Abstract

In this paper, we determine conditions for the existence of an epimorphism between two finite 4-valued modal algebras and state a method to obtain it. Furthermore, we obtain formulas which generalize those indicated by R. Sikorski for finite Boolean algebras [1] , and by M. Abad and A. V. Figallo for finite 3-valued ukasiewicz algebras [2] .

Keywords:Four Valued Modal Algebras, Tetravalent Modal Algebras, De Morgan Algebra, Epimorphisms, Automorphisms

1. Introduction

Tetravalent modal algebras were first considered by A. Monteiro and studied mainly by I. Loureiro, A. V. Figallo, A. Ziliani and P. Landini (see [3] -[10] ). Later, in 2000, J. M. Font and M. Rius [11] showed their interest in the logics originated by the lattice aspects of these algebras. Following A. V. Figallo’s terminology, we call them four-valued modal algebras (or 4-valued modal algebras). In addition, the interesting 2013 work by M. Coniglio and M. Figallo (see [12] ) has given, in our opinion, a new impulse to the profound study and development of the tetravalent modal algebras. For further information the reader interested in this subject is referred to the bibliography recommended below.

Let us now recall that An algebra of type is a De Morgan algebra [13] if is a bounded distributive lattice with least element 0, greatest element 1, and satifies the equations and  .

It is known that a finite De Morgan algebra is determined by its determinant system, where is the ordered set of all prime elements of and is an antitone involution of

. Thus, has the following properties:

(i), for each(ii) if and, then.

Moreover, the operation is given in the following way:

In 1978 A. Monteiro defined the 4-valued modal algebras (or -algebras) as algebras of type, where is a De Morgan algebra and the properties

are verified.

We assume the reader to be familiar with the theory of -algebras as it is given in [3] [4] . In particular, in [4] is the demonstration of the following properties, which are used in this paper:

If the operator is defined by the formula, then it is easy to see that.

In what follows, we only consider finite -algebras.

From [13] me have that the determinant system of an -algebra has connected components of the three following types:

The ordered set has the following diagram:

We will denote by, , and the ordered sets, , , respectively.

Frequently we will write instead of.

On the other hand, the operator is given by the formula

Then,

2. M₄-Epimorphisms and M₄-Functions

We will denote the sets and by and, respectively.

The lemma given below will be used in the proof of Lemma 2.5 Lemma 2.1 If, then either or

Proof. If, then, so with. Then,

,

.

Thus, , that is.                                                  In what follows, we will denote with the set of all -epimorphisms from the -algebra into the -algebra.

Definition 2.1 A mapping is called an -function if and only if the following conditions are satisfied:

(M1) is one to one(M2) for all and(M3) for all.

We will denote by the set of all -functions from into.

Lemma 2.2 If, and, then:

(i) is an isotone map,

(ii).

Proof.

(i) If and, then. Thus, , and by (M1) we obtain.

(ii) Let, , ,. It is easy to see that. We shall show that. Indeed, if, then there exists such that. Hence, and. Since, we obtain and, with, that is. Hence,; thus,. Finally,.

Definition 2.2 Let. is an -function associated with if and only if is defined by:

We denote by the set of all -functions.

The following lemma is used to prove that every -function is an epimorphism.

Lemma 2.3

(I) If and, then.

(II) If, then or.

(III)If, then.

Proof.

(I) Assume that. Since and, it follows that there exists, such that. From this we obtain; as a consequence, , which contradicts the hypothesis.

(II) Suppose. There are two possibilities for:

(a) is minimal in. As, , then there exists such that. Since and is minimal, then must hold, and from (M1), results. Thus, with.

(b) is maximal and not minimal in. Then, with and. From there exists such that. If we get, and then. If, as in (a), there exists a unique such that. Moreover, (i) because if, then, which is not possible. Consequently, (ii).

From (i) it follows that is Type II or Type III; in the latter case and therefore

, which is not possible. Then, is Type II, so that, then, and thus (iii). From (ii) and (iii) it results. Therefore, , so.

(III) Let.

(a) If, then from (II), with. Thus,; then,.

(b) If, then. There are two cases:

(b₁). Then,.

(b₂) and, then:

(b₂₁) if, then with, and; therefore, by (M1). Moreover, it is clear that is a minimal prime, because if, then, which is not possible. Then, is minimal, and as a consequence,. Thus, , which is a contradiction. Then, holds; therefore,

(b₂₂) if, then is Type III; therefore, and .

Lemma 2.4 If, then.

Proof. Let and be the -function associated by.

(a) It is obvious that and.

(b). Actually, it is easy to see that; moreover, if and only if. Since, or must hold; hence, or, that is; then. Finally,

.

(c) for all. Indeed,

(c₁).

It is sufficient to show that, if, then.

If, then and. Hence, there exists such that, and. Then, holds. Taking into account that, by Lemma 2.3(I), we have that; then.

(c₂).

Let. The proof is a consequence of the fact that the following conditions are pairwise equivalent:

(1)   (2)   (3)

(4)    (5)    (6)

(d), for all, which is inmediate consequence of (III), and (b).

(e) is clearly onto.

Lemmas 2.5 and 2.6 are necessary to show that all -epimorphisms are an -function M.

Lemma 2.5 If and, then there exists a unique such that.

Proof.

(1) (a) If, then there is such that Indeed, let; since is onto, there exists such that; therefore, and then,. If, then, and. If, then, and; since

, or must hold. If, then; therefore for some; hence, and so. Then or. Thus,.

(b) If and, then is unique. Indeed, if, then and. Accordlingly, is the smallest such that,. If, then; by Lemma 2.1, must hold. Furthermore, , and then . On the other hand, , which is a contradiction.

(2) If, then there is such that; by 1. there is a unique such that. Let; it is easy to prove that and are of the same type. So.

In addition, the following holds:, giving the result that follows:

(i)

Moreover,;; then

(ii)

From (i) and (ii), it follows that. It is evident that is unique and.

Definition 2.3 Let. The function is induced by if is defined by, if and only if.

Lemma 2.6 Every function induced by an -epimorphism is an -function.

Proof. be the function induced by the -epimorphism.

(M1) It is easy to prove that is one to one.

(M3) Let, with. Therefore,; consequently, , i.e..

As a consequence of (M1) and (M3), is an order-preserving function.

(M2) Let, with.

(1) If is Type I, then, and

(a) if, since and is one to one on the set of all prime that, must hold; then, which is a contradiction.

(b) if, then. If, then; since f es one to one must be, which is a contradiction. So and consequently and then, which is not possible.

This shows that in this case and; as a result, so that either, or and are not comparable, which is not possible since .

(2) If is Type II, then either or. In the first case and therefore. A similar argument is valid if.

(3) If is Type III, then is Type III, and it is not comparable with; consequently,.

Lemma 2.7.

Proof. If, the proof is obvious. Let, the function induced by, and the -function associated to. It is clear that. Let,. Then, and consequently,. For each, , let. Then. Thus,; as a consequence,. Teorema 2.1.

Proof. It follows from Lemmas 2.4 and 2.7. Teorema 2.2.

Remarks 2.1 Let,. Then(i) if only if, , and.

(ii) The following holds

where

is the set of all injective functions, is the set of all injective functions

such that for each, is the set of all injective functions such that for all, and for all.

Teorema 2.3 If, , and, then

Proof. If, , and, then the map is a bijective correspondence between and. From this, Theorem 2.1 and Remarks 2.1 this theorem follows.

An immediate consequence of theorem 2.3 is

Teorema 2.4

,

References

AMS 2000 Subject Classification

Primary 08A35, 06D30. Secondary 03G25.