International Journal of Mathematics and Mathematical Sciences

Volume 2014 (2014), Article ID 428940, 12 pages

http://dx.doi.org/10.1155/2014/428940

Research Article

Folding Theory Applied to Residuated Lattices

Albert Kadji,¹ Celestin Lele,² Jean B. Nganou,³ and Marcel Tonga¹

¹Departmen of Mathematics, University of Yaounde 1, P.O. Box 812, Yaounde, Cameroon

²Departmen of Mathematics, University of Dschang, P.O. Box 67, Dschang, Cameroon

³Departmen of Mathematics, University of Oregon, Eugene, OR 97403, USA

Received 14 February 2014; Revised 20 April 2014; Accepted 20 May 2014; Published 25 June 2014

Academic Editor: Radko Mesiar

Copyright © 2014 Albert Kadji et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

Residuated lattices play an important role in the study of fuzzy logic based on -norms. In this paper, we introduce some notions of -fold filters in residuated lattices, study the relations among them, and compare them with prime, maximal and primary, filters. This work generalizes existing results in BL-algebras and residuated lattices, most notably the works of Lele et al., Motamed et al., Haveski et al., Borzooei et al., Van Gasse et al., Kondo et al., Turunen et al., and Borumand Saeid et al., we draw diagrams summarizing the relations between different types of -fold filters and -fold residuated lattices.

1. Introduction

Since Hájek introduced his basic fuzzy logics, (BL-logics in short) in 1998 [1], as logics of continuous -norms, a multitude of research papers related to the algebraic counterparts of BL-logics, has been published. In [2–5], the authors defined the notions of -fold (implicative, positive implicative, Boolean, fantastic, obstinate, and normal) filters in BL-algebras and studied the relation among them.

A close analysis of the situation reveals that the main drive in all the previously mentioned works resides in the existence of an adjoint pair of operations. Just as the foldness theory for filters in BL-algebras generalizes filters introduced by Hájek, our foldness theory for filters in residuated lattices builds on recently published works on filters in residuated lattices by Haveshki et al. [6], Van Gasse et al. [7], Kondo and Dudek [8], Kondo and Turunen [9], Borumand Saeid and Pourkhatoun in [10], and Zahiri and Farahani in [11].

More specifically, we introduce the notions of -fold (implicative, positive implicative, Boolean, fantastic, normal, integral, and involutive) filters and -fold Boolean filters of the second kind in residuated lattices, notions that naturally generalize the corresponding ones previously studied in BL-algebras. Concurrently, we introduce the same foldness concepts on residuated lattices. In each folding class, we tie together the two concepts by characterizing the corresponding residuated lattices using their filters. For instance, it is shown (Proposition 20) that a residuated lattice is -fold implicative if and only if its trivial filter is -fold implicative if and only if all its filters are -fold implicative. Examples are included not only to illustrate the newly introduced concepts but also to differentiate them from the existing ones. Finally, diagrams summarizing all the relationships between the above classes of filters and residuated lattices are given (see Figures 1 and 2) for quick referencing. It should be noted that when restricted to BL-algebras these diagrams contain previously discovered relationships and also some newly found ones.

Figure 1: Diagram of -fold filters.

Figure 2: Diagrams of -fold residuated lattices.

2. Preliminaries

A residuated lattice is a nonempty set with four binary operations , and two constants satisfying the following properties:

 L-1: is a bounded lattice. L-2: is a commutative monoid. L-3: if and only if (residuation).

A MTL-algebra is a residuated lattice which satisfies the following condition:

 L-4: (prelinearity).

A BL-algebra is a MTL-algebra which satisfies the following condition:

 L-5: (divisibility).

A MV-algebra is a BL-algebra which satisfies the following condition:

 L-6: , where . Alternatively, a MV-algebra can be defined as a residuated lattice which satisfies the following condition: L-7: .

A Heyting lattice is a residuated lattice which satisfies the following condition:

 L-8: .

A Boolean lattice is a residuated lattice which satisfies the following condition:

 L-9: .

In this work, unless mentioned otherwise, will be a residuated lattice, which will often be referred to by its support set .

For any element , we define and for any integer .

Proposition 1 (see [7, 9, 10, 12]). The following properties hold in a residuated lattice.

(1) if and only if ; .(2). (3).(4)If then and . (5). (6), and   . (7)  if and only if . (8)If   ,  then     and    . (9). (10)implies   and   for  every   . (11). (12). (13). (14). (15).

Let and be a nonempty subset of . Then, is called a filter if it satisfies the following two conditions.

 (F1): for every , . (F2): for every , if and , then .

is called a deductive system  if , and for all , and implies .

It is known that in a residuated lattice, filters and deductive systems coincide [7].

The residuated lattices listed below are not BL-algebras and will be used to illustrate the concepts treated in the paper.

Example 2. Let be a lattice such that , , where and are not comparable. Define the operations and by the two tables shown in (1). Then is a residuated lattice which is not a MTL-algebra since :

; ; are the proper nontrivial filters of .

Example 3. Let be a lattice such that and , where and are not comparable. Define the operations and by the two tables shown in (2).  Then is a residuated lattice which is not a MTL-algebra, since :

is the only nontrivial filter of .

Example 4. Let be a lattice such that . Define the operations and by the two tables shown in (3). Then is a residuated lattice which is not a BL-algebra, since :

, , are filters of .

Example 5.   Let be a lattice such that , , where and are not comparable. Define the operations and by the two tables shown in (4). Then is a residuated lattice which is not a BL-algebra, since :

, , , are filters of .

Example 6. Let be a lattice such that , , where and are not comparable. Define the operations and by the two tables shown in (5). Then is a residuated lattice which is not a MTL-algebra, since :

is the only proper filter of .

Remark 7 (see [13]). It is well known that the class of residuated lattices is a variety. So from the above examples, we may obtain infinite residuated lattices which are not -algebras.

Definition 8 (see [14]). Let be a residuated lattice.

(i) is said to be locally finite if for every , there exists an integer such that.(ii) is said to be local if it has a unique maximal filter.

Clearly, a locally finite residuated lattice is local.

Given a filter of a residuated lattice , there is a well-known congruence on defined by if and only if and ;  the quotient structure is also a residuated lattice where;  ; ; .

One can easily verify the following result.

Proposition 9 (see [14]). For any filter of a residuated lattices , the following conditions are equivalent.

(i) is a maximal filter of .(ii)For any , if and only if for some .(iii) is a locally finite residuated lattice.

Consequently, it is straightforward to see that is locally finite if and only if is a maximal filter,   and is local if and only if ; is a proper filter.

Definition 10. Let and be two residuated lattices. Then a map is called a residuated lattice homomorphism if it satisfies the following conditions.

(i).(ii), for every .(iii), for every .

If is bijective, the homomorphism is called a residuated lattice isomorphism. In this case we write .

Definition 11 (see [7, 9, 14, 15]). A proper filter is said to be:

(i)prime if, for all , or ;(ii)prime of the second kind if,  for all , implies or ;(iii)prime of the third kind if,  for all , ;(iv)boolean if,  for all , ;(v)boolean of the second kind if,  for all , or ;(vi)primary if,  for all : implies or , for some integer ;(vii)a semimaximal filter if , where is the intersection of all maximal filters of which contain .

Remark 12 (see [7, 9, 14, 15]). (i)  Prime filters are prime filters of the second kind. The converse is true if is a MTL-algebra.

(ii)  Prime filters are prime filters of the third kind.

(iii)  Boolean filters of the second kind are boolean filters.

(iv)  If is a MTL-algebra, then maximal filters are prime filters.

(v)  If a filter is prime of the second kind and boolean, then it is boolean of the second kind.

(vi)  Maximal filters are semi maximal filters.

Lemma 13. Maximal filters are prime filters of the second kind.

Proof. Assume that is a maximal filter of and let , be two proper filters of such that . Then and ; hence and since is a maximal filter and , proper filters. From [14, Lemma 3], it follows that is prime of the second kind.

Proposition 14. Any prime filter of is a primary filter of .

Proof. Using Definition 11(vi), the proof is as shown in the case of pseudo-BL algebra [16].

Now, unless mentioned otherwise, will be an integer and .

A class of filters of will be said to be closed under extension if, for any filters and of , and imply .

3. -Fold Implicative Filter

Definition 15. Let be a residuated lattice.

(i) is said to be -fold implicative if for all .(ii)A subset is called an -fold implicative filter if and for all , and imply .

In particular, a 1-fold implicative filter is an implicative filter [12].

By taking in the definition, we see that any -fold implicative filter is a filter.

Like in the case of -algebras [8], a 1-fold implicative residuated lattice may be called a Gödel residuated lattice.

Example 16. Let . The residuated lattice of Example 5 is -fold implicative. But the residuated lattice of Example 6 is not 1-fold implicative, since .

Lemma 17. A filter of is -fold implicative if and only if, for all , ; is a filter of .

Proof. :  Let be an -fold implicative filter of . Since , we have . Let be such that ; then and . So, , and , showing that is a filter of .

:   Conversely, suppose that is a filter of , for all . Let such that and . Then, and thus , so, .

Proposition 18. The following conditions are equivalent for a filter and .

(i) is -fold implicative.(ii)For all , .(iii)For all , imply .

Moreover, the class of -fold implicative filters is closed under extension.

Proof. : Let ; by Proposition 1, we have and . Since is an -fold implicative filter of , we have .

:  Let such that and . By Proposition 1 we have , so .

Since , we obtain that . Now, by Proposition 1(14), . So, by the hypothesis and , we obtain .

:  Let be such that . By Proposition 1, we have and ; since is an -fold implicative filter of we have .

:

 Firstly, for ,; from , we have . Secondly, for ,;  from we have ; that is. From again, we obtain . Finally for ,  note that . From we obtain ; that is . From again, we also have .

By repeating the process times, we obtain .

Using Propositions 18 and 1, we can easily see that an -fold implicative filter is -fold implicative. But the converse is not true, as illustrated by Example 6, where the filter is 2-fold implicative but not 1-fold implicative, since .

Given a filter of and , the congruence class of in will sometimes be denoted by .

Proposition 19. The following conditions are equivalent for a filter of and .

(i) is -fold implicative. (ii) is an -fold implicative residuated lattice.

Proof. : Let and . From Proposition 18 it follows that ,   or equivalently ,  or equivalently . But . Hence for all ; that is, for all .

:  Let such that ; then, or equivalently ; that is, . Since is an -fold implicative residuated lattice, we have ; thus , and.

From Propositions 19 and 18, we have the following.

Proposition 20. The following conditions are equivalent for a residuated lattice .

(i) is -fold implicative.(ii) is an -fold implicative filter of .(iii)Every filter of is -fold implicative.(iv) for all .

Corollary 21. Any 1-fold implicative residuated lattice is a Heyting lattice, and any Heyting lattice is an -fold implicative residuated lattice.

Proof. Assume that is a 1-fold implicative residuated lattice.  If , we have . Hence .

An -fold implicative residuated lattice may not be a Heyting lattice, as illustrated by Example 6, for .

4. -Fold Boolean Filter

Definition 22. A filter of is said to be -fold Boolean if for all .

In particular, any 1-fold Boolean filter is a Boolean filter. [7]

From this definition, the class of -fold boolean filters is closed under extension. Moreover, in Example 4, is an -fold boolean filter and is not.

Proposition 23. The following conditions are equivalent for any filter and any .

(i) is -fold Boolean.(ii)For all , implies .(iii)For all , implies .

Proof. : Suppose that is -fold Boolean. Let be such that ; we must show that . By Proposition 1(12), we have ; since , we have , as is a filter.

: Let ; we must show that . Since , we have and then . So ; by , we get that .

:  Let such that ; we must show that . By Proposition 1, we have , and then . Since , we have and by .

:  This follows by setting in condition .

Remark 24. A filter satisfying Proposition  23(iii) is also said to be -fold positive implicative  [5].

Proposition 25. Every filter of which is -fold Boolean is also -fold implicative.

Proof. Let be an -fold Boolean filter of and such that . We have . By Proposition 1, we have ,  since ,  so , and is -fold implicative.

The converse of Proposition 25 is not true, as Example 5 shows. Indeed, for , is an -fold implicative filter which is not -fold Boolean because .

Definition 26. A residuated lattice is called -fold Boolean  (or -fold positive implicative) if for each .

Remark 27. Let be a residuated lattice.

(i)Since , it is clear that every -fold Boolean filter is ()-fold Boolean, and -fold Boolean residuated lattices are ()-fold Boolean.(ii)It is easy to check that the residuated lattice of Example 6 is -fold boolean for . But the residuated lattice of Example 5 is not, since .

As in the case of -fold implicativeness, one easily verifies that for a filter of a residuated lattice , is -fold boolean if and only if is an -fold boolean residuated lattice.

Proposition 28. The following conditions are equivalent for a residuated lattice .

(i) is -fold Boolean.(ii)Every filter of is -fold Boolean. (iii) is an -fold Boolean filter.

So, -fold Boolean residuated lattices are -fold implicative residuated lattices. But the converse is not true.

To end this section, we note that any 1-fold Boolean residuated lattice is a Boolean lattice, and any Boolean lattice is an -fold Boolean residuated lattice, for .

However, for the residuated lattice of Example 6 and , and are -fold Boolean residuated lattices which are not Boolean lattices since (here, is any nonempty set). Also, and are not BL-algebras.

5. -Fold Normal Filter

Definition 29. A filter is -fold normal if, for all , implies .

Example 30. Let .  In Example 5, simple computations prove that is an -fold normal filter,  but the filter is not -fold normal, since and.

Definition 31. A residuated lattice is said to satisfy the weak double negation if the following conditions hold for all, .

(i).(ii).(iii).

Note.  Any BL-algebra satisfies the weak double negation [17, Proposition 2.6(17)].

Example 32. (i)   By simple computation, one can show that the lattice of Example 2 satisfies the weak double negation and is not a BL-algebra.

(ii)  The lattice of Example 3 does not satisfy the weak double negation because.

It is easy to prove the following result.

Remark 33. If satisfies the weak double negation, then, for all .

Proposition 34. Let be a residuated lattice which satisfies the weak double negation and let be a filter of . The following conditions are equivalent.

(i) is -fold normal.(ii)For every , if   , then .(iii), where is the operator defined by .

Proof. :  Suppose that is -fold normal, and let such that . Then , and since is -fold normal, then ; that is, .

:  Suppose that (ii) holds, and let . Since , we have and . Thus, , since satisfies the weak double negation. From this and the fact that is a filter, we have . Thus we obtain that .

:  Suppose that (ii) holds, and let , that is, , then , and by (ii), . Thus as needed.

:  Suppose , and let such that . Since satisfies the weak double negation, we have , so . Hence, .

Note.    From the definition, it is clear that an -fold normal filter is -fold normal.

In [18, Theorem 8], the authors state a result (in the case ) that was meant to be (i)  (ii) above, but there seems to be a typo in the statement of their result. However, we have a correct proof, and (iii) provides an answer to their open problem. So, under the assumption of the weak double negation, the class of -fold normal filters is closed under extension.

Now we give the definition of an -fold normal residuated lattice.

Definition 35. is an -fold normal residuated lattice if it satisfies the following condition. For all , implies .

As in the case of -fold implicativeness, one easily sees that, given a filter of , is -fold normal if and only if is an -fold normal residuated lattice.

Note that the residuated lattice of Example 6 is -fold normal, for .

6. -Fold Fantastic Filter

Definition 36. Let . A subset of is an -fold fantastic filter if and, for all , implies .

In particular a 1-fold fantastic filter is a fantastic filter.

Example 37. Let .

(1)Let be the residuated lattice of Example 6. It is easy to check that is an -fold fantastic filter.(2)For the residuated lattice of Example 2. is not an -fold fantastic filter since , but .

The following result gives a simple characterization of -fold fantastic filters.

Proposition 38. Let and let be a filter. is an -fold fantastic filter if and only if , for all .

Thus the class of -fold fantastic filters is closed under extension.

Proof. Assume that is -fold fantastic. Since , from the hypothesis, we obtain .

Since , we have and . Thus .

Conversely, assume that , for .

Let be such that .   By Proposition 1, we have ,  so  , and the latter is; it follows that .

Proposition 39. Let . Then any -fold Boolean filter is -fold fantastic.

Proof. Assume that is -fold Boolean. Let be such that .

We have ,  so and .

We also have

Since is -fold positive implicative, we obtain .  Hence is an -fold fantastic filter.

Proposition 40. Let . Any -fold fantastic filter is -fold normal.

Proof. Assume that is an -fold fantastic filter. Let be such that .   We must show that .

Now, implies , and implies .  Since is -fold fantastic, we have   .

Moreover, , so by Proposition 1 (14), we have, so the latter is in , and by we obtain .

Let us note that if , then ; thus

Theorem 41. Let . A filter of is -fold Boolean if and only if it is -fold fantastic and -fold implicative.

Proof. : This follows from Propositions 25 and 39.

: Let be -fold fantastic and -fold implicative, and let be such that . We must show that .

Since is -fold fantastic and , we have.

From the observation above, we have, = . So . Thus , and , since is -fold implicative.

Definition 42. A residuated lattice is said to be -fold fantastic if for all , .

Example 43. Let .

(1)The residuated lattice of Example 6 is -fold fantastic.(2)The residuated lattice of Example 2 is not -fold fantastic,  since .

Here is a characterization of -fold fantastic residuated lattices.

Proposition 44. The residuated lattice is -fold fantastic if and only if the inequality holds, for all .

Proof. :  Assume that is an -fold fantastic residuated lattice. Let .

We have . So , by Proposition 1.

: Suppose conversely that the inequality holds for all . Then .  So .  Now, implies that, and whence the equality.

Proposition 45. The following conditions are equivalent for a residuated lattice .

(i) is -fold fantastic.(ii)Every filter of is -fold fantastic. (iii) is an -fold fantastic filter of .

Proof. : Follows from the definitions of -fold fantastic filter and -fantastic residuated lattice.

:  Follows from the fact that is a filter of .

:  Assume that is -fold fantastic.  Let and . By Proposition 1, . So and by the hypothesis, we have ;  that is, .

On the other hand, implies , and hence .  Then, it follows from that . Hence by Proposition 44, is -fold fantastic.

So, a filter of is -fold fantastic if and only if is an -fold fantastic residuated lattice.

Combining Propositions 45, 28, and 20 and Theorem 41, we have the following result.

Corollary 46. Let . A residuated lattice is -fold Boolean if and only if it is -fold fantastic and -fold implicative.

To end this section, we note that:

(i)-fold fantastic residuated lattices are -fold normal;(ii)1-fold fantastic residuated lattices are -algebras, and any -algebra is an -fold fantastic residuated lattice, for ;(iii)however, an -fold fantastic residuated lattice may not be a -algebra, as illustrated by Example 6 for .

7. -Fold Obstinate Filter

Definition 47. Let . A proper filter of is said to be -fold obstinate if for all , implies and .

In particular a 1-fold obstinate filter is an obstinate filter.

The following result gives a characterization of -fold obstinate filter.

Lemma 48. Let ; a proper filter is -fold obstinate if and only if, for all , implies .

Proof. :  Let with . By setting in the definition, we have ; that is, .

:  Conversely, let ;  then .  But , so, .  Similarly, .

Remark 49. (i)  An -fold obstinate filter is also called an -fold Boolean filter of the second kind.

(ii)  Since , any -fold obstinate filter is -fold obstinate.

Example 50. Let be the lattice of Example 2 and .

(i)The filter is -fold obstinate.(ii)The filter is not -fold obstinate, since and .

Proposition 51. The following conditions are equivalent for any proper filter and any .

(i) is -fold obstinate.(ii) is maximal and -fold Boolean.(iii) is maximal and -fold implicative.(iv) is prime of the second kind and -fold Boolean.

Proof. :  Assume that is an -fold obstinate filter. We first show that is maximal. Let ; since is -fold obstinate, by Lemma 48,. Let such that ; since , it is clear that and then . From this, we get that, for all , if and only if ; hence by Proposition 9, is a maximal filter.

On the other hand, let such that .  If , then , a contradiction, since is -fold obstinate. Thus .

:  Follows from Proposition 25.

:  Assume that is a maximal and -fold implicative filter of . Let be such that . By Lemma 17, is a filter of and so is .

Let ; since , we have , and hence and we obtain . On the other hand, , since , and hence . By hypothesis, . So . Since is a maximal filter of , we get . Therefore, , or equivalently . Similarly, we get . Hence is an -fold obstinate filter of .

:  Follows from Lemma 13.

:  Assume that is a prime filter of the second kind and is -fold Boolean. Let be such that . Since is -fold boolean, we have . Since is a prime filter of the second kind and , we have . Hence is an -fold obstinate filter.

We note that this is an improvement of [11, Theorem 4.14].

Definition 52. A residuated lattice is said to be -fold obstinate (or -fold Boolean of the second kind) if, for all , .

This means that is an -nilpotent commutative semigroup.

Proposition 53. The following conditions are equivalent for any proper filter and .

(i) is an -fold obstinate residuated lattice.(ii) is an -fold obstinate filter of .

Proof. : Assume that is an -fold obstinate residuated lattice. Let be such that ;  then, . Since is an -fold obstinate residuated lattice, it follows that . This implies .

: Assume that is an -fold obstinate filter of . Let be such that ; then, . It follows that , or equivalently , or equivalently ; that is, .

Since , from Proposition 53, we have the following result.

Proposition 54. Let . Then is an -fold obstinate residuated lattice if and only if is an -fold obstinate filter of .

Thus, an -fold obstinate residuated lattice is locally finite.

Example 55. Let .

(i)The lattice of Example 6 is -fold obstinate.(ii)The lattice of Example 3 is not -fold obstinate, since and .

8. -Fold Integral Filter

Definition 56. Let be a filter of the residuated lattice , and .

(i) is said to be -fold integral if,  for all , implies or . In particular, 1-fold integral filters are integral filters.(ii) is said to be -fold integral if, for all , implies or .  In particular, a -fold integral residuated lattice is an integral residuated lattice.

Example 57. Let .

(i)In Example 5, the filter is -fold integral, but the filter is not, because , whereas and .(ii)The residuated lattice of Example 4 is -fold integral, but that of Example 5 is not, since and.

Proposition 58. Let be a filter of . The following conditions are equivalent.

(i) is an -fold integral filter.(ii) is an -fold integral residuated lattice.

Proof. : Assume that is an -fold integral filter. Then, let be such that, or equivalently ; then . Since is -fold integral, it follows that or , so or . From this we have or .

: Let be such that . We have , or equivalently . Since is an -fold integral residuated lattice, we have or ; that is, or .

From Proposition 58 and the fact that , it is clear that the residuated lattice is -fold integral if and only if is an -fold integral filter.

Also, any -fold integral filter is a primary filter. But the converse is not true: in Example 3, simple computations prove that is a primary filter, but not a 1-fold integral filter, because and .

Proposition 59. Any proper filter of which is -fold obstinate is also -fold integral.

Proof. Assume that is a proper -fold obstinate filter and let such that . Since is a proper filter, it follows that , so or . Since is a proper -fold obstinate filter, this implies or , and hence is an -fold integral filter of .

The converse of the above proposition is not true, since in Example 4, is an -fold integral filter which is not -fold obstinate because and .

It follows that any -fold obstinate residuated lattice is -fold integral.

Definition 60. A filter satisfies -fold double negation if,  for all , implies .

Example 61. Let .

(i)In Example 2, we see by simple computations that the filter satisfies -fold double negation.(ii)In Example 3, the filter does not satisfy -fold double negation since but .

Proposition 62. Let be a proper filter which satisfies -fold double negation. If is -fold integral and -fold fantastic, then is -fold obstinate.

Proof. Assume that is -fold integral and -fold fantastic. Let such that . By Proposition 1, we have . Since is -fold integral, we have or .

Now, by Proposition 38, as is -fold fantastic; thus by -fold double negation. Since , we also have ,  so.

Definition 63. A residuated lattice satisfies -fold double negation if,  for all , implies .

One easily verifies that a residuated lattice satisfies -fold double negation if and only if so does filter .

Example 64. Let . The residuated lattice of Example 2 satisfies -fold double negation. But the lattice of Example 3 does not, since but .

From Propositions 62 and 45, we obtain the following corollary.

Corollary 65. Let be a residuated lattice satisfying -fold double negation.  If is -fold integral and -fold fantastic, then is -fold obstinate.

9. -Fold Involutive filter

In [11], Zahiri and Farahani introduced the notion of -fold involutive filter of -algebra. In this section, we follow their idea and give the corresponding definition on residuated lattices.

Definition 66. Let be a filter of the residuated lattice .

(i) is called an -fold involutive filter of (or -fold IRL filter)  if   , for all .(ii) is called an -fold involutive residuated lattice (or -fold IRL)  if   , for all .

Remark 67. (i)  It is easy to verify that is an -fold involutive filter if and only if is an -fold involutive residuated lattice.

(ii)  The class of -fold involutive filters is closed under extension.

These two notions are weakenings of the corresponding ones for -fold fantastic, by setting in Definitions 36 and 42. So, -fold fantastic filters are -fold involutive

In [11], they also study the notion of extended involutive filter of -algebra, by taking in Definition 29.

Definition 68. (i) is called an -fold extended involutive filter of the residuated lattice (or -fold EIRL filter)  if implies , for all .

(ii) is called an -fold extended involutive residuated lattice (or -fold EIRL)   if implies, for all .

Clearly, is an -fold extended involutive filter if and only if is an -fold extended involutive residuated lattice. Moreover, an -fold involutive filter (an -fold normal filter, resp.) is -fold extended involutive.

Now, let us consider some statements about .

   ,.   ,.   , (-fold involutive filter).   ,implies(-fold double negation).

Then we have the following implications.

implies ,  because   .   Moreover, by repeated use of Proposition 1 (14) and (15), we have ; and ; and .  So, and .  Then, we easily obtain the following result.

Proposition 69. Let be a filter of the residuated lattice

(i)If satisfies -fold double negation, then is -fold involutive if and only if satisfies .(ii)If is -fold involutive, then satisfies -fold double negation if and only if satisfies .

Now, it is easy to see that an -fold Boolean filter satisfies -fold double negation if and only if it satisfies . So it needs not to satisfy any one of them, as can be observed on filter of Example 6.

We may restate [11, Theorem 3.14, Theorem 3.16] as follows.

Theorem 70. Let be a filter of the residuated lattice .

(i)If is -fold Boolean, then is -fold implicative and -fold extended involutive.  The converse holds if satisfies -fold double negation.(ii)If is -fold Boolean, then is -fold implicative and -fold extended involutive. The converse holds if satisfies -fold double negation.

Now, here are some consequences of .

Theorem 71. Let be a filter satisfying -fold double negation (or ). Then is -fold obstinate if and only if is -fold integral and -fold involutive.

Proof. This follows from Propositions 39, 51, 59, and 69 and the proof of Proposition 62.

Theorem 72. Let be a filter of satisfying . Then is -fold obstinate if and only if is -fold integral and -fold extended involutive.

Proof. :  This follows from Theorem 71, the fact that implies , and the fact that an -fold involutive filter is -fold extended involutive.

:  Let such that .  Since , we have or .

If , then by we would have, and as is -fold EIRL; a contradiction. So, ,  and   .

Conflict of Interests 

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors wish to thank the referees for their excellent suggestions that improved the presentation and the readability of the paper.

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