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Revista de la Unión Matemática Argentina
Print version ISSN 00416932
Rev. Unión Mat. Argent. vol.50 no.1 Bahía Blanca June 2009
The subvariety of QHeyting algebras generated by chains
Laura A. Rueda
Abstract. The variety of Heyting algebras with a quantifier [14] corresponds to the algebraic study of the modal intuitionistic propositional calculus without the necessity operator. This paper is concerned with the subvariety of generated by chains. We prove that this subvariety is characterized within by the equations and . We investigate free objects in .
1. Introduction and Preliminaries
Distributive lattices with a quantifier were considered as algebras for the first time by Cignoli in [7] who studied them under the name of distributive lattices. A distributive lattice is an algebra of type such that is a bounded distributive lattice and the unary operation satisfies the following conditions, for any , : , , and . These conditions were introduced by Halmos [9] as an algebraic counterpart of the logical notion of an existential quantifier.
Various further investigations have been carried out since [7] (see R. Cignoli [8], H. Priestley [13], M. Adams and W. Dziobiak [4], M. Abad and J. P. Díaz Varela [2] and A. Petrovich [11]). As a natural generalization, the operation of quantification was considered for Heyting algebras in [3] and [15]. A Heyting algebra is an algebra of type for which is a bounded distributive lattice and for , , is the relative pseudocomplement of with respect to , i.e., if and only if . It is known that the class of Heyting algebras forms a variety. An important subvariety of Heyting algebras is the class of linear Heyting algebras [5]. A linear Heyting algebra is a Heyting algebra that satisfies the equation . Throughout this paper will denote the category of Heyting algebras and Heyting algebra homomorphisms and will denote the subcategory of linear Heyting algebras.
A Heyting algebra is an algebra such that is an object of and is a quantifier on , that is, is a unary operation defined as for distributive lattices. Monadic Boolean algebras are the simplest examples of Heyting algebras. The class of Heyting algebras forms a variety, which we denote . The subvariety of characterized within by the equation , that is, the subvariety of linear Heyting algebras will be denoted by . Heyting algebras were first introduced in [14] and have been investigated in [14, 15, 3].
In this paper we investigate the subvariety of the variety of Heyting algebras generated by chains. We characterize by identities in Section 2 and we investigate free objects in this variety in Section 3.
We will usually use the same notation for a variety and for the algebraic category associated with it. And, similarly we will use the same notation for a structure and for its universe.
Recall that Heyting algebras are algebraic models of the intuitionistic propositional logic and that the study of extensions of Intuitionistic Propositional Calculus (IPC) reduces to the study of subvarieties of the variety . The language of intuitionistic modal logic (MIPC) is the language of IPC enriched with two modal unary operators of necessity and of posibility . The algebraic models of MIPC are the monadic Heyting algebras.
Now, in MIPC the operators and are independent from each other, that is and are not theorems in MIPC. Hence, the set of theorems of the propositional calculus without of the necessity operator , called the free fragment of MIPC is different from that of MIPC. Similarly, the set of theorems of the propositional calculus without of the possibility operator , called the free fragment of MIPC is different from that of MIPC.
It turns out that the behaviour of the free fragment of MIPC is very much similar to that of MIPC. However, surprisingly enough, the free fragment of MIPC behaves pretty diferent from MIPC.
Heyting algebras are the algebraic models of the free fragment of MIPC, that is, Heyting algebras are the free reducts of monadic Heyting algebras.
For a poset and , let and . We write , instead of , respectively. We say that is decreasing if , increasing if and convex if . A mapping is order preserving if whenever .
In order to describe the dual category of we recall that a Priestley space is a triple such that is a partially ordered set, is a compact topological space, and the triple is totally orderdisconnected (that is, for , , if then there exists a clopen increasing such that and ). Priestley showed that the category of bounded distributive lattices and lattice homomorphisms is dually equivalent to the category of Priestley spaces and order preserving continuous functions (see the survey paper [12]).
A Heyting space (see [7, 14, 15]) is a Priestley space together with an equivalence relation defined on such that is clopen for every convex clopen , for each , where and is the lattice of clopen increasing subsets of , and the blocks of are closed in . For , let denote the clopen increasing set that represents , where is the set of prime filters of , ordered by set inclusion and with the topology having as a subbasis the sets and for . If then, under the duality, corresponds to the clopen increasing set .
For Heyting spaces and , a Heyting morphism is a continuous orderpreserving mapping such that and , for each .
It can be proved in the usual way that the category of Heyting algebras and homomorphisms is dually equivalent to the category of Heyting spaces and Heyting morphisms [14, 15]. For each Heyting algebra the corresponding Heyting space is , where . Conversely, if is a Heyting space, the corresponding Heyting algebra is , where is defined as in .
In this section we will study the subvariety generated by chains within . Observe that if , then , that is, . Consequently, .
Recall that in the variety of Heyting algebras, congruences are determined by filters. Precisely, if and is a filter of , then is a congruence on , and the correspondence establishes an isomorphism from the lattice of filters of on , the lattice of congruences of . If is generated by an element , , we write .
Observe that if is a Heyting chain and is a filter of , if and only if or . Then, . As a consequence of this, we have that if is a chain, is a subdirectly irreducible algebra in if and only if is a subdirectly irreducible algebra in , that is, has a unique dual atom.
A quantifier on an algebra is said to be multiplicative if , for every .
Let be the subvariety of characterized by the equation .
Proof Let and , i.e., . As , then , that is, . So . Thus , so . Therefore, .
Observe that , that is, . Let us see that .
Lemma 2.2. Let be a subdirectly irreducible algebra in . Then is a chain.
Proof Let be a subdirectly irreducible algebra. Then and hence is subdirectly irreducible in , that is, has a unique dual atom. Since for every , , then or , that is, or . So is a chain.
As a consequence of this corollary we have that is characterized within by the identities and .
The following theorem characterizes the dual space of an algebra in .
Theorem 2.4. Let be a Heyting algebra, let be the associated Heyting space and the partition of determined by . Then, if and only if each has exactly one maximal element.
Proof Suppose that and there exists such that has two maximal elements , , . Let be such that and . For each , we have that . Thus there exists such that and . Consequently
As is closed, by a compacteness argument
and . So and . This implies that and consequently . On the other hand, , which contradicts that .
Conversely, we know that . Let us see that . Since is a quantifier, . Let us prove the other inclusion. Let and such that . Since and , if , then . Therefore and so
Lemma 2.5. is the greatest subvariety of such that every filter determines a congruence.
Proof Let such that . We are going to construct a filter in which does not determine a congruence. From , there exist such that . Then there exists a prime ideal such that and . Since is an ideal we have that and . Consider the filter . Then , being that . Let us see that . Suppose on the contrary that . Thus , which implies that (*) since the image of is closed under implication. On the other hand, , so . Since is a prime ideal and we have that . This, together with (*), implies that , which is a contradiction.
In this section we characterize the free algebra in with generators. Following a path analogous to that of M. Abad and L. Monteiro in [1], we will provide a method to construct the order set of all joinirreducible elements of the free algebra, and as a consequence, we will obtain a formula to compute .
It is clear that for any subset of a chain , the subalgebra of generated by is . Thus, every generated subalgebra of a chain of has at most elements, that is, the class of all chains in is uniformly locally finite. So is generated by a uniformly locally finite class, and consequently, is a variety locally finite [6, Theorem 3.7].
If is a finite algebra, the Heyting space has the discrete topology and is antiisomorphic to the ordered set of joinirreducible elements of . In this section we will use the set instead of and we will consider the relation defined on , that is we consider . If is the partition determined by in , we say that if and only if . This is an order relation.
Theorem 3.1. [10] A Heyting algebra is linear if and only if the family of prime filters which contain a prime filter is a chain.
Definition 3.2. Let be a finite algebra. Let and let , such that , , where . We say that has coordinates , if the chain is of length and if , .
Notice that the set of the previous definition is considered within .
Let be a non negative integer. Let be the chain with elements. Let , , with for . Let be the interval in consisting of the elements such that . We denote the algebra , where , .
Observe that if , is a chain. More precisely, is of length if and only if is a chain with elements [1, p. 7]. If is the natural homomorphism and are such that , then there exist joinirreducible elements in such that and , . Moreover, taking into account that en if and only if en , it follows that has coordinates , if and only if . Since is the trivial relation, we have that is a subdirect product of the chains .
Let be the free algebra with a finite set of generators of cardinal . For the sake of simplicity we will write instead of .
We know that every generated subalgebra of a chain of has at most elements. Since is a chain generated by at most elements, we have the following
If then from Lemma 3.3, has coordinates , for some , and such that .
Consider the following sets:
and
For a subset of to generate the algebra , every non constant element must be contained in , that is, . Besides, every constant can be obtained from , except the constants of . So we have that and consequently, .
For every , , consider the sets
We will denote instead of . Observe that , and for every , that is, consists of one tuple whose first coordinates are equal to 2. Moreover, if , we have that , but if , .
Let . It is clear that
and that for .
Let be the set of all functions from the set of free generators of into such that . Observe that every is nonempty, as if and only if , that is .
Recall that a filter in a finite Heyting algebra is prime if and only if , where is joinirreducible element.
If , can be extended to a unique homomorphism from onto . If is the kernel of , it is well known that is a prime filter in , so , with . Thus, for each , we have a function
defined by .
Lemma 3.4. The following holds .
Proof Let us see that is onto. For , consider the natural homomorphism from onto , and the restriction of to . Then and therefore . Let be the extension of . Since , then and therefore .
Let us prove that the function , is onetoone. If the functions , satisfy then there is an automorphism of such that . But the only automorphism of is the identity, then and then .
If is the number of functions from a set with elements onto a set with elements, then:
Let . Then, for each ,
In particular, for , and then . For , , and for , . So in both cases, , then . And for , , , , then and
Consequently,
Consider the set
If , there is a unique and a unique such that . If we put we have a onetoone mapping from onto .
The following lemma is immediate (recall that ).
Lemma 3.6. has coordinates if and only if for all .
As a consequence, the set has minimal elements.
Lemma 3.7. For , and , has coordinates if and only if .
Proof From the proof of Lemma 3.4, has coordinates if and only if , and from the comment preceding that lemma, this is equivalent to .
Remark 3.8. We know that if , the extension homomorphism and the natural homomorphism from into satisfy . Then if in , , we have
The proof of the following lemma will be omitted since it is an adaptation of that of [1, Lemma 3.13].
We say covers if and implies .
Lemma 3.9. If , , covers if and only if the following conditions hold:

(i) ,

(ii) , , .

(iii) or , , and .
In the following theorem we denote .
Theorem 3.10. Let , . Then covers if and only if , or , , , and for the following conditions hold:

(I) if and only if , .

(II) if and only if or .
Proof Suppose that covers . The first part of the theorem is an immediate consequence of Lemma 3.9
Since in , , we have
if and only if
if and only if
if and only if
and if and only if
In particular, we have the conditions and .
Conversely, let , be such that
and satisfying and . Then,
From Lemma 3.9 , we must prove that .
Consider in
and
the chains and respectively and consider the following sets:
Then
We have that is a filter, is an ideal and , , are nonempty sets, being that , , . is also nonempty. Indeed, if , since , there is such that , then from and , and , that is . If , there is such that , then from and , and , that is .
It is clear that the sets , , are pairwise disjoint. Observe that , and so it is a filter. Using these remarks it is a routine matter to show that the set is a subalgebra of .
Let us see that . If , .
If , and from , , that is, . Then .
If , and from , , that is . So .
If , , then and from , , that is, . Then .
Therefore, and consequently .
Then we can write, .
Since , we have
The previous theorem allows us to construct the ordered set of joinirreducible elements of the free algebra . By virtue of Lemma 3.6, there exists a onetoone correspondence between the set of minimal elements of and the set of functions from into . Since is a chain, for every , then is also a chain for . So, the orderedconnected components of are , where is minimal, that is, the orderconnected components of are the sets , where .
We have constructed the Heyting space . The free algebra with a finite set of generators of cardinality , is the algebra obtained from considering the decreasing subsets of with the quantifier given by , for each decreasing set of .
Example 3.11. In the next figure we give the free algebra generated by an element , and the ordered set of its joinirreducible elements, with the equivalence relation which determines the quantifier.
Where , , , and . We denote .
In the rest of this section we investigate the poset in order to obtain a recursive formula for the number of elements of .
Let be the family of orderconnected components , with minimal, such that , . It is clear that , and if , then . In general, .
For a given , all the orderconnected components in have the same number of elements. So if and for , then
We are going to determine .
Consider . From Theorem 3.10, we know that covers if and only if or and

(I) if and only if .

(II) if and only if .
In particular, there are funtions in covering , and similarly there are funtions in covering . So, there are functions covering , of which satisfy , .
In these conditions we have the following result.

(1) If , there exists with such that and are orderisomorphic.

(2) If and covers , with and , for every , then there exists , such that , and are orderisomorphic.
Proof

(1) If is the function defined by:
,
is clearly a minimal element of , and .
Let us see that and are orderisomorphic. Observe that if , then , where and . We define by means of , where
, .
Clearly is an isomorphism.

(2) Observe that, if , , then , , and . If is the function defined by , then and are orderisomorphic. Indeed, if we define by means of , where and defined as in , and it can be proved that is an isomorphism.
Finally, consider defined by , where , , , and
,
Clearly is an isomorphism.
From the previous proposition,
Therefore
and then
Acknowledgment: I gratefully acknowledge helpful comments of the referees. In particular, one of them suggested the algebraic proof of Lemma 2.5.
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Laura A. Rueda
Departamento de Matemática,
Universidad Nacional del Sur
Bahía Blanca, Argentina
larueda@criba.edu.ar
Recibido: 12 de noviembre de 2007
Aceptado: 2 de junio de 2009