Hyperstructure Theory Applied to BF-Algebras

Abstract

This study applies the hyperstructure theory to BF-algebra, which is an algebraic structure. In fact, we define hyper-BF-algebras and hyper-BF ideals and investigate several of their related characteristics. BF-algebra and hyper-BF ideal characteristics are taken into account, and supported examples are built. Here, we also develop new concepts known as hyper-B-algebra, hyper-BG-algebra, and hyper-BH algebra as generalizations of other classes of hyper-BCK-/BCI-algebras. Additionally, we demonstrate that each hyper-BF is a weak hyper-BF in hyper-BF-algebra, but the opposite is not true. It is further established that the intersection of the weak hyper-BF ideal family is weak.

1. Introduction

Imai and Iseki presented the BCK- and BCI-algebra ideas in 1966 [1,2]. BCK-algebra and BCI-algebra are mathematical plans of the BCK and BCI frameworks in combinatory logic. Various articles on the hypothesis of “BCK/BCI-algebras” have been distributed from that point forward, with an emphasis on the “BCK/BCI ideals”.

Neggers and Kim [3] originally introduced the concept of B-algebra by extending the notion of a BCK-/BCI-algebras. Jun et al. presented BH-algebra, which is a development of B-algebra, in [4]. BG-algebra was considered by Kim and Kim [5]. They also presented the concept of BN-algebra [6] and provided some equivalent conditions related to BN-algebra. Hwang et al. defined BV-algebra and showed that it is logically equivalent to BT-algebra, BM-algebra, 0-commutative B-algebra, and BV-algebra in [7]. The term BP-algebra was created by Ahn and Han [8], who also demonstrated that the quadratic BP-algebra is identical to multiple quadratic algebras. Andrzej Walendziak [9] developed the concept of BF-algebra, as well as the concepts of ideals and normal ideals of BF-algebras.

In 1934, Marty [10] created the theory of a hyperstructure and provided a description of a hypergroup. A set in an algebraic hyperstructure is formed by multiplying two elements, whereas an element in a classical algebraic structure is formed by multiplying two elements. Several authors later investigated hyperstructure in various algebraic structures, such as hyperfields [11], n-ary semihypergroups [12], ordered semihypergroups [13], Γ-semihypergroups [14], etc. More hyperstructure theories are studied in [15,16,17,18,19,20,21,22,23].

In Reference [24], Jun et al. applied a hyperstructure to BCK algebra and introduced the concept of hyper-BCK algebra, which is the generalization of BCK algebra. They then applied the same methodology to BCC-algebra in [25]. Muhiuddin et al., as well as other authors, adapted the hyperstructure theory to a variety of algebraic structures [26,27,28,29,30,31,32,33].

This article is organized as follows: Section 2 presents the concepts that are essential to these findings. In Section 3, we define hyper-BF-algebras and hyper-BF ideals, as well as provide theorems pertaining to these concepts. We extensively discuss some of their related properties and provide numerous examples to support these new ideas. As a generalization of other classes of hyper-BCK/BCI-algebras, we introduce brand-new concepts known as hyper-B-algebra, hyper-BG-algebra, and hyper-BH-algebra, and provide detailed examples of each. Additionally, in the case of hyper-BF-algebra, every hyper-BF ideal is a weak hyper-BF ideal, but the opposite is not true. Additionally, we demonstrate that the intersection of a family of weak hyper-BF ideals yields a weak hyper-BF ideal. Section 4 concludes the investigation.

2. Preliminaries

Here, we revisit a few definitions that are essential to our study.

A hyperoperation “∘” is a mapping from $𝔉\times 𝔉$ to the set of all nonempty subsets of $(\varnothing \ne )𝔉$. In this case, 𝔉 is referred to as a hypergroupoid. Allow 𝔉 to be a hypergroupoid, with $A,B\subseteq 𝔉$. The following definition applies to $A\circ B$:

$$A\circ B=\bigcup _{\genfrac{}{}{0pt}{}{{\kappa }_0\in A}{q_0\in B}}{\kappa }_0\circ q_0.$$

Note that ${\kappa }_0\circ q_0$ is used rather than ${\kappa }_0\circ \left\{q_0\right\},\left\{{\kappa }_0\right\}\circ \left\{q_0\right\}$, and $\left\{{\kappa }_0\right\}\circ q_0$ for any ${\kappa }_0,q_0\in 𝔉$. We define two hyperorders, “≪” and “<”, on a hypergroupoid $(𝔉,\circ )$, as follows:

$$(\forall {\kappa }_0,q_0\in 𝔉)({\kappa }_0\ll q_0⟺0\in {\kappa }_0\circ q_0),$$

$$(\forall A,B\subseteq 𝔉)(A\ll B⟺(\forall {\kappa }_0\in A)(\exists q_0\in B)({\kappa }_0\ll q_0)),$$

and

$$(\forall {\kappa }_0,q_0\in 𝔉)({\kappa }_0<q_0⟺0\in {\kappa }_0\circ q_0),$$

$$(\forall A,B\subseteq 𝔉)(A<B⟺(\exists a\in A\mathrm{and}\exists q_0\in B)({\kappa }_0<q_0)),$$

respectively. If the non-empty set 𝔉 meets the following assumptions, it is called hyper-BCK-algebra, which contains “∘” and “$0$” for all ${\kappa }_0,q_0,\kappa \in 𝔉$.

(h1)

$({\kappa }_0\circ \kappa )\circ (q_0\circ \kappa )\ll {\kappa }_0\circ q_0.$

(h2)

$({\kappa }_0\circ q_0)\circ \kappa =({\kappa }_0\circ \kappa )\circ q_0.$

(h3)

${\kappa }_0\circ 𝔉\ll \left\{{\kappa }_0\right\}.$

(h4)

${\kappa }_0\ll q_0$ and $q_0\ll {\kappa }_0$ ⟹${\kappa }_0=q_0.$

Proposition 1

([24]). In a hyper-BCK-algebra 𝔉, the following assertions are valid for all ${\kappa }_0,q_0,\kappa \in 𝔉$

(a1)

$0\circ 0=\left\{0\right\}.$

(a2)

$0\ll {\kappa }_0.$

(a3)

${\kappa }_0\ll {\kappa }_0.$

(a4)

$0\circ {\kappa }_0=\left\{0\right\}$ and ${\kappa }_0\circ 0=\left\{{\kappa }_0\right\}.$

(a5)

${\kappa }_0\circ 0\ll q_0⟹{\kappa }_0\ll q_0.$

(a6)

${\kappa }_0\ll q_0⟹\kappa \circ q_0\ll \kappa \circ {\kappa }_0.$

(a7)

${\kappa }_0\circ q_0=\left\{0\right\}⟹({\kappa }_0\circ \kappa )\circ (q_0\circ \kappa )=\left\{0\right\}$ and ${\kappa }_0\circ \kappa \ll q_0\circ \kappa .$

Definition 1

([9]). A B-algebra is an algebra $(X;\ast ,0)$ of type $(2,0)$, satisfying the following conditions:

(B1)

${\kappa }_0\ast {\kappa }_0=0$,

(B2)

${\kappa }_0\ast 0={\kappa }_0$,

(B3)

$({\kappa }_0\ast q_0)\ast \kappa ={\kappa }_0\ast (\kappa \ast (0\ast q_0))$, $\forall {\kappa }_0,q_0,\kappa \in X$.

Definition 2

([9]). BF-algebra acts on sets X with two operations: ∗ and 0. BF-algebra must meet the following criteria:

(B1)

${\kappa }_0\ast {\kappa }_0=0$;

(B2)

${\kappa }_0\ast 0={\kappa }_0$;

(BF)

$0\ast ({\kappa }_0\ast q_0)=q_0\ast {\kappa }_0$, $\forall {\kappa }_0,q_0\in X$.

Definition 3

([9]). BF${}_1$-algebra is BF-algebra satisfying the following condition:

(BG)

${\kappa }_0=({\kappa }_0\ast q_0)\ast (0\ast q_0)$, $\forall {\kappa }_0,q_0\in X$.

Definition 4

([9]). BF${}_2$-algebra is BF-algebra satisfying the following condition:

(BH)

${\kappa }_0\ast q_0=0$ and $q_0\ast {\kappa }_0=0$ imply ${\kappa }_0=q_0$, $\forall {\kappa }_0,q_0\in X$.

3. Hyper-BF-Algebras

In this section, we will discuss our main findings. For a hypergroupoid $(𝔉,\circ )$, we consider the following assertions for all ${\kappa }_0,q_0,\kappa \in 𝔉$:

(c1)

${\kappa }_0\ll {\kappa }_0.$

(c2)

${\kappa }_0\circ 0=\left\{{\kappa }_0\right\}.$

(c3)

$0\circ ({\kappa }_0\circ q_0)=q_0\circ {\kappa }_0.$

(c4)

${\kappa }_0\ll q_0$ and $q_0\ll {\kappa }_0$ $({\kappa }_0=q_0)$.

(c5)

$({\kappa }_0\circ q_0)\circ (0\circ q_0)=\left\{{\kappa }_0\right\}.$

(c6)

${\kappa }_0\in ({\kappa }_0\circ q_0)\circ (0\circ q_0).$

(c7)

$({\kappa }_0\circ q_0)\circ \kappa ={\kappa }_0\circ (\kappa \circ (0\circ q_0)).$

Now, in support of our new notions, we need to construct the following examples:

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1) and (c2).

Example 1.

Let $𝔉=\{0,1\}$. Consider the following table:

$$\begin{array}{|ccc|}\hline \circ & 0& 1\\ 0& \left\{0\right\}& \{0,1\}\\ 1& \left\{1\right\}& \{0,1\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1) and (c2).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1) and (c2), but not (c3).

Example 2.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \{0,1\}& \{1,2\}\\ 1& \left\{1\right\}& \{0,1\}& \{1,2\}\\ 2& \left\{2\right\}& \{1,2\}& \{0,1,2\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1) and (c2), but not (c3).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1), (c2), and (c3).

Example 3.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}\\ 1& \left\{1\right\}& \{0,1\}& \{1,2\}\\ 2& \left\{2\right\}& \{1,2\}& \{0,1,2\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1), (c2), and (c3).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1) and (c2), but not (c4).

Example 4.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \{0,1\}& \{1,2\}\\ 1& \left\{1\right\}& \{0,1\}& \{0,1,2\}\\ 2& \left\{2\right\}& \{0,1,2\}& \{0,1,2\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1) and (c2), but not (c4).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1), (c2), and (c4).

Example 5.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}\\ 1& \left\{1\right\}& \{0,1\}& \left\{1\right\}\\ 2& \left\{2\right\}& \left\{2\right\}& \{0,2\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1), (c2), and (c4).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1) and (c2), but not (c5).

Example 6.

We define a hyperoperation “∘” on $𝔉:=[0,\infty )$ by

$${\kappa }_0\circ q_0:=\left\{\begin{array}{cc}\left(0,{\kappa }_0\right]\hfill & \mathrm{if}{q}_0>{\kappa }_0\ne 0\hfill \\ \left\{q_0\right\}\hfill & \mathrm{if}{\kappa }_0=0\hfill \\ \left\{{\kappa }_0\right\}\hfill & \mathrm{if}{q}_0=0\hfill \\ \left(0,q_0\right]\hfill & \mathrm{if}{\kappa }_0>q_0\ne 0\hfill \\ \left[0,{\kappa }_0\right]\hfill & \mathrm{if}{\kappa }_0=q_0\hfill \end{array}\right.$$

$\forall {\kappa }_0,q_0\in 𝔉$. Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1) and (c2), but not (c5).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1), (c2), and (c5).

Example 7.

If $(𝔉,\ast ,0)$ is an algebra of $B{F}_1$, we can define the hyperoperation “∘” in 𝔉 and specify the $({\kappa }_0\circ q_0=\{{\kappa }_0\ast q_0\})$$(\forall {\kappa }_0,q_0\in 𝔉)$. Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1), (c2), and (c5).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1) and (c2), but not (c6).

Example 8.

The hypergroupoid $(𝔉,\circ )$ in Example 6 satisfies (c1) and (c2), but not (c6).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1), (c2), and (c6).

Example 9.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}\\ 1& \left\{1\right\}& \{0,1\}& \{1,2\}\\ 2& \left\{2\right\}& \{1,2\}& \{0,1,2\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1), (c2), and (c6).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1) and (c2), but not (c7).

Example 10.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}\\ 1& \left\{1\right\}& \left\{0\right\}& \{1,2\}\\ 2& \left\{2\right\}& \{1,2\}& \left\{0\right\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1) and (c2), but not (c7).

We define a hypergroupoid $(𝔉,\circ )$ that satisfies (c1), (c2), and (c7).

Example 11.

Let $𝔉=\{0,1,2\}$. Consider the following table:

$$\begin{array}{|cccc|}\hline \circ & 0& 1& 2\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}\\ 1& \left\{1\right\}& \{0,1\}& \{0,1,2\}\\ 2& \left\{2\right\}& \{0,1,2\}& \{0,1,2\}\\ \hline\end{array}$$

Then $(𝔉,\circ )$ is a hypergroupoid that satisfies (c1), (c2), and (c7).

Proposition 2.

Suppose that 𝔉 is furnished with a hyperoperation denoted by “∘” and a fixed value of “0”.

$\left(1\right)$

If $(𝔉,\circ ,0)$ satisfies (c5), then ${\kappa }_0\circ q_0$ is a singleton set $\forall {\kappa }_0,q_0\in 𝔉$.

$\left(2\right)$

If $(𝔉,\circ ,0)$ fulfills the equation $(0\in {\kappa }_0\circ q_0,({\kappa }_0\circ q_0)\circ (0\circ q_0)=\left\{q_0\right\}),$ then $0\circ q_0$ is a singleton set $(\forall {\kappa }_0,q_0\in 𝔉)\in 𝔉$.

Proof. 

(1).

Using (c5), we have

$$\left\{{\kappa }_0\right\}=({\kappa }_0\circ q_0)\circ (0\circ q_0)=\bigcup _{\genfrac{}{}{0pt}{}{a\in {\kappa }_0\circ q_0}{b\in 0\circ q_0}}a\circ b$$

$\forall {\kappa }_0,q_0\in 𝔉.$ Let $a_1,a_2\in {\kappa }_0\circ q_0$ and $b\in 0\circ {\kappa }_0.$ Then $a_1\circ b=\left\{{\kappa }_0\right\}$ and $a_2\circ b=\left\{{\kappa }_0\right\}$. Thus,

$$\left\{a_1\right\}=(a_1\circ b)\circ (0\circ b)=(a_2\circ b)\circ (0\circ b)=\left\{a_2\right\}.$$

It follows that $a_1=a_2$$\forall a_1,a_2\in {\kappa }_0\circ q_0$, so that ${\kappa }_0\circ q_0$ is a singleton set.

(2).

By assumption, $\forall {\kappa }_0,q_0\in 𝔉$, $\left\{q_0\right\}=({\kappa }_0\circ q_0)\circ (0\circ q_0)={\displaystyle \bigcup _{\genfrac{}{}{0pt}{}{a\in {\kappa }_0\circ q_0}{b\in 0\circ q_0}}}a\circ b$. Let $a\in {\kappa }_0\circ q_0$ and $b_1,b_2\in 0\circ q_0.$ Then $a\circ b_1=\left\{q_0\right\}$ and $a\circ b_2=\left\{q_0\right\}$. Then $\left\{b_1\right\}=(a\circ b_1)\circ (0\circ b_1)=(a\circ b_2)\circ (0\circ b_2)=\left\{b_2\right\}$. Thus, $\forall b_1,b_2\in 0\circ q_0$, $b_1=b_2$. Hence, $\forall {\kappa }_0,q_0\in 𝔉$, $0\circ q_0$ is a singleton set.

□

Proposition 2 (1) shows that condition (c5) is not a proper condition in hyper-theory. Thus, we provide the following definition.

Definition 5.

If “∘” is a hyperoperation and 0 is a constant in a set $(\varnothing \ne )𝔉$ with conditions (c1), (c2), and (c7), then 𝔉 is known as the hyper-B-algebra related to the hyperorder ≪ (i.e., ≪-hyper-B-algebra).

Example 12.

It is straightforward to verify that hypergroupoid $(𝔉,\circ )$ in Example 1 is the hyper-B-algebra.

Definition 6.

A set $(\varnothing \ne )𝔉$ equipped with a hyperoperation “∘” and a constant “0” satisfying conditions (c1), (c2), (c3), and (c4) is known as a hyper-BH-algebra related to the hyperorder ≪ (i.e., ≪-hyper-BH-algebra).

Example 13.

Hypergroupoid $(𝔉,\circ )$ in Example 5 satisfies (c1), (c2), (c3), and (c4). Thus, it is a hyper-BH-algebra.

Definition 7.

If “∘” is a hyperoperation and 0 is a constant in a set $(\varnothing \ne )𝔉$ with conditions (c1), (c2), (c3), and (c6), then 𝔉 is known as a hyper-BG-algebra related to the hyperorder ≪ (i.e., ≪-hyper-BG-algebra).

Example 14.

Hypergroupoid $(𝔉,\circ )$ in Example 9 satisfies (c1), (c2), (c3), and (c6). Thus, it is a hyper-BG-algebra.

Definition 8.

A set $(\varnothing \ne )𝔉$ equipped with a hyperoperation “∘” and a constant “0”, satisfying conditions (c1), (c2), and (c3), is known as a hyper-BF-algebra related to the hyperorder ≪ (i.e., ≪-hyper-BF-algebra).

Example 15.

Let $𝔉=\{0,1,2,3\}$. Consider the following table:

$$\begin{array}{|ccccc|}\hline \circ & 0& 1& 2& 3\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}& \left\{3\right\}\\ 1& \left\{1\right\}& \{0,1\}& \left\{3\right\}& \left\{2\right\}\\ 2& \left\{2\right\}& \left\{3\right\}& \{0,2\}& \left\{1\right\}\\ 3& \left\{3\right\}& \left\{2\right\}& \left\{1\right\}& \{0,3\}\\ \hline\end{array}$$

Clearly, $(𝔉,\circ )$ is a hypergroupoid. As 𝔉 satisfies (c1), (c2), and (c3), it is straightforward to show that it is a hyper-BF-algebra.

Proposition 3.

If 𝔉 is a ≪-hyper-BF-algebra, then $(\forall {\kappa }_0,q_0\in 𝔉)$

(1)

$0\circ (0\circ {\kappa }_0)=\left\{{\kappa }_0\right\}.$

(2)

$0\circ {\kappa }_0=0\circ q_0⟹{\kappa }_0=q_0.$

(3)

${\kappa }_0\circ q_0=\left\{0\right\}⟹q_0\circ {\kappa }_0=\left\{0\right\}.$

(4)

${\kappa }_0\in ({\kappa }_0\circ {\kappa }_0)\circ (0\circ {\kappa }_0)$.

Proof. 

(1).

By Definition of hyper-BF-algebras, $\forall {\kappa }_0,q_0\in 𝔉,0\circ ({\kappa }_0\circ q_0)=q_0\circ {\kappa }_0.$ Taking ${\kappa }_0=0$ and $q_0={\kappa }_0$, $0\circ (0\circ {\kappa }_0)={\kappa }_0\circ 0=\left\{{\kappa }_0\right\}$, by condition (c2).

(2).

Assume that $0\circ {\kappa }_0=0\circ q_0$. By Proposition 3 (1) and the assumption, $\left\{{\kappa }_0\right\}=0\circ (0\circ {\kappa }_0)=0\circ (0\circ q_0)=\left\{q_0\right\}$. Therefore, ${\kappa }_0=q_0$.

(3).

Assume that ${\kappa }_0\circ q_0=\left\{0\right\}$. By condition (c3), the assumption, and condition (c2), $q_0\circ {\kappa }_0=0\circ ({\kappa }_0\circ q_0)=0\circ \left\{0\right\}=\left\{0\right\}$. Henceforth, $q_0\circ {\kappa }_0=\left\{0\right\}$.

(4).

$\forall {\kappa }_0\in 𝔉$, $({\kappa }_0\circ {\kappa }_0)\circ (0\circ {\kappa }_0)\supseteq 0\circ (0\circ {\kappa }_0)={\kappa }_0\circ 0=\left\{{\kappa }_0\right\}\ni {\kappa }_0$, by (c1), (c3), and (c2). Hence, ${\kappa }_0\in ({\kappa }_0\circ {\kappa }_0)\circ (0\circ {\kappa }_0)$.

□

Proposition 4.

Let $(𝔉,\circ )$ be a hyper-BF-algebra. For every $(\varnothing \ne )T_1,J_1\subseteq 𝔉,$ we have the following assertions.

(1)

$T_1\ll T_1.$

(2)

$T_1\circ 0=T_1.$

(3)

$0\circ (T_1\circ J_1)=J_1\circ T_1.$

(4)

$T_1\subseteq J_1$⟹$T_1\ll J_1.$

Proof. 

(1).

By condition (c1), $\forall {\kappa }_0\in 𝔉$, ${\kappa }_0\ll {\kappa }_0$. Then $\forall {\kappa }_0\in T_1\exists {\kappa }_0\in T_1$, such that ${\kappa }_0\ll {\kappa }_0$. Hence, $T_1\ll T_1$.

(2).

By condition (c2), $\forall (\varnothing \ne )A\subseteq 𝔉$, we have

$$\begin{array}{cc}{T}_1\circ 0\hfill & =\bigcup \{a\circ 0\mid a\in T_1\}\hfill \\ & =\bigcup \{\left\{a\right\}\mid a\in T_1\}\hfill \\ & =T_1.\hfill \end{array}$$

(3).

By condition (c3), $\forall (\varnothing \ne )T_1,J_1\subseteq 𝔉$, we have

$$\begin{array}{cc}0\circ (T_1\circ J_1)\hfill & =\bigcup \{0\circ {\kappa }_0\mid {\kappa }_0\in T_1\circ J_1\}\hfill \\ & =\bigcup \{0\circ (a\circ b)\mid a\in T_1,b\in J_1\}\hfill \\ & =\bigcup \{b\circ a\mid a\in T_1,b\in J_1\}\hfill \\ & =J_1\circ T_1.\hfill \end{array}$$

(4).

Assume that $T_1\subseteq J_1$. Then $\forall {\kappa }_0\in T_1\exists {\kappa }_0\in J_1$, such that ${\kappa }_0\ll {\kappa }_0$, by condition (c1). Hence, $T_1\ll J_1$.

□

Remark 1.

The converse of Proposition 4 (4) is not true. In Example 11, let $A=\{0,1\}$ and $B=\{0,2\}$. Then $A\ll B$, but $A⊈B$.

Proposition 5.

Let $(𝔉,\circ )$ be a hyper-BF-algebra with $\mid 𝔉\mid \le 3$. Then $0\circ q_0$ is a singleton set.

Proof. 

Case 1. $\mid 𝔉\mid =1$. By $𝔉=\left\{0\right\}$ and (c2), $\forall q_0\in 𝔉$, $0\circ q_0=0\circ 0=\left\{0\right\}$. Thus, $0\circ q_0$ is a singleton set.

• Case 2. $\mid 𝔉\mid =2$. Assume that $\exists q_0\in 𝔉$, such that $0\circ q_0$ is not a singleton set. Then $\exists q_0\in 𝔉$, such that $0\circ q_0=\{{{\kappa }_0}_1,{{\kappa }_0}_2\}$, where ${{\kappa }_0}_1\ne {{\kappa }_0}_2,{{\kappa }_0}_1,{{\kappa }_0}_2\in 𝔉$. By Proposition 3, $0\circ {{\kappa }_0}_1=\left\{q_0\right\}$ and $0\circ {{\kappa }_0}_2=\left\{q_0\right\}$. Since $q_0={{\kappa }_0}_1$ or $q_0={{\kappa }_0}_2$, $0\circ q_0=\left\{q_0\right\}$. Thus, $0\circ q_0$ is a singleton set.
• Case 3. $\mid 𝔉\mid =3$. If $q_0=0$, then by (c2), $0\circ q_0=0\circ 0=\left\{0\right\}$. Thus, $0\circ q_0$ is a singleton set. If $q_0\ne 0$, then $𝔉=\{0,{\kappa }_0,q_0\}$ with $x\ne q_0$ and ${\kappa }_0\ne 0$. Suppose that $0\circ q_0$ is not a singleton set. Then there exists ${q_0}_1,{q_0}_2\in 0\circ q_0$, such that ${q_0}_1\ne {q_0}_2$. By Proposition 3, $0\circ {q_0}_1=\left\{q_0\right\}$ and $0\circ {q_0}_2=\left\{q_0\right\}$. If ${q_0}_1=q_0$ or ${q_0}_2=q_0$, then $0\circ q_0=\left\{q_0\right\}$ is a singleton set. Therefore, ${q_0}_1=0$ or ${q_0}_2=0$, because $H=\{0,{\kappa }_0,q_0\}$ and ${q_0}_1={q_0}_2$. Without loss of generality, ${q_0}_1=0$ and ${q_0}_2={\kappa }_0$. Then $\left\{q_0\right\}=0\circ {q_0}_1=0\circ 0=\left\{0\right\}$. Therefore, $q_0=0$, which is a contradiction. Hence, $0\circ q_0$ is a singleton set. □

Proposition 6.

If $(𝔉,\circ )$ is a hyper-B-algebra, then $(𝔉,\circ )$ is a hyper-BG-algebra.

Proof. 

Let $(𝔉,\circ )$ be a hyper-B-algebra. Then $\forall {\kappa }_0,q_0\in 𝔉$,

$$\begin{array}{cc}({\kappa }_0\circ q_0)\circ (0\circ q_0)\hfill & =\bigcup _{b\in 0\circ q_0}({\kappa }_0\circ q_0)\circ b\hfill \\ & =\bigcup _{b\in 0\circ q_0}{\kappa }_0\circ (b\circ (0\circ q_0))\hfill \\ & =\bigcup _{\genfrac{}{}{0pt}{}{b\in 0\circ q_0}{a\in 0\circ q_0}}{\kappa }_0\circ (b\circ a)\hfill \\ & \supseteq {\kappa }_0\circ 0\hfill \\ & =\left\{{\kappa }_0\right\}\ni {\kappa }_0\hfill \end{array}$$

by (c7), (c1), and (c2). Hence, $(𝔉,\circ )$ is a hyper-BG-algebra. □

Definition 9.

A hyper-BF-algebra is called a hyper-BG-algebra (resp., a hyper-BH-algebra), if it satisfies (c6) (resp., (c4)).

Definition 10.

Let $(𝔉,\circ )$ represent a hyper-BF-algebra, and let S represent the subset of 𝔉 that includes the value 0. If S is a hyper-BF-algebra for the hyperoperation “$\circ$” on 𝔉, it is a hyper-subalgebra of 𝔉.

Proposition 7.

Let $(\varnothing \ne )S\subseteq 𝔉$ (hyper-BF-algebra) be such that ${\kappa }_0\circ {\kappa }_0\subseteq S$ for all ${\kappa }_0\in S$. In this case, the element 0 belongs to S.

Proof. 

Let ($\forall {\kappa }_0\in S$ and $a\in S$), ${\kappa }_0\circ {\kappa }_0\subseteq S$. As $a\ll a$, we have $0\in a\circ a\subseteq S$, as required. □

Proposition 8.

Let $(\varnothing \ne )S\subseteq 𝔉$ (hyper-BF-algebra) be such that ${\kappa }_0\circ q_0\subseteq S$ for all ${\kappa }_0,q_0\in S$. In this case, element 0 belongs to S.

Proof. 

Let ${\kappa }_0\circ q_0\subseteq S\forall {\kappa }_0,q_0\in S$ and $a\in S$. As $a\ll a$, so $0\in a\circ a\subseteq S$, as required. □

Theorem 1.

Let $(𝔉,\circ )$ be a hyper-BF-algebra and $(\varnothing \ne )S\subseteq 𝔉$. Then S is a hyper-subalgebra of $H\iff {\kappa }_0\circ q_0\subseteq S,\forall {\kappa }_0,q_0\in S$.

Proof. 

$(⟹)$ is obvious.

$(⟸)$ Let ${\kappa }_0\circ q_0\subseteq S\forall {\kappa }_0,q_0\in S$. Then $0\in S$ by Proposition 8. For any ${\kappa }_0,q_0\in S$, we have ${\kappa }_0\circ q_0\subseteq S$. Hence,

$$0\circ ({\kappa }_0\circ q_0)=\bigcup _{a\in {\kappa }_0\circ q_0}0\circ a\subseteq S$$

and so (c3) is valid in S. We may also show that axioms (c1) and (c2) are valid in S. Thus, S is a hyper-subalgebra of 𝔉. □

Example 16.

(1)

We define a hyperoperation “∘” on 𝔉 by ${\kappa }_0\circ q_0={\kappa }_0\ast q_0,\forall {\kappa }_0,q_0\in 𝔉$, using $(𝔉,\ast ,0)$ as the BF-algebra. Then $(𝔉,\circ )$ is a hyper-BF-algebra. If S is a subalgebra of a BF-algebra $(H,\ast ,0)$, then it is a hyper-subalgebra of $(𝔉,\circ )$.

(2)

In Example 6, let $(𝔉,\circ )$ be the hyper-BF-algebra, and let $S=\left[0,a\right],\forall a\in \left[0,\infty \right)$. Then S is a $(𝔉,\circ )$ hyper-subalgebra.

(3)

Let $(𝔉,\circ )$ be the hyper-BF-algebra in Example 3, and $S_1=\{0,1\}$ and $S_2=\{0,2\}$, respectively. Thus, $S_1$ is a hyper-subalgebra of 𝔉, but $S_2$ is not a hyper-subalgebra of 𝔉 because $2\circ 2=\{0,1,2\}⊈S_2$.

Definition 11.

Let $(𝔉,\circ )$ be a hyper-BF-algebra and $(\varnothing \ne )I\subseteq 𝔉$. Then I is called a hyper-BF ideal of 𝔉 if:

(HI1) 

$0\in I$,

(HI2) 

${\kappa }_0\circ q_0\ll I$ and $q_0\in I$ imply ${\kappa }_0\in I$, $\forall {\kappa }_0,q_0\in 𝔉$.

Example 17.

Let $H=\{0,1,2,3\}$. Consider the following table:

$$\begin{array}{|ccccc|}\hline \circ & 0& 1& 2& 3\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}& \left\{3\right\}\\ 1& \left\{1\right\}& \{0,1\}& \left\{3\right\}& \{2,3\}\\ 2& \left\{2\right\}& \left\{3\right\}& \{0,2\}& \{1,3\}\\ 3& \left\{3\right\}& \{2,3\}& \{1,3\}& H\\ \hline\end{array}$$

Clearly, $(𝔉,\circ )$ is a hyper-BF-algebra. Verifying that the set $I=\{0,1\}$ is the hyper-BF ideal of 𝔉 is routine.

Proposition 9.

Let $(𝔉,\circ )$ be a hyper-BF-algebra. Then I is a hyper-BF ideal ⇔

(HFI1) 

$0\in I$;

(HFI2) 

$\forall {\kappa }_0\in 𝔉\backslash I$, $\forall q_0\in I$, $0\notin {\kappa }_0\circ q_0$;

(HFI3) 

($\forall {\kappa }_0\in 𝔉\backslash I$), $\forall q_0\in I$, $\exists (\kappa \in {\kappa }_0\circ q_0$), for some $\kappa \in 𝔉\backslash I$.

Proof. 

(⟹) Suppose that I is a hyper-BF ideal. Then trivially, (HFI1) holds. Moreover, ${\kappa }_0\circ q_0\ll I$ and $q_0\in I$ implies ${\kappa }_0\in I$. By contrapositive, ${\kappa }_0\notin I$⇒$q_0\notin I$ or ${\kappa }_0\circ q_0\nless I$. Then

$$\forall {\kappa }_0\in 𝔉\backslash Iand\forall q_0\in I\Rightarrow {\kappa }_0\circ q_0\nless I.$$

(1)

Since $0\in I$, so ${\kappa }_0\circ 0\nless I$$\forall {\kappa }_0\in 𝔉\backslash I$. It follows from (c2) that $\left\{{\kappa }_0\right\}\nless I$$\forall {\kappa }_0\in 𝔉\backslash I$. Therefore, $\forall {\kappa }_0\in 𝔉\backslash I$, $\nexists q_0\in I$ such that ${\kappa }_0\ll q_0$. Then ${\kappa }_0\circ 0\nless I$. It follows that $\forall {\kappa }_0\in 𝔉\backslash I$, $\nexists q_0\in I$ such that $0\in {\kappa }_0\circ q_0$. Thus, $\forall {\kappa }_0\in 𝔉\backslash I$, $\forall q_0\in I$, $0\notin {\kappa }_0\circ q_0$ and, hence, (HFI2) holds. By (1), $\exists \kappa \in {\kappa }_0\circ q_0$, such that $\kappa \nless I$. Then $\kappa \notin I$, by (c1). Therefore, $\forall {\kappa }_0\in 𝔉\backslash I$, $\forall q_0\in I$, $\exists \kappa \in {\kappa }_0\circ q_0$, for some $\kappa \in 𝔉\backslash I$. Thus, (HFI3) holds.

(⟸) This is enough to show (1). By (HFI3), $\forall {\kappa }_0\in 𝔉\backslash I$, $\forall q_0\in I$, $\exists \kappa \in {\kappa }_0\circ q_0$, for some $\kappa \in 𝔉\backslash I$. Moreover, by (HFI2), we have $0\notin {\kappa }_0\circ I$, $\forall {\kappa }_0\in 𝔉\backslash I$. Then $\forall {\kappa }_0\in 𝔉\backslash I$ and $\forall q_0\in I$, $\exists \kappa \in {\kappa }_0\circ q_0$, such that $0\notin \kappa \circ I$. It follows that $\forall {\kappa }_0\in 𝔉\backslash I$ and $\forall q_0\in I$, $\exists \kappa \in {\kappa }_0\circ q_0$, such that $\kappa \nless I$. Hence, ${\kappa }_0\circ q_0\nless I$$\forall {\kappa }_0\in H\backslash I$ and $\forall q_0\in I$. □

Proposition 10.

If $\{I_{\lambda }\mid \lambda \in \Lambda \}$ is a family of hyper-BF ideals of a hyper-BF-algebra $(𝔉,\circ )$, then $\bigcap \{I_{\lambda }\mid \lambda \in \Lambda \}$ is a hyper-BF ideal of 𝔉.

Proof. 

Let $I=\bigcap \{I_{\lambda }\mid \lambda \in \Lambda \}$. Clearly, $0\in I$. Assume that ${\kappa }_0\circ q_0\ll I$ and $q_0\in I$. Then ${\kappa }_0\circ q_0\ll I_{\lambda }$ and $q_0\in I_{\lambda }$$\forall \lambda \in \Lambda$. From (HI2), ${\kappa }_0\in I_{\lambda }$$\forall \lambda \in \Lambda$ so that ${\kappa }_0\in I$, ending the proof. □

Definition 12.

Let $(𝔉,\circ )$ be a hyper-BF-algebra and let $A\subseteq 𝔉$. By the hyper-BF ideal generated by A, written $\left[A\right]$, we mean the intersection of all hyper-BF ideals, which contain A.

Definition 13.

Let $(𝔉,\circ )$ be a hyper-BF-algebra and $(\varnothing \ne )I\subseteq 𝔉$. Then I is called a weak hyper-BF ideal of 𝔉 if:

(HI1) 

$0\in I$,

(WHI1) 

${\kappa }_0\circ q_0\subseteq I$ and $q_0\in I$ imply ${\kappa }_0\in I$ $\forall {\kappa }_0,q_0\in 𝔉$.

Example 18.

(1)

Let $(𝔉,\circ )$ be the hyper-BF-algebra in Example 16 (1). Then every ideal I of a BF-algebra $(H,\ast ,0)$ in Example 16 (1) is a weak hyper-BF ideal of 𝔉.

(2)

Let $(𝔉,\circ )$ be the hyper-BF-algebra in Example 16 (2) and let $I=\{0,a\}$ for every $a\in \left[0,\infty \right)$. Then I is a weak hyper-BF ideal of 𝔉.

(3)

Let $(𝔉,\circ )$ be the hyper-BF-algebra in Example 16 (3). Then $I_1=\{0,1\}$ and $I_2=\{0,2\}$ are weak hyper-BF ideals of 𝔉.

Theorem 2.

Every hyper-BF ideal of hyper-BF-algebra $(𝔉,\circ )$ is a weak hyper-BF ideal of 𝔉.

Proof. 

Assume that $\forall {\kappa }_0,q_0\in 𝔉$, ${\kappa }_0\circ q_0\subseteq I$ and $q_0\in I$. Then $\forall {\kappa }_0,q_0\in 𝔉$, ${\kappa }_0\circ q_0\ll I$, and $q_0\in I$, by Proposition 4 (4). Then, by the definition of the hyper-BF ideal, we have ${\kappa }_0\in I$ and $0\in I$. □

Remark 2.

In general, weak hyper-BF ideals of hyper-BF-algebra 𝔉 may not be hyper-BF ideals of 𝔉. The following illustration validates this statement.

Example 19.

Let $𝔉=\{0,1,2,3\}$. Consider the following table:

$$\begin{array}{|ccccc|}\hline \circ & 0& 1& 2& 3\\ 0& \left\{0\right\}& \left\{1\right\}& \left\{2\right\}& \left\{3\right\}\\ 1& \left\{1\right\}& \{0,1\}& \{0,1,2\}& \{0,1,2,3\}\\ 2& \left\{2\right\}& \{0,1,2\}& \{0,1,2\}& \{0,1,2,3\}\\ 3& \left\{3\right\}& \{0,1,2,3\}& \{0,1,2,3\}& \{0,1,2,3\}\\ \hline\end{array}$$

Clearly, $(𝔉,\circ )$ is a hyper-BF-algebra. It is straightforward to verify that the set $I=\{0,1\}$ is a weak hyper-BF ideal but is not a hyper-BF ideal as

$$2\circ 1\le Iand1\in Ibut2\notin I.$$

Proposition 11.

If $\{I_{\lambda }\mid \lambda \in \Lambda \}$ is a class of weak hyper-BF ideals of a hyper-BF-algebra $(𝔉,\circ )$, then $\bigcap \{I_{\lambda }\mid \lambda \in \Lambda \}$ is a weak hyper-BF ideal of 𝔉.

Proof. 

Let $I=\bigcap \{I_{\lambda }\mid \lambda \in \Lambda \}$. Clearly, $0\in I$. Assume that ${\kappa }_0\circ q_0\subseteq I$ and $q_0\in I$. Then ${\kappa }_0\circ q_0\subseteq I_{\lambda }$ and $q_0\in I_{\lambda }$$\forall \lambda \in \Lambda$. It follows from (WHI1) that ${\kappa }_0\in I_{\lambda }$$\forall \lambda \in \Lambda$ so that ${\kappa }_0\in I$, ending the proof. □

4. Conclusions

In this research, we went into great detail about various characteristics of hyper-BF ideals and hyper-BF-algebras. Numerous cases back up these innovative theories. The unique ideas, known as hyper-B-algebra, hyper-BG-algebra, and hyper-BH-algebra, were also demonstrated through examples. Additionally, it was shown that each hyper-BF ideal in a hyper-BF-algebra is a weak hyper-BF ideal, even though the opposite is not true. The intersection of a family of weak hyper-BF ideals with another weak hyper-BF ideal was also found.