Groups of Negations on the Unit Square

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 G 2 (or G 3 or G 4, correspondingly). All these four kinds of mappings form a group G 5. And all the groups G i, i = 2,3, 4 are normal subgroups of G 5. Moreover, for G 5, 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 L I, and the other group contains all the isomorphisms and all the strict negations on L I, 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.


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 [0, 1] 2 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][4][5]. And the unit square [0, 1] 2 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.
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 intervalvalued sets. At the end of this paper, a section of conclusion is shown.
The composition of two negations , is defined as ( ∘ )( 1 , 2 ) = ( ( 1 , 2 )). 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.
Obviously, each involutive negation is strict, but the converse is not valid.
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.
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.
is an involutive negation of the first kind. The composition of it and the second standard negation 2, , is a strict but not involutive negation of the third kind. Obviously, each monotonic isomorphism is continuous.

Similar for
= ,2 or = ,3 . Therefore, Φ ∘ and ∘ Φ are strict negations and of the same kind as , for = , , = 1, 2, 3. 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 Φ. As a result, for any that is, = Φ ∘ .

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] 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. (2) For every monotonic isomorphism Φ on the unit square, there exist four involutive negations 1 , 2 , 3 , and 4 of the second, s.
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 Proof. Obviously, for any element in , = 1, . . . , 5, id∘ = ∘ id = , that is, id is the unit element of , = 1, . . . , 5.
For any two monotonic isomorphisms Φ 1 , Φ 2 , the composition of them is still a monotonic isomorphism. Thus 1 is closed under the operator "∘". For two strict negations 1 , 2 of the same kind, the composition of them is also a monotonic isomorphism. The composition of a strict negation and a monotonic isomorphism Φ is still a strict negation of the same kind with . Therefore, 2 , 3 , and 4 are closed under the operator "∘". For any two strict negations 1 , 2 of different kind, the composition is a strict negation of the other kind. So 5 is closed. Now, let us prove the associativity of 5 . This proof also shows the associativity of , = 1, 2, 3, 4.
For any monotonic isomorphism Φ and any strict negation , obviously Φ −1 and −1 exist. Moreover, Φ −1 is also a monotonic isomorphism and −1 is a negation of the same kind as , that is, the inverse about the composition operator "∘" exists. Therefore, , = 1, . . . , 5 are groups.
From Theorem 13, we immediately get the following result.
Theorem 15. The set ,2 of all the involutive negations of the second kind could generate the group 3 , that is, for any ∈ 3 , there is some involutive negations , = 1, . . . , 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 3 . However, the following problem is still unproven.
Problem 16. Could 2 (or 4 ) be generated by the set of all the involutive negations of the first (or the third) kind?
A generator set of 5 is given in the following theorem.
Theorem 17. The set ,2 ∪{ ,1 } is a generator set of the group 5 . And ,2 ∪ { ,3 } is also a the generator set of 5 .
Proof. By Lemma 11, for any strict negation of the first or the third kind, there exists a monotonic isomorphism Φ, s.t.

Negations on the Interval-Valued Set
The interval-valued set is a sublattice of the unit square [0, 1] 2 , 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, The Scientific World Journal 5 in [6], it is proven that for each involutive interval-valued negation , it keeps the diagonal Δ = {( 1 , 2 ) ∈ : 1 = 2 }, that is, Definition 20. A mapping Φ on the interval-valued set is an isomorphism, if it is a bijection and keeps the natural order.
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 1 : It is not hard to check that 1 is a strict negation on the interval-valued set.
It seems that there is a mapping 1 ( 1 , 2 ) = (√1 − 2 2 , 1− 1 ), s.t. 1 is its inverse. However, not all of the points of the interval-valued set are well defined under 1 , 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 2 is also not the inverse of 1 , because it is not an injection, thus not a strict negation on . Consider Now, let us give the proof of Example 21. Suppose 2 is the inverse negation of 1 . Clearly, 1 maps the interval-valued set to the set = {( 1 , 2 ) : 0 ≤ 1 − 2 ≤ (1 − 1 ) 2 ≤ 1}. Then ( 2 )| is a surjection to . Since for the points in \ , their images under 2 also should be in , we can get that 2 is not an injection, thus not a strict negation on . So 1 has no inverse.
Denote that is a strict negation on and keeps the diagonal Δ.} Then we have the following theorem. Proof. Obviously, the unit element is the identity mapping id, and the operation is closed and associative.
Let be a strict negation on , which keeps the diagonal Δ. From Lemma 6, there exist some strict negations that is, ( ) −1 is the inverse of . Thus, 2 is a group.
Similar to the proof of Theorem 14, 1 ⊲ 2 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 6 .
(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 7 .
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 .
(2) The set of all the involutive negations of the first kind on the unit square, which keep the diagonal, generates the group 7 .

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.