Some remarks on hyper MV -algebras

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1 Journal of Intelligent & Fuzzy Systems 27 (2014) DOI: /IFS IOS Press 2997 Some remarks on hyper MV -algebras R.A. Borzooei a, W.A. Dudek b,, A. Radfar c and O. Zahiri a a Department of Mathematics, Shahid Beheshti University, Tehran, Iran b Institute of Mathematics and Computer Science, Wroclaw University of Technology, Wroclaw, Poland c Department of Mathematics, Payam Noor University, Tehran, Iran Abstract. We describe relations between hyper MV -algebras and hyper K-algebras and prove that a finite hyper MV -algebra satisfying the condition 0 x ={x}, used by many authors, is in fact an ordinary MV -algebra. We also characterize relations between the main types of deductive systems of hyper MV -algebras. Keywords: (proper) hyper MV -algebra, hyper K-algebra, deductive system 1. Introduction MV -algebras were introduced by Chang in [3] in order to show Lukasiewicz logic to be standard complete, i.e., complete with respect to evaluations of propositional variables in the real unit interval [0,1]. In [13], Mundici has proved that MV -algebras are categorically equivalent to Abelian lattice ordered groups with strong unit (see also [4]). The theory of hyperstructures which is a generalization of the concept of algebraic structures was first introduced by Marty [11] and since then many researchers have worked on and developed this new field of modern algebra. Some connections between a deterministic finite automaton and hyper algebraic structures called K-algebras are described in [8]. Many other interesting applications of hyper structures one can find in the book [5]. There are applications to the following subjects: geometry, binary relations, lattices, fuzzy sets and rough sets, automata, combinatorics, codes, artificial intelligence, and probabilities. Seems that one of important applications is in logic where the logical operations is not uniquely determined, i.e., they give some (set) possibilities. Corresponding author. W.A. Dudek, Institute of Mathematics and Computer Science, Wroclaw University of Technology, Wyb. Wyspianskiego 27, Wroclaw, Poland. Fax: ; s: wieslaw.dudek@pwr.edu.pl (W.A. Dudek); borzooei@ sbu.ac.ir (R.A. Borzooei), ateferadfar@yahoo.com (A. Radfar), om.zahiri@gmail.com (O. Zahiri). Recently, Borzooei et al. [1] introduced and studied hyper K-algebras and Ghorbani et al. [7], applied the hyperstructures to MV -algebras and introduced the concept of hyper MV -algebras which are a generalization of MV -algebras. They also, investigated the congruence relations to define quotient hyper MV -algebras, introduced the category of hyper MV - algebras, proved that this category has a terminal object and a co-equalizer, used Jia s construction to provide a free hyper MV -algebra by a set and shown that the monomorphisms in this category are exactly the one-to-one homomorphisms. Rasouli and Davvaz, [14], consider their results and studied homomorphisms, dual homomorphisms, strong homomorphisms and fundamental relations on hyper MV -algebras and gave some results about the connections between hyper MV - algebras and fundamental MV -algebras. In [6] the concept of (strong) prime hyper MV filter was introduced and the uniform topology on hyper MV -algebras was described. Moreover, Jun et al. [9, 10], introduced some new types of deductive systems on hyper MV - algebras and investigated their relations. Now, many authors study various types of hyper MV -algebras. Unfortunately, in mamy papers (see for example [7] and [9]) was used the condition 0 x ={x} which leading to some confusions. In this paper, we describe relations between hyper MV -algebras and hyper K-algebras and prove that there is no finite proper hyper MV -algebra in which 0 x = {x} is valid for all its elements. All such hyper MV /14/$ IOS Press and the authors. All rights reserved

2 2998 R.A. Borzooei et al. / Some remarks on hyper MV -algebras algebras are in fact classical MV -algebras. Finally, we consider relations between the main types of hyper MV - deductive systems. 2. Preliminaries Let M be a nonempty set, P(M) the family its nonempty subsets. By a binary hyper operation we mean each map from M M into P(M), i.e., any map which for each pair of elements from M asign some subset of M. Definition 2.1. A hyper MV -algebra is a nonempty set M endowed with a binary hyper operation, a unary operation and a constant 0 satisfying the following conditions: for all x, y, z M (hmv 1) x (y z) = (x y) z, (hmv 2) x y = y x, (hmv 3) (x ) = x, (hmv 4) (x y) y = (y x) x, (hmv 5) 0 x 0, (hmv 6) 0 x x, (hmv 7) if x y and y x, then x = y, where x y means that 0 x y. For every nonempty subsets A, B of M we define A B = {a b a A, b B} and A B if and only if there exist a A and b B such that a b. Also, we put 0 = 1 and A ={a a A}. If0 /= 0, then we say that a hyper MV -algebra (M,,, 0) is non-trivial. A hyper MV -algebra in which for some x, y M the set x y has at least two elements is called proper. The following proposition proved in [7] is almost obvious. Proposition 2.2. In a hyper MV -algebra (M,,, 0) for all x, y M the following statements hold: (i) x y y x, (ii) 0 x 1, (iii) 0 0 ={0}, (iv) x 0 x, (v) y 0 x = y x. Let (M,,, 0) be a hyper MV -algebra. Following [9] and [10] we say that a subset D M containing 0 is its hyper MV -deductive system if (x y) D and y D imply x D, for all x, y M, weak hyper MV -deductive system if (x y) D and y D imply x D, for all x, y M, (, ; )-hyper MV -deductive system if ((x y) z) D and (y z) D imply (x z) D, for all x, y, z M, (, ; )-hyper MV -deductive system if ((x y) z) D and (y z) D imply (x z) D, for all x, y, z M, (, ; )-hyper MV -deductive system if ((x y) z) D and (y z) D imply (x z) D, for all x, y, z M, (, ; )-hyper MV -deductive system if ((x y) z) D and (y z) D imply (x z) D, for all x, y, z M. Hyper MV -deductive systems have the following useful characterization proved in [10]. Proposition 2.3. A subset D of a hyper MV -algebra (M,,, 0) is its hyper MV -deductive system if and only if 0 D and (x y) D/= y D = x D for all x, y M. 3. Relation between hyper MV -algebras and hyper K-algebras Definition 3.1. A nonempty set H with a binary hyper operation and a constant 0 is called a hyper K-algebra if for all x, y, z H hold: (HK1) (x z) (y z) <x y, (HK2) (x y) z = (x z) y, (HK3) x<x, (HK4) x<yand y<ximply x = y, (HK5) 0 <x, where x<ymeans that 0 x y. If there exists an element e H such that x<e, for all x H, then a hyper K-algebra (H,, 0) is called bounded and e is its unit. A hyper K-algebra in which each set x y has only one element is a classical K-algebra. K-algebras, studied by many authors, are an important generalization of BCK-algebras strongly connected with BCK-logic [2]. Below we present one of possible definitions of BCK-algebras (see for example [12]).

3 R.A. Borzooei et al. / Some remarks on hyper MV -algebras 2999 Definition 3.2. A nonempty set H with a binary operation and a constant 0 is called a BCK-algebra if for all x, y, z H hold: (BCK1) (x z) (y z) x y, (BCK2) x (x y) y, (BCK3) x x, (BCK4) x y and y x imply x = y, (BCK5) 0 x, where x y means that x y = 0. The study of hyper K-algebras was initiated in [1] and continued in [7] and [17]. Below we present three theorems proved in [7]. Theorem 3.3. On any hyper MV -algebra (M,,, 0) we can define a bounded hyper K-algebra (M,, 0) by putting x y = (x y) for all x, y M. Theorem 3.4. If (H,, 0) is a bounded hyper K-algebra with a unit e such that (i) e (e x) ={x}, (ii) e ((e (x y)) y) = e ((e (y x)) x) hold for all x, y H, then the set e x has only one element. Moreover, (H,,, 0), where x = e x and x y = (x y), is a hyper MV -algebra. Theorem 3.5. If in a hyper MV -algebra (H,,, 0) for every x H we have x 0 ={x}, then (H,, 0), where x y = (x y), is a bounded hyper K-algebra satisfying the conditions of Theorem 3.4. Conversely, any bounded hyper K-algebra (H,, 0) satisfying the conditions of Theorem 3.4 becomes a hyper MV -algebra (H,,, 0) with respect to the operations x = 1 x and x y = (x y). Moreover, in this hyper MV - algebra x 0 ={x} and x y = (x y). In fact, in [7] it was proved that hyper MV -algebras in which 0 x ={x} and hyper K-algebras satisfying the conditions (i) and (ii) of Theorem 3.4, are equivalent. This situation is illustrated in [7] by the following example. Example 3.6. Let H ={0,b,1}. Then (H,, 0), where is defined by table: 0 b 1 0 {0} H {0} b {b} H H 1 {1} {b} {0} is a bounded hyper K-algebra and e = 1 is its unit. Further, the authors of [7] state that (H,, 0) satisfies the conditions of the Theorem 3.4, hence the set H with the operations 0 b 1 0 {0} {b} {1} b {b} H H 1 {1} H {1} is a hyper MV -algebra. x x 0 1 b b 1 0 Unfortunately, (H,,, 0) is not a hyper MV - algebra because for b H we have (0 b) b = (1 b) b = H b = H b = H /={b} =b 0 = (b 0) 0, which contradicts the axiom (hmv 4). On the other hand, for a hyper K-algebra defined in this example we have 1 ((1 (0 b)) b) = 1 ((1 H) b) = 1 (H b) = 1 H = H and 1 ((1 (b 0)) 0) = 1 ((1 {b}) 0) = 1 ({b} 0) = 1 {b} ={b}, which means that this hyper K-algebra does not satisfy the second condition of Theorem 3.4. This incorrect example of a hyper MV -algebra was used to show that there exists a hyper MV -algebra satisfying the following theorem and corollary proved in [9]. Theorem 3.7. If in a hyper MV -algebra (M,,, 0) for every x M the set (x 0) has only one element, then every (, ; )-hyper MV -deductive system of M is its weak hyper MV -deductive system. Corollary 3.8. If in a hyper MV -algebra (M,,, 0) for every x M the set (x 0) has only one element, then every (, ; )-hyper MV -deductive system of M is its hyper MV -deductive system. Below we prove that there are no finite proper hyper MV -algebras satisfying the assumption of Theorem 3.7 and Corollary 3.8. Consequently, there are no finite proper hyper K-algebras satisfying the assumption of Theorem 3.4.

4 3000 R.A. Borzooei et al. / Some remarks on hyper MV -algebras To prove this fact observe first that by (hmv 3) the map : M M is a bijection. So, the set (x 0) has one element if and only if the set x 0 has one element. But, by Proposition 2.2 (iv), for every x M we have x x 0. Hence x 0 ={x } for every x M. Thus x 0 ={x}. So, the set (x 0) has one element if and only if x 0 ={x}. This, by (hmv 2), means that the assumption of Theorem 3.7 is equivalent to the condition 0 x ={x} for all x M. (1) Sometimes the element 0 with this property is called skalar zero. Lemma 3.9. In a hyper MV -algebra M satisfying (1) we have (i) (1 x) x ={x}, (ii) (1 x) 1 = x x, (iii) x 1 x implies x = 1 for all x M. Proof. By (hmv 4), we have (1 x) x = (x 0) 0 = x 0 ={x}. Moreover, (1 x) 1 = (x 1) 1 = (1 x ) x = (0 x ) x = x x = x x so (i) and (ii) hold. (iii). Let x 1 x. Then by (i) we obtain that x x (1 x) x ={x}. From this, by (hmv 2) and (hmv 6), we conclude that 1 x x ={x} and so x = 1. Lemma In a hyper MV -algebra satisfying (1) holds 1 1 ={1}. Proof. Let x 1 1. Then, by (hmv 4) and Proposition 2.2 (ii), x 1 (1 1) 1 = (0 0) 0 = 0 0 ={1}. Thus 1 x ={1}. But by Lemma 3.9(ii), we have (1 x ) 1 = x x and so x x = (x 1) 1 = 0 1 ={1}. This together with (hmv 1) gives 1 1 = (x x ) 1 = x (1 x ) = x 1. Hence 1 1 = x 1 for every x 1 1. Thus, x x 1, so x = 1, by Lemma 3.9(iii). Therefore, 1 1 = {1}. The above lemma (in another way) is also proved in [16] as Theorem Lemma In a finite hyper MV -algebra M satisfying (1) we have 1 x ={1} for all x M. Proof. Let x M. By Lemma 3.10 and (hmv 1), we get 1 (1 x) = (1 1) x = 1 x. Hence we have 1 (1 x) = 1 x, for any x M. Obviously, 1 1 x. Suppose that there exists a 1 x such that a/= 1. Case 1. If 1 x ={1,a}, then a 1 x = 1 (1 x) = (1 1) (1 a). Since 1 1 ={1}, must be a a 1, which contradicts to Lemma 3.9 (iii). So, this case is impossible. Case 2. Let 1 x ={1,a,b}. Since a/ 1 a, b/ 1 b and {a, b} 1 x = 1 (1 x) = (1 1) (1 a) (1 b), we get a 1 b and b 1 a. From this, by applying 1 (1 x) = 1 x, we obtain 1 a 1 (1 b) = 1 b 1 (1 a) = 1 a. Hence 1 a = 1 b. Thus, a 1 a and b 1 b, which is a contradiction. So, this case is also impossible. Case 3. If 1 x ={1,a,b,c}, then {1,a,b,c}=1 x = 1 (1 x) = (1 1) (1 a) (1 b) (1 c). Since a/ 1 1 and a/ 1 a (Lemma 3.9), we have a 1 b or a 1 c. Without loss of generality, let a 1 b. Then b 1 a or b 1 c. Ifb 1 a, then, similarly as in the case of 1 x ={1,a,b}, we obtain a contradiction. If b 1 c, then 1 b 1 (1 c) = 1 c. Analogously a 1 b implies 1 a 1 b. Consequently, 1 a 1 b 1 c. Using the same method, from the fact that c 1 a or c 1 b, we deduce 1 c 1 a or 1 c 1 b. In the first case we obtain 1 a = 1 b = 1 c,in the second 1 b = 1 c. Thus, in both cases we get c 1 c, which is a contradiction. Hence this case is also impossible. Case 4. Suppose now that 1 x has at least five elements, i.e., 1 x ={1,a 2,a 3,...,a n } for some n 5. Then

5 R.A. Borzooei et al. / Some remarks on hyper MV -algebras 3001 {1,a 2,a 3,...,a n }=1 x = 1 (1 x) = (1 1) (1 a 2 ) (1 a 3 )... (1 a n ). From the above (Case 2) it follows that there is no 2 i<j n such that a i 1 a j and a j 1 a i. Also there is no 2 i<j<k n for which a i 1 j, a j 1 a k and a k 1 a i (Case 3). There remains the case when the elements of the set 1 x can be ordered in the following way a 2 1 a 3, a 3 1 a 4, a 4 1 a 5,..., a n 1 1 a n, a n 1 a k for some 2 k<n. Then, similarly as in the Case 3, 1 a k 1 (1 a k+1 ) = 1 a k+1 1 a k a n 1 1 a n and 1 a n 1 a k. Thus, 1 a k = 1 a n. Hence a n 1 a n, which is a contradiction. Therefore, the set 1 x has only one element, i.e., 1 x ={1} for all x M. Corollary In a finite hyper MV -algebra M satisfying (1) we have 1 A ={1} for any subset A of M. Lemma In a finite hyper MV -algebra M satisfying (1) we have x x ={1} for all x M. Proof. Indeed, by Lemmas 3.9 (ii) and 3.11, we get for all x M. x x = (1 x) 1 = 1 1 ={1} Theorem 3.14 There are no finite proper hyper MV - algebras satisfying (1). Proof. We show that a finite hyper MV -algebra M satisfying (1) is a MV -algebra, i.e., x y is singleton for any x, y M. Let x, y M and a, b be arbitrary elements of x y. Then (a b ) (a b) (x y) (x y). On the other hand, by (hmv 1), (hmv 4) and Lemma 3.11, we conclude that (x y) (x y) = x (y (x y) ) = x ((y x ) x ) = (x x ) (y x ) = 1 (y x ) ={1}. Hence a b ={1} and b a ={1}. From this, by (hmv 7), we obtain a = b. This means that x y has only one element. Corollary [16] A finite hyper MV -algebra satisfying (1) is an MV -algebra. Problem 1. Is an infinite proper hyper MV -algebra satisfying (1)? Theorem A finite bounded hyper K-algebra satisfying the conditions of Theorem 3.4 is a BCK-algebra. Proof. Let (H,, 0) be a finite bounded hyper K-algebra satisfying the conditions of Theorem 3.4. We show that for all x, y H the set x y has only one element. Let x, y H and the operations and will be defined as in Theorem 3.4. Then (H,,, 0) is a hyper MV -algebra and, by Theorem 3.5, (H,,, 0) satisfies (1). So, by Theorem 3.14, the set x y has only one element. In particular, x y and (x y) also have one element. Therefore, x y = (x y) has one element. Moreover, in this algebra x 0 = x = x. This means that (H,, 0) is a BCK-algebra. Corollary There are no finite proper bounded hyper K-algebras satisfying the conditions of Theorem 3.4. Problem 2. Is a proper bounded hyper K-algebra satisfying the conditions of Theorem 3.4? 4. Deductive systems of hyper MV -algebras According to [15] a nonempty subset D of a hyper MV -algebra (M,,, 0) is called S-reflexive if (x y) D/= implies (x y) D. Lemma 4.1. Let D be a nonempty subset of a hyper MV -algebra (M,,, 0). Then D is S-reflexive if and only if for all x, y M (x y) D/= = (x y) D. (2) Proof. Suppose that D is S-reflexive and x, y D are arbitrary elements such that (x y) D/=. Then

6 3002 R.A. Borzooei et al. / Some remarks on hyper MV -algebras (x y) D /=, and consequently (x y) D. Thus (x y) D. Hence (x y) D/= implies (x y) D. The converse statement is obvious. Theorem 4.2. In a hyper MV -algebra (M,,, 0) (a) each (, ; )-hyper MV -deductive system D such that x y and y D imply x D, (b) each (, ; )-hyper MV -deductive system, (c) each (, ; )-hyper MV -deductive system D satisfying (2) is a hyper MV -deductive system of M. Proof. (a). Let D bea(, ; )-hyper MV -deductive system in M. Assume that (x y) D and y D. By Proposition 2.2 (iv), (x y) ((x y) 0) and so (x y) ((x y) 0). Since (x y) D, then also ((x y) 0) D. In a similar way, we can show that y (y 0) and y D imply (y 0) D. Hence, ((x y) 0) D and (y 0) D, which, by the assumption on D, gives (x 0) D. Thus, there exists u (x 0) such that u D. So, u D and u x 0. This, by Proposition 2.2 (v), gives u x. Consequently, x u, which implies x D (since u D). Therefore, D is a hyper MV - deductive system of M. (b). Let D bea (, ; )-hyper MV -deductive system in M. If(x y) D and y D, then similarly as in the proof of (a), ((x y) 0) D and (y 0) D, which, by the assumption on D and Proposition 2.2 (iv), gives x = x (x 0) D. Therefore, D is a hyper MV -deductive system in M. (c). Let D bea(, ; )-hyper MV -deductive system in M such that D is S-reflexive. If (x y) D and y D, for some x, y M, then, in the same manner as in the proof of (a), we can show that ((x y) 0) D and y (y 0) D. Thus (y 0) D by Lemma 4.1. Hence, by the assumption on D, we obtain x (x 0) D. This means that D is a hyper MV -deductive system of M. Theorem 4.3. Each (, ; )-hyper MV -deductive system D satisfying (2) is a weak hyper MV -deductive system. Proof. Suppose that a (, ; )-hyper MV -deductive system D satisfies (2). Let (x y) D and y D. We will show that ((x y) 0) D and (y 0) D. For each u ((x y) 0) there is a u x y such that u (a u 0). Since a u D and a u a u 0, we have (a u 0) D /=. This by Lemma 4.1 gives (a u 0) D. Hence u (a u 0) D and consequently ((x y) 0) D. Analogously y D implies (y 0) D. From the above, by the assumption on D, we conclude x (x 0) D, which completes the proof. The following example shows that the converse of Theorem 4.2 and 4.3 is not true. Example 4.4. The set M ={0,a,b,1} with the operations and defined by: 0 a b 1 0 {0} {0,a} {b} {1,b} a {0,a} {a} {b, 1} {1} b {b} {b, 1} {b, 1} {b, 1} 1 {1,b} {1} {b, 1} {1} x x 0 1 a b b a 1 0 is a hyper MV -algebra and D ={0,a} is its hyper MV - deductive system. Since and ((1 a) b) = ((0 a) b) = ({0,a} b) ={1,b} ={0,a}, (a b) = (b b) ={1,b} ={0,a} (1 a) = (0 a) ={1,b}, we have ((1 a) b) D, ((1 a) b) D, (a b) D and (a b) D. But (1 a) / D and (1 a) / D. Thus D is not a (, ; )-hyper MV -deductive system nor a (, ; )-hyper MV - deductive system nor a (, ; )-hyper MV -deductive system nor a (, ; )-hyper MV -deductive system. In general, it is not true that a weak hyper MV - deductive system is a (, ; )-hyper MV -deductive system. But in some cases when a hyper MV -algebra satisfies some additional conditions it is true. One of such conditions is given in the following theorem proved in [9]. Theorem 4.5. If a hyper MV -algebra (M,,, 0) satisfies the identity ((x z) (y z) ) = ((x y) z) (3) then each its weak hyper MV -deductive system is a (, ; )-hyper MV -deductive system. Unfortunately there are no proper finite hyper MV - algebras satisfying this identity. To prove this fact

7 R.A. Borzooei et al. / Some remarks on hyper MV -algebras 3003 observe first that the identity (3) is equivalent to the identity (x z) (y z) = (x y) z (4) and the following lemma takes place. Lemma 4.6. In a hyper MV -algebra (M,,, 0) satisfying the identity (3) we have (i) (x y) z = (x y) (z y), (ii) 0 x = 0 (0 x), (iii) 0 x ={x} for all x, y, z M. Proof. (i). By (hmv 4) and (hmv 2), z (y z) = (y z) z = (z y) y = y (z y). From this, by applying (3), we obtain (x y) z = (x z) (y z) = x (z (y z) ) = x (y (z y) ) = (x y) (z y), which proves (i). (ii). Putting x = 1, y = 0 and z = x in (i), we get (1 0) x = (1 0) ((x ) 0) and so (0 0) x = (0 0) (x 0). This, by Proposition 2.2 (iii), implies (ii). (iii). Let x be an arbitrary element of M and let t 0 x. Then t x and x t. Since t 0 x implies t (0 x), we have 0 t 0 (0 x), which, by just proved (ii), gives 0 t 0 x. Thus, t 0 t 0 x and t x, by Proposition 2.2. From this, applying (hmv 7), we conclude t = x. Therefore, 0 x ={x} for all x M. Theorem 4.7. Any finite hyper MV -algebra satisfying the identity (3) is an MV -algebra. Proof. It is a consequence of Lemma 4.6 and Corollary Problem 3. Is an infinite proper hyper MV -algebra satisfying (3) or (4)? Another relation between weak hyper MV -deductive systems and (, ; )-hyper MV -deductive system is presented below. Theorem 4.8. Let D be a nonempty subset of a hyper MV -algebra (M,,, 0). If for every a M the set D(a, ) ={x M (x a) D} is a weak hyper MV -deductive system of M, then D is a (, ; )-hyper MV -deductive system of M. Proof. Let each D(a, ) be a weak hyper deductive system of M. Then obviously 0 D(a, ) for each a M. Thus, 0 (0 a) D, by(hmv 5). Hence 0 D. Further, if ((x y) z) D and (y z) D, for some x, y, z M, then (x y) D(z, ) and y D(z, ). This implies x D(z, ), so (x z) D. Therefore, D is a (, ; )-hyper deductive system of M. Theorem 4.9. If D is a (, ; )-hyper MV -deductive system of a hyper MV -algebra (M,,, 0), then each subset D(a, ) containing 0 is a weak hyper MV - deductive system of M. Proof. Let 0 D(a, ). If (x y) D(a, ) and y D(a, ), then for each u (x y) we have (u a) D and (y a) D, so ((x y) a) = ((x y) a) D. Since D isa(, ; )- hyper deductive system, the above implies (x a) D. Thus x D(a, ), which shows that D(a, ) isa weak hyper deductive system of M. Theorem Let D be a nonempty subset of a hyper MV -algebra (M,,, 0) containing 0. If for every a M the set D(a, ) ={x M (x a) D} is a hyper MV -deductive system of M, then D is a (, ; )-hyper MV -deductive system of M. Proof. By the assumption 0 D. Let for some x, y, z M will be ((x y) z) D and (y z) D. Then (x y) D(z, ) and y D(z, ). So, there exists u (x y) such that (u z) D. Thus u D(z, ). But u (x y) means also that u (x y). Hence u (x y) D(z, ). This, together with y D(z, ) and Proposition 2.3, gives x D(z, ). Therefore, (x z) D(z, ), which completes the proof. Theorem If D is a (, ; )-hyper MV - deductive system of a hyper MV -algebra (M,,, 0), then each subset D(a, ) is a hyper MV -deductive system of M.

8 3004 R.A. Borzooei et al. / Some remarks on hyper MV -algebras Proof. Since 0 D and 0 (0 a) for each a M, we have (0 a) D. Thus, 0 D(a, ) for every a M. Now let (x y) D(a, ) /= and y D(a, ). Then obviously (u a) D for some u (x y) D(a, ). On the other hand (u a) ((x y) a) = ((x y) a), hence ((x y) a) D. Since also (y a) D, by the assumption on D, we conclude (x a) D. Consequently, x D(a, ). Proposition 2.3 completes the proof. Analogously we can prove the following theorem. Theorem If D is a (, ; )-hyper or a (, ; )-hyper MV -deductive system of a hyper MV -algebra (M,,, 0), then each subset D(a, ) containing 0 is a hyper MV -deductive system of M. In the next theorem, we attempt to find the real relation between proper hyper K-algebras and proper hyper MV -algebras. Theorem Let (H,, 0) be a bounded hyper K-algebra with 1 as the greatest element such that: (i) each 1 x has the least element with respect to the relation <, (ii) (x ) = x, where x = min(1 x), (iii) (x y) = (y x), (iv) ((y x) x) = ((x y) y) for all x, y H. Then (H,,, 0), where x y = (x y), is a hyper MV -algebra. Conversely, if (M,,, 0) is a hyper MV -algebra, then (M,, 0), where (x y) = (x y), is a bounded K-algebra satisfying the conditions (i) (iv). Proof. Let (H,, 0) be a bounded hyper K-algebra satisfying the conditions (i) (iv). By (HK3) we have 0 1 1, so 1 = min(1 1) = 0, by (HK5). Thus, 1 = 0 and 0 = (1 ) = 1, by (ii). From (HK3) and (HK5) we get 0 x x and 0 (0 x), whence, applying (iii), we obtain (hmv 5) and (hmv 6). (hmv 2), (hmv 3) and (hmv 4) are a consequence of (iii), (ii) and (iv), respectively. Since x<y 0 x y 0 (x y) = x y x y, we see that (HK4) implies (hmv 7). To prove (hmv 1) observe that for any x, y, z H we have (x y) z = (x y) z = ((x y) z) = ((x y) z) by (ii) = ((y x) z) by (HK2) = ((y x) z) by (ii) = ((y z) x) by (HK2) = ((y z) x) = (y z) x = x (y z). Therefore, (H,,, 0) is a hyper MV -algebra. On the other hand, if (M,,, 0) is a hyper MV - algebra, then, by Theorem 3.3, (M,, 0) with the operation x y = (x y) is a bounded K-algebra. Moreover, for any x M and u 1 x we have u (1 x) = 0 x. Thus, u x and x u, by Proposition 2.2. But x 0 x = 1 x means that x (1 x) = 1 x. Hence x = min(1 x). The other conditions follows directly from the axioms of a hyper MV -algebra. Theorem For a bounded hyper K-algebra (H,, 0) with the greatest element 1 the conditions of Theorem 4.13 are equivalent to the conditions: (A1) 1 x has a least element with respect to <, (A2) x<yimplies y <x, where x = min(1 x), (A3) x y = y x, (A4) y (y x) = x (x y), (A5) x = max{u u 1 x} for all x, y H. Proof. Assume that a hyper K-algebra (H,, 0) is bounded and satisfies the conditions of Theorem We show that it satisfies (A1) (A5). (A1) and (i) are identical. (A2) Let x<yfor some x, y H. Then 0 x y and so 0 x y, by (ii). From this, by applying (iii), we obtain 0 (x y) = (y x ). Thus, 0 (y x ) = y x, whence y <x. So, (A2) is satisfied. (A3) is a consequence of (iii) and (ii). (A4) For any x, y H, using (ii) and (iii), we have y (y x) = y (y x) = {y a a y x} = {a y a y x} =(y x) y = (y x) y = (x y ) y.

9 R.A. Borzooei et al. / Some remarks on hyper MV -algebras 3005 Thus y (y x) = (x y ) y. Hence, by (iv), we get y (y x) = y (y x) = ((x y ) y ) = ((y x ) x ) = (x (x y)) = x (x y). This proves (A4). (A5) Let x H. Then x = min(1 x) <y for all y 1 x. So, min(1 y) < min(1 min(1 x)). Since min(1 x) 1 x, by (ii), we get x = x = max{min(1 y) y 1 x}. Therefore (H,, 0) satisfies (A5) and consequently it satisfies the conditions (A1) (A5). Conversely, let (H,, 0) be a bounded hyper K-algebra satisfying the conditions (A1) (A5). Obviously it satisfies (i). To prove (ii) consider an arbitrary element u 1 x. Then x <u and so u <x.on the other hand, x 1 x implies x {u u 1 x}. Hence, by (A5), x = max{u u 1 x} =x, which proves (ii). (iii) is a consequence of (A3). By (A3), for any x, y H, we also have (x y) y = y (x y) = y (x y) = y (y x ). This, by (A4), gives (x y) y = (y x) x. Hence ((x y) y) = ((y x) x), i.e., (iv) holds. 5. Conclusion We show that any finite hyper MV -algebra in which 0 x ={x} is valid for all its elements is MV -algebra and the corresponding hyper K-algebra is a BCKalgebra. This means that results proved in [7] and [9] are valid only for infinite hyper MV -algebras if such infinite hyper MV -algebras exist. References [1] R.A. Borzooei, A. Hasankhani, M.M. Zahedi and Y.B. Jun, On hyper K-algebras, Math Japon 52 (2000), [2] M.W. Bunder, Simpler axioms for BCK algebras and the connection between the axioms and the combinators B, C and K, Math Japon 26 (1981), [3] C.C. Chang, Algebraic analysis of many valued logics, Trans Amer Math Soc 88 (1958), [4] W.M. Cornish, Lattice-ordered groups and BCK-algebras, Math Japon 25 (1980), [5] P. Corsini and V. Leoreanu, Applications of hyperstructures theory, Adv in Math, Kluwer Acad Publ, [6] S. Ghorbani, E. Eslami and A. Hasankhani, Uniform topology and spectral topology on hyper MV-algebras, Quasigroups and Related Systems 17 (2009), 39-ŰU54. [7] S. Ghorbani, A. Hasankhani and E. Eslami, Hyper MV-algebras, Set-Valued Math Appl 1 (2008), [8] M. Golmohamadian and M.M. Zahedi, Hyper K-algebras induced by a deterministic finite automaton, Italian J pure Appl Math 27 (2010), [9] Y.B. Jun, M.S. Kang and H.S. Kim, New type of hyper MVdeductive system in hyper MV-algebras, Math Log Quart 56 (2009), [10] Y.B. Jun, M.S. Kang and H.S. Kim, Hyper MV-deductive systems of hyper MV-algebras, Commun Korean Math Soc 25 (2010), [11] F. Marty, Sur une generalization de la notation de group, 8th Congress Math Scandianaves, Stockholm, (1934), [12] J. Meng and Y.B. Jun BCK-algebras, Kyung Moon SA Co, Seoul, [13] D. Mundici, Interpretation of AF C*-algebras in Lukasiewicz sentential calculus, J Funct Anal 65 (1986), [14] S. Rasouli and B. Davvaz, Homomorphisms, ideals and binary relations on hyper-mv algebras, Multiple-valued Logic and Soft Computing 17 (2011), [15] L. Torkzadeh and A. Ahadpanah, Hyper MV-ideals in hyper MV-algebras, Math Log Quart 56 (2010), [16] L. Torkzadeh and S. Ghorbani, Some characterization of hyper MV-algebras, J Mahani Math Research Center 1 (2012), [17] M.M. Zahedi, R.A. Borzooei, Y.B. Jun and A. Hasankhani, Some results on hyper K-algebras, Scientiae Math 3 (2000),