RATIO MATHEMATICA 22 (2012) 13-35 ISSN:1592-7415

Classifications of Hyper Pseudo
BCK-algebras of Order 3

R. A. Borzooei, A. Rezazadeh, R. Ameri
Department of Mathematics, Shahid Beheshti University, Tehran, Iran

Department of Mathematics, Payam Noor University, Tehran, Iran

Department of Mathematics, Tehran University, Tehran, Iran

borzooei@sbu.ac.ir, Rezazade2008@gmail.com, rez−ameri@yahoo.com

Abstract

In this paper by considering the notion of hyper pseudo BCK-
algebra, we classify the set of all non-isomorphic hyper pseudo BCK-
algebras of order 3. For this, we define the notion of simple and normal
condition and we characterize the all of hyper pseudo BCK-algebras
of order 3 that satisfies these conditions.

Key words: Hyper pseudo BCK-algebra, Simple condition, Nor-
mal condition.

2000 AMS subject classifications: 97U99.

1 Introduction

The study of BCK-algebras was initiated at 1966 by Y. Imai and K.
Iséki in [5] as a generalization of the concept of set-theoretic difference and
propositional calculi. In order to extend BCK-algebras in a noncommutative
form, Georgescu and Iorgulescu [4] introduced the notion of pseudo BCK-
algebras and studied their properties. The hyperstructure theory (called also
multialgebra) was introduced in 1934 by F. Marty [10] at the 8th Congress
of Scandinavian Mathematicians. Since then many researchers have worked
on algebraic hyperstructures and developed it. A recent book [3] contains a
wealth of applications. Via this book, Corsini and Leoreanu presented some
of the numerous applications of algebraic hyperstructures, especially those
from the last fifteen years, to the following subjects: geometry, hypergraphs,

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R. A. Borzooei, A. Rezazadeh and R. Ameri

binary relations, lattices, fuzzy sets and rough sets, automata, cryptography,
codes, median algebras, relation algebras, artificial intelligence and proba-
bilities. Hyperstructures have many applications to several sectors of both
pure and applied sciences. In [1, 9], R. A. Borzooei et al. applied the hy-
perstructures to (pseudo) BCK-algebras and introduced the notion of hyper
(pseudo) BCK-algebra which is a generalization of (pseudo) BCK-algebra
and investigated some related properties. In [2], R. A. Borzooei et al. clas-
sified all hyper BCK-algebras of order 3. Now, in this paper we classify the
set of all non-isomorphic hyper pseudo BCK-algebras of order 3.

2 Preliminaries

Definition 2.1. [9] By a hyper BCK-algebra we mean a nonempty set H
endowed with a hyperoperation ”◦” and a constant 0 satisfy the following
axioms:

(HK1) (x ◦ z) ◦ (y ◦ z) ≪ x ◦ y,

(HK2) (x ◦ y) ◦ z = (x ◦ z) ◦ y,

(HK3) x ◦ H ≪ {x},

(HK4) x ≪ y and y ≪ x imply x = y.

for all x, y, z ∈ H, where x ≪ y is defined by 0 ∈ x ◦ y and for every
A, B ⊆ H, A ≪ B is defined by ∀a ∈ A, ∃b ∈ B such that a ≪ b. In such
case, we call ” ≪ ” the hyperorder in H.

Definition 2.2. [1] A hyper pseudo BCK-algebra is a structure (H, ◦, ∗, 0)
where ”∗” and ”◦” are hyper operations on H and ”0” is a constant element,
that satisfies the following: for all x, y, z ∈ H,

(PHK1) (x ◦ z) ◦ (y ◦ z) ≪, x ◦ y, (x ∗ z) ∗ (y ∗ z) ≪ x ∗ y,

(PHK2) (x ◦ y) ∗ z = (x ∗ z) ◦ y,

(PHK3) x ◦ H ≪ {x}, x ∗ H ≪ {x},

(PHK4) x ≪ y and y ≪ x imply x = y.

where x ≪ y ⇔ 0 ∈ x ◦ y ⇔ 0 ∈ x ∗ y and for every A, B ⊆ H, A ≪ B is
defined by ∀a ∈ A, ∃b ∈ B such that a ≪ b.

Theorem 2.1. [2, 9] Any BCK-algebra and hyper BCK-algebra is a hyper
pseudo BCK-algebra.

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Classifications of Hyper Pseudo BCK-algebras of Order 3

Proposition 2.1. [1] In any hyper pseudo BCK-algebra H, the following
hold:

(i) 0 ◦ 0 = {0}, 0 ∗ 0 = {0}, x ◦ 0 = {x}, x ∗ 0 = {x},

(ii) 0 ≪ x, x ≪ x, A ≪ A,

(iii) 0 ◦ x = {0}, 0 ∗ x = {0}, 0 ◦ A = {0}, 0 ∗ A = {0}.

for all x, y, z ∈ H and for all nonempty subsets A and B of H.

Theorem 2.2. [2] There are 19 non-isomorphic hyper BCK-algebras of or-
der 3.

Note: From now on, in this paper H = {0, a, b} is a hyper pseudo BCK-
algebra of order 3, unless otherwise state.

3 Characterization of Hyper Pseudo BCK-

algebras of Order 3

Definition 3.1. [2] We say that H satisfies the normal condition if one of
the conditions a ≪ b or b ≪ a holds. If no one of these conditions hold, then
we say that H satisfies the simple condition.

Definition 3.2. Let (H1, ◦1, ∗1, 01) and (H2, ◦2, ∗2, 02) be two hyper pseudo
BCK-algebras and f : H1 → H2 be a function. Then f is said to be a
homomorphism iff

(i) f(01) = 02

(ii) f(x ◦1 y) = f(x) ◦2 f(y), ∀x, y ∈ H1

(iii) f(x ∗1 y) = f(x) ∗2 f(y), ∀x, y ∈ H1.

If f is one to one (onto) we say that f is a monomorphism (epimorphism)
and if f is both one to one and onto, we say that f is an isomorphism.

Definition 3.3. Let I ⊆ H. Then we say that I is a proper subset of H if
I ̸= {0} and I ̸= H.

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R. A. Borzooei, A. Rezazadeh and R. Ameri

3.1 Characterization of hyper pseudo BCK-algebras of
order 3 that satisfy the simple condition

Theorem 3.1. There are only 10 hyper pseudo BCK-algebras of order 3,
that satisfy the simple condition.

Proof. Let H satisfy the simple condition. Now, we prove the following
statements:

(i) For all x, y ∈ H which x ̸= y, then x ̸∈ y ◦ x and x ̸∈ y ∗ x.

(ii) a ◦ b = a ∗ b = {a} and b ◦ a = b ∗ a = {b}.

(iii) a ◦ a and a ∗ a are equal to {0} or {0, a} and b ◦ b and b ∗ b are equal
to {0} or {0, b}.

For the proof of (i), let x ̸= y and x ∈ y ◦ x, by the contrary. Clearly x ̸= 0.
Because if x = 0, then y ̸= 0 and 0 ∈ y ◦ 0 = {y}, which is impossible.
Moreover, since y ◦ x ≤ y, then x ≤ y which is impossible by the simplicity
of H. By the similar way, we can prove that x ̸∈ y ∗ x.

(ii) Since a ̸≤ b, then 0 ̸∈ a ◦ b and 0 ̸∈ a ∗ b. Hence a ◦ b and a ∗ b can
not be equal to {0}, {0, a}, {0, b} or {0, a, b}. Since by (i), we have b ̸∈ a ◦ b
and b ̸∈ a ∗ b, we conclude that a ◦ b and a ∗ b can not be equal to {b} or
{a, b}. Thus a ◦ b = a ∗ b = {a}. By the similar way, we can prove that
b ◦ a = b ∗ a = {b}.

(iii) Since a ≪ a, the only cases for a◦a and a∗a are {0}, {0, a}, {0, b} or
{0, a, b}. Also we have a ◦ a ≤ a and a ∗ a ≤ a. Thus b ̸∈ a ◦ a and b ̸∈ a ∗ a.
Hence the only cases for a ◦ a and a ∗ a are {0} or {0, a}. By the similar way,
we can prove that b ◦ b and b ∗ b are equal to {0} or {0, b}.

Therefore, by (i), (ii) and (iii) we conclude that there are 16 hyper pseudo
BCK-algebras of order 3, which satisfy the simple condition. But some of
them are isomorphic under the map f : H → H which is defined by f(0) = 0,
f(a) = b and f(b) = a. Hence there are 10 hyper pseudo BCK-algebras of
order 3, that satisfy the simple condition. Now, we give these hyper pseudo
BCK-algebras:

◦1 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0}

∗1 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0}

◦2 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0}

∗2 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

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Classifications of Hyper Pseudo BCK-algebras of Order 3

◦3 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0}

∗3 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0,b}

◦4 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

∗4 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0}

◦5 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

∗5 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

◦6 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

∗6 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0,b}

◦7 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

∗7 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0}

◦8 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0,b}

∗8 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0}

◦9 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0,b}

∗9 0 a b
0 {0} {0} {0}
a {a} {0} {a}
b {b} {b} {0,b}

◦10 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0,b}

∗10 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {b} {b} {0,b}

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R. A. Borzooei, A. Rezazadeh and R. Ameri

3.2 Characterization of hyper pseudo BCK-algebras of
order 3 that satisfy the normal condition

Note: From now on, in this section we let H = {0, a, b} satisfies the
normal condition. Since in this condition, a ≤ b or b ≤ a, so without loss of
generality we let a ≤ b and b ̸≤ a i.e., 0 ∈ a ◦ b ∩ a ∗ b, 0 ̸∈ b ◦ a and 0 ̸∈ b ∗ a.
Lemma 3.1. Only one of the following cases hold for b ∗ a and b ◦ a.

(NPHB1) b ◦ a = b ∗ a = {a},

(NPHB2) b ◦ a = {a}, b ∗ a = {a, b},

(NPHB3) b ◦ a = b ∗ a = {b},

(NPHB4) b ◦ a = {a, b}, b ∗ a = {a},

(NPHB5) b ◦ a = b ∗ a = {a, b}.

Proof. Since 0 ̸∈ b ◦ a and 0 ̸∈ b ∗ a, then b ◦ a and b ∗ a are equal to
one of the sets {a}, {b} or {a, b}.

If b◦a = {a}, then b∗a ̸= {b}. Since if b∗a = {b}, then (b◦a)∗a ̸= (b∗a)◦a.
Hence b ∗ a = {a} or {a, b}.

If b ◦ a = {b}, then b ∗ a ̸= {a, b} and {a}. Since if b ∗ a = {a, b} or {a},
then (b ◦ a) ∗ a ̸= (b ∗ a) ◦ a. Hence b ∗ a = {b}.

If b ◦ a = {a, b}, then b ∗ a ̸= {b}. Since if b ∗ a = {b}, then (b ◦ a) ∗ a ̸=
(b ∗ a) ◦ a. Hence b ∗ a = {a} or {a, b}. Therefore, we have the above cases.
Lemma 3.2. Only one of the following cases hold for a ∗ b and a ◦ b.

(NPHA1) a ◦ b = a ∗ b = {0}.

(NPHA2) a ◦ b = a ∗ b = {0, a}.

(NPHA3) a ◦ b = {0}, a ∗ b = {0, a}.

(NPHA4) a ◦ b = {0, a}, a ∗ b = {0}.

Proof. Since 0 ∈ a ◦ b ∩ a ∗ b, then a ◦ b and a ∗ b are equal to one of the
sets {0}, {0, a}, {0, b} or {0, a, b}. Moreover, since a ◦ b ≤ a and a ∗ b ≤ a,
then b ̸∈ a ◦ b and b ̸∈ a ∗ b. Hence the only cases for a ◦ b and a ∗ b are {0}
or {0, a}. Therefore, we have the above cases.
Lemma 3.3. Only one of the following cases hold for a ∗ a and a ◦ a.

(i) a ◦ a = a ∗ a = {0},

(ii) a ◦ a = {0}, a ∗ a = {0, a},

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Classifications of Hyper Pseudo BCK-algebras of Order 3

(iii) a ◦ a = {0, a}, a ∗ a = {0},

(iv) a ◦ a = a ∗ a = {0, a}.

Proof. Since 0 ∈ a ◦ a ∩ a ∗ a, then a ◦ a and a ∗ a can be equal to
the one of cases {0}, {0, a}, {0, b} or {0, a, b}. Moreover, since a ◦ a ≤ a and
a ∗ a ≤ a, then b ̸∈ a ◦ a and b ̸∈ a ∗ a. Hence the only cases for a ◦ a and
a ∗ a are {0} or {0, a}. Therefore, we have the above cases.
Theorem 3.4. There are only 5 non-isomorphic hyper pseudo BCK-algebras
of order 3, that satisfy the normal condition and condition (NPHB1).

Proof. Since H satisfies the condition (NPHB1), then b◦a = b∗a = {a}.
Case (NPHA1): We have a ∗ b = a ◦ b = {0}. If a ◦ a = {0, a} or

a ∗ a = {0, a}, then (a ◦ a) ◦ (b ◦ a) ̸≪ a ◦ b and (a ∗ a) ∗ (b ∗ a) ̸≪ a ∗ b.
Therefore, a◦a = a∗a = {0}. Moreover, if b◦b and b∗b are equal to the one
of sets {0, b} or {0, a, b}, then (b◦b)◦(a◦b) ̸≪ b◦a and (b∗b)∗(a∗b) ̸≪ b∗a.
Therefore, in this case b◦b and b∗b are equal to the one of sets {0} or {0, a}.
Now, we consider the following cases:

(1) b ◦ b = b ∗ b = {0}. Thus in this case, we have the following tables:

∗1 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0}

◦1 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0}

(2) b◦b = {0} and b∗b = {0, a}. Thus in this case, we have the following
tables:

∗2 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0}

◦2 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0,a}

(3) b ◦ b = {0, a} and b ∗ b = {0}. Thus similar to (2), we have one hyper
pseudo BCK-algebra in this case.

(4) b ◦ b = b ∗ b = {0, a}. Thus in this case, we have the following tables:

∗4 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0,a}

◦4 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0,a}

Case (NPHA2): We have a ∗ b = a ◦ b = {0, a}. If a ◦ a = {0} or a ∗ a = {0},
then (a ◦ b) ◦ (a ◦ b) ̸≪ a ◦ a and (a ∗ b) ∗ (a ∗ b) ̸≪ a ∗ a. Therefore, a ◦ a =
a ∗ a = {0, a}. Moreover, if b ◦ b and b ∗ b are equal to the one of sets {0, b}

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R. A. Borzooei, A. Rezazadeh and R. Ameri

or {0, a, b}, then (b ◦ b) ◦ (a ◦ b) ̸≪ b ◦ a and (b ∗ b) ∗ (a ∗ b) ̸≪ b ∗ a and if
b◦b = {0} or b∗b = {0}, then (b◦a)◦(b◦a) ̸≪ b◦b and (b∗a)∗(b∗a) ̸≪ b∗b.
Therefore, b ◦ b = b ∗ b = {0, a}. Thus we have the following case:

∗5 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a} {0,a}

◦5 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a} {0,a}

Case (NPHA3): We have a ∗ b = {0, a} and a ◦ b = {0}. Since (a ∗ b) ∗
(a ∗ b) ≪ a ∗ a, then a ∗ a = {0, a}. Moreover, since a ◦ b = {0}, then
a ◦ a ̸= {0, a}, because (a ◦ a) ◦ (b ◦ a) ̸≪ a ◦ b. Thus a ◦ a = {0} and in this
case (b ∗ a) ◦ a ̸= (b ◦ a) ∗ a.

Case (NPHA4): We have a ∗ b = {0} and a ◦ b = {0, a}. In this case we
can show that (b ∗ a) ◦ a ̸= (b ◦ a) ∗ a. Therefore, we have not any hyper
pseudo BCK-algebras.

We can check that all of the these 5 cases are hyper pseudo BCK-algebras
and each of them are not isomorphic together.
Theorem 3.5. There are only 6 non-isomorphic hyper pseudo BCK-algebras
of order 3, that satisfy the normal condition and condition (NPHB2).

Proof. Since H satisfies the condition (NPHB2), then b ◦ a = {a} and
b ∗ a = {a, b}.

Case (NPHA1): We have a ∗ b = a ◦ b = {0}. If one of the a ◦ a or a ∗ a
are equal to {0, a}, then (a ◦ a) ◦ (b ◦ a) ̸≪ a ◦ b and (a ∗ a) ∗ (b ∗ a) ̸≪ a ∗ a.
Hence we have a ∗ a = a ◦ a = {0}. But in this case (b ∗ a) ◦ a ̸= (b ◦ a) ∗ a.
Therefore, we have not any hyper pseudo BCK-algebras in this case.

Case (NPHA2): We have a ∗ b = a ◦ b = {0, a}. If a ◦ a = {0} or
a∗a = {0}, then (a◦b)◦(a◦b) ̸≪ a◦a and (a∗b)∗(a∗b) ̸≪ a∗a. Therefore,
a ◦ a = a ∗ a = {0, a}. Moreover, if b ∗ b is equal to the one of sets {0} or
{0, a}, then (b∗a)∗(b∗a) ̸≪ b∗b. Therefore, in this case b∗b is equal to the
one of sets {0, b} or {0, a, b} and if b ◦ b is equal to the one of sets {0, b} or
{0, a, b}, then (b ◦ b) ◦ (a ◦ b) ̸≪ b ◦ a. Hence in this case b ◦ b is equal to the
one of sets {0} or {0, a}. Moreover, since a ◦ a = {0, a}, if b ◦ b = {0}, then
(b ◦ a) ◦ (b ◦ a) ̸≪ b ◦ b. Thus b ◦ b = {0, a}. Now, we consider the following
cases:

(1) b ◦ b = {0, a} and b ∗ b = {0, b}. Thus in this case, we have the
following tables:

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

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

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Classifications of Hyper Pseudo BCK-algebras of Order 3

(2) b ◦ b = {0, a} and b ∗ b = {0, a, b}. Thus in this case, we have the
following tables:

◦2 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a} {0,a}

∗2 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

Case (NPHA3): We have a ∗ b = {0, a} and a ◦ b = {0}. If a ∗ a = {0}
and a ◦ a is equal to {0} or {0, a}, then (a ∗ b) ∗ (a ∗ b) ̸≪ a ∗ a. Thus
a ∗ a = {0, a}. Moreover, if b ∗ b is equal to the one of sets {0} or {0, a}, then
(b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b. Therefore, in this case b ∗ b is equal to the one of
sets {0, b} or {0, a, b} and if b ◦ b is equal to the one of sets {0, b} or {0, a, b},
then (b ◦ b) ◦ (a ◦ b) ̸≪ b ◦ a. Thus in this case b ◦ b is equal to the one of sets
{0} or {0, a}. Now, we consider the following cases:

(1) b◦b = {0} and b∗b = {0, b}. If a◦a = {0, a}, then (b◦a)◦(b◦a) ̸≪ b◦b.
Hence a ◦ a = {0}. Thus in this case, we have the following tables:

◦3 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0}

∗3 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

(2) b◦b = {0} and b∗b = {0, a, b}. If a◦a = {0, a}, then (b◦a)◦(b◦a) ̸≪
b ◦ b. Hence a ◦ a = {0}. Thus in this case, we have the following tables:

◦4 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0}

∗4 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

(3) b◦b = {0, a} and b∗b = {0, b}. If a◦a = {0, a}, then (a◦a)◦(b◦a) ̸≪
a ◦ b. Hence a ◦ a = {0}. Thus in this case, we have the following tables:

◦5 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0,a}

∗5 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

(4) b◦b = {0, a} and b∗b = {0, a, b}. If a◦a = {0, a}, then (a◦a)◦(b◦a) ̸≪
a ◦ b. Hence a ◦ a = {0}. Thus in this case, we have the following tables:

◦6 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a} {0,a}

∗6 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

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R. A. Borzooei, A. Rezazadeh and R. Ameri

Case (NPHA4): We have a ∗ b = {0} and a ◦ b = {0, a}. If a ◦ a = {0}
and a ∗ a is equal to {0} or {0, a}, then (a ◦ b) ◦ (a ◦ b) ̸≪ a ◦ a. Thus
a ◦a = {0, a}. Moreover, if b ∗b is equal to the one of sets {0} or {0, a}, then
(b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b. Therefore, in this case b ∗ b is equal to the one of sets
{0, b} or {0, a, b}. Moreover, if a ∗ a = {0, a}, then (a ∗ a) ∗ (b ∗ a) ̸≪ a ∗ b.
Hence a ∗ a = {0}. But in this case (b ◦ a) ∗ b ̸= (b ∗ b) ◦ a. Therefore, we have
not any hyper pseudo BCK-algebras.

We can check that all of the these 6 cases are hyper pseudo BCK-algebras
and each of them are not isomorphic together.
Theorem 3.6. There are only 70 non-isomorphic hyper pseudo BCK-
algebras of order 3, that satisfy the normal condition and condition (NPHB3).

Proof. Since H satisfies the condition (NPHB3), then b◦a = b∗a = {b}.
Case (NPHA1): We have a ∗ b = a ◦ b = {0}. Now, we consider the

following cases:
(1) b ◦ b = b ∗ b = {0}. In this case, we have the following tables:

◦1 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

∗1 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

◦2 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

∗2 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

◦3 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗3 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

◦4 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗4 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

(2) b ◦ b = {0} and b ∗ b = {0, b}. In this case, we have the following tables:

◦5 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

∗5 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

22



Classifications of Hyper Pseudo BCK-algebras of Order 3

◦6 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

∗6 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

◦7 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗7 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

◦8 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗8 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

(3) b ∗ b = {0} and b ◦ b = {0, b}. Similar to (2), we have four hyper
pseudo BCK-algebras in this case.

(4) b◦b = {0} and b∗b = {0, a}. If a◦a = {0}, then (b◦a)∗b ̸= (b∗b)◦a.
Thus in this case, we have the following tables:

◦13 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗13 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,a}

◦14 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗14 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

(5) b∗b = {0} and b◦b = {0, a}. If a∗a = {0}, then (b∗a)◦b ̸= (b◦b)∗a.
Thus similar to (4), we have two hyper pseudo BCK-algebras in this case.

(6) b◦b = {0} and b∗b = {0, a, b}. If a◦a = {0}, then (b◦a)∗b ̸= (b∗b)◦a.
Thus in this case, we have the following tables:

◦17 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗17 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,a,b}

◦18 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗18 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

23



R. A. Borzooei, A. Rezazadeh and R. Ameri

(7) b∗b = {0} and b◦b = {0, a, b}. If a∗a = {0}, then (b∗a)◦b ̸= (b◦b)∗a.
Thus similar to (6), we have two hyper pseudo BCK-algebras in this case.

(8) b ◦ b = b ∗ b = {0, b}. Thus in this case, we have the following tables:

◦21 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

∗21 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

◦22 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

∗22 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

◦23 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

∗23 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

◦24 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

∗24 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

(9) b ◦ b = {0, b} and b ∗ b = {0, a}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(10) b ∗ b = {0, b} and b ◦ b = {0, a}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(11) b ◦ b = {0, b} and b ∗ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(12) b ∗ b = {0, b} and b ◦ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(13) b ◦ b = b ∗ b = {0, a}. If a ◦ a = {0}, then (b ◦ a) ∗ b ̸= (b ∗ b) ◦ a and

if a ∗ a = {0}, then (b ∗ a) ◦ b ̸= (b ◦ b) ∗ a. Therefore, a ◦ a = a ∗ a = {0, a}.
Thus in this case, we have the following tables:

◦25 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

∗25 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

(14) b ◦ b = {0, a} and b ∗ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(15) b ∗ b = {0, a} and b ◦ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(16) b ◦ b = b ∗ b = {0, a, b}. If a ◦ a = {0}, then (b ◦ a) ∗ b ̸= (b ∗ b) ◦ a and

if a ∗ a = {0}, then (b ∗ a) ◦ b ̸= (b ◦ b) ∗ a. Therefore, a ◦ a = a ∗ a = {0, a}.
Thus in this case, we have the following tables:

24



Classifications of Hyper Pseudo BCK-algebras of Order 3

◦26 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

∗26 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

Case (NPHA2): We have a ∗ b = a ◦ b = {0, a}. If one of the a ∗ a or a ◦ a
are equal to {0}, then (a∗b)∗(a∗b) ̸≪ a∗a or (a◦b)◦(a◦b) ̸≪ a◦a. Hence
a ∗ a = a ◦ a = {0, a}.

(1) b ◦ b = b ∗ b = {0}. Thus in this case, we have the following tables:

◦27 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

∗27 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

(2) b ◦ b = {0} and b ∗ b = {0, b}. Thus in this case, we have the following
tables:

◦28 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

∗28 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

(3) b ∗ b = {0} and b ◦ b = {0, b}. Thus similar to (2), we have one hyper
pseudo BCK-algebra in this case.

(4) b ◦ b = {0} and b ∗ b = {0, a}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(5) b ∗ b = {0} and b ◦ b = {0, a}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(6) b ◦ b = {0} and b ∗ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(7) b ∗ b = {0} and b ◦ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(8) b ◦ b = b ∗ b = {0, b}. Thus in this case, we have the following tables:

◦30 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗30 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

(9) b ◦ b = {0, b} and b ∗ b = {0, a}. Thus in this case, we have the
following tables:

◦31 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗31 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a}

25



R. A. Borzooei, A. Rezazadeh and R. Ameri

(10) Let b ∗ b = {0, b} and b ◦ b = {0, a}. Thus similar to (9), we have one
hyper pseudo BCK-algebra in this case.

(11) b ◦ b = {0, b} and b ∗ b = {0, a, b}. Thus in this case, we have the
following tables:

◦33 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗33 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

(12) b ∗ b = {0, b} and b ◦ b = {0, a, b}. Thus similar to (11), we have one
hyper pseudo BCK-algebra in this case.

(13) b ◦ b = b ∗ b = {0, a}. Thus in this case, we have the following tables:

◦35 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a}

∗35 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a}

(14) b ◦ b = {0, a} and b ∗ b = {0, a, b}. Thus in this case, we have the
following tables:

◦36 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a}

∗36 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

(15) b ∗ b = {0, a} and b ◦ b = {0, a, b}. Thus similar to (14), we have one
hyper pseudo BCK-algebra in this case.

(16) b ◦ b = b ∗ b = {0, a, b}. Thus in this case, we have the following
tables:

◦38 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

∗38 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

Case (NPHA3): We have a∗b = {0, a} and a◦b = {0}. If a∗a = {0} and
a◦a is equal to {0} or {0, a}, then (a∗b)∗(a∗b) ̸≪ a∗a. Thus a∗a = {0, a}.

(1) b ◦ b = b ∗ b = {0}. Thus in this case, we have the following tables:

◦39 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

∗39 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

26



Classifications of Hyper Pseudo BCK-algebras of Order 3

◦40 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗40 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

(2) b ◦ b = {0} and b ∗ b = {0, b}. Thus in this case, we have the following
tables:

◦41 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

∗41 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

◦42 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗42 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

(3) b ∗ b = {0} and b ◦ b = {0, b}. Thus similar to (2), we have two hyper
pseudo BCK-algebras in this case.

(4) b◦b = {0} and b∗b = {0, a}. If a◦a = {0} then (b◦a)∗b ̸= (b∗b)◦a.
Thus in this case, we have the following tables:

◦45 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗45 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a}

(5) b ∗ b = {0} and b ◦ b = {0, a}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(6) b◦b = {0} and b∗b = {0, a, b}. If a◦a = {0}, then (b◦a)∗b ̸= (b∗b)◦a.

Thus in this case, we have the following tables:

◦46 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

∗46 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

(7) b ∗ b = {0} and b ◦ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(8) b ◦ b = b ∗ b = {0, b}. Thus in this case, we have the following tables:

◦47 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

∗47 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

27



R. A. Borzooei, A. Rezazadeh and R. Ameri

◦48 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

∗48 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

(9) b ◦ b = {0, b} and b ∗ b = {0, a}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(10) b ∗ b = {0, b} and b ◦ b = {0, a}. Thus in this case, we have the

following tables:

◦49 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,a}

∗49 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

◦50 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

∗50 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

(11) b ◦ b = {0, b} and b ∗ b = {0, a, b}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(12) b ∗ b = {0, b} and b ◦ b = {0, a, b}. Thus in this case, we have the

following tables:

◦51 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,a,b}

∗51 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

◦52 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

∗52 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

(13) b ◦ b = b ∗ b = {0, a}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(14) b ◦ b = {0, a} and b ∗ b = {0, a, b}. If a ◦ a = {0}, then (b ◦ a) ∗ b ̸=

(b ∗ b) ◦ a. Thus in this case, we have the following tables:

◦53 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

∗53 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

28



Classifications of Hyper Pseudo BCK-algebras of Order 3

(15) b ∗ b = {0, a} and b ◦ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(16) b ◦ b = b ∗ b = {0, a, b}. If a ◦ a = {0}, then (b ◦ a) ∗ b ̸= (b ∗ b) ◦ a.

Thus in this case, we have the following tables:

◦54 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

∗54 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

Case (NPHA4): We have a◦b = {0, a} and a∗b = {0}. If a◦a = {0} and
a∗a is equal to {0} or {0, a}, then (a◦b)◦(a◦b) ̸≪ a◦a. Thus a◦a = {0, a}.

(1) b ◦ b = b ∗ b = {0, b}. Thus in this case, we have the following tables:

◦55 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

∗55 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0}

◦56 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

∗56 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

(2) b ◦ b = {0} and b ∗ b = {0, b}. Thus in this case, we have the following
tables:

◦57 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

∗57 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

◦58 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0}

∗58 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

(3) b ∗ b = {0} and b ◦ b = {0, b}. Thus similar to (2), we have two hyper
pseudo BCK-algebras in this case.

(4) b ◦ b = {0} and b ∗ b = {0, a}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(5) b∗b = {0} and b◦b = {0, a}. If a∗a = {0}, then (b∗a)◦b ̸= (b◦b)∗a.

Thus in this case, we have the following tables:

29



R. A. Borzooei, A. Rezazadeh and R. Ameri

◦61 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a}

∗61 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

(6) b ◦ b = {0} and b ∗ b = {0, a, b}. Then (b ◦ b) ∗ b ̸= (b ∗ b) ◦ b.
(7) b∗b = {0} and b◦b = {0, a, b}. If a∗a = {0}, then (b∗a)◦b ̸= (b◦b)∗a.

Thus in this case, we have the following tables:

◦62 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

∗62 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0}

(8) b ◦ b = b ∗ b = {0, b}. Thus in this case, we have the following tables:

◦63 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗63 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,b}

◦64 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗64 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,b}

(9) b ◦ b = {0, b} and b ∗ b = {0, a}. Thus in this case, we have the following
tables:

◦65 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗65 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,a}

◦66 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗66 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

(10) b ∗ b = {0, b} and b ◦ b = {0, a}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(11) b ◦ b = {0, b} and b ∗ b = {0, a, b}. Thus in this case, we have the

following tables:

30



Classifications of Hyper Pseudo BCK-algebras of Order 3

◦67 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗67 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {b} {0,a,b}

◦68 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,b}

∗68 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

(12) b ∗ b = {0, b} and b ◦ b = {0, a, b}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(13) b ◦ b = b ∗ b = {0, a}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(14) b ◦ b = {0, a} and b ∗ b = {0, a, b}. Then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.
(15) b ∗ b = {0, a} and b ◦ b = {0, a, b}. If a ∗ a = {0}, then (b ∗ a) ◦ b ̸=

(b ◦ b) ∗ a. Thus in this case, we have the following tables:

◦69 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

∗69 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a}

(16) b ◦ b = b ∗ b = {0, a, b}. If a ∗ a = {0}, then (b ∗ a) ◦ b ̸= (b ◦ b) ∗ a.
Thus in this case, we have the following tables:

◦70 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {b} {0,a,b}

∗70 0 a b
0 {0} {0} {0}
a {a} {0,a} {0}
b {b} {b} {0,a,b}

We can check that all of the these 70 cases are hyper pseudo BCK-
algebras and each of them are not isomorphic together.
Theorem 3.7. There are only 6 non-isomorphic hyper pseudo BCK-algebras
of order 3, that satisfy the normal condition and condition (NPHB4).

Proof. The proof is the similar to the proof of Theorem 3.5, by the
some modification.
Theorem 3.8. There are only 9 non-isomorphic hyper pseudo BCK-algebras
of order 3, that satisfy the normal condition and condition (NPHB5).

Proof. Since H satisfies the condition (NPHB5), then b ◦ a = b ∗ a =
{a, b}.

Case (NPHA1): We have a∗b = a◦b = {0}. If one of the b◦b or b∗b are
equal to {0}, {0, a} or {0, b}, then (b◦a)◦(b◦a) ̸≪ b◦b or (b∗a)∗(b∗a) ̸≪ b∗b.
Therefore, we have only the following case:

31



R. A. Borzooei, A. Rezazadeh and R. Ameri

(1) b◦b = b∗b = {0, a, b}. If a◦a = {0, a}, then (a◦a)◦(b◦a) ̸≪ a◦b and
if a ∗ a = {0, a}, then (a ∗ a) ∗ (b ∗ a) ̸≪ a ∗ b. Therefore, a ◦ a = a ∗ a = {0}.
Thus in this case, we have the following tables:

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

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

Case (NPHA2): We have a ∗ b = a ◦ b = {0, a}. If one of the a ∗ a or a ◦ a
are equal to {0}, then (a∗b)∗(a∗b) ̸≪ a∗a or (a◦b)◦(a◦b) ̸≪ a◦a. Hence
a ∗ a = a ◦ a = {0, a}.

If b◦b = {0}, but b∗b = {0}, {0,b}, {0,a} or {0,a,b}, then (b◦a)◦(b◦a) ̸≪
b ◦ b.

If b ◦ b = {0, b}, but b ∗ b = {0} or {0, a}, then (b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b.
Hence b ∗ b = {0, b} or {0, a, b}.

If b ◦ b = {0, a}, but b ∗ b = {0}, {0, b}, {0, a} or {0, a, b}, then (b ◦ a) ◦
(b ◦ a) ̸≪ b ◦ b.

If b ◦ b = {0, a, b}, but b ∗ b = {0} or {0, a}, then (b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b.
Hence b ∗ b = {0, b} or {0, a, b}. Therefore, we have the following cases:

(1) b ◦ b = b ∗ b = {0, b}. Thus we have the following tables:

◦2 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

∗2 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

(2) b ◦ b = {0, b} and b ∗ b = {0, a, b}. Thus we have the following tables:

◦3 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

∗3 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

(3) b ◦ b = {0, a, b} and b ∗ b = {0, b}. Thus similar to (2), we have one
hyper pseudo BCK-algebra in this case.

(4) Let b ◦ b = b ∗ b = {0, a, b}. Thus we have the following tables:

◦5 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

∗5 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

32



Classifications of Hyper Pseudo BCK-algebras of Order 3

Case (NPHA3): We have a∗b = {0, a} and a◦b = {0}. If a∗a = {0} and
a◦a is equal to {0} or {0, a}, then (a∗b)∗(a∗b) ̸≪ a∗a. Thus a∗a = {0, a}.

If b◦b = {0}, but b∗b = {0}, {0,b}, {0,a} or {0,a,b}, then (b◦a)◦(b◦a) ̸≪
b ◦ b.

If b ◦ b = {0, b}, but b ∗ b = {0} or {0, a}, then (b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b.
If b ◦ b = b ∗ b = {0, b}, then (b ∗ a) ◦ b ̸= (b ◦ b) ∗ a. If b ◦ b = {0, b} and
b ∗ b = {0, a, b}, then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.

If b◦b = {0, a}, but b∗b = {0}, {0, b} or {0, a, b}, then (b◦a)◦(b◦a) ̸≪ b◦b.
If b ◦ b = b ∗ b = {0, a}, then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.

If b ◦ b = {0, a, b}, but b ∗ b = {0} or {0, a}, then (b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b.
Hence b ∗ b = {0, b} or {0, a, b}. Therefore, we have the following cases:

(1) b◦b = {0, a, b} and b∗b = {0, b}. If a◦a = {0, a}, then (a◦a)◦(b◦a) ̸≪
a ◦ b. Thus in this case, we have the following tables:

◦6 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a,b} {0,a,b}

∗6 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

(2) b ◦ b = b ∗ b = {0, a, b}. If a ◦ a = {0, a}, then (a ◦ a) ◦ (b ◦ a) ̸≪ a ◦ b.
Thus in this case, we have the following tables:

◦7 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a,b} {0,a,b}

∗7 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

Case (NPHA4): We have a◦b = {0, a} and a∗b = {0}. If a◦a = {0} and
a∗a is equal to {0} or {0, a}, then (a◦b)◦(a◦b) ̸≪ a◦a. Thus a◦a = {0, a}.

If b◦b = {0}, but b∗b = {0}, {0,b}, {0,a} or {0,a,b}, then (b◦a)◦(b◦a) ̸≪
b ◦ b.

If b ◦ b = {0, b}, but b ∗ b = {0} or {0, a}, then (b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b. If
b ◦ b = b ∗ b = {0, b}, then (b ◦ a) ∗ b ̸= (b ∗ b) ◦ a. Hence b ∗ b = {0, a, b}.

If b◦b = {0, a}, but b∗b = {0}, {0, b} or {0, a, b}, then (b◦a)◦(b◦a) ̸≪ b◦b.
If b ◦ b = b ∗ b = {0, a}, then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b.

If b ◦ b = {0, a, b}, but b ∗ b = {0} or {0, a}, then (b ∗ a) ∗ (b ∗ a) ̸≪ b ∗ b.
If b ◦ b = {0, a, b}, but b ∗ b = {0, b}, then (b ∗ b) ◦ b ̸= (b ◦ b) ∗ b. Hence
b ∗ b = {0, a, b}. Therefore, we have the following cases:

(1) b◦b = {0, b} and b∗b = {0, a, b}. If a∗a = {0, a}, then (a∗a)∗(b∗a) ̸≪
a ∗ b. Thus in this step, we have the following case:

33



R. A. Borzooei, A. Rezazadeh and R. Ameri

◦8 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,b}

∗8 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a,b} {0,a,b}

(2) b ◦ b = b ∗ b = {0, a, b}. If a ∗ a = {0, a}, then (a ∗ a) ∗ (b ∗ a) ̸≪ a ∗ b.
Thus in this step, we have the following case:

◦9 0 a b
0 {0} {0} {0}
a {a} {0,a} {0,a}
b {b} {a,b} {0,a,b}

∗9 0 a b
0 {0} {0} {0}
a {a} {0} {0}
b {b} {a,b} {0,a,b}

We can check that all of the these 9 cases are hyper pseudo BCK-algebras
and each of them are not isomorphic together.
Theorem 3.9. There are 96 hyper pseudo BCK-algebras of order 3, that
satisfies the normal condition.

Proof. By Theorems 3.4, 3.5, 3.6, 3.7 and 3.8, the proof is clear.

4 Conclusion

Theorem 4.1. There are 106 hyper pseudo BCK-algebras of order 3 up to
isomorphism.

Proof. The proof follows by Theorems 3.1 and 3.9.

Definition 4.1. We say that H is a proper hyper pseudo BCK-algebra if H
is not a hyper BCK-algebra.

Theorem 4.2. There are 87 proper hyper pseudo BCK-algebras of order 3
up to isomorphism.

Proof. The proof follows by Theorems 2.5 and 4.1.

References

[1] R. A. Borzooei, A. Rezazadeh and R. Ameri, On Hyper Pseudo BCK-
algebras, submitted.

[2] R. A. Borzooei, M. M. Zahedi and H. Rezaei, Classification of hyper
BCK-algebras of order 3, Ital. J. Pure Appl. Math., 12(2002), 175-184.

[3] P. Corsini and V. Leoreanu, Applications of hyperstructure theory,
Kluwer Academic Publications, (2003).

34



Classifications of Hyper Pseudo BCK-algebras of Order 3

[4] G. Georgescu and A. Iorgulescu, Pseudo BCK-algebras: an extension
of BCK-algebras, Proceedings of DMTCS 01: Combinatorics, Com-
putability and Logic, Springer, London, (2001), 97-114.

[5] Y. Imai and K. Iseki, On axiom systems of prepositional calculi, XIV.
Proc Japan Acad, 42(1996), 26-29.

[6] A. Iorgulescu, Classes of pseudo-BCK algebras - Part I, Journal of
Multiple-Valued Logic and Soft Computing, 12(2006), 71-130.

[7] A. Iorgulescu, Classes of pseudo-BCK algebras - Part II, Journal of
Multiple-Valued Logic and Soft Computing, 12(2006), 575-629.

[8] Y. B. Jun, M. Kondo and K. H. Kim, Pseudo-ideals of pseudo-BCK
algebras, Scientiae Mathematicae Japonicae, 8(2003), 87-91.

[9] Y. B. Jun, M. M. Zahedi, X. L. Xin and R. A. Borzooei, On hyper BCK-
algebras, Italian Journal of Pure and Applied Mathematics, 10(2000),
127-136.

[10] F. Marty, Sur une generalization de la notion degroups, 8th Congress
Math. Scandinaves, Stockholm, (1934), 45-49.

35