# A Bijection between Lattice-Valued Filters and Lattice-Valued Congruences in Residuated Lattices

Authors: Wei Wei; Yan Qiang; Jing Zhang
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

The aim of this paper is to study relations between lattice-valued filters and lattice-valued congruences in residuated lattices. We introduce a new definition of congruences which just depends on the meet∧ and the residuum →. Then it is shown that each of these congruences is automatically a universal-algebra-congruence. Also, lattice-valued filters and lattice-valued congruences are studied, and it is shown that there is a one-to-one correspondence between the set of all (lattice-valued) filters and the set of all (lattice-valued) congruences.


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## Body

## 1. Introduction and Preliminaries

The interest in lattice-valued logic has been rapidly growing recently. Several algebras playing the role of structures of true values have been introduced and axiomatized [1–3]. The most general structure considered in this paper is that of a residuated lattice [4].In a narrow sense, a residuated lattice is an algebraL=(L,∧,∨,⊗,→,0,1) of type (2,2,2,2,0,0) satisfying the following: (i) (L,∧,∨) is a bounded lattice with 0,1 as the bottom element, and the top element respectively; (ii) (L,⊗,1) is a commutative monoid and monotone at both arguments; (iii) a⊗b≤c if and only if a≤b→c(foralla,b,c∈L). The operations ⊗,→ are called the multiplication and residuum, respectively. A residuated lattice in this paper is generally called a bounded, integral, and commutative residuated lattice in [4].Residuated lattices were first introduced as a generalization of ideal lattices of rings in 1939 by Ward and Dilworth [5]. In their original definition, a residuated lattice was what we would call an integral commutative one.For a residuated latticeL, the negation operation ¬:L→L is defined by ¬x=x→0(forallx∈L).Residuated lattices are very common in mathematical science and a lot of lattices and algebras are residuated lattices firstly. For example, an integral commutative Girard-monoid [2] is a residuated lattice satisfying the law of double negation: x=¬¬x; a Heyting algebra [6] is a residuated lattice with ⊗=∧; an MV-algebra [7] is a residuated lattice where x∨y=(x→y)→y holds; an MTL-algebra [1] is a residuated lattice satisfying (x→y)∨(y→x)=1; a BL-algebra [8] is an MTL-algebra satisfying x∧y=x⊗(x→y); a product algebra (or Π-algebra) [8] is a BL-algebra satisfying ¬¬z≤((x⊗z)→(y⊗z))→(x→y) and x∧¬x=0; a G-algebra (Gödel algebra) [2] is both a Heyting algebra and an MTL-algebra; an R0-algebra [3] is a residuated lattice where x⊗y=¬(x→¬y); a lattice implication algebra [9] is a residuated lattice with a⊗b=(a→b′)′ (where ′: L→L is an order-reversing involution).Since the class of all residuated lattices is a variety of algebras (Proposition  2 in [10]), we can study them as universal algebras. Now, consider a residuated lattice L=(L,∧,∨,⊗,→,0,1) as a universal algebra; a congruence ~ on L is an equivalence relation which preserves all operators on L; that is, (a,b),(c,d)∈~ implies that (a∧c,b∧d),(a∨c,b∨d),(a⊗c,b⊗d),(a→c,b→d)∈~.The aim of this paper is to study the relation between lattice-valued filters and lattice-valued congruences in residuated lattices. We will introduce a new definition of congruences just depending on the meet∧ and the residuum →. Then it is shown that each of these congruences is automatically a universal-algebra-congruence. Also, lattice-valued filters and lattice-valued congruences are studied, and it is shown that there is a one-to-one correspondence between the set of all (lattice-valued) filters and the set of all (lattice-valued) congruences.


## 2. Filters and Congruences

In pure mathematics, (lattice-valued) filters (or ideals) and (lattice-valued) congruences are useful tools in investigating the structure of the corresponding algebras.The definition of a residuated lattice (in a narrow sense) has been given in Section1. In the following discussion, L always denotes a residuated lattice.Proposition 1 (see [3, 8, 10, 11]).
LetL be a residuated lattice. Then(R1)
a⊗b≤a∧b;(R2)
a→(b∧c)=(a→b)∧(a→c);(R3)
(a∨b)→c=(a→c)∧(b→c);(R4)
a⊗(b∨c)=(a⊗b)∨(a⊗c);(R5)
b→c≤(a→b)→(a→c);(R6)
a=1→a;(R7)
a≤b  if and only if  a→b=1;(R8)
a≤b→c  if and only if  b≤a→c;(R9)
a→(b→c)=b→(a→c)=(a⊗b)→c;(R10)
a→b≤(a⊗c)→(b⊗c);(R11)
a→b≥b;(R12)
b≥a⊗(a→b);(R13)
b≤a→(a⊗b);(R14)
a→(a⊗b)≤a→b;(R15)
a≤(a→b)→b;(R16)
(a→b)⊗(b→c)≤a→c.Definition 2 (see [8]).
A nonempty subsetF of L is called a filter if(F1)F is an upper set; that is, x≤y and x∈F imply y∈F for all x,y∈L;(F2)F is closed under ⊗; that is, x⊗y∈F holds for all x,y∈F.Proposition 3 (see [8]).
LetF be a nonempty subset of  L. Then the following three are equivalent:(1)
F  is a filter;(2)
1∈F  and  x,x→y∈F  imply  y∈F  for all  x,y∈F;(3)
F  is closed under  ⊗  and  x∨y∈F  for all  x∈F  and  y∈L.DenoteF(L) as the set of all filters of  L. Then F(L) is a complete lattice under the partial order of set inclusion with the largest element L and the least element {1}. Furthermore, the meets in F(L) are the usual intersection of sets.For simplification of congruence relation in algebraic structures, related attempts have been made in [12–14].Definition 4.
A nonempty subset~ of L×L is called a {∧,→}-congruence on L if the following conditions hold:(ER)~ is an equivalence;
for anyx,y,z∈L,(C1) if(x,y)∈~, then (x∧z,y∧z)∈~;(C2) if(x,y)∈~, then (x→z,y→z)∈~.
Obviously, a congruence is always a{∧,→}-congruence. Let Con(L) denote the set of all congruences on L. It is easy to verify that Con(L) is a complete lattice, where the meets are the usual intersection of sets and L×L, {(1,1)} are the largest and the least elements, respectively.
Let~ be a congruence on L and L/~={[x]|x∈L}, where [x] is the congruence class of x with respect to ~. Define [x]∧[y]=[x∧y], [x]∨[y]=[x∨y], [x]⊗[y]=[x⊗y], [x]→[y]=[x→y](forallx,y∈L). It is easy to verify that (L/~, ∧,∨,⊗,→,[0],[1]) is also a residuated lattice.Proposition 5 (see [15]).
LetF be a filter of L. Then ~F={(x,y)∈L×L∣x→y,y→x∈F} is a {∧,→}-congruence on L.Proof.
(ER) Obviously,~F is reflexive and symmetric. To show the transitivity of ~F, suppose that (x,y),(y,z)∈~F; we have x→y,y→x,y→z,z→y∈F. Then
(1)(x⟶y)⊗(y⟶z),(z⟶y)⊗(y⟶x)∈F.
By (R16)
(2)(x⟶y)⊗(y⟶z)≤x⟶z,(z⟶y)⊗(y⟶x)≤z⟶x,
and F is an upper set; we have x→z,z→x∈F.
Suppose that(x,y)∈~F and z∈L. Then x→y,y→x∈F.
(C1) First,(3)(x∧z)⟶(y∧z)=((x∧z)⟶y)∧((x∧z)⟶z)≥(x⟶y)∧1=x⟶y.
Then (x∧z)→(y∧z)∈F since F is an upper set. Similarly, we have (y∧z)→(x∧z)∈F. Hence (x∧z,y∧z)∈~F.
(C2) By (R16), we have(4)(y⟶x)⊗(x⟶z)≤(y⟶z),y⟶x≤(x⟶z)⟶(y⟶z).
Thus (x→z)→(y→z)∈F since F is an upper set. Similarly, we have (y→z)→(x→z)∈F. Hence (x→z,y→z)∈~F.Proposition 6.
LetF be a filter of L. If (a,b),(c,d)∈~F, then (a∧c,b∧d),(a∨c,b∨d),(a⊗c,b⊗d),(a→c,b→d)∈~F.Proof.
Suppose that(a,b),(c,d)∈~F. Then a→b,b→a,c→d,d→c∈F.
(1)(a∧c,b∧d)∈~F. In fact, by (R1) and (R2),
(5)(a∧c)⟶(b∧d)=((a∧c)⟶b)∧((a∧c)⟶d)≥(a⟶b)∧(c⟶d)≥(a⟶b)⊗(c⟶d)∈F.
It follows that (a∧c)→(b∧d)∈F. Similarly, (b∧d)→(a∧c)∈F. Hence (a∧c,b∧d)∈~F.
(2)(a∨c,b∨d)∈~F. In fact, by (R1) and (R3),
(6)(a∨c)⟶(b∨d)=(a⟶(b∨d))∧(c⟶(b∨d))≥(a⟶b)∧(c⟶d)≥(a⟶b)⊗(c⟶d)∈F.
It follows that (a∨c)→(b∨d)∈F. Similarly, (b∨d)→(a∨c)∈F. Hence (a∨c,b∨d)∈~F.
(3)(a⊗c,b⊗d)∈~F. In fact, by (R10), (a⊗c)→(b⊗c)≥a→b∈F, which implies that (a⊗c)→(b⊗c)∈F. Similarly, (b⊗c)→(a⊗c)∈F. Thus (a⊗c,b⊗c)∈~F. Similarly, (c⊗b,d⊗b)∈~F. Hence (a⊗c,b⊗d)∈~F by the transitivity of ~F.
(4)(a→c,b→d)∈~F. In fact, by (R16), we have
(7)(b⟶a)⊗(a⟶c)⊗(c⟶d)≤b⟶d,
which implies that
(8)(b⟶a)⊗(c⟶d)≤(a⟶c)⟶(b⟶d).
Thus, (a→c)→(b→d)∈F. Similarly, (b→d)→(a→c)∈F. Hence (a→c,b→d)~F.Proposition 7.
Let~  be a {∧,→}-congruence on L. Then F~={x∈L∣(x,1)∈~} is a filter of L.Proof.
Obviously,1∈F~. Suppose that x,x→y∈F~; that is, (x,1),(x→y,1)∈~. By (R6) and (C2), we have (x→y,y)=(x→y,1→y)∈~ and by the transitivity of  ~, we have (y,1)∈~. Thus y∈F~. Hence F~ is a filter of L.Lemma 8.
Let~ be a {∧,→}-congruence on L. Then (x,y)∈~ if and only if (x→y,1)∈~ and (y→x,1)∈~.Proof.
Suppose that(x,y)∈~. Then (x→y,1)=(x→y,y→y)∈~ and similarly (y→x,1)∈~. Conversely, suppose that (x→y,1)∈~ and (y→x,1)∈~. Then
(9)((x⟶y)⟶y,y)=((x⟶y)⟶y,1⟶y)∈~.
By (C1) and (R15),
(10)(x,x∧y)=(((x⟶y)⟶y)∧x,y∧x)∈~.
Similarly, we have (y,x∧y)∈~. Hence (x,y)∈~ by the transitivity of ~.Theorem 9.
LetF,~ be a filter of L and a {∧,→}-congruence on L, respectively. Then ~F~=~ and F~F=F. Thus there is a bijection between F(L) and Con(L).Proof.
(1) By Lemma8, (x,y)∈~F~ if and only if x→y∈F~ and y→x∈F~ if and only if (x→y,1)∈~ and (y→x,1)∈~ if and only if (x,y)∈~. Hence ~F~=~.
(2)x∈F~F if and only if (x,1)∈~F if and only if x→1∈F and 1→x∈F if and only if x∈F. Hence F~F=F.Remark 10.
(1) By Proposition6 and Theorem 9, if ~ is a {∧,→}-congruence on L and (a,b),(c,d)∈~, then (a∧c,b∧d),(a∨c,b∨d),(a⊗c,b⊗d),(a→c,b→d)∈~. That is to say, a {∧,→}-congruence and a (universal) congruence are equivalent to each other, and so are the symbols Con(L).
(2) In [16], Pavelka firstly showed that there is a one-to-one correspondence between all filters and all congruences in a residuated lattice. And a binary relation is a universal-algebra-congruence if and only if it is an equivalence relation that preserves both ⊗ and → (that is, it just depends on the operations ⊗,→; the other two operations ∨,∧ are automatically preserved).


## 3.M-Filters

In the following part of this paper, unless otherwise stated,M always denotes a lattice with a greatest element 1. In a lattice M, an element a is called prime (resp., coprime) if b∧c≤a (resp., a≤b∨c) always implies b≤a or c≤a (resp., a≤b or a≤c) for all b,c∈M. The set of all prime (resp., coprime) elements of M is denoted by J(M) (resp., P(M)). A complete lattice M is called a spatial frame [6] if a=∧{r∈P(M)∣a≤r} and M is called a closed set lattice [17] if a=∨{r∈J(M)∣r≤a}.In this section, we will studyM-filters and their properties in the residuated lattice L.Definition 11.
We call a mappingA:L→M a lattice-valued filter of L if(FF1)A(1)=1;(FF2)A(y)≥A(x)&A(x→y) for all x,y∈L.Remark 12.
The definition of a lattice-valued filter [13] is a lattice-valued set A of L satisfying (FF2) and(FF1′) for all  x∈L,A(1)≥A(x),which is different from Definition 11. It is easy to see that a lattice-valued filter in this paper is always a lattice-valued filter in [13]. In a common sense, a lattice-valued filter should be equivalent to a crisp one if we replaced M by {0,1}. Thus, the lattice-valued filter in [13] is not a direct generalization of a crisp one since 0L (the constant map valued at 0) is a lattice-valued filter of L while ∅ (the crisp counterpart) is not a crisp one.DenoteFF(L) as the set of all lattice-valued filters of L.Proposition 13.
LetA:L→M be a mapping with A(1)=1. The following two are equivalent:(1)
A∈FF(L);(2)
A  is monotone with respect to the order on  L  and  A(x)∧A(y)≥A(x⊗y)≥A(x)&A(y)   for all  x,y∈L.Proof.
(
1
)
⇒
(
2
): for any x,y∈L with x≤y, we have x→y=1 and
(11)A(y)≥A(x⟶y)&A(x)=A(1)&A(x)=1&A(x)=A(x).
Thus A is monotone. By (FF2) and (R13),
(12)A(x⊗y)≥A(x)&A(x⟶(x⊗y))≥A(x)&A(y)
since A is monotone. Also, A(x⊗y)≤A(x)∧A(y) since A is monotone. Therefore, A(x)∧A(y)≥A(x⊗y)≥A(x)&A(y).
(
2
)
⇒
(
1
): by (R12), forallx,y∈L,A(y)≥A(x⊗(x→y))=A(x)&A(x→y).Corollary 14.
If&=∧ in M, then for each A∈FF(L), A(x∧y)=A(x⊗y)=A(x)∧A(y).Proof.
By Proposition13 and (R1), A(x∧y)≤A(x)∧A(y)=A(x⊗y)≤A(x∧y). Then A(x∧y)=A(x⊗y)=A(x)∧A(y).LetA:L→M be a mapping. For any r∈M, define
(13)A[r]={x∈L∣A(x)≥r},A(r)={x∈L∣A(x)≰r}.Proposition 15.
A
∈
F
F
(
L
) if and only if A[r]∈F(L) for any r∈M.Proof.
(
1
)
⇒
(
2
): clearly, for any r∈M,1∈A[r]. If x,x→y∈A[r], then A(x),A(x→y)≥r. Then A(y)≥A(x)∧A(x→y)≥r. Thus, y∈A[r]. Hence A[r]∈F(L).
(
2
)
⇒
(
1
): clearly, A(1)=1 since A[1]∈F(L). For any x,y∈L, suppose that A(x)∧A(x→y)=r. Then x,x→y∈A[r]. Thus y∈A[r] and A(y)≥r=A(x)∧A(x→y). Hence A∈FF(L).Proposition 16.
(1) IfM is a closed set lattice, then A∈FF(L) if and only if A[r]∈F(L) for any r∈J(M).
(2) IfM is a spatial frame, then A∈FF(L) if and only if A(r)∈F(L) for any r∈P(M).Proof.
(1) The necessity is from Proposition15. Sufficiency: clearly, A(1)=1 since A[r]∈F(L) for any r∈J(L). For any x,y∈L, suppose that r∈J(L) and r≤A(x)∧A(x→y). Then x,x→y∈A[r]. Thus y∈A[r] and A(y)≥r, By the arbitrariness of r∈J(M) and A∈FF(L).
(2) Necessity: clearly, for anyr∈P(M),1∈A(r) since 1∉P(L). If x,x→y∈A(r), then A(x),A(x→y)≰r and A(y)≥A(x)∧A(x→y). Then A(y)≰r and y∈A(r). Hence A(r)∈F(L). Sufficiency: if A(1)≠1, then there exists r∈P(M) such that A(1)≤r. Then 1∉A(r), which contradicts A(r)∈F(L). Thus A(1)=1. For any x,y∈L, for any r∈P(M) such that A(x)∧A(x→y)≰r, we have A(x)≰r and A(x→y)≰r and then x,x→y∈A(r), which implies that y∈A(r) and A(y)≰r. By the arbitrariness of r, we have A(y)≥A(x)∧A(x→y). Hence A∈FF(L).


## 4. Lattice-Valued Congruences

In this section, we will study lattice-valued congruences and the relations among filters, congruences, lattice-valued filters, and lattice-valued congruences in residuated lattices.Definition 17.
A mappingθ:L×L→M is called a lattice-valued congruence on L if it satisfies the following, for any x,y,z∈L:(FC1)θ(x,x)=1;(FC2)θ(x,y)=θ(y,x);(FC3)θ(x,z)≥θ(x,y)&θ(y,z);(FC4)θ(x∧z,y∧z)≥θ(x,y);(FC5)θ(x→z,y→z)≥θ(x,y).DenoteFCon(L) as the set of all lattice-valued congruences on L.Definition 18.
Letθ be a lattice-valued congruence on L. Define θx:L→M by θx(y)=θ(x,y)(forally∈L). θx is called the lattice-valued congruence class of x with respect to θ on L.Proposition 19.
Letθ be a lattice-valued congruence on L. Then θ1 is a lattice-valued filter on L, called the lattice-valued filter induced by θ, denoted by Aθ.Proof.
(FF1) Clearly,θ1(1)=θ(1,1)=1. (FF2) forallx,y∈L, by (FC3),
(14)θ1(y)=θ(1,y)≥θ(1,x⟶y)&θ(x⟶y,y),
and by (FC5),
(15)θ(x⟶y,y)=θ(x⟶y,1⟶y)≥θ(x,1)=θ1(x).
Thus θ1(y)≥θ1(x→y)&θ1(x). Hence θ1∈FF(L).Proposition 20.
LetA be a lattice-valued filter on L and θA(x,y)=A(x→y)&A(y→x)(forallx,y∈L). Then θA is a lattice-valued congruence on L, called the lattice-valued congruence induced by A.Proof.
(FC1) and (FC2) are obvious and omitted. For anyx,y,z∈L, (FC3) by Proposition 13 and (R16),
(16)θA(x,y)&θA(y,z)=A(x⟶y)&A(y⟶x)&A(z⟶y)&A(y⟶z)≤A((x⟶y)⊗(y⟶z))&A((z⟶y)⊗(y⟶x))≤A(x⟶z)&A(z⟶x)=θA(x,z);
(FC4) by Proposition 13, (R2), and (R7),
(17)θA(x∧z,y∧z)=A((x∧z)⟶(y∧z))&A((y∧z)⟶(x∧z))=A(((x∧z)⟶y)mmm∧((x∧z)⟶z))&A(((y∧z)⟶x)mmmmmmmmmmmmmmm∧((y∧z)⟶z))=A((x∧z)⟶y)&A((y∧z)⟶x)≥A(x⟶y)&A(y⟶x)=θA(x,y);
(FC5) by Proposition 13, (R9), and (R15),
(18)θA(x⟶z,y⟶z)=A((x⟶z)⟶(y⟶z))&A((y⟶z)⟶(x⟶z))=A(y⟶((x⟶z)⟶z))&A(x⟶((y⟶z)⟶z))≥A(y⟶x)&A(x⟶y)=θA(x,y).Theorem 21.
Letθ,A be a lattice-valued congruence and a lattice-valued filter on L, respectively. Then(1)
θAθ=θ;(2)
AθA=A.
Thus there is a bijection betweenFF(L) and FCon(L).Proof.
(1)Forallx,y∈L, by (FC2)–(FC5), (R6), and (R15),
(19)θAθ(x,y)=Aθ(x⟶y)&Aθ(y⟶x)=θ(1,x⟶y)&θ(1,y⟶x)≤θ(1⟶y,(x⟶y)⟶y)&θ(1⟶x,(y⟶x)⟶x)=θ(y,(x⟶y)⟶y)&θ(x,(y⟶x)⟶x)≤θ(y∧x,((x⟶y)⟶y)mmm∧x((x⟶y)⟶y))&θ(x∧y,((y⟶x)⟶x)∧y)=θ(x∧y,x)&θ(x∧y,y)≤θ(x,y).
And by (FC5) and (R7),
(20)θAθ(x,y)=θ(1,x⟶y)&θ(1,y⟶x)=θ(y⟶y,x⟶y)&θ(x⟶x,y⟶x)≥θ(y,x)&θ(x,y)=θ(x,y).
(2) For allx∈L,AθA(x)=θA(1,x)=A(1→x)&A(x→1)=A(x)&A(1)=A(x).Lemma 22.
Letθ be a lattice-valued congruence on L. Then(1)
for anyr∈M, one has θ[r]∈Con(L);(2)
ifM is a spatial frame, then for any r∈P(M), θ(r)∈Con(L).Proof.
This proof is trivial by the definitions of congruences and lattice-valued congruences.Proposition 23.
For anyA∈FF(L),θ∈FCon(L), one has forallr∈M,(1)
(θA)[r]=~(A[r]);(2)
(Aθ)[r]=F(θ[r]).Proof.
(1) Consider the following:(21)(θA)[r]={(x,y)∈L×L∣θA(x,y)≥r}={(x,y)∈L×L∣A(x⟶y)∧A(y⟶x)≥r}={(x,y)∈L×L∣x⟶y,y⟶x∈A[r]}=~(A[r]).
(2) Consider(22)(Aθ)[r]={x∈L∣Aθ(x)≥r}={x∈L∣θ(1,x)≥r}={x∈L∣(1,x)∈θ[r]}=F(θ[r]).Replacing“≥" by “≰" in Proposition 23, we have the following.Theorem 24.
LetM be a spatial frame. Then A∈FF(L),θ∈FCon(L), and one has for any r∈P(M),(1)
(θA)(r)=~(A(r));(2)
(Aθ)(r)=F(θ(r)).By Theorem9, Proposition 16, Theorem 21, Proposition 23 and Theorem 24, we have the following.Corollary 25.
(1)θ∈FCon(L) if and only if for any r∈M,θ[r]∈Con(L).
(2)θ∈FCon(L) if and only if for any r∈P(M),θ(r)∈Con(L).By Corollary25 and Remark 10, we have the following.Corollary 26.
Letθ be a lattice-valued congruence on L. Then each of θ(a∧c,b∧d),θ(a∨c,b∨d),θ(a⊗c,b⊗d), and θ(a→c,b→d) is larger than or equal to θ(a,b)∧θ(c,d).At last, we will give some properties of lattice-valued congruence classes of lattice-valued congruences.Lemma 27.
Letθ be a lattice-valued congruence on L. Then for any x,y∈L,θ(1,x→y)∧θ(1,y→x)=θ(x,y).Proof.
It is a corollary of Theorem21(1).Proposition 28.
Letθ be a lattice-valued congruence on L and x,y∈L. Then the following four are equivalent.(1)
θx=θy.(2)
θ(x,y)=1.(3)
θ(1,x→y)=θ(1,y→x)=1.(4)
θx→y=θy→x=θ1.Proof.
Clearly,(2) is equivalent to (3) by Lemma 27. (1)⇒(2): θ(x,y)=θx(y)=θy(y)=θ(y,y)=1.(2)⇒(1): for all  z∈L,θx(z)=θ(x,z)≥θ(x,y)∧θ(y,z)=1∧θ(y,z)=θy(z).Similarly, θy(z)≥θx(z) and so θx=θy.
Similar to(1)⇔(2), we can show that (3)⇔(4).



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