Hyper MV -ideals in hyper MV -algebras

Math. Log. Quart. 56, No. 1, 51 – 62 (2010) / DOI 10.1002/malq.200810035

Hyper MV-ideals in hyper MV-algebras

Lida Torkzadeh∗ and Afsaneh Ahadpanah∗∗

Department of Mathematics, Isalamic Azad University, Kerman Branch, Kerman, Iran

Received 20 October 2008, revised 18 January 2009, accepted 3 March 2009Published online 18 January 2010

Key words Hyper MV -algebra, hyper MV -ideal, hyper K-algebra.MSC (2000) 06D35, 06F35, 03G25

In this paper we define the hyper operations ⊗, ∨ and ∧ on a hyper MV-algebra and we obtain some re-lated results. After that by considering the notions of hyper MV-ideals and weak hyper MV-ideals, we provesome theorems. Then we determine relationships between (weak) hyper MV-ideals in a hyper MV-algebra(M,⊕,∗ , 0) and (weak) hyper K-ideals in a hyper K-algebra (M, ◦, 0). Finally we give a characterization ofhyper MV-algebras of order 3 or 4 based on the (weak) hyper MV-ideals.

c© 2010 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

MV-algebras were introduced by C. C. Chang [2] in 1958 to provide an algebraic proof of completeness theoremof infinite valued Lukasewicz propositional calculus. The hyper structure theory was introduced by F. Marty [9]at the 8th congress of Scandinavian Mathematicians in 1934. Since then many researches have worked in theseareas. Recently in [4], Sh. Ghorbani, A. Hasankhni and E. Eslami applied the hyper structure to MV-algebrasand introduced the concept of hyper MV-algebra which is a generalization of MV-algebra and investigated somerelated results. Now we follow [4] and [5] and obtain some results as mentioned in the abstract.

2 Preliminaries

Definition 2.1 [4] A hyper MV-algebra is a non-empty set M endowed with a hyper operation “⊕”, a unaryoperation “∗” and a constant “0” satisfying the following axioms for all x, y, z ∈ M :(hMV1) x⊕ (y ⊕ z) = (x⊕ y)⊕ z,(hMV2) x⊕ y = y ⊕ x,(hMV3) (x∗)∗ = x,(hMV4) (x∗ ⊕ y)∗ ⊕ y = (y∗ ⊕ x)∗ ⊕ x,(hMV5) 0∗ ∈ x⊕ 0∗,(hMV6) 0∗ ∈ x⊕ x∗,(hMV7) if x � y and y � x, then x = y, where x � y is defined by 0∗ ∈ x∗ ⊕ y.

For every A,B ⊆ M , we define: A � B if and only if there exist a ∈ A and b ∈ B, such that a � b, andA⊕B =

⋃a∈A

b∈B

a⊕ b . Also, we define 0∗ = 1 and A∗ = {a∗ : a ∈ A}.

∗ Corresponding author: e-mail: [email protected]∗∗ e-mail: a−[email protected]

c© 2010 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

52 L. Torkzadeh and A. Ahadpanah: Hyper MV-ideals in hyper MV-algebras

Proposition 2.2 [4] Let (M,⊕,∗ , 0) be a hyper MV-algebra. Then for all x, y, z ∈ M and for all non-emptysubsets A, B and C of M the following statements hold:

(1) (A⊕B)⊕ C = A⊕ (B ⊕ C);(2) 0 � x;(3) x � x;(4) if x � y, then y∗ � x∗, and A � B implies B∗ � A∗;(5) x � 1;(6) A � A;(7) A ⊆ B implies A � B;(8) x � x⊕ y and A � A⊕B;(9) (A∗)∗ = A;(10) 0⊕ 0 = {0};(11) x ∈ x⊕ 0;(12) if y ∈ x⊕ 0, then y � x;(13) if y ⊕ 0 = x⊕ 0, then x = y.

A hyper MV-algebra (M,⊕,∗ , 0) is called nontrivial if M �= {0}. It is clear that a hyper MV-algebra isnontrivial if and only if 0 �= 1. In this paper, we consider nontrivial hyper MV-algebras.

Definition 2.3 [4] Let (M,⊕,∗ , 0) be a hyper MV-algebra and S be a non-empty subset of M . If S is a hyperMV-algebra with respect to the hyper operation “⊕” and the unary operation “∗” on M , we say that S is a hyperMV-subalgebra of M .

Theorem 2.4 [4] Let S be a non-empty subset of a hyper MV-algebra (M,⊕,∗ , 0). S is a hyper MV-subal-gebra of M if and only if x∗ ∈ S and x⊕ y ⊆ S for all x, y ∈ S.

Definition 2.5 [1] Let H be a non-empty set and “◦” be a hyper operation on H . Then H is called a hyperK-algebra if it contains a constant “0” and satisfies the following axioms for all x, y, z ∈ H , where x < y isdefined by 0 ∈ x ◦ y:(HK1) (x ◦ z) ◦ (y ◦ z) < x ◦ y,(HK2) (x ◦ y) ◦ z = (x ◦ z) ◦ y,(HK3) x < x,(HK4) if x < y and y < x, then x = y,(HK5) 0 < x.

For every A,B ⊆ H we write A < B if there exist a ∈ A and b ∈ B such that a < b.Theorem 2.6 [1] Let (H, ◦, 0) be a hyper K-algebra. Then for all x, y, z ∈ H and for all non-empty subsets

A, B and C of H the following hold:(i) x ◦ y < z if and only if x ◦ z < y; (ii) (x ◦ z) ◦ (x ◦ y) < y ◦ z;

(iii) x ◦ (x ◦ y) < y; (iv) x ◦ y < x;(v) A ⊆ B implies A < B; (vi) x ∈ x ◦ 0;

(vii) (A ◦ C) ◦ (A ◦B) < B ◦ C; (viii) (A ◦ C) ◦ (B ◦ C) < A ◦B;(ix) A ◦B < C if and only if A ◦ C < B; (x) A ◦B < A.Definition 2.7 [1] Let I be a non-empty subset of a hyper K-algebra (H, ◦, 0) and 0 ∈ I .(i) I is called a weak hyper K-ideal of H if x ◦ y ⊆ I and y ∈ I imply that x ∈ I , for all x, y ∈ H .(ii) I is called a hyper K-ideal of H if x ◦ y < I and y ∈ I imply that x ∈ I , for all x, y ∈ H .Theorem 2.8 [4] Let (M,⊕,∗ , 0) be a hyper MV-algebra. Then (M, ◦, 0) is a bounded hyper K-algebra and

also for all x, y ∈ M , x < y if and only if x � y, where x ◦ y = (x∗ ⊕ y)∗.

Theorem 2.9 [4] Let (M1,⊕1,∗1 , 01) and (M2,⊕2,

∗2 , 02) be two hyper MV-algebras and M = M1 ×M2.Then (M,⊕,∗ , 0) is a hyper MV-algebra and (x1, y1) � (x2, y2) if and only if x1 � x2 and y1 � y2, where(a1, b1)⊕ (a2, b2) = (a1 ⊕1 a2, b1 ⊕2 b2), (a, b)∗ = (a∗1 , b∗2) and 0 = (01, 02).

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Math. Log. Quart. 56, No. 1 (2010) / www.mlq-journal.org 53

3 Some results on hyper MV-algebras

In the rest of this paper, by M we denote a hyper MV-algebra. On M we define the hyper operations “⊗” and“” as follows: x⊗ y = (x∗ ⊕ y∗)∗ and x y = x⊗ y∗ = (x∗ ⊕ y)∗ (we consider the operation ∗ more bindingthan any other hyper operations and the hyper operation ⊗ more binding than ⊕ and ).

Lemma 3.1 For x, y ∈ M the following conditions are equivalent:(i) 1 ∈ x∗ ⊕ y;(ii) 0 ∈ x⊗ y∗.

P r o o f. By hypothesis we have 1 ∈ x∗ ⊕ y ⇔ 1∗ ∈ (x∗ ⊕ y)∗ ⇔ 0 ∈ x⊗ y∗.

Lemma 3.2 Let 1 ∈ x∗ ⊕ y for x, y ∈ M . Then y ∈ x⊕ (y x).P r o o f. By Proposition 2.2(11) and (hMV4) we have

y ∈ 0⊕ y = 1∗ ⊕ y ⊆ (x∗ ⊕ y)∗ ⊕ y = (y∗ ⊕ x)∗ ⊕ x = (y x)⊕ x = x⊕ (y x).

Note that by the above lemma we have: if x � y, then there exists z ∈ M such that y ∈ x ⊕ z. The followingexample shows that the converse of the above lemma may not be true.

Example 3.3 Let M = {0, a, b, c, 1}. Consider the following tables on M :

⊕ 0 a b c 1

0 {0} {0, a} {0, a, b} {0, c} {0, a, b, c, 1}

a {0, a} {0, a} {0, a, b, c, 1} {0, a, c} {0, a, b, c, 1}

b {0, a, b} {0, a, b, c, 1} {0, a, b, 1} {0, a, b, c} {0, a, b, c, 1}

c {0, c} {0, a, c} {0, a, b, c} {0, a, b, c, 1} {0, a, b, c, 1}

1 {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1}

∗ 0 a b c 1

1 b a c 0

Then (M,⊕,∗ , 0) is a hyper MV-algebra. Also

a⊕ (c a) = a⊕ (c∗ ⊕ a)∗ = a⊕ (c⊕ a)∗ = a⊕ {0, a, c}∗ = a⊕ {1, b, c} = {0, a, b, c, 1}

and so c ∈ a⊕ (c a) but 1 �∈ {0, a, b, c} = b⊕ c = a∗ ⊕ c.Note that A⊗B =

⋃a∈A

b∈B

a⊗ b. If A has just one element x, then we write x⊗B instead of {x} ⊗B.

Theorem 3.4 If x, y, z ∈ M , then the following hold:(a1) x⊗ (y ⊗ z) = (x⊗ y)⊗ z;(a2) x⊗ y = y ⊗ x;(a3) 0 ∈ x⊗ x∗;(a4) 0 ∈ x⊗ 0;(a5) x ∈ x⊗ 1;(a6) x⊗ y � x, y;(a7) if x ∈ x⊕ x, then x � x⊗ x;(a8) if x = x∗, then x ∈ x⊕ x if and only if x ∈ x⊗ x;(a9) if x ∈ x⊗ x, then x � x⊕ x;(a10) x � y implies x⊕ z � y ⊕ z and x⊗ z � y ⊗ z;(a11) z ⊗ x � y ⇔ z � x∗ ⊕ y;

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(a12) (x y)⊕ y = (y x)⊕ x;(a13) x ∈ x 0, 0 ∈ 0 x, 0 ∈ x x, x∗ ∈ 1 x, 0 ∈ x 1;(a14) x y � x, y∗;(a15) x � y implies x z � y z and z y � z x;(a16) (x⊕ y) x � y.

P r o o f.(a1) We have

x⊗ (y ⊗ z) = x⊗ (y∗ ⊕ z∗)∗

=⋃{x⊗ t : t ∈ (y∗ ⊕ z∗)∗}

=⋃{(x∗ ⊕ t∗)∗ : t ∈ (y∗ ⊕ z∗)∗}

=⋃{(x∗ ⊕ t∗)∗ : t∗ ∈ y∗ ⊕ z∗}

=⋃{a∗ : a ∈ x∗ ⊕ t∗, t∗ ∈ y∗ ⊕ z∗}

=⋃{a∗ : a ∈ x∗ ⊕ (y∗ ⊕ z∗)}

=⋃{a∗ : a ∈ (x∗ ⊕ y∗)⊕ z∗}

=⋃{a∗ : a ∈ (x⊗ y)∗ ⊕ z∗}

=⋃{a∗ : a ∈ k∗ ⊕ z∗, k ∈ x⊗ y}

=⋃{a∗ : a ∈ (k ⊗ z)∗, k ∈ x⊗ y}

=⋃{k ⊗ z : k ∈ x⊗ y}

= (x⊗ y)⊗ z.

(a2) By (hMV2) we have x⊗ y = (x∗ ⊕ y∗)∗ = (y∗ ⊕ x∗)∗ = y ⊗ x.(a3) Since 0∗ ∈ x⊕ x∗, then 0 = (0∗)∗ ∈ (x⊕ x∗)∗ = x∗ ⊗ x.(a4) Since 0∗ ∈ x∗ ⊕ 0∗, then 0 = (0∗)∗ ∈ (x∗ ⊕ 0∗)∗ = x⊗ 0.(a5) By Proposition 2.2(11), x∗ ∈ x∗ ⊕ 0, so x ∈ (x∗ ⊕ 0)∗ = x⊗ 0∗ = x⊗ 1.(a6) By Proposition 2.2.(8) we have x∗ � x∗ ⊕ y∗ = (x ⊗ y)∗. So x ⊗ y � x, by Proposition 2.2(4).

Similarly, x⊗ y � y.(a7) By (hMV4), we get that x∗ ⊕ (x ⊗ x) = x∗ ⊕ (x∗ ⊕ x∗)∗ = x ⊕ (x ⊕ x)∗. Since x ∈ x ⊕ x, then

0∗ ∈ x ⊕ x∗ ⊆ x ⊕ (x ⊕ x)∗ = x∗ ⊕ (x ⊗ x). So there exists z ∈ x ⊗ x such that 0∗ ∈ x∗ ⊕ z. Thereforex � x⊗ x.

(a8) Let x = x∗. Then x ∈ x⊕ x ⇔ x∗ ∈ (x⊕ x)∗ = x∗ ⊗ x∗ ⇔ x ∈ x⊗ x.(a9) The proof is similar to the proof of (a7).(a10) By Proposition 2.2(1) and (hMV4) we have

(x⊕ z)∗ ⊕ (y ⊕ z) = ((x⊕ z)∗ ⊕ z)⊕ y = ((z∗ ⊕ x∗)∗ ⊕ x∗)⊕ y = (z∗ ⊕ x∗)∗ ⊕ (x∗ ⊕ y).

Since x � y, then 0∗ ∈ x∗ ⊕ y. So 0∗ ∈ (z∗ ⊕x∗)∗ ⊕ (x∗ ⊕ y) = (x⊕ z)∗ ⊕ (y⊕ z). Therefore x⊕ z � y⊕ z.Also x � y implies y∗ � x∗, so (y ⊗ z)∗ = y∗ ⊕ z∗ � x∗ ⊕ z∗ = (x⊗ z)∗. Therefore by Proposition 2.2(4),x⊗ z � y ⊗ z.

(a11) z ⊗ x � y ⇔ 0∗ ∈ (z ⊗ x)∗ ⊕ y = (z∗ ⊕ x∗)⊕ y = z∗ ⊕ (x∗ ⊕ y) ⇔ z � x∗ ⊕ y.(a12) (x y)⊕ y = (x∗ ⊕ y)∗ ⊕ y = (y∗ ⊕ x)∗ ⊕ x = (y x)⊕ x.(a13) The proof is easy.(a14) By (a6), x y = x⊗ y∗ � x, y∗.(a15) x � y implies that y∗ � x∗. Then by (a10) we get that x z = x ⊗ z∗ � y ⊗ z∗ = y z and

z y = z ⊗ y∗ � z ⊗ x∗ = z x.(a16) We have

0∗ ∈ (x∗ ⊕ y∗)∗ ⊕ 0∗ ⊆ (x∗ ⊕ y∗)∗ ⊕ (y∗ ⊕ y) = ((x∗ ⊕ y∗)∗ ⊕ y∗)⊕ y)

= ((x⊕ y)∗ ⊕ x)⊕ y = ((x⊕ y)⊗ x∗)∗ ⊕ y = ((x⊕ y) x)∗ ⊕ y.

Therefore (x⊕ y) x � y.



The following example shows that b ∈ b⊕ b but b �∈ b⊗ b and also the converse of (a7) is not true in general.Example 3.5 Consider the following tables on M = {0, a, b, 1}:

⊕ 0 a b 1

0 {0} {0, a, b} {0, b} {0, a, b, 1}

a {0, a, b} {0, 1} {0, a, b, 1} {0, a, b, 1}

b {0, b} {0, a, b, 1} {b} {0, a, b, 1}

1 {0, a, b, 1} {0, a, b, 1} {0, a, b, 1} {0, a, b, 1}

∗ 0 b a 1

1 a b 0

Then (M,⊕,∗ , 0) is a hyper MV-algebra and b ∈ b⊕ b, but b �∈ {0, 1} = (a⊕ a)∗ = (b∗ ⊕ b∗)∗ = b⊗ b. Alsoa⊗ a = {a} and so a � a⊗ a but a �∈ {0, 1} = a⊕ a.

Lemma 3.6 Let x � y and y∗ ⊕ z = {1}, for x, y, z ∈ M . Then x � z.

P r o o f. Let x � y. Then y∗ � x∗ and so {1} = y∗ ⊕ z � x∗ ⊕ z. Thus there exists t ∈ x∗ ⊕ z such that1 � t. Since t � 1, then t = 1. Therefore x � z.

The following example shows that it is possible that a � b and b � c but a �� c. Hence the relation “�” onM is not transitive in general.

Example 3.7 Let M = {0, a, b, c, 1}. Consider the following tables on M :

⊕ 0 a b c 1

0 {0} {0, a} {0, b} {0, c} {0, a, b, c, 1}

a {0, a} {0, a} {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1}

b {0, b} {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c} {0, a, b, c, 1}

c {0, c} {0, a, b, c, 1} {0, a, b, c} {0, a, b, c, 1} {0, a, b, c, 1}

1 {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1} {0, a, b, c, 1}

∗ 0 a b c 1

1 b a c 0

Then (M,⊕,∗ , 0) is a hyper MV-algebra. By 1 �∈ {0, a, b, c} = b⊕ c = a∗ ⊕ c we have also a � b and b � c

but a �� c.Lemma 3.8 The following hold for x, y ∈ M :

(a17) if x⊕ y = {0}, then x = 0 and y = 0;(a18) if x⊗ y = {1}, then x = 1 and y = 1.

P r o o f.(a17) By Proposition 2.2(8) we get that x � 0 and y � 0. Since, by Proposition 2.2(2), 0 � x and 0 � y,

then x = 0 and y = 0.(a18) By (a6) we get that 1 � x and 1 � y. Also by Proposition 2.2(5) we have x � 1 and y � 1. Therefore

x = y = 1.

Lemma 3.9 Let A,B ⊆ M and y ∈ M . Then(i) A � B if and only if 0∗ ∈ A∗ ⊕B;(ii) 0∗ ∈ A⊕A∗;(iii) A⊗B = (A∗ ⊕B∗)∗;(iv) (A∗ ⊕ y)∗ ⊕ y ⊆ (y∗ ⊕A)∗ ⊕A.

P r o o f. The proof is easy.



The following example shows that the reverse inclusion in the Lemma 3.9(iv) may not be true.Example 3.10 The following tables show a hyper MV-algebra structure on M = {0, a, b, 1}:

⊕ 0 a b 1

0 {0} {0, a} {b} {b, 1}

a {0, a} {0, a} {b, 1} {b, 1}

b {b} {b, 1} {b, 1} {b, 1}

1 {b, 1} {b, 1} {b, 1} {b, 1}

∗ 0 a b 1

1 b a 0

Consider A = {a, 1}. Then we can check that (A∗ ⊕ b)∗ ⊕ b = {b, 1} and (b∗ ⊕ A)∗ ⊕ A = {0, a, b, 1}. So(A∗ ⊕ b)∗ ⊕ b �= (b∗ ⊕A)∗ ⊕A.

Lemma 3.11 For x, y, z ∈ M , we have x∗ ⊕ y � (x∗ ⊕ z)⊕ (y⊗ z∗) and so (x∗ ⊕ y)⊗ x � z ⊕ (y⊗ z∗).

P r o o f. Using (hMV4) we get that

(x∗ ⊕ y)∗ ⊕ (x∗ ⊕ z)⊕ (y ⊗ z∗) = (x∗ ⊕ y)∗ ⊕ x∗ ⊕ (z ⊕ (y∗ ⊕ z)∗)

= (x∗ ⊕ y)∗ ⊕ x∗ ⊕ (y ⊕ (y ⊕ z∗)∗)

= (x∗ ⊕ y)∗ ⊕ (x∗ ⊕ y)⊕ (y ⊕ z∗)∗.

Since, by Lemma 3.9(ii), 0∗ ∈ (x∗ ⊕ y)∗ ⊕ (x∗ ⊕ y), then

0∗ ∈ 0∗ ⊕ (y∗ ⊗ z) ⊆ (x∗ ⊕ y)∗ ⊕ (x∗ ⊕ y)⊕ (y ⊕ z∗)∗ = (x∗ ⊕ y)∗ ⊕ (x∗ ⊕ z)⊕ (y ⊗ z∗).

So, by Lemma 3.9(i) we get that x∗ ⊕ y � (x∗ ⊕ z)⊕ (y ⊗ z∗).

On M we define the hyper operations “∨” and “∧” as follows:

x ∨ y = x⊗ y∗ ⊕ y and x ∧ y = (x∗ ∨ y∗)∗ = (x⊕ y∗)⊗ y.

Theorem 3.12 If x, y, z ∈ M , then the following hold:(a19) x ∨ y = y ∨ x and x ∧ y = y ∧ x;(a20) x⊗ y � x ∧ y � x, y;(a21) x, y � x ∨ y � x⊕ y;(a22) 0 ∈ x ∧ 0 and 1 ∈ x ∨ 1;(a23) x ∈ (x ∨ 0) ∩ (x ∧ 1);(a24) x ∈ (x ∧ x) ∩ (x ∨ x);(a25) x � y implies x ∧ z � y ∧ z and x ∨ z � y ∨ z;(a26) x ∈ (x ∧ (x ∨ y)) ∩ (x ∨ (x ∧ y));(a27) x � y implies x ∈ x ∧ y and y ∈ x ∨ y;(a28) x⊕ y ⊆ (x ∨ y)⊕ (x ∧ y).

P r o o f.(a19) By definition of hyper operations ” ∨ ” and ” ∧ ” and (hMV4) the proof is clear.(a20) By Proposition 2.2(8) we have x � x⊕ y∗, so x⊗ y � (x⊕ y∗)⊗ y by (a10). Hence x⊗ y � x ∧ y.

By (a6) we get that y ⊗ (y∗ ⊕ x) � y and x⊗ (x∗ ⊕ y) � x. Therefore x ∧ y � x, y.(a21) Since, by Proposition 2.2(8), y � (x⊗ y∗)⊕ y and x � x⊕ (y ⊗ x∗), then x, y � x ∨ y. Using (a6),

we have x⊗ y∗ � x. Thus x⊗ y∗ ⊕ y � x⊕ y. Therefore x ∨ y � x⊕ y.(a22) By (a20) we have x ∧ 0 � 0. Thus there exists t ∈ x ∧ 0 such that t � 0. Also by Proposition 2.2(2),

0 � t. Then t = 0 and so 0 ∈ x ∧ 0. By (a21) we get that 1 � x ∨ 1. So there is h ∈ x ∨ 1 such that 1 � h.Therefore 1 = h ∈ x ∨ 1.

(a23) By (a5) we get that x ∈ 1⊗ x ⊆ 1⊗ (0⊕ x) = 1⊗ (1∗ ⊕ x) = x ∧ 1 and similar we also havex ∈ x⊕ 0 ⊆ (x⊗ 1)⊕ 0 = (x⊗ 0∗)⊕ 0 = x ∨ 0. Therefore x ∈ (x ∨ 0) ∩ (x ∧ 1).



(a24) Using (a3), (a5) and Proposition 2.2(11) we have

x ∈ x⊕ 0 ⊆ x⊕ (x⊗ x∗) = x ∨ x and x ∈ x⊗ 1 = x⊗ 0∗ ⊆ x⊗ (x∗ ⊕ x) = x ∧ x

Therefore x ∈ (x ∧ x) ∩ (x ∨ x).(a25) Let x � y. Then x⊗ z∗ � y⊗ z∗ and so x⊗ z∗⊕ z � y⊗ z∗⊕ z. Therefore x∨ z � y∨ z. Similarly

x ∧ z � y ∧ z.(a26) By (a5) and (hMV5) we have

x ∈ 1⊗ x = 0∗ ⊗ x ⊆ ((y ⊗ x∗)⊕ 0∗)⊗ x

⊆ ((y ⊗ x∗)⊕ x⊕ x∗)⊗ x

= (x ∨ y ⊕ x∗)⊗ x = x ∧ (x ∨ y).

Similarly, we can get that x ∈ x ∨ (x ∧ y).(a27) Let x � y. Then 0∗ ∈ x∗ ⊕ y. So

x ∈ x⊗ 0∗ ⊆ x⊗ (x∗ ⊕ y) = x ∧ y and y ∈ y ⊕ 0 ⊆ y ⊕ (x∗ ⊕ y)∗ = y ⊕ (y∗ ⊗ x) = x ∨ y.

Therefore x ∈ x ∧ y and y ∈ x ∨ y.(a28) It holds

x⊕ y ⊆ (x⊕ y)⊕ 0 ⊆ (x⊕ y)⊕ (x⊕ y∗ ⊕ y)∗ = x⊕ (y ⊕ ((x⊕ y∗)⊕ y)∗

⊆ x⊕ (x⊕ y∗)∗ ⊕ ((x⊕ y∗)∗ ⊕ y∗)∗ = (x ∨ y)⊕ (x ∧ y).

Open problem: Prove or disprove, the hyper operation ∨(∧) is associative, i.e. x ∨ (y ∨ z) = (x ∨ y) ∨ z forall x, y, z ∈ M .

Theorem 3.13 Let x⊕ z = y ⊕ z, x � z∗ and y � z∗, for x, y, z ∈ M . Then {x, y} ⊆ (x ∧ z∗) ∩ (y ∧ z∗).

P r o o f. By (a27), x � z∗ implies x ∈ x∧ z∗, similarly, y � z∗ implies y ∈ y ∧ z∗. Now by hypothesis and(a5) we have

x ∈ x⊗ 1 ⊆ x⊗ (x∗ ⊕ z∗) = x ∧ z∗ = z∗ ⊗ (z ⊕ x) = z∗ ⊗ (z ⊕ y) = y ∧ z∗.

Similarly we can get that y ∈ x ∧ z∗. Therefore {x, y} ⊆ (x ∧ z∗) ∩ (y ∧ z∗).

The following example shows that x⊕ z = y ⊕ z, x � z∗ and y � z∗ but x �= y.Example 3.14 Consider the following tables on M = {0, a, b, 1}:

⊕ 0 a b 1

0 {0} {0, a} {0, b} {0, a, b, 1}

a {0, a} {0, a} {0, a, b, 1} {0, a, b, 1}

b {0, b} {0, a, b, 1} {0, a, b, 1} {0, a, b, 1}

1 {0, a, b, 1} {0, a, b, 1} {0, a, b, 1} {0, a, b, 1}

∗ 0 a b 1

1 b a 0

Then (M,⊕,∗ , 0) is a hyper MV-algebra. Also a⊕ a = {0, a} = 0⊕ a, 0 � a∗ and a � a∗ but a �= 0.Theorem 3.15(i) If x ∈ x⊗ y, then 1 ∈ x∗ ∨ y for all x, y ∈ M .(ii) If x ∈ x⊕ y, then 0 ∈ x∗ ∧ y for all x, y ∈ M .

P r o o f.(i) By hypothesis we get that 0∗ ∈ x⊕ x∗ ⊆ (x⊗ y)⊕ x∗ = x∗ ∨ y. Thus 1 ∈ x∗ ∨ y.(ii) The proof is similar to the proof of (i).

Consider Example 3.5, we can see that 1 ∈ b∗ ∨ b but b �∈ b ⊗ b, also 0 ∈ a∗ ∧ a, but a �∈ a ⊕ a. Hence theconverse of Theorem 3.14 (i), (ii) is not true in general.



4 Ideals in hyper MV-algebras

Definition 4.1 A hyper MV-ideal of a hyper MV-algebra is a non-empty subset I of M satisfying the fol-lowing conditions:(I0) if x ∈ I , y ∈ M and y � x, then y ∈ I;(I1) if x, y ∈ I , then x⊕ y ⊆ I .

We denote by HId(M) the set of all hyper MV-ideals of M and by HI(M) the set

HI(M) = {I ⊆ M : if x, y ∈ M, y � x and x ∈ I , then y ∈ I}.

Remark 4.2 Clearly HId(M) ⊆ HI(M) and if {Ii}i∈Λ ⊆ HI(M), then⋃

i∈ΛIi,

⋂i∈Λ

Ii ∈ HI(M). Also,if ∅ �= I ∈ HI(M), then 0 ∈ I .

Example 4.3 Consider the hyper MV-algebra in Example 3.10. Then I1 = {0, a} is a hyper MV-ideal, whileI2 = {0, b} is not a hyper MV-ideal, because b⊕ b = {b, 1} �⊆ I2.

Definition 4.4 Let I be a non-empty subset of a hyper MV-algebra M . Then I is called a weak hyperMV-ideal of M , if I satisfies (I0) and

(I2) if x, y ∈ I , then x⊕ y � I .We denote by WHId(M) the set of all weak hyper MV-ideals of M .Example 4.5 Consider the following tables on M = {0, b, 1}:

⊕ 0 b 1

0 {0} {0, b} {0, b, 1}

b {0, b} {1} {0, b, 1}

1 {0, b, 1} {0, b, 1} {0, b, 1}

∗ 0 b 1

1 b 0

Then (M,⊕,∗ , 0) is a hyper MV-algebra. Also I1 = {0} is a weak hyper MV-ideal, while I2 = {0, b} is not aweak hyper MV-ideal, because b⊕ b = {1} �� I2.

Theorem 4.6 Every hyper MV-ideal is a weak hyper MV-ideal. The converse is not true in general.

P r o o f. It is proved by Proposition 2.2(7) and Definition 4.4. In Example 4.3, I = {0, a, b} is a weak hyperMV-ideal, but it is not a hyper MV-ideal, because a, b ∈ I , but a⊕ b = {b, 1} �⊆ I .

Note that M and {0} are two (weak) hyper MV-ideals of M . A (weak) hyper MV-ideal I of M such thatI �= M is called proper.

Theorem 4.7 A (weak) hyper MV-ideal I of M is proper if and only if 1 �∈ I .

P r o o f. Let I be a proper hyper MV-ideal of M . Then I �= M . We show that 1 �∈ I . On the contrary, let1 ∈ I . Since x � 1, for all x ∈ M , then x ∈ I , by (I0). So I = M , which is a contradiction.

Conversely, let 1 �∈ I . Then it is clear that I �= M .

Theorem 4.8 Let I be a non-empty subset of M . Then the following conditions are equivalent:(i) I ∈ WHId(M);(ii) I satisfies (I0) and if x, y ∈ I , then (x⊕ y) ∩ I �= ∅.

P r o o f.(i) → (ii). Since I ∈ WHId(M), it is clear that (I0) holds. Now let x, y ∈ I . Then x ⊕ y � I and so there

exist t ∈ x⊕ y and s ∈ I such that t � s. Thus t ∈ I . Therefore (x⊕ y) ∩ I �= ∅.(ii) → (i). Let x, y ∈ I . Then by hypothesis we have (x⊕ y) ∩ I �= ∅ and so there is h ∈ (x⊕ y) ∩ I . Hence

x⊕ y � I . Therefore I ∈ WHId(M).



Theorem 4.9 Let I be a proper (weak) hyper MV-ideal of M . Then I is not a hyper MV-subalgebra of M .

P r o o f. On the contrary, let I be a hyper MV-subalgebra of M . Since I �= ∅, then there is x ∈ I , and sox∗ ∈ I , by Theorem 2.4. Hence 1 = 0∗ ∈ x ⊕ x∗ ⊆ I , which is a contradiction. Therefore I is not a hyperMV-subalgebra of M .

Theorem 4.10 Let ∅ �= I ∈ HI(M). If x ∈ I and y ∈ M , then (x⊗ y) ∩ I �= ∅ and (x ∧ y) ∩ I �= ∅.

P r o o f. Since x⊗ y � x, then there is t ∈ x⊗ y such that t � x and so x ∈ I implies that t ∈ I . Therefore(x⊗ y) ∩ I �= ∅. Similarly, we can prove that (x ∧ y) ∩ I �= ∅.

Theorem 4.11 Let I ∈ HId(M) ∪WHId(M). If x, y ∈ I , then (x ∨ y) ∩ I �= ∅.

P r o o f. Let I ∈ HId(M). Since x ∈ I , then (x⊗ y∗)∩ I �= ∅ by Theorem 4.10. So there is t ∈ (x⊗ y∗)∩ I .Hence by hypothesis we get that t ⊕ y ⊆ I . Also t ⊕ y ⊆ (x ⊗ y∗) ⊕ y = x ∨ y. Therefore (x ∨ y) ∩ I �= ∅.If I ∈ WHId(M), then similar to above argument there exists t ∈ (x ⊗ y∗) ∩ I . Thus, by Theorem 4.8(ii),(t⊕ y) ∩ I �= ∅. So t⊕ y ⊆ x ∨ y implies that (x ∨ y) ∩ I �= ∅.

Theorem 4.12 Let I be a hyper MV-ideal of hyper MV-algebra (M,⊕,∗ , 0). Then I is a weak hyper K-idealof hyper K-algebra (M, ◦, 0), where x ◦ y = (x∗ ⊕ y)∗.

P r o o f. Let I be a hyper MV-ideal of M . Then 0 ∈ I . Let x ◦ y ⊆ I and y ∈ I . We have

x ∨ y = (x⊗ y∗)⊕ y = (x∗ ⊕ y)∗ ⊕ y = (x ◦ y)⊕ y.

Since x � x ∨ y, then there is t ∈ x ∨ y such that x � t. So there exists h ∈ x ◦ y such that t ∈ h ⊕ y. Sinceh ∈ x ◦ y ⊆ I and y ∈ I , then t ∈ h ⊕ y ⊆ I . Hence x � t and t ∈ I imply that x ∈ I . Therefore I is a weakhyper K-ideal of (M, ◦, 0).

Definition 4.13 A non-empty subset I of (M,⊕,∗ , 0) is called S-reflexive if (x⊕y)∩I �= ∅ implies x⊕y ⊆ I

for all x, y ∈ M .Note that M is S-reflexive.Consider Example 4.3, we can see that {b, 1} and {0, a} are S-reflexive, but {0}, {0, a, b} are not S-reflexive.Theorem 4.14 Every S-reflexive weak hyper MV-ideal is a hyper MV-ideal.

P r o o f. The proof follows from Theorem 4.8 and Definition 4.13.

Theorem 4.15 Let I be a non-empty subset of M and I and I∗ be S-reflexive subsets of (M,⊕,∗ , 0). If I isa weak hyper MV-ideal of (M,⊕,∗ , 0), then I is a hyper K-ideal of (M, ◦, 0), where x ◦ y = (x∗ ⊕ y)∗ .

P r o o f. Since I is a weak hyper MV-ideal, then 0 ∈ I . Let x ◦ y < I and y ∈ I . Then by Theorem 2.18(x∗⊕y)∗ � I , and so (x∗⊕y)∩I∗ �= ∅. Thus by hypothesis we get that x∗⊕y ⊆ I∗. Hence x◦y = (x∗⊕y)∗ ⊆ I .Since x � x ∨ y = (x ◦ y)⊕ y, then there exists t ∈ (x ◦ y)⊕ y such that x � t. So there exists h ∈ x ◦ y ⊆ I

such that t ∈ h⊕ y. Hence h, y ∈ I implies that h⊕ y � I . Then by Theorem 4.14 we get that t ∈ h⊕ y ⊆ I .So x � t and t ∈ I imply that x ∈ I . Therefore I is a hyper K-ideal of (M, ◦, 0).

The following example shows that the condition “I and I∗ are S-reflexive” is necessary in the above theorem.Example 4.16 The following tables show a hyper MV-algebra structure on M = {0, a, b, 1}.

⊕ 0 a b 1

0 {0} {0, a} {0, b} {0, a, b, 1}

a {0, a} {0, a} {0, a, b, 1} {0, a, b, 1}

b {0, b} {0, a, b, 1} {0, b} {0, a, b, 1}

1 {0, a, b, 1} {0, a, b, 1} {0, a, b, 1} {0, a, b, 1}

∗ 0 a b 1

1 b a 0



We can construct a hyper K-algebra (M, ◦, 0) as follow:

⊕ 0 a b 1

0 {0, a, b, 1} {0, a, b, 1} {0, a, b, 1} {0, a, b, 1}

a {a, 1} {0, a, b, 1} {a, 1} {0, a, b, 1}

b {b, 1} {b, 1} {0, a, b, 1} {0, a, b, 1}

1 {1} {b, 1} {a, 1} {0, a, b, 1}

It is clear that I = {0, a, b} is a weak hyper MV-ideal in (M,⊕,∗ , 0), but it is not hyper K-ideal in (M, ◦, 0),because 1 ◦ b = {1, a} < I and b ∈ I but 1 �∈ I . Also I and I∗ are not S-reflexive.

Theorem 4.17 Let (M1,⊕1,∗1 , 01) and (M2,⊕2,

∗1 , 01) be two hyper MV-algebras and consider the hyperMV-algebra M = M1 ×M2. Then it holds:

(i) if I1 ∈ HId(M1) (WHId(M1)) and I2 ∈ HId(M2) (WHId(M2)), then I1 × I2 ∈ HId(M) (WHId(M));(ii) if I ∈ HId(M), then there are two unique I1 ∈ HId(M1) and I2∈ HId(M2) such that I = I1 × I2;(iii) if I is S-reflexive and I ∈ WHId(M), then there are two unique I1 ∈ WHId(M1) and I2 ∈ WHId(M2)

such that I = I1 × I2.

P r o o f.(i) We prove (I0). Let (x1, y1) ∈ M1×M2, (x2, y2) ∈ I1×I2 and (x1, y1) � (x2, y2). Then by Theorem 2.9

x1 � x2 and y1 � y2. So x2 ∈ I1 and y2 ∈ I2 implies that x1 ∈ I1 and y1 ∈ I2. Therefore (x1, y1) ∈ I1 × I2.Now let (x1, y1) and (x2, y2) be in ∈ I1 × I2. Then it holds x1 ⊕1 x2 ⊆ I1 and y1 ⊕2 y2 ⊆ I2 and so(x1, y1)⊕ (x2, y2) = (x1 ⊕1 x2, y1 ⊕2 y2) ⊆ I1 × I2. Therefore I1 × I2 ∈ HId(M).

(ii) Let I ∈ HId(M). Suppose that

I1 = {a ∈ M1 : (a, b) ∈ I for some b ∈ M2} and I2 = {b ∈ M2 : (a, b) ∈ I for some a ∈ M1}.

Now we show that I1 ∈ HId(M1). The proof for I2 is similar. Let a1 ∈ M1, a2 ∈ I1 and a1 � a2. Then thereis b2 ∈ M1 such that (a2, b2) ∈ I . Since 02 � b2 and a1 � a2, then (a1, 02) � (a2, b2) ∈ I , so (a1, 02) ∈ I .Thus a1 ∈ I1.

Let a1, a2 ∈ I1. Then there are b1, b2 ∈ M2 such that (a1, b1), (a2, b2) ∈ I . Thus

(a1 ⊕1 a2, b1 ⊕2 b2) = (a1, b1)⊕ (a2, b2) ⊆ I.

Hence a1 ⊕1 a2 ⊆ I1. Therefore I1 ∈ HId(M1). We show that I = I1 × I2. Let x ∈ I . Then there area ∈ M1 and b ∈ M2 such that x = (a, b). By definition of I1 and I2 we get that a ∈ I1 and b ∈ I2. Thusx ∈ I1 × I2 and hence I ⊆ I1 × I2. On the other hand let (a, b) ∈ I1 × I2. Then there are b1 ∈ M2

and a1 ∈ M1 such that (a1, b), (a, b1) ∈ I . Since (01, b) � (a1, b) ∈ I and (a, 02) � (a, b1) ∈ I , then(a, b) ∈ (a⊕1 01, b⊕2 02) = (a, 02)⊕(01, b) ⊆ I , so (a, b) ∈ I . Therefore I = I1×I2. The proof of uniquenessis clear.

(iii) The proof is similar to the proof of (ii).

5 Hyper MV-ideals in hyper MV-algebras of order 3 and 4

Theorem 5.1 Let |M | > 2. Then I = M − {1} is not a hyper MV-ideal.

P r o o f. On the contrary, let I be a hyper MV-ideal. Consider a ∈ I−{0}. Then there exists b ∈ M−{1} = I

such that a∗ = b. Thus by hypothesis we have 1 = 0∗ ∈ a∗⊕ a = b⊕ a ⊆ I , which is a contradiction. ThereforeI is not a hyper MV-ideal.

Corollary 5.2 Let M = {0, b, 1} be a hyper MV-algebra. Then the only proper hyper MV-ideal of M isI = {0}.



Theorem 5.3 Let a ∈ M , 1 ∈ a⊕ a and I be a non-empty subset of M . If {0, a} ⊆ I , then I is not a properhyper MV-ideal.


Theorem 5.4 Let M = {0, a, b, 1} be a hyper MV-algebra and I = {0, a}. Then it holds:(i) if a = a∗, then I is not a hyper MV-ideal of M ;(ii) if a = b∗, then I is a hyper MV-ideal of M if and only if a⊕ a ⊆ I .

P r o o f.(i) The proof follows from Theorem 5.3.(ii) Let a ⊕ a ⊆ I . We show that 0 ⊕ a ⊆ I . It is clear that 1 �∈ 0 ⊕ a, because if 1 ∈ 0 ⊕ a, then we

get that 1 � a, which is not true. Also b �∈ 0 ⊕ a, since if b ∈ 0 ⊕ a, then by Proposition 2.2(12), b � a, so0∗ ∈ b∗ ⊕ a = a ⊕ a ⊆ I , it is impossible. Thus 0 ⊕ a ⊆ I . Therefore I is a hyper MV-ideal of M . The proofof the converse is clear.

Theorem 5.5 Let a ∈ M , a ⊕ a = {1} and I be a non-empty subset of M . If {0, a} ⊆ I , then I is not aproper weak hyper MV-ideal of M .


Theorem 5.6 Let M = {0, b, 1} be a hyper MV-algebra. Then I = {0, b} is a weak hyper MV-ideal of Mif and only if b⊕ b �= {1}.

P r o o f. Let b⊕b �= {1}. Then 0 ∈ b⊕b or b ∈ b⊕b and so b⊕b � I . Also b ∈ 0⊕b implies that 0⊕b � I .Therefore I is a weak hyper MV-ideal of M .

Conversely, the proof follows from Theorem 5.5.

Theorem 5.7 Let M = {0, a, b, 1} be a hyper MV-algebra and I = {0, a}. Then it holds:(i) if a∗ = b, then I is a weak hyper MV-ideal of M if and only if 1 �∈ a⊕ a and a⊕ a �= {b};(ii) if a∗ = a, then I is a weak hyper MV-ideal of M if and only if a⊕ a �⊆ {1, b} and b �� a.

P r o o f.(i) Let I be a weak hyper MV-ideal of M . If 1 ∈ a ⊕ a, then b � a, and so by hypothesis we have b ∈ I ,

which is not true. If a⊕a = {b}, since a⊕a � I , then we get that b � 0 or b � a, which is not true. Therefore1 �∈ a⊕ a and a⊕ a �= {b}. Conversely, let 1 �∈ a⊕ a and a⊕ a �= {b}. Then a⊕ a = {0} or {a} or {0, a} or{0, b} or {0, a, b} and so a⊕ a � I . Also a ∈ a⊕ 0 implies that a⊕ 0 � I . Since 1 �∈ a⊕ a, then b �� a andhence I0 holds. Therefore I is a weak hyper MV-ideal.

(ii) Since a∗ = a, then 1 ∈ a ⊕ a. Let I be a weak hyper MV-ideal of M . Since a ∈ I , then it is clear thatb �� a. Now we show that a ⊕ a �⊆ {1, b}, on the contrary, let a ⊕ a ⊆ {1, b}. Then by hypothesis we get that1 � 0 or b � 0 or 1 � a or b � a, which is a contradiction. Thus a⊕a �⊆ {1, b}. Conversely, let a⊕a �⊆ {1, b}and b �� a. Then I0 holds. Also a⊕ a �⊆ {1, b} implies that a⊕ a = {0, 1} or {1, a} or {1, a, b} or {0, a, b} or{0, 1, a} or {0, 1, a, b}, hence a⊕ a � I . Also we have a⊕ 0 � I . Therefore I is a weak hyper MV-ideal.

Theorem 5.8 Let M = {0, a, b, 1} be a hyper MV-algebra. Then I = {0, a, b} is a weak hyper MV-ideal ifand only if a⊕ a �= {1}, b⊕ b �= {1} and a⊕ b �= {1}.

P r o o f. Let I be a weak hyper MV-ideal. Then a ⊕ a � I , a ⊕ b � I and b ⊕ b � I , so a ⊕ a �= {1},b⊕ b �= {1} and a⊕ b �= {1}. Conversely, let a⊕ a �= {1}, b⊕ b �= {1} and a⊕ b �= {1}. Then (a⊕ a)∩ I �= ∅,(b ⊕ b) ∩ I �= ∅ and (a ⊕ b) ∩ I �= ∅ and also it is clear that I0 holds. Therefore by Theorem 4.8, I is a weakhyper MV-ideal.



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Hyper MV -ideals in hyper MV -algebras