In this paper, we investigated the number of hyper MV -algebrasof order 3. In fact, we prove that there are 33 hyper MV -algebras oforder 3, up to isomorphism

Borzooei, R.A.

Radfar, A.

English

In this paper, we investigated the number of hyper MV -algebras
of order 3. In fact, we prove that there are 33 hyper MV -algebras of
order 3, up to isomorphism

R.A. Borzooei

A. Radfar

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RATIO MATHEMATICA 25 (2013), 15–28 ISSN:1592-7415Classification of Hyper MV -algebrasof Order 3R. A. Borzooei∗, A. Radfar∗∗∗Department of Mathematics, Shahid Beheshti University, G. C.,Tehran, Iran∗∗Department of Mathematics, Payame Noor University, Tehran, Iranborzooei@sbu.ac.ir, Ateferadfar@yahoo.comAbstractIn this paper, we investigated the number of hyper MV -algebrasof order 3. In fact, we prove that there are 33 hyper MV -algebras oforder 3, up to isomorphism.Key words: hyper MV -algebraMSC 2010: 97U99.1 IntroductionThe concept of MV -algebras was introduced by Chang in [1] in order toshow Lukasiewicz logic to be standard complete, i.e. complete with respectto evaluations of propositional variables in the real unit interval [0, 1]. In[6], Mundici showed that any MV -algebra is an interval of an Abelian lat-tice ordered group with a strong unit. Also, he introduced the concept ofstate on MV -algebra. Georgescu and Iorgulescu [2] introduced a new non-commutative algebraic structures, which were called pseudo MV -algebras.It can be obtained by dropping commutative axioms in MV -algebras, whichare a generalization of MV -algebras. The hyper structure theory was intro-duced by F. Marty [5] at the 8th congress of Scandinavian Mathematiciansin 1934. Since then many researches have worked in these areas. Recently in[4], Sh. Ghorbani, A. Hasankhni and E. Eslami applied the hyper structureto MV -algebras and introduced the concept of a hyper MV -algebra whichis a generalization of an MV -algebra and investigated some related results.Now, in this paper we find all hyper MV -algebras of order 3.15R. A. Borzooei, A. Radfar2 PreliminaryDefinition 2.1. [1] An MV -algebra (X,⊕,∗ , 0) is a set X equipped witha binary operation ⊕, a unary operation ∗ and a constant 0 satisfying thefollowing equations:(MV1) x⊕ (y ⊕ z) = (x⊕ y)⊕ z,(MV2) x⊕ y = y ⊕ x,(MV3) x⊕ 0 = x,(MV4) (x∗)∗ = x,(MV5) x⊕ 0∗ = 0∗,(MV6) (x∗ ⊕ y)∗ ⊕ y = (y∗ ⊕ x)∗ ⊕ x,for all x, y, z ∈ X.Definition 2.2. [3]A hyperalgebra (M,⊕,∗ , 0) with a hyperoperation ⊕ : M × M −→P∗(M), a unary operation ∗ : M −→ M and a constant 0, is said to bea hyper MV -algebra if and only if satisfies the following axioms, for allx, y, z ∈M :(HMV1) x⊕ (y ⊕ z) = (x⊕ y)⊕ z,(HMV2) x⊕ y = y ⊕ x,(HMV3) (x∗)∗ = x,(HMV4) 0∗ ∈ x⊕ 0∗,(HMV5) (x∗ ⊕ y)∗ ⊕ y = (y∗ ⊕ x)∗ ⊕ x,(HMV6) 0∗ ∈ x⊕ x∗,(HMV7) If x 6 y and y 6 x, then x = y,where x 6 y is defined by 0∗ ∈ x∗ ⊕ y. For every X, Y ⊆ M , X 6 Y ifthere exist x ∈ X and y ∈ Y such that x 6 y. We define 1 = 0∗Theorem 2.3. [3] Let (M,⊕,∗ , 0) be a hyper-MV algebra. Then for allx, y, z ∈ M and for all non-empty subsets A, B and C of M the followinghold:(i) (A⊕B)⊕ C = A⊕ (B ⊕ C),(ii) 0 6 x 6 1, x 6 x and A 6 A,(iii) If x 6 y then y∗ 6 x∗ and A 6 B implies B∗ 6 A∗,(iv) If x 6 0 or 1 6 x, then x = 0 or x = 1, respectively,(v) 0⊕ 0 = {0},(vi) x ∈ x⊕ 0,(vii) If x⊕ 0 = y ⊕ 0, then x = y.16Classification of Hyper MV -algebras of Order 33 Classification of hyper MV -algebras of or-der 3In this section we try to find all hyper MV -algebras of order 3, up toisomorphism.Theorem 3.1. Let M be a hyper MV -algebra and x be an element of Msuch that 0⊕ x = {x} and x∗ = x. Then the following statements hold:(i) (1⊕ x)∗ ⊕ x = {x},(ii) (1⊕ x)∗ ⊕ 1 = x⊕ x,(iii) x 6∈ 1⊕ x and 0 6∈ 1⊕ x.Proof. Since 0∗ = 1, then by hypothesis and (HMV 5);(1⊕x)∗⊕x = (0∗⊕x)∗⊕x = (x∗⊕0)∗⊕0 = (x⊕0)∗⊕0 = x∗⊕0 = x⊕0 = {x}(1⊕ x)∗ ⊕ 1 = (x⊕ 1)∗ ⊕ 1 = ((x∗)∗ ⊕ 1)∗ ⊕ 1 == (1∗ ⊕ x∗)∗ ⊕ x∗ = (0⊕ x)∗ ⊕ x∗ = x∗ ⊕ x∗ = x⊕ xand so (i) and (ii) hold.(iii) If x ∈ 1⊕x, then x = x∗ ∈ (1⊕x)∗ and so x⊕x = x∗⊕x ⊆ (1⊕x)∗⊕x.By (i), x⊕ x ⊆ {x}. Hence x⊕ x = {x}. Now, since by (HMV6), 1 = 0∗ ∈x ⊕ x∗ = x ⊕ x = {x}, then x = 1 and so 0 = 1∗ = x∗ = x = 1, which is acontradiction. Hence x /∈ 1⊕ x. Now, let 0 ∈ 1⊕ x. Then 1 = 0∗ ∈ (1⊕ x)∗and so 1⊕ x ⊆ (1⊕ x)∗ ⊕ x. By (i), 1⊕ x ⊆ {x}. Thus 1⊕ x = {x}, whichis a contradiction. Hence 0 /∈ 1⊕ x.Note. From now one in this paper, we let M = {0, a, 1} be a hyperMV -algebra of order 3.Theorem 3.2. (i) 1 ≤ 1, 0 ≤ 0, a ≤ a, 0 ≤ 1 and 0 ≤ a,(ii) a 6≤ 0,(iii) a∗ = a,(iv) 1 ∈ 1⊕ a.Proof. (i). By Theorem 2.3(ii), the proof is clear.(ii). By Theorem 2.3(iv), the proof is clear.(iii). By Definition 2.2, 0∗ = 1 and by (HMV3), 0 = (0∗)∗ = 1∗. Now, ifa∗ = 1, then 0 = 1∗ = (a∗)∗ = a, which is a contradiction. By similar way, ifa∗ = 0, then 1 = 0∗ = (a∗)∗ = a, which is a contradiction. Hence, a∗ = a.(iv). By (HMV4), 1 = 0∗ ∈ 0∗ ⊕ a = 1⊕ a.Theorem 3.3. If 0⊕ a = {a} or 1⊕ a = {1}, then M is an MV -algebra.17R. A. Borzooei, A. RadfarProof. Let 0 ⊕ a = {a}. Since a∗ = a, then by Theorem 3.1(iii), a 6∈ 1 ⊕ aand 0 6∈ 1⊕ a and so 1⊕ a = {1}.Moreover, By Theorem 3.1(iii) and (i), 0 6∈ 1⊕ 0 and (1⊕ 0)∗⊕ 0 = {0}.Since 0 6∈ {a} = 0⊕ a and 0 6∈ 1⊕ 0, then (1⊕ 0)∗ = {0} and so 1⊕ 0 = {1}.By Theorem 3.1(i) and (ii), 0 ⊕ 1 = {1} = (1 ⊕ a)∗ ⊕ 1 = a ⊕ a. Hencea⊕ a = {1}. Now, by (HMV1),1⊕ 1 = (a⊕ a)⊕ 1 = a⊕ (1⊕ a) = a⊕ 1 = {1}.Therefore, x⊕y is singleton for all x, y ∈M and so M is an MV -algebra.Now, if 1⊕a = {1}, then {0} = {1∗} = (1⊕a)∗ and so 0⊕a = (1⊕a)∗⊕a.By (HMV5),0⊕ a = (1⊕ a)∗ ⊕ a = 0⊕ (0⊕ a)∗.By Theorem 3.2, a 6< 0, 1 6∈ 0⊕ a. If 0 ∈ 0⊕ a, then 0⊕ a = {0, a} and{0, a} = 0⊕ a = 0⊕ (0⊕ a)∗ = 0⊕ {0, a}∗ == 0⊕ {1, a} = (0⊕ 1) ∪ (0⊕ a) = (0⊕ 1) ∪ {0, a}.Hence 0 ⊕ 1 ⊆ {0, a}. By (HMV 4), 1 ∈ 0 ⊕ 1. Thus 1 ∈ {0, a}, which is acontradiction. Thus 0 6∈ 0 ⊕ a and so 0 ⊕ a = {a}. Therefore, M is a sameMV -algebra, which is as follows:⊕1 0 a 10 {0} {a} {1}a {a} {1} {1}1 {1} {1} {1}Definition 3.4. We call a hyper MV -algebra is proper, if it is not an MV -algebra.Lemma 3.5. Let M = {0, a, 1} be a proper hyper MV -algebra of order 3.Then(i) 0⊕ a = {0, a},(ii) 0⊕ 1 = {1}, {0, 1} or M ,(iii) a⊕ a = {1}, {0, 1}, {1, a} or M ,(iv) 1⊕ a = {0, 1}, {1, a} or M ,(v) 1⊕ 1 = {1}, {0, 1} {1, a} or M ,(vi) If a⊕ a = {1}, then 0⊕ 1 = M .18Classification of Hyper MV -algebras of Order 3Proof. (i). Since a 6< 0, then 1 6∈ 0⊕a. By Theorem 2.3 (vi), a ∈ 0⊕a.Thus 0⊕ a = {a} or {0, a}. If 0⊕ a = {a}, then by Theorem 3.3, M is notproper. Thus 0⊕ a = {0, a}(ii). Since 0 ≤ 0, then 1 = 0∗ ∈ 0∗ ⊕ 0 = 1 ⊕ 0 = 0 ⊕ 1. Hence it issufficient to show that 0 ⊕ 1 6= {1, a}. Let 0 ⊕ 1 = {1, a}, by the contrary.Then by (HMV1),{1, a} = 0⊕ 1 = (0⊕ 0)⊕ 1 = (0⊕ 1)⊕ 0 = {1, a} ⊕ 0 = {0, a, 1},which is impossible. Therefore, 0⊕ 1 6= {1, a} and so 0⊕ 1 = {1}, {0, 1} orM .(iii), (v). Since a ≤ a and 0 ≤ 1, then 1 ∈ a⊕ a and 1 ∈ 1⊕ 1 and so (v)and (iii) are hold.(iv). Since 0 ≤ a, then 1 ∈ 1⊕ a. By Theorem 3.3, if a⊕ 1 = {1}, thenM is an MV algebra which is impossible. Hence 1 ⊕ a = {0, 1}, {1, a} orM .(vi). Let a⊕ a = {1}. Then by (HMV1),0⊕ 1 = 0⊕ (a⊕ a) = (0⊕ a)⊕ a = {0, a} ⊕ a = (0⊕ a) ∪ (a⊕ a) = M.By Lemma 3.5 (ii), we know that 0 ⊕ 1 = {1}, {0, 1} or M . So, for theclassification of all hyper MV -algebras of order 3, we consider the followingthree cases.Case 1: 0⊕ 1 = {1}Lemma 3.6. Let M = {0, a, 1} be a proper hyper MV -algebra of order 3and 0⊕ 1 = {1}. Then(i) a⊕ a = {1, a} or M ,(ii) 1⊕ 1 = {1},(iii) 1⊕ a = M .Proof. (i). By Lemma 3.5 (i) and (iii), 0⊕a = {0, a} and 1 ∈ a⊕a. Hence(0⊕ a)⊕ a = {0, a} ⊕ a = (0⊕ a) ∪ (a⊕ a) = {0, a} ∪ (a⊕ a) = M.Since by (HMV1), (0 ⊕ a) ⊕ a = 0 ⊕ (a ⊕ a), then 0 ⊕ (a ⊕ a) = M . ByLemma 3.5(iii), a ⊕ a = {1}, {0, 1}, {1, a} or M . If a ⊕ a = {1}, then0⊕ (a⊕ a) = 0⊕ 1 = {1}, which is a contradiction.If a ⊕ a = {0, 1}, then by Theorem 2.3(v), 0 ⊕ (a ⊕ a) = 0 ⊕ {0, 1} =(0 ⊕ 0) ∪ (0 ⊕ 1) = {0, 1}, which is a contradiction. Hence, a ⊕ a = {1, a}or M .19R. A. Borzooei, A. Radfar(ii). By (HMV5), and Theorem 2.3(v),(1⊕ 1)∗ ⊕ 1 = (0∗ ⊕ 1)∗ ⊕ 1 = (1∗ ⊕ 0)∗ ⊕ 0 = (0⊕ 0)∗ ⊕ 0 = 1⊕ 0 = {1}.If 0 ∈ 1 ⊕ 1, then 1 ⊕ 1 ⊆ (1 ⊕ 1)∗ ⊕ 1 = {1} and so 0 /∈ 1 ⊕ 1, which is acontradiction. If a ∈ 1⊕1, then a⊕1 ⊆ (1⊕1)∗⊕1 = {1}. Thus a⊕1 = {1}and so by Theorem 3.3, M is an MV -algebra, which is a contradiction.Hence, 1⊕ 1 = {1}.(iii). By Lemma 3.5, 1 ⊕ a = {0, 1}, {1, a} or M . If 1 ⊕ a = {0, 1},since by (HMV1), 1⊕ (1⊕ a) = (1⊕ 1)⊕ a = 1⊕ a, then 1⊕ (1⊕ a) = {1},which is a contradiction. If 1⊕ a = {1, a}, since by (HMV1), 0⊕ (1⊕ a) =(0 ⊕ 1) ⊕ a = 1 ⊕ a, then 0 ⊕ (1 ⊕ a) = (0 ⊕ 1) ∪ (0 ⊕ a) = M , which is acontradiction. Hence, 1⊕ a = M .Theorem 3.7. There are two non-isomorphic proper hyper MV -algebras oforder 3 such that 0⊕ 1 = {1}.Proof. According Theorem 3.6, if M is a proper hyper MV -algebraof order 3 and 0 ⊕ 1 = {1}, then we must investigate two following tables,which both of them are non-isomorphic hyper MV -algebras.⊕2 0 a 10 {0} {0, a} {1}a {0, a} {1, a} {0, a, 1}1 {1} {0, a, 1} {1}⊕3 0 a 10 {0} {0, a} {1}a {0, a} {0, a, 1} {0, a, 1}1 {1} {0, a, 1} {1}Case 2: 0⊕ 1 = {0, 1}Lemma 3.8. Let M = {0, a, 1} be a proper hyper MV -algebra of order 3and 0⊕ 1 = {0, 1}. Then(i) (a⊕ a) ∪ (1⊕ a) = M ,(ii) a⊕ 1 = {a, 1} or M ,(iii) a⊕ a = {a, 1} or M ,(iv) 1⊕ 1 = {0, 1} or {1}.Proof. (i). Let 0⊕ 1 = {0, 1}. By Theorem 3.5(iv), since 1 ∈ 1⊕ a, by(HMV1),(0⊕a)⊕1 = (0⊕1)⊕a = {0, 1}⊕a = (0⊕a)∪(1⊕a) = {0, a}∪(1⊕a) = M.On the other hands(0⊕ a)⊕ 1 = {0, a} ⊕ 1 = (0⊕ 1) ∪ (a⊕ 1) = {0, 1} ∪ (a⊕ 1).20Classification of Hyper MV -algebras of Order 3Thus {0, 1} ∪ (a ⊕ 1) = M and so a ∈ a ⊕ 1. New, we consider two cases0 ∈ a ⊕ 1 or 0 6= a ⊕ 1. If 0 ∈ a ⊕ 1, since by Theorem 3.5, 1 ∈ a ⊕ 1, thena⊕1 = M and so (a⊕a)∪ (1⊕a) = M . Now, if 0 6= a⊕1, then by Theorem3.5, a ∈ a⊕ 1. Hence by Theorem 3.2(iv), {1, a} ⊆ a⊕ 1. ThusM = (0⊕ 1) ∪ (a⊕ 1) = {0, a} ⊕ 1 = {1, a}∗ ⊕ 1 ⊆ (a⊕ 1)∗ ⊕ 1 ⊆Mand so (a ⊕ 1)∗ ⊕ 1 = M . On the other hands, by (HMV5), (a ⊕ 1)∗ ⊕ 1 =(0⊕ a)∗ ⊕ a. Hence (0⊕ a)∗ ⊕ a = M . Since 0⊕ a = {0, a}, thenM = (0⊕ a)∗ ⊕ a = {1, a} ⊕ a = (1⊕ a) ∪ (a⊕ a).(ii). By Lemma 3.5(iv), it is enough to show that 1 ⊕ a = {0, 1}. Let0 ∈ a ⊕ 1, by the contrary. Since by Lemma 3.5(iv) and (i), 0 ⊕ a = {0, a}and 1 ∈ 1⊕ a, then(0⊕ 1)⊕ a = {0, 1} ⊕ a = (0⊕ a) ∪ (1⊕ a) = M.Thus by (HMV1),M = (0⊕ 1)⊕ a = (0⊕ a)⊕ 1 = {0, 1} ∪ (1⊕ a).and so a ∈ 1⊕a. Hence a⊕1 6= {0, 1} and so by lemma 3.5(iv), a⊕1 = {a, 1}or M .(iii). By Lemma 3.5(i), 0⊕ a = {0, a}. Now, since 1 ∈ a⊕ a, then(0⊕ a)⊕ a = {0, a} ⊕ a = (0⊕ a) ∪ (a⊕ a) = M.Hence, by (HMV1), 0 ⊕ (a ⊕ a) = (0 ⊕ a) ⊕ a = M . Since a 6∈ 0 ⊕ 0 anda 6∈ 0⊕ 1, then a ∈ a⊕ a. Hence a⊕ a = {a, 1} or M .(iv). Let a ∈ 1⊕ 1. By (HMV5),a⊕ 1 = a∗ ⊕ 1 ⊆ (1⊕ 1)∗ ⊕ 1 = (0⊕ 0)∗ ⊕ 0 = {0, 1}.which is a contradiction by (i). Hence a 6∈ 1 ⊕ 1 and so by Lemma 3.5(v),1⊕ 1 = {0, 1} or {1}.Theorem 3.9. There are 6 non-isomorphic proper hyper MV -algebras oforder 3 such that 0⊕ 1 = {0, 1}.Proof. By Lemma 3.8 (iii), a⊕a = {a, 1} or M . If a⊕a = {a, 1}, thenby Lemma 3.8 (ii), a⊕1 = {a, 1} or M . By Lemma 3.8 (i), if a⊕a = {a, 1},21R. A. Borzooei, A. Radfarthen a⊕1 6= {a, 1}. Hence we must investigate 2 following tables which bothof them are hyper MV -algebras.⊕4 0 a 10 {0} {0, a} {0, 1}a {0, a} {a, 1} {0, a, 1}1 {0, 1} {0, a, 1} {1}⊕5 0 a 10 {0} {0, a} {0, 1}a {0, a} {a, 1} {0, a, 1}1 {1} {0, a, 1} {0, 1}Now, if a⊕a = M , then by Lemma 3.8 (ii) and (iv), a⊕1 = {a, 1} or M and1⊕ 1 = {0, 1} or {1}. Thus we must investigate 4 following tables, which allof them are hyper MV -algebras.⊕6 0 a 10 {0} {0, a} {0, 1}a {0, a} {0, a, 1} {a, 1}1 {0, 1} {a, 1} {1}⊕7 0 a 10 {0} {0, a} {0, 1}a {0, a} {0, a, 1} {a, 1}1 {1} {a, 1} {0, 1}⊕8 0 a 10 {0} {0, a} {0, 1}a {0, a} {0, a, 1} {0, a, 1}1 {0, 1} {0, a, 1} {1}⊕9 0 a 10 {0} {0, a} {0, 1}a {0, a} {0, a, 1} {0, a, 1}1 {0, 1} {0, a, 1} {0, 1}Case 3: 0⊕ 1 = MLemma 3.10. Let M = {0, a, 1} be a proper hyper MV -algebra of order 3such that 0⊕ 1 = M . Then(i) (a⊕ a) ∪ (1⊕ a) = M ,(ii) If a⊕ a = {1}, then a⊕ 1 = 1⊕ 1 = M ,(iii) If a ⊕ a = {0, 1}, then a ⊕ 1 = {a, 1} or M and if a ⊕ 1 = {a, 1},then 1⊕ 1 = {1}, {0, 1} or M ,(iv) If a ⊕ a = {a, 1}, then a ⊕ 1 = {0, 1} or M and if a ⊕ 1 = {0, 1},then 1⊕ 1 = {a, 1} or M ,(v) If a⊕ a = M and a⊕ 1 = {1, a}, then 1⊕ 1 = {1}, {0, 1} or M ,(vi) If a⊕ a = M and a⊕ 1 = {0, 1}, then 1⊕ 1 = {0, 1}, {a, 1} or M .Proof.(i). Since by Lemma 3.5(iv), 1 ∈ 1 ⊕ a, then M = 0 ⊕ 1 = 1∗ ⊕ 1 ⊆(a ⊕ 1)∗ ⊕ 1 and so (a ⊕ 1)∗ ⊕ 1 = M . Hence by (HMV5), (0 ⊕ a)∗ ⊕ a =(a⊕ 1)∗ ⊕ 1 = M and so by Lemma 3.5(i),M = (0⊕ a)∗ ⊕ a = {0, a}∗ ⊕ a = {1, a} ⊕ a = (1⊕ a) ∪ a⊕ a.22Classification of Hyper MV -algebras of Order 3(ii). Let a ⊕ a = {1}. Since 1 ∈ 1 ⊕ a, then by (HMV5) and Lemma3.5(i),1⊕ a = (1⊕ a) ∪ (a⊕ a) = {1, a} ⊕ a = {0, a}∗ ⊕ a = (0⊕ a)∗ ⊕ a= (a⊕ 0)∗ ⊕ 0 = {1, a} ⊕ 0 = (1⊕ 0) ∪ (a⊕ 0)= MNow, since a⊕ a = {1} and 1⊕ a = M , then by (HMV1),1⊕ 1 = (a⊕ a)⊕ (a⊕ a) = a⊕ (a⊕ (a⊕ a))= a⊕ (a⊕ 1) = a⊕M = (a⊕ 1) ∪ (a⊕ a) ∪ (a⊕ 0) = M.(iii). If a⊕ a = {0, 1}, then by (i) and Lemma 3.5(iv), a⊕ 1 = {a, 1} orM . Let a⊕ 1 = {a, 1}. If 1⊕ 1 = {a, 1}, then by (HMV1) and (i),M = (a⊕ a) ∪ (1⊕ a) = {a, 1} ⊕ a = (1⊕ 1)⊕ a= 1⊕ (1⊕ a) = 1⊕ {1, a} = (1⊕ 1) ∪ (1⊕ a)= (1⊕ 1) ∪ {1, a}Hence 0 ∈ 1⊕ 1 = {a, 1}, which is a contradiction. Thus 1⊕ 1 6= {a, 1} andso by Lemma 3.5(v), 1⊕ 1 = {1}, {0, 1} or M .(iv). By (i), if a⊕ a = {a, 1}, then a⊕ 1 = {0, 1} or M .If a⊕ 1 = {0, 1}, then by (HMV1),M = {0, a} ∪ (1⊕ a) = {0, 1} ⊕ a = (1⊕ a)⊕ a= 1⊕ (a⊕ a) = 1⊕ {a, 1} = (1⊕ a) ∪ (1⊕ 1)= {0, 1} ∪ (1⊕ 1)Hence a ∈ 1⊕ 1. By Lemma 3.5(v), 1⊕ 1 = {1, a} or M .(v). Let a⊕a = M and 1⊕a = {1, a}. If 1⊕1 = {a, 1}, then by (HMV1),M = (a⊕ a) ∪ (1⊕ a) = {1, a} ⊕ a = (1⊕ 1)⊕ a = 1⊕ (1⊕ a)= 1⊕ {1, a} = (1⊕ 1) ∪ (1⊕ a)= (1⊕ 1) ∪ {1, a}Hence 0 ∈ 1 ⊕ 1 = {a, 1}, which is impossible. Thus 1 ⊕ 1 6= {1, a} and soby Lemma 3.5(v), 1⊕ 1 = {1}, {0, 1} or M .(vi). Let a⊕ a = M and 1⊕ a = {0, 1}. Then by (HMV1),(1⊕ 1)⊕ a = 1⊕ (1⊕ a) = 1⊕ {0, 1} = (0⊕ 1) ∪ (1⊕ 1) = M.Now, if 1⊕ 1 = {1}, then 1⊕ a = (1⊕ 1)⊕ a = M , which is a contradiction.Hence 1⊕ 1 6= {1} and so by Theorem 3.5(v), 1⊕ 1 = {0, 1}, {a, 1} or M23R. A. Borzooei, A. RadfarTheorem 3.11. There are 24 non-isomorphic proper hyper MV -algebras oforder 3 such that 0⊕ 1 = M .Proof. By Lemma 3.5 (iii), a ⊕ a = {1}, {0, 1}, {1, a} or M . Ifa⊕ a = {1} , then by Lemma 3.10 (ii), a⊕ 1 = 1⊕ 1 = M and so we mustinvestigate the following table, which is a hyper MV -algebra.⊕10 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, a, 1}If a ⊕ a = {0, 1}, then by Lemma 3.10 (iii), a ⊕ 1 = {a, 1} or M and ifa ⊕ 1 = {a, 1}, then 1 ⊕ 1 = {1}, {0, 1} or M . Thus we must investigatethe following 3 cases which all of them are hyper MV -algebras.⊕11 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {a, 1}1 {0, a, 1} {a, 1} {1}⊕12 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {a, 1}1 {0, a, 1} {a, 1} {0, 1}⊕13 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {a, 1}1 {0, a, 1} {a, 1} {0, a, 1}If a ⊕ 1 = M , then by Lemma 3.5 (v), 1 ⊕ 1 = {1}, {0, 1}, {1, a} or M .Hence we must investigate the following 4 cases which all of them are hyperMV -algebras.⊕14 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {1}⊕15 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, 1}⊕16 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {a, 1}⊕17 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, a, 1}24Classification of Hyper MV -algebras of Order 3Now, if a ⊕ a = {a, 1}, then by Lemma 3.10 (iv), a ⊕ 1 = {0, 1} or M andif a ⊕ 1 = {0, 1}, then 1 ⊕ 1 = {a, 1} or M . Hence we must investigate thefollowing 2 cases which both of them are hyper MV -algebras.⊕18 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {a, 1} {0, 1}1 {0, a, 1} {0, 1} {a, 1}⊕19 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {a, 1} {0, 1}1 {0, a, 1} {0, 1} {0, a, 1}If a ⊕ 1 = M , then by Lemma 3.5 (v), 1 ⊕ 1 = {1}, {0, 1}, {a, 1} or Mand so we must investigate the following 4 cases which all of them are hyperMV -algebras.⊕20 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {1}⊕21 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, 1}⊕22 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {a, 1}⊕23 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, a, 1}Now, let a⊕ a = M . Then by Lemma 3.10 (v), a⊕ 1 = {1, a}, {0, 1} or M .If a ⊕ 1 = {1, a}, then 1 ⊕ 1 = {1}, {0, 1} or M . Thus we must investigatethe following 3 cases which all of them are hyper MV -algebras.⊕24 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {a, 1}1 {0, a, 1} {a, 1} {1}⊕25 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {a, 1}1 {0, a, 1} {a, 1} {0, 1}⊕26 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {a, 1}1 {0, a, 1} {a, 1} {0, a, 1}Also by Lemma 3.10 (v), if a⊕ 1 = {0, 1}, then 1⊕ 1 = {0, 1}, {a, 1} or M .Hence we must investigate the following 3 cases which all of them are hyper25R. A. Borzooei, A. RadfarMV -algebras.⊕27 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, 1}1 {0, a, 1} {0, 1} {0, 1}⊕28 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, 1}1 {0, a, 1} {0, 1} {a, 1}⊕29 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, 1}1 {0, a, 1} {0, 1} {0, a, 1}Finally, if a⊕1 = M , then by Lemma 3.5 (v), 1⊕1 = {1}, {0, 1}, {a, 1} or M .Hence we must investigate the following 4 cases which all of them are hyperMV -algebras.⊕30 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {1}⊕31 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, 1}⊕32 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {a, 1}⊕33 0 a 10 {0} {0, a} {0, a, 1}a {0, a} {0, a, 1} {0, a, 1}1 {0, a, 1} {0, a, 1} {0, a, 1}Corolary 3.12. There are 33 non-isomorphic hyper MV -algebras of order 3.Proof. By Theorems 3.3, 3.7, 3.9 and 3.11, we have 33 non-isomorphic hyperMV -algebras of order 3.References[1] C. C. Chang, Algebraic analysis of many valued logics, Trans. Amer.Math. Soc, 88 (1958), 467–490.[2] G. Georgescu, A. Iorgulescu, Pseudo-MV algebras, Multi Valued Logic,6, (2001), 95-135.26Classification of Hyper MV -algebras of Order 3[3] S. Ghorbani, E. Eslami and A. Hasankhani, Quotient hyper MV-algebras, Scientiae Mathematicae Japonicae, 3 (2007) 371–386.[4] Sh. Ghorbani, A. Hasankhani, and E. Eslami, Hyper MV-algebras, Set-Valued Math. Appl, 1 (2008), 205–222.[5] F. Marty, Sur une generalization de la notion de groupe, 8th CongressMath. Scandin aves, Stockholm (1934), 45–49.[6] D. Mundici, Interpretation of AFC∗-algebras in Lukasiewicz sententialcalculus, J. Funct. Anal, 65, (1986), 15–63.2728

Classification of Hyper MV -algebras of Order 3

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Classification of Hyper MV -algebras of Order 3

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