∨ (x∧z) for everyx, y, z∈L,
  • (2)x∨ (y∧z) = (x∨y) ∧ (x∨z) for everyx, y, z∈L.
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
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This implies
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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 ,
  • (Q1)x(yz) =y(xz),
  • (Q2)xx= 1,
  • (Q3) (xy)y= (yx)x,
  • (Q4)xy= 1and yx= 1imply x=y,
  • (B) 0x= 1.
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 1 x = x for every x L in the same way as the proof of 2.2(1). Let x′ = x 0. Then we have x′′ = ( x 0)0 = (0 x ) x = 1 x = x by (Q3) and (B). So ′ is an involution, and it satisfies the following.
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Lager Image
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 ,
  • (F1) 1 ∈F,
  • (F2)x∈Fandxy∈Fimplyy∈F.
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:
  • (1)x∈F and x≤y imply y∈F,
  • (2)x, y∈F implies x∧y∈F.
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:
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If we define x′ = x 0 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
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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 Φ F on L by
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for any x, y L .
Lemma 3.3. Let F be a filter of L. Then Φ F is a congruence relation.
Proof . For any x L , it is clear that x Φ Fx , since xx = 1 ∈ F , and that x Φ Fy implies y Φ Fx .
To show the transitivity of Φ F , let x Φ Fy and y Φ Fz . 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 x Φ Fz . Thus Φ F is an equivalence relation in L .
To show that Φ F is a congruence relation on L , let x Φ Fy 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 F ( zy ), and Φ F is left compatible. In the similar way, we can show ( xz )( yz ), ( yz )( xz ) ∈ F by 2.4(4). That is ( xz F ( yz ), and Φ F is right compatible. Hence Φ F is a congruence relation on L . □
We will call this relation Φ F 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 x Θ1 and xy Θ1 imply xy Θ1 y and xy Θ1, hence y Θ1 since 1 y = 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 = Φ F .
Let x Θ y . Then xy Θ yy and yx Θ yy , since Θ is a congruence relation. This implies xy Θ1 and yx Θ1, and xy, yx ∈ [1] Θ = F . Thus x Φ Fy . Also, let x Φ Fy . Then xy , yx F = [1] Θ . This implies xy Θ1 and yx Θ1, and ( xy ) y Θ1 y and ( yx ) x Θ1 x , i.e., ( xy ) y Θ y and ( yx ) x Θ x . Hence x Θ y since ( xy ) y = ( yx ) x .
This mean Θ = Φ F , 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
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for arbitrary subset { Fα | α ∈ Λ} of Fil ( L ) (resp. {Θ α | 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:
  • (1) [1]ΦF=F for every F∈Fil(L),
  • (2) Φ[1]Θ= Θfor everyΘ ∈Con(L),
  • (3)F⊆G if and only ifΦF⊆ ΦGfor any F,G∈FIl(L),
  • (4) Θ ⊆ Ψif and only if[1]Θ⊆ [1]Ψfor anyΘ,Ψ ∈Con(L).
Proof . (1) Let F Fil ( L ). Then we have
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Hence [1] ΦF = F .
(2) It was proved in the proof of 3.5.
(3) Let F G in Fil ( L ) and x Φ Fy . Then xy, yx F , and xy, yx G since F G . This implies x Φ Gy . Hence Φ F ⊆ Φ G . Conversely, let Φ F ⊆ Φ G and x F . Then x 1 = 1 ∈ F , since F is a filter, and 1 x = x F . This implies x Φ F 1, and x Φ G 1 since Φ F ⊆ Φ G . This implies 1 x = 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
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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 ) = Φ F for every F Fil ( L ). Then it satisfies the following:
  • (1)ϕ is order-isomorphism,
  • (2)ϕ preserves arbitrary join and arbitrary meet.
Proof . (1) Let ϕ : Fil ( L ) → Con ( L ) be a map defined by ϕ ( F ) = Φ F for every F Fil ( L ). Then F G if and only if ϕ( F ) = Φ F ⊆ Φ G = 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 ) = Φ Fβ ⊆ 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
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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.
BIO
Yong Ho Yon received M.Sc. from Chungbuk National University, and Ph.D. from Chungbuk National University. He is currently a professor at Mokwon University since 2011. His research interests are lattice theory and algebras with implication.
Division of Information and Communication Convergence Engineering, Mokwon University, Daejeon 302-729, Korea.
e-mail: yhyon@mokwon.ac.kr
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