RATIO MATHEMATICA 25 (2013), 15–28 ISSN:1592-7415

Classification of Hyper M V -algebras
of Order 3

R. A. Borzooei∗, A. Radfar∗∗

∗Department of Mathematics, Shahid Beheshti University, G. C.,Tehran, Iran
∗∗Department of Mathematics, Payame Noor University, Tehran, Iran

borzooei@sbu.ac.ir, Ateferadfar@yahoo.com

Abstract

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

Key words: hyper M V -algebra

MSC 2010: 97U99.

1 Introduction

The concept of M V -algebras was introduced by Chang in [1] 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
[6], Mundici showed that any M V -algebra is an interval of an Abelian lat-
tice ordered group with a strong unit. Also, he introduced the concept of
state on M V -algebra. Georgescu and Iorgulescu [2] introduced a new non-
commutative algebraic structures, which were called pseudo M V -algebras.
It can be obtained by dropping commutative axioms in M V -algebras, which
are a generalization of M V -algebras. The hyper structure theory was intro-
duced by F. Marty [5] at the 8th congress of Scandinavian Mathematicians
in 1934. Since then many researches have worked in these areas. Recently in
[4], Sh. Ghorbani, A. Hasankhni and E. Eslami applied the hyper structure
to M V -algebras and introduced the concept of a hyper M V -algebra which
is a generalization of an M V -algebra and investigated some related results.
Now, in this paper we find all hyper M V -algebras of order 3.

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R. A. Borzooei, A. Radfar

2 Preliminary

Definition 2.1. [1] An M V -algebra (X,⊕,∗ , 0) is a set X equipped with
a binary operation ⊕, a unary operation ∗ and a constant 0 satisfying the
following equations:

(M V1) x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z,

(M V2) x ⊕ y = y ⊕ x,

(M V3) x ⊕ 0 = x,

(M V4) (x
∗)∗ = x,

(M V5) x ⊕ 0∗ = 0∗,

(M V6) (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 be
a hyper M V -algebra if and only if satisfies the following axioms, for all
x, y, z ∈ M :

(HM V1) x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z,
(HM V2) x ⊕ y = y ⊕ x,
(HM V3) (x

∗)∗ = x,
(HM V4) 0

∗ ∈ x ⊕ 0∗,
(HM V5) (x

∗ ⊕ y)∗ ⊕ y = (y∗ ⊕ x)∗ ⊕ x,
(HM V6) 0

∗ ∈ x ⊕ x∗,
(HM V7) 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 if

there 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-M V algebra. Then for all
x, y, z ∈ M and for all non-empty subsets A, B and C of M the following
hold:

(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.

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Classification of Hyper M V -algebras of Order 3

3 Classification of hyper M V -algebras of or-

der 3

In this section we try to find all hyper M V -algebras of order 3, up to
isomorphism.

Theorem 3.1. Let M be a hyper M V -algebra and x be an element of M
such 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 (HM V 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 ⊕ x

and 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 (HM V6), 1 = 0∗ ∈
x ⊕ x∗ = x ⊕ x = {x}, then x = 1 and so 0 = 1∗ = x∗ = x = 1, which is a
contradiction. 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}, which
is a contradiction. Hence 0 /∈ 1 ⊕ x.

Note. From now one in this paper, we let M = {0, a, 1} be a hyper
M V -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 (HM V3), 0 = (0

∗)∗ = 1∗. Now, if
a∗ = 1, then 0 = 1∗ = (a∗)∗ = a, which is a contradiction. By similar way, if
a∗ = 0, then 1 = 0∗ = (a∗)∗ = a, which is a contradiction. Hence, a∗ = a.

(iv). By (HM V4), 1 = 0
∗ ∈ 0∗ ⊕ a = 1 ⊕ a.

Theorem 3.3. If 0 ⊕ a = {a} or 1 ⊕ a = {1}, then M is an M V -algebra.

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R. A. Borzooei, A. Radfar

Proof. Let 0 ⊕ a = {a}. Since a∗ = a, then by Theorem 3.1(iii), a 6∈ 1 ⊕ a
and 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. Hence
a ⊕ a = {1}. Now, by (HM V1),

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 M V -algebra.

Now, if 1⊕a = {1}, then {0} = {1∗} = (1⊕a)∗ and so 0⊕a = (1⊕a)∗⊕a.
By (HM V5),

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 (HM V 4), 1 ∈ 0 ⊕ 1. Thus 1 ∈ {0, a}, which is a
contradiction. Thus 0 6∈ 0 ⊕ a and so 0 ⊕ a = {a}. Therefore, M is a same
M V -algebra, which is as follows:

⊕1 0 a 1
0 {0} {a} {1}
a {a} {1} {1}
1 {1} {1} {1}

Definition 3.4. We call a hyper M V -algebra is proper, if it is not an M V -
algebra.

Lemma 3.5. Let M = {0, a, 1} be a proper hyper M V -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 .

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Classification of Hyper M V -algebras of Order 3

Proof. (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 not
proper. Thus 0 ⊕ a = {0, a}

(ii). Since 0 ≤ 0, then 1 = 0∗ ∈ 0∗ ⊕ 0 = 1 ⊕ 0 = 0 ⊕ 1. Hence it is
sufficient to show that 0 ⊕ 1 6= {1, a}. Let 0 ⊕ 1 = {1, a}, by the contrary.
Then by (HM V1),

{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} or
M .

(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}, then
M is an M V algebra which is impossible. Hence 1 ⊕ a = {0, 1}, {1, a} or
M .

(vi). Let a ⊕ a = {1}. Then by (HM V1),

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 the
classification of all hyper M V -algebras of order 3, we consider the following
three cases.

Case 1: 0 ⊕ 1 = {1}

Lemma 3.6. Let M = {0, a, 1} be a proper hyper M V -algebra of order 3
and 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 (HM V1), (0 ⊕ a) ⊕ a = 0 ⊕ (a ⊕ a), then 0 ⊕ (a ⊕ a) = M . By
Lemma 3.5(iii), a ⊕ a = {1}, {0, 1}, {1, a} or M . If a ⊕ a = {1}, then
0 ⊕ (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 .

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R. A. Borzooei, A. Radfar

(ii). By (HM V5), 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 a
contradiction. If a ∈ 1⊕1, then a⊕1 ⊆ (1⊕1)∗⊕1 = {1}. Thus a⊕1 = {1}
and so by Theorem 3.3, M is an M V -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 (HM V1), 1 ⊕ (1 ⊕a) = (1 ⊕ 1) ⊕a = 1 ⊕a, then 1 ⊕ (1 ⊕a) = {1},
which is a contradiction. If 1 ⊕ a = {1, a}, since by (HM V1), 0 ⊕ (1 ⊕ a) =
(0 ⊕ 1) ⊕ a = 1 ⊕ a, then 0 ⊕ (1 ⊕ a) = (0 ⊕ 1) ∪ (0 ⊕ a) = M , which is a
contradiction. Hence, 1 ⊕ a = M .

Theorem 3.7. There are two non-isomorphic proper hyper M V -algebras of
order 3 such that 0 ⊕ 1 = {1}.

Proof. According Theorem 3.6, if M is a proper hyper M V -algebra
of order 3 and 0 ⊕ 1 = {1}, then we must investigate two following tables,
which both of them are non-isomorphic hyper M V -algebras.

⊕2 0 a 1
0 {0} {0, a} {1}
a {0, a} {1, a} {0, a, 1}
1 {1} {0, a, 1} {1}

⊕3 0 a 1
0 {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 M V -algebra of order 3
and 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
(HM V1),

(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).

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Classification of Hyper M V -algebras of Order 3

Thus {0, 1}∪ (a ⊕ 1) = M and so a ∈ a ⊕ 1. New, we consider two cases
0 ∈ a ⊕ 1 or 0 6= a ⊕ 1. If 0 ∈ a ⊕ 1, since by Theorem 3.5, 1 ∈ a ⊕ 1, then
a⊕1 = M and so (a⊕a)∪(1⊕a) = M . Now, if 0 6= a⊕1, then by Theorem
3.5, a ∈ a ⊕ 1. Hence by Theorem 3.2(iv), {1, a}⊆ a ⊕ 1. Thus

M = (0 ⊕ 1) ∪ (a ⊕ 1) = {0, a}⊕ 1 = {1, a}∗ ⊕ 1 ⊆ (a ⊕ 1)∗ ⊕ 1 ⊆ M

and so (a ⊕ 1)∗ ⊕ 1 = M . On the other hands, by (HM V5), (a ⊕ 1)∗ ⊕ 1 =
(0 ⊕ a)∗ ⊕ a. Hence (0 ⊕ a)∗ ⊕ a = M . Since 0 ⊕ a = {0, a}, then

M = (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}. Let
0 ∈ 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 (HM V1),

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 (HM V1), 0 ⊕ (a ⊕ a) = (0 ⊕ a) ⊕ a = M . Since a 6∈ 0 ⊕ 0 and
a 6∈ 0 ⊕ 1, then a ∈ a ⊕ a. Hence a ⊕ a = {a, 1} or M .

(iv). Let a ∈ 1 ⊕ 1. By (HM V5),

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 M V -algebras of
order 3 such that 0 ⊕ 1 = {0, 1}.

Proof. By Lemma 3.8 (iii), a⊕a = {a, 1} or M . If a⊕a = {a, 1}, then
by Lemma 3.8 (ii), a⊕1 = {a, 1} or M . By Lemma 3.8 (i), if a⊕a = {a, 1},

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R. A. Borzooei, A. Radfar

then a⊕1 6= {a, 1}. Hence we must investigate 2 following tables which both
of them are hyper M V -algebras.

⊕4 0 a 1
0 {0} {0, a} {0, 1}
a {0, a} {a, 1} {0, a, 1}
1 {0, 1} {0, a, 1} {1}

⊕5 0 a 1
0 {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 and
1⊕1 = {0, 1} or {1}. Thus we must investigate 4 following tables, which all
of them are hyper M V -algebras.

⊕6 0 a 1
0 {0} {0, a} {0, 1}
a {0, a} {0, a, 1} {a, 1}
1 {0, 1} {a, 1} {1}

⊕7 0 a 1
0 {0} {0, a} {0, 1}
a {0, a} {0, a, 1} {a, 1}
1 {1} {a, 1} {0, 1}

⊕8 0 a 1
0 {0} {0, a} {0, 1}
a {0, a} {0, a, 1} {0, a, 1}
1 {0, 1} {0, a, 1} {1}

⊕9 0 a 1
0 {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 = M

Lemma 3.10. Let M = {0, a, 1} be a proper hyper M V -algebra of order 3
such 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 (HM V5), (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.

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Classification of Hyper M V -algebras of Order 3

(ii). Let a ⊕ a = {1}. Since 1 ∈ 1 ⊕ a, then by (HM V5) and Lemma
3.5(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)
= M

Now, since a ⊕ a = {1} and 1 ⊕ a = M , then by (HM V1),

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} or
M . Let a ⊕ 1 = {a, 1}. If 1 ⊕ 1 = {a, 1}, then by (HM V1) 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} and
so 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 (HM V1),

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 (HM V1),

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 so
by Lemma 3.5(v), 1 ⊕ 1 = {1}, {0, 1} or M .

(vi). Let a ⊕ a = M and 1 ⊕ a = {0, 1}. Then by (HM V1),

(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 M

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R. A. Borzooei, A. Radfar

Theorem 3.11. There are 24 non-isomorphic proper hyper M V -algebras of
order 3 such that 0 ⊕ 1 = M .

Proof. By Lemma 3.5 (iii), a ⊕ a = {1}, {0, 1}, {1, a} or M . If
a ⊕ a = {1} , then by Lemma 3.10 (ii), a ⊕ 1 = 1 ⊕ 1 = M and so we must
investigate the following table, which is a hyper M V -algebra.

⊕10 0 a 1
0 {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 if
a ⊕ 1 = {a, 1}, then 1 ⊕ 1 = {1}, {0, 1} or M . Thus we must investigate
the following 3 cases which all of them are hyper M V -algebras.

⊕11 0 a 1
0 {0} {0, a} {0, a, 1}
a {0, a} {0, 1} {a, 1}
1 {0, a, 1} {a, 1} {1}

⊕12 0 a 1
0 {0} {0, a} {0, a, 1}
a {0, a} {0, 1} {a, 1}
1 {0, a, 1} {a, 1} {0, 1}

⊕13 0 a 1
0 {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 hyper
M V -algebras.

⊕14 0 a 1
0 {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 1
0 {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 1
0 {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 1
0 {0} {0, a} {0, a, 1}
a {0, a} {0, 1} {0, a, 1}
1 {0, a, 1} {0, a, 1} {0, a, 1}

24



Classification of Hyper M V -algebras of Order 3

Now, if a ⊕ a = {a, 1}, then by Lemma 3.10 (iv), a ⊕ 1 = {0, 1} or M and
if a ⊕ 1 = {0, 1}, then 1 ⊕ 1 = {a, 1} or M . Hence we must investigate the
following 2 cases which both of them are hyper M V -algebras.

⊕18 0 a 1
0 {0} {0, a} {0, a, 1}
a {0, a} {a, 1} {0, 1}
1 {0, a, 1} {0, 1} {a, 1}

⊕19 0 a 1
0 {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 M
and so we must investigate the following 4 cases which all of them are hyper
M V -algebras.

⊕20 0 a 1
0 {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 1
0 {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 1
0 {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 1
0 {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 investigate
the following 3 cases which all of them are hyper M V -algebras.

⊕24 0 a 1
0 {0} {0, a} {0, a, 1}
a {0, a} {0, a, 1} {a, 1}
1 {0, a, 1} {a, 1} {1}

⊕25 0 a 1
0 {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 1
0 {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 hyper

25



R. A. Borzooei, A. Radfar

M V -algebras.

⊕27 0 a 1
0 {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 1
0 {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 1
0 {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 hyper
M V -algebras.

⊕30 0 a 1
0 {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 1
0 {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 1
0 {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 1
0 {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 M V -algebras of order 3.

Proof. By Theorems 3.3, 3.7, 3.9 and 3.11, we have 33 non-isomorphic hyper
M V -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-M V algebras, Multi Valued Logic,
6, (2001), 95-135.

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Classification of Hyper M V -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 Congress
Math. Scandin aves, Stockholm (1934), 45–49.

[6] D. Mundici, Interpretation of AF C∗-algebras in Lukasiewicz sentential
calculus, J. Funct. Anal, 65, (1986), 15–63.

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