Research Article | Open Access

# Groups of Negations on the Unit Square

**Academic Editor:**Jesús Medina

#### Abstract

The main results are about the groups of the negations on the unit square, which is considered as a bilattice. It is proven that all the automorphisms on it form a group; the set, containing the monotonic isomorphisms and the strict negations of the first (or the second or the third) kind, with the operator “composition,” is a group (or or , correspondingly). All these four kinds of mappings form a group . And all the groups are normal subgroups of . Moreover, for , a generator set is given, which consists of all the involutive negations of the second kind and the standard negation of the first kind. As a subset of the unit square, the interval-valued set is also studied. Two groups are found: one group consists of all the isomorphisms on , and the other group contains all the isomorphisms and all the strict negations on , which keep the diagonal. Moreover, the former is a normal subgroup of the latter. And all the involutive negations on the interval-valued set form a generator set of the latter group.

#### 1. Introduction

Negations, as a basic operation, play important roles in logic. In [1], the groups of the negations on the unit interval are studied. And in this paper, groups of negations on the bilattice and on the interval-valued set , which can be seen as a sublattice of the unit square, will be studied.

Bilattices, introduced by Ginsberg [2] as a uniform framework for inference in artificial intelligence, are algebraic structures that proved useful in many fields [3–5]. And the unit square is a very special bilattice, which is a subset of the real plane. Thus it is of particular interest. In [6, 7], three kinds of negations on bilattices are introduced. Then one problem arises: are the properties of these negations similar to the negations on [0, 1]? In the third section of this paper, some results will be given.

All the vague sets, the interval-valued sets, the intuitionisitic fuzzy sets and the grey sets are isomorphic to sublattice , with the natural order if and only if , . Their properties could be found in [3, 7–23], and so forth. For sometimes, the sets are characterized as , with the order if and only if , . Both and are sublattices of the bilattice , with orders and . Therefore, similar to [1], the groups of negations on the interval-valued sets can be discussed.

The contents will be arranged as follows. In the next section, some basic notions will be given. In the third section, the groups of the strict negations will be studied. And in the fourth section, we will study the negations on the interval-valued sets. At the end of this paper, a section of conclusion is shown.

#### 2. Negations on the Bilattice

In [6, 7], three kinds of negations on the bilattice are given. As a particular bilattice, the unit square also has these kinds of negations. These negations are based on two kinds of orders on the unit square , and , defined as, for any , , in which is the natural order on [0, 1]. Actually, is the natural order on the unit square.

*Definition 1. *The first kind negation , called reflection, is a unary operator on the square, satisfying the following properties: , ,(1)if , then ;(2)if , then ;(3) and .

The second kind negation , named conflation, is a unary operator satisfying(1′)if , then ;(2′)if , then ;(3′) and .

The last kind negation is a unary operator satisfying(1′′)if , then ;(2′′)if , then ;(3′′) and .

For convenience, in this paper, these three kinds of negations are collectedly called negations on the unit square, as a bilattice .

The composition of two negations , is defined as . Then, it is not hard to check that and are negations of the third kind; and are the second kind negations, and and are of the first kind.

Similar to the definition of strict interval-valued negations in [11], the strict negations on the unit square are given as follows.

*Definition 2. *A negation on the unit square is called strict, if it is continuous and both and are strict, that is, in Definition 1, if both and are replaced by and in the premise, then the conclusion will be changed to and correspondingly.

*Example 3. *Let and , with
in which , are constants in [0, 1]. Then both and are negations, but neither of them is strict.

*Definition 4. *If a negation satisfies , then it is called an involutive negation.

The mapping defined as , for all , is the standard negation of the first kind and named as the first standard negation. The mapping defined as , for all , is called the second standard negation. And is called the third standard kind negation. It is obvious that , , and are involutive.

Obviously, each involutive negation is strict, but the converse is not valid.

*Example 5. *The following negation is strict but not involutive:

Lemma 6 (see [6]). *Each of the first kind strict negations could be characterized as , with , being strict negations on [0, 1], and every second kind strict negation could be characterized as , with and being strict negations on [0, 1].**Each of the first kind involutive negations could be characterized as , with , negations on [0, 1], and every second kind involutive negation could be characterized as , with and being involutive negations on [0, 1].*

Similarly, each of the third kind negations could be characterized as follows.

Lemma 7. * is a strict negation of the third kind, if and only if there are two isomorphisms and on [0, 1], s.t. *

*Proof. *Since each could be represented as , in which is the first kind and is the second kind. Then from Lemma 6, the result could be obtained.

From Lemmas 6 and 7, since and , are bijections on [0, 1], we can know each strict negation is a bijection on .

Note that, the composition of two involutive negations of different kinds is still a negation, but it may not be involutive. The following example shows it.

*Example 8. * is an involutive negation of the first kind. The composition of it and the second standard negation ,
is a strict but not involutive negation of the third kind.

*Definition 9. *A continuous mapping is a monotonic isomorphism on the unit square, if it is bijective and preserves both the orders and .

Obviously, each monotonic isomorphism is continuous. Define and as and .

Lemma 10. *If is a monotonic isomorphism on the unit square, then there are isomorphisms and on the unit interval [0, 1] such that
*

*Proof. *Firstly, let us show , , , and . For any , . Since preserves , we can know . Thus, the greatest element of is . Because is a bijection, we have . Similarly, , , and can be proven.

Now, let us show that is a strict negation of the first kind.

Suppose , then . Because preserves , we have
It shows that reverses the order . Similarly, since both and keep the order , we can get that preserves the order . Since is a bijection, strictly preserves the orders and . Similarly, it can be proven that strictly preserves the order and strictly reverses the order .

Since both and are continuous, is continuous.

Since , , and , , the following hold:
These two formulas, together with the facts that is continuous, strictly reverses the order , and strictly preserves the order , show that is a strict negation of the first kind, denoted by .

Then for any , , that is, . From Lemma 6, , with , being negations on the unit interval [0, 1]. Since , we could get . Because and are isomorphisms on the unit interval, can be characterized as , in which , are isomorphisms on the unit interval.

Lemma 11. *If is a monotonic isomorphism and is a standard negation, then and are strict negations and of the same kind as . Conversely, for each strict negation , there exists a monotonic isomorphism and a standard negation of the same kind with , s.t. .*

*Proof. *If , the proof of Lemma 10 has already shown that is a strict negation. And by Lemma 10, . It shows that is a continuous bijection on , which strictly preserves the order and strictly reverses and satisfies and . Therefore, is a strict negation of the first kind.

Similar for or . Therefore, and are strict negations and of the same kind as , for .

Next, let us show the second part. Suppose is a negation and is the standard negation of the same kind as . Then, is a monotonic isomorphism . As a result, for any ,
that is, .

#### 3. Groups of the Negations on the Unit Square

Based on the notions in the above section, the groups of the negations on the unit square will be discussed.

In [1], the following theorem is obtained.

Theorem 12 (see [1, Theorem 4]). *(**1**)* For every strict negation on the unit interval, there exist three involutive negations , and s.t. .*(**2**)* For every isomophism on the unit interval, there exist four involutive negations , , and s.t. .

This theorem shows that under the operator “composition,” all the involutive negations and isomorphisms on the unit interval cannot form a group, since it is not closed under the operator “composition.” But the set of all the strict negations and isomorphisms, together with the operator “composition,” is a group [1].

From this theorem and Lemmas 6 and 10, we could get the following result on the unit square.

Theorem 13. *(**1**)* For every strict negation of the second kind on the unit square, there exist three involutive negations , , and of the second kind, s.t. .*(**2**)* For every monotonic isomorphism on the unit square, there exist four involutive negations , , , and of the second, s.t. .

For the other two kinds of negations, we have not got similar results, since the composition operator is not commutative.

Next, let us discuss the groups of the negations on the unit square. The following sets are denoted by * * is a monotonic isomorphism on the unit square.} * * is a first kind strict negation on the unit square.} * * is a second kind strict negation on the unit square.} * * is a third kind strict negation on the unit square.} * *.

Theorem 14. *, are groups, with the same unit element id. Moreover, , that is, is a normal subgroup of , , and are normal subgroups of .*

*Proof. *Obviously, for any element in , , , that is, id is the unit element of , .

For any two monotonic isomorphisms , , the composition of them is still a monotonic isomorphism. Thus is closed under the operator “”. For two strict negations , of the same kind, the composition of them is also a monotonic isomorphism. The composition of a strict negation and a monotonic isomorphism is still a strict negation of the same kind with . Therefore, , , and are closed under the operator “”. For any two strict negations , of different kind, the composition is a strict negation of the other kind. So is closed.

Now, let us prove the associativity of . This proof also shows the associativity of , .

From Lemmas 6, 7, and 10, each of the negations and the monotonic isomorphisms can be characterized as or , in which both of and are isomorphisms on [0, 1] or both are strict negations on [0, 1]. Let , , be negations or monotonic isomorphisms on . Then,
that is, , which shows the associativity of .

For any monotonic isomorphism and any strict negation , obviously and exist. Moreover, is also a monotonic isomorphism and is a negation of the same kind as , that is, the inverse about the composition operator “” exists. Therefore, , are groups.

Since , , we only need to prove that they are normal subgroups.

Firstly, let us show . For any , obviously, , because is a group. For , from Lemma 6, we know and . For any ,
Because both and are isomorphisms on the unit interval [0, 1], is in . Thus, , that is, is a normal subgroup of .

Similarly, we could prove . Since, , we could know .

Now, let us show . Because is a group, for all ( is a strict negation of the first kind or a monotonic isomorphism on ), we have . For any , is a negation of the second kind or the third kind. If is of the second kind, then is also of the second kind. Thus for all , , and for all , is of the first kind, since is of the third kind. If is of the third kind, then is also of the third kind. Thus for all , , and for all , is of the first kind, since is of the second kind. Therefore, for all , , that is, is a normal subgroup of . Similarly, , could be proven.

From Theorem 13, we immediately get the following result.

Theorem 15. *The set of all the involutive negations of the second kind could generate the group , that is, for any , there is some involutive negations , of the second kind, such that . And if is a negation, could be 3; if is an isomorphism, could be 4.*

This theorem shows the relation between the involutive negations of the second kind and the group . However, the following problem is still unproven.

*Problem 16. *Could (or ) be generated by the set of all the involutive negations of the first (or the third) kind?

A generator set of is given in the following theorem.

Theorem 17. *The set is a generator set of the group . And is also a the generator set of .*

*Proof. *By Lemma 11, for any strict negation of the first or the third kind, there exists a monotonic isomorphism , s.t. or . Since , from Theorem 15, generates the group .

Because , is also a the generator set of .

#### 4. Negations on the Interval-Valued Set

The interval-valued set is a sublattice of the unit square , so the negations of the interval-valued set could be defined as the restriction of the negations on the unit square.

*Definition 18. *A negation on the interval-valued set is the restriction of some negation of the first kind on the unit square, and satisfies that

*Definition 19. *An interval-valued negation is strict, if it is the restriction of some strict negations of the first kind on the unit square and satisfies (12).

An interval-valued negation is involutive, if it satisfies

From this definition, each strict interval-valued negation is an injection and each involutive negation is strict [6]. Also, in [6], it is proven that for each involutive interval-valued negation , it keeps the diagonal , that is,

*Definition 20. *A mapping on the interval-valued set is an isomorphism, if it is a bijection and keeps the natural order.

Actually, if is an isomorphism on the interval-valued set, then keeps both the orders and [6]. Thus, there exist some isomorphisms on the unit square, s.t. . Moreover, every keeps the diagonal [6].

Different from the unit square, all the strict negations and the isomorphisms on the interval-valued set, together with the composition operator, do not consist of a group. The following is a counter example.

*Example 21. *There is no strict negation on the interval-valued set, which is an inverse of the following negation :

It is not hard to check that is a strict negation on the interval-valued set.

It seems that there is a mapping , s.t. is its inverse. However, not all of the points of the interval-valued set are well defined under , such as the point (0.5, 0.5). The “image” of it is (0.75, 0.5), which is out of the interval-valued set.

Also, the following mapping is also not the inverse of , because it is not an injection, thus not a strict negation on . Consider

Now, let us give the proof of Example 21.

Suppose is the inverse negation of . Clearly, maps the interval-valued set to the set . Then is a surjection to . Since for the points in , their images under also should be in , we can get that is not an injection, thus not a strict negation on . So has no inverse.

Denote that * * * * is a strict negation on and keeps the diagonal Then we have the following theorem.

Theorem 22. * and are groups, with the composition of the mappings. Moreover, .*

*Proof. *Obviously, the unit element is the identity mapping id, and the operation is closed and associative.

Suppose is an isomorphism in . Then can be represented as , with an isomorphism on the unit interval [0, 1], because keeps the diagonal. Define the mapping as . Then is also an isomorphism on , that is, , and
that is, is the inverse of . Thus, is a group.

Let be a strict negation on , which keeps the diagonal . From Lemma 6, there exist some strict negations on [0, 1], s.t. . From (14), . Then is also a strict negation on and keeps the diagonal. Also we can check that
that is, is the inverse of . Thus, is a group.

Similar to the proof of Theorem 14, could be proven.

This theorem could be extended to the unit square.

Theorem 23. *(**1**)* All the monotonic isomorphisms on the unit square, which keep the diagonal, form a group, called .*(**2**)* All the strict negations of the first kind and the monotonic isomorphisms on the unit square, which keep the diagonal, form a group, called .

The proof is similar to Theorem 22.

From Theorems 12, 22, and 23, we could obtain the following theorem.

Theorem 24. *(**1**)* The set of all the involutive negations on generates the group .*(**2**)* The set of all the involutive negations of the first kind on the unit square, which keep the diagonal, generates the group .

#### 5. Conclusion

In this paper, we firstly study the negations on the unit square. The main results are Theorems 14 and 23, which show the groups that are formed by the strict negations and the monotonic isomorphisms. Then we discuss the negations on the interval-valued set. The main result is Theorem 22, that is, all the strict negations and isomorphisms on , which keep the diagonal, form a group. Moreover, some generator sets of the groups are given.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This project is supported by the Tianyuan special funds of the National Natural Science Foundation of China (Grant no. 11226265) and Promotive research fund for excellent young and middle-aged scientisits of Shandong Province (Grant no. 2012BSB01159).

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Copyright © 2014 Jiachao Wu. 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.