Elitable GE-filters of bordered GE-algebras

Abstract

In this paper, the notions of maximal GE-filter and prime GE-filter of a GE-algebra are introduced and the relation between them is given. Some characterizations of prime GE-filters of a transitive GE-algebra are given in terms of the GE-filter generated by a subset of a transitive GE-algebra. We generalized Stone’s theorem to transitive GE-algebras. The notion of elitable GE-filter of a bordered GE-algebra is introduced and investigated its properties. We observed that the class of all elitable GE-filters of a transitive bordered GE-algebra is a complete distributive lattice. Equivalent conditions for a GE-filter of a transitive bordered GE-algebra to be elitable GE-filter are given. We provided conditions for a subset of a transitive bordered GE-algebra to be elitable GE-filter.

1 Introduction

Imai and Iséki introduced BCK-algebras (see Imai and Iséki 1966; Iséki 1966) in 1966 as the algebraic semantics for a non-classical logic with only implication. Various researchers have examined the generalised concepts of BCK-algebras since then. Henkin and Skolem introduced Hilbert algebras in the 1950 s to investigate intuitionistic and other non-classical logics. A. Diego established that Hilbert algebras form a locally finite variety (see Diego 1966). Later several researchers extended the theory on Hilbert algebras. S. Celani given a representation theorem for Hilbert algebras by means of ordered sets and characterized the homomorphisms of Hilbert algebras in terms of applications defined between the sets of all irreducible deductive systems of the associated algebras (see Celani 2002). Chajda et al. considered the properties of deductive systems in Hilbert algebras which are upper semi-lattices as posets and shown that every maximal deductive system is prime. They have given a condition for a deductive system to be prime and shown that the annihilator of any non-empty subset of a Hilbert algebra is a deductive system which is an annihilator of the induced upper semilattice(see Chajda et al. 2002). Hong et al. introduced the concept of maximal deductive systems and shown that every bounded Hilbert algebra with at least two elements contains at least one maximal deductive system (see Hong and Jun 1996). Jun et al. introduced the concept of Hilbert filter in Hilbert algebras and studied how to generate a Hilbert filter by a set (see Jun and Kim 2005). The notion of BE-algebra was introduced by H.S. Kim and Y.H. Kim as a generalization of a dual BCK-algebra (see Kim and Kim 2006). Rezaei et al. discussed relations between Hilbert algebras and BE-algebras (see Rezaei et al. 2013). Borumand Saeid et al. introduced the notions of implicative filter, positive implicative filter, normal filter, fantastic filter, obstinate filter and maximal filter in a BE-algebra and obtained the related properties (see Borumand Saeid et al. 2013). Dudek et al. introduced the notion of poor and crazy filters in BCK-algebras and studied their properties in different types of BCK-algebras (see Dudek and Jun 2007).

The generalisation process is another important topic in the study of algebraic structures. Bandaru et al. introduced the concept of GE-algebras as a generalisation of Hilbert algebras and studied several properties (see Bandaru et al. 2021). Rezaei and colleagues introduced and discussed the concept of prominent GE-filters in GE-algebras (see Rezaei et al. 2021). Bandaru et al. introduced and investigated the concept of bordered GE-algebra (see Bandaru et al. 2021). Later, Ozturk et al. introduced and investigated the concept of Strong GE-filters, GE-ideals of bordered GE-algebras (see Ozturk et al. 2021). Song et al. introduced and discussed the concept of Imploring GE-filters of GE-algebras (see Song et al. 2021).

In this paper, we introduce and investigate the concepts of maximal GE-filter and prime GE-filter of a GE-algebra. We define prime GE-filter as the GE-filter produced by a subset of a transitive GE-algebra. We define and investigate the properties of an elitable GE-filter of a bordered GE-algebra. The class of all elitable GE-filters of a transitive bordered GE-algebra is a complete distributive lattice, as we observe. We describe an elitable GE-filter in a transitive bordered GE-algebra. In addition, we define the conditions under which a subset of a transitive bordered GE-algebra is an elitable GE-filter.

2 Preliminaries

Definition 2.1

(Bandaru et al. 2021) A GE-algebra is a non-empty set $R$ with a constant 1 and a binary operation $*$ satisfying the following axioms:

• (GE1) $\mu *\mu = 1$,

• (GE2) $1 *\mu =\mu$,

• (GE3) $\mu *(\nu *\tau )=\mu *(\nu *(\mu *\tau ))$

for all $\mu ,\nu ,\tau \in R$.

In a GE-algebra $R$, a binary relation “$\le$” is defined by

$$\begin{aligned} (\forall \beta , \gamma \in R) \left( \beta \le \gamma ~\Leftrightarrow ~\beta *\gamma =1\right) . \end{aligned}$$

(2.1)

Definition 2.2

(Bandaru et al. 2021) A GE-algebra $R$ is said to be

• transitive if it satisfies:

  $$\begin{aligned} (\forall \beta ,\gamma ,\alpha \in R)\left( \beta *\gamma \le (\alpha *\beta )*(\alpha *\gamma )\right) . \end{aligned}$$

  (2.2)

• antisymmetric if the binary relation “$\le$” is antisymmetric.

• commutative if it satisfies:

  $$\begin{aligned} (\forall \beta ,\gamma \in R)\left( (\beta *\gamma )*\gamma = (\gamma *\beta )*\beta \right) . \end{aligned}$$

  (2.3)

Theorem 2.3

(Bandaru et al. 2021) Every self-distributive BE-algebra is a GE-algebra.

The following proposition gives the equivalent conditions for a GE-algebra to be implication algebra, dual implicative BCK-algebra and commutative Hilbert algebra.

Proposition 2.4

(Bandaru et al. 2021) Let $(R, *, 1)$ be a GE-algebra. Then, the following are equivalent.

• (i) $R$ is commutative,

• (ii) $R$ is implication algebra,

• (iii) $R$ is dual implicative BCK-algebra,

• (iv) $R$ is commutative Hilbert algebra.

Definition 2.5

(Bandaru et al. 2021) If a GE-algebra $R$ has a special element, say $0$, that satisfies $0\le \beta$ for all $\beta \in R$, we call $R$ the bordered GE-algebra.

For every element $\beta$ of a bordered GE-algebra $R$, we denote $\beta *0$ by , and is denoted by .

Definition 2.6

(Bandaru et al. 2021) If a bordered GE-algebra $R$ satisfies the condition (2.2), we say that $R$ is a transitive bordered GE-algebra.

Definition 2.7

(Bandaru et al. 2021) A bordered GE-algebra $R$ is said to be antisymmetric if the binary operation “$\le$” is antisymmetric.

Proposition 2.8

(Bandaru et al. 2021) Every GE-algebra $R$ satisfies the following items.

$$\begin{aligned}&(\forall \beta \in R) \left( \beta *1=1\right) . \end{aligned}$$

(2.4)

$$\begin{aligned}&(\forall \beta ,\gamma \in R) \left( \beta *(\beta *\gamma )=\beta *\gamma \right) . \end{aligned}$$

(2.5)

$$\begin{aligned}&(\forall \beta ,\gamma \in R) \left( \beta \le \gamma *\beta \right) . \end{aligned}$$

(2.6)

$$\begin{aligned}&(\forall \beta , \gamma , \alpha \in R)\left( \beta *(\gamma *\alpha )\le \gamma *(\beta *\alpha )\right) . \end{aligned}$$

(2.7)

$$\begin{aligned}&(\forall \beta \in R)\left( 1\le \beta ~\Rightarrow ~\beta =1\right) . \end{aligned}$$

(2.8)

$$\begin{aligned}&(\forall \beta ,\gamma \in R)\left( \beta \le (\gamma *\beta )*\beta \right) . \end{aligned}$$

(2.9)

$$\begin{aligned}&(\forall \beta ,\gamma \in R)\left( \beta \le (\beta *\gamma )*\gamma \right) . \end{aligned}$$

(2.10)

$$\begin{aligned}&(\forall \beta ,\gamma ,\alpha \in R)\left( \beta \le \gamma *\alpha ~\Leftrightarrow ~\gamma \le \beta *\alpha \right) . \end{aligned}$$

(2.11)

If $R$ is transitive, then

$$\begin{aligned}&(\forall \beta ,\gamma ,\alpha \in R)\nonumber \\&\quad \left( \beta \le \gamma ~\Rightarrow ~\alpha *\beta \le \alpha *\gamma , ~\gamma *\alpha \le \beta *\alpha \right) . \end{aligned}$$

(2.12)

$$\begin{aligned}&(\forall \beta ,\gamma ,\alpha \in R)\left( \beta *\gamma \le (\gamma *\alpha )*(\beta *\alpha )\right) . \end{aligned}$$

(2.13)

Lemma 2.9

(Bandaru et al. 2021) The following are equivalent to each other in a GE-algebra $R$.

$$\begin{aligned}&(\forall \beta , \gamma , \alpha \in R)\left( \beta *\gamma \le (\alpha *\beta )*(\alpha *\gamma )\right) . \end{aligned}$$

(2.14)

$$\begin{aligned}&(\forall \beta , \gamma , \alpha \in R)\left( \beta *\gamma \le (\gamma *\alpha )*(\beta *\alpha )\right) . \end{aligned}$$

(2.15)

Definition 2.10

(Bandaru et al. 2021) A subset $K$ of a GE-algebra $R$ is called a GE-filter of $R$ if it satisfies:

$$\begin{aligned}&1\in K, \end{aligned}$$

(2.16)

$$\begin{aligned}&(\forall \beta , \gamma \in R) (\beta *\gamma \in K, ~\beta \in K~\Rightarrow ~\gamma \in K). \end{aligned}$$

(2.17)

Lemma 2.11

(Bandaru et al. 2021) In a GE-algebra $R$, every GE-filter $K$ of $R$ satisfies:

$$\begin{aligned} (\forall \beta , \gamma \in R) \left( \beta \le \gamma , ~\beta \in K~\Rightarrow ~\gamma \in K\right) . \end{aligned}$$

(2.18)

Proposition 2.12

(Bandaru et al. 2021) The following assertions are true in a bordered GE-algebra $R$.

(2.19)

(2.20)

(2.21)

(2.22)

(2.23)

If $R$ is a transitive bordered GE-algebra, then

(2.24)

(2.25)

If $R$ is an antisymmetric bordered GE-algebra, then

(2.26)

If $R$ is a transitive and antisymmetric bordered GE-algebra, then

(2.27)

Definition 2.13

(Bandaru et al. 2021) A duplex bordered element in a bordered GE-algebra $R$ is defined as an element $\beta$ of $R$ that satisfies .

The set of all duplex bordered elements of a bordered GE-algebra $R$ is denoted by ${0}^{2}(R)$ and is referred to as the $R$ duplex bordered set. It is clear that $0, 1\in {0}^{2}(R)$.

Definition 2.14

(Bandaru et al. 2021) A bordered GE-algebra $R$ is said to be duplex if every element of $R$ is a duplex bordered element, that is, $R= {0}^{2}(R)$.

Definition 2.15

(Ozturk et al. 2021) Let $R$ be a bordered GE-algebra. If a subset $G$ of $R$ meets the following conditions for all $\beta , \gamma \in R$, it is termed a GE-ideal of $R$:

• (1) $0\in G$,

• (2) $\beta \in G$ and imply that $\gamma \in G$.

Clearly, $\{0\}$ is a GE-ideal of $R$.

Proposition 2.16

(Ozturk et al. 2021) Let $G$ be a GE-ideal of $R$. Then, for any $\eta ,\zeta \in R$, we have

• (1) $\eta \in G$ and $\zeta \le \eta$ imply $\zeta \in G.$

• (2) .

3 Maximal and prime GE-filters

Definition 3.1

(Ozturk et al. 2021) Let $K$ be a subset of a GE-algebra $R$. The GE-filter of $R$ generated by $K$ is denoted by $\langle K\rangle$ and is defined to be the intersection of all GE-filters of $R$ containing $K$.

Example 3.2

Let $R=\{1, e, f, g, h, a, b\}$ be a set with the binary operation “$*$” in the following Cayley Table.

$$\begin{aligned}\begin{array}{c|ccccccc} *&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h&{}\quad a&{}\quad b\\ \hline 1&{} 1&{} e&{} f&{} g&{} h&{} a&{} b\\ e&{} 1&{} 1&{} 1&{} g&{} a&{} a&{} 1\\ f&{} 1&{} e&{} 1&{} h&{} h&{} h&{} b\\ g&{} 1&{} 1&{} f&{} 1&{} 1&{} 1&{} 1\\ h&{} 1&{} e&{} 1&{} 1&{} 1&{} 1&{} b\\ a&{} 1&{} e&{} f&{} 1&{} 1&{} 1&{} 1\\ b&{} 1&{} e&{} f&{} a&{} h&{} a&{} 1\end{array} \end{aligned}$$

Then, $(R, *, 1)$ is a GE-algebra. If we take a subset $G=\{1,e\}$ of $R$, then the GE-filter of $R$ generated by $G$ is $\langle G\rangle =\{1,e,f,b\}$.

The next theorem shows how the elemental structure of $\langle K\rangle$ is constructed.

Theorem 3.3

Let $K$ be a non-empty subset of a transitive GE-algebra $R$. Then, $\langle K\rangle$ consists of $\beta$’s that satisfies the following condition:

$$\begin{aligned}&(\exists \beta _1, \beta _2, \cdots , \beta _n\in K)\nonumber \\&\quad \left( \beta _n*(\cdots *(\beta _2*(\beta _1*\beta ))\cdots )=1\right) , \end{aligned}$$

(3.1)

that is, $\langle K\rangle =\{\beta \in R\mid \beta _n*(\cdots *(\beta _2*(\beta _1*\beta ))\cdots )=1\text {for some} ~\beta _1, \beta _2, \cdots , \beta _n\in K\}$.

Proof

Let

$$\begin{aligned} G&:=\{\beta \in R\mid \beta _n*(\cdots *(\beta _2*(\beta _1*\beta ))\cdots )\\&=1\text {for some} ~\beta _1, \beta _2, \cdots , \beta _n\in K\}. \end{aligned}$$

Obviously, $1\in G$. Let $\mu , \nu \in R$ be such that $\mu *\nu \in G$ and $\mu \in G$. Then, there is $\beta _1, \beta _2, \cdots , \beta _m, \gamma _1, \gamma _2,\cdots , \gamma _n\in K$ such that

$$\begin{aligned}&\beta _m*(\cdots *(\beta _2*(\beta _1*(\mu *\nu )))\cdots )=1, \end{aligned}$$

(3.2)

$$\begin{aligned}&\gamma _n*(\cdots *(\gamma _2*(\gamma _1*\mu ))\cdots )=1. \end{aligned}$$

(3.3)

By (GE3), (3.2) and (GE1), we can observe that,

$$\begin{aligned} \mu *(\beta _m*(\cdots *(\beta _2*(\beta _1*\nu ))\cdots ))=1, \end{aligned}$$

that is,

$$\begin{aligned} \mu \le \beta _m*(\cdots *(\beta _2*(\beta _1*\nu ))\cdots ). \end{aligned}$$

(3.4)

By (2.12), we get

$$\begin{aligned} \gamma _1*\mu \le \gamma _1*(\beta _m*(\cdots *(\beta _2*(\beta _1*\nu ))\cdots )). \end{aligned}$$

(3.5)

If we repeat this process n times, we have

$$\begin{aligned} 1&=\gamma _n*(\cdots *(\gamma _2*(\gamma _1*\mu ))\cdots )\\&\le \gamma _n*(\cdots *(\gamma _1*(\beta _m*(\cdots *(\beta _2*(\beta _1*\nu ))\cdots )))\cdots ). \end{aligned}$$

By (2.8), it follows that

$$\begin{aligned} \gamma _n*(\cdots *(\gamma _1*(\beta _m*(\cdots *(\beta _2*(\beta _1*\nu ))\cdots )))\cdots )=1. \end{aligned}$$

Hence, $\nu \in G$. Thus, $G$ is a GE-filter of $R$. It is obvious that $K\subseteq G$. Let $M$ be a GE-filter for $R$ that includes $K$. If $\beta \in G$, then $\alpha _n*(\cdots *(\alpha _2*(\alpha _1*\beta ))\cdots )=1\in M$ for some $\alpha _1, \alpha _2,\cdots ,\alpha _n\in K\subseteq M$. It follows that $\beta \in M$. Therefore, $G\subseteq M$. This shows that $G= \langle K\rangle$. $\square$

Corollary 3.4

For every element $\gamma$ in a transitive GE-algebra $R$, we have

$$\begin{aligned} \langle \gamma \rangle =\{\beta \in R\mid \gamma *\beta =1\}, \end{aligned}$$

that is, $\langle \gamma \rangle =\{\beta \in R\mid \gamma \le \beta \}$ which is called principal GE-filter generated by $\gamma$.

We construct the smallest GE-filter containing $K$ and $\eta$, given a GE-filter $K$ and an element $\eta$ in a transitive GE-algebra $R$.

Theorem 3.5

Let $K$ be a GE-filter of a transitive GE-algebra $R$ and $\eta$ be any element of $R$. Then,

$$\begin{aligned} \langle K\cup \{\eta \}\rangle =\{\zeta \in R\mid \eta *\zeta \in K\}. \end{aligned}$$

(3.6)

Proof

Let $K_{\eta }:=\{\zeta \in R\mid \eta *\zeta \in K\}.$ Since $\eta *\eta =1\in K$, $\eta \in K_{\eta }$. For any $\zeta \in K$, we have $\zeta \le \eta *\zeta$ by (2.6). Hence, $\eta *\zeta \in K$ by Lemma 2.11. Thus, $\zeta \in K_{\eta }$, which shows that $K\subseteq K_{\eta }$. Hence, $K\cup \{\eta \}\subseteq K_{\eta }$. Since $\eta *1= 1\in K,$ we get $1\in K_{\eta }.$ Let $\zeta , \alpha \in R$ be such that $\zeta \in K_{\eta }$ and $\zeta *\alpha \in K_{\eta }$. Then, $\eta *(\zeta *\alpha ) \in K$ and $\eta *\zeta \in K.$ Using (2.7), (2.12) and (2.5), we have

$$\begin{aligned} \eta *(\zeta *\alpha )&\le \zeta *(\eta *\alpha ) \le (\eta *\zeta ) *(\eta *(\eta *\alpha ))\\&= (\eta *\zeta ) *(\eta *\alpha ), \end{aligned}$$

and so $\eta *\alpha \in K$ by Lemma 2.11 and (2.17). Hence, $\alpha \in K_{\eta }$, and thus, $K_{\eta }$ is a GE-filter of $R$. Let $G$ be a GE-filter of $R$ containing $K\cup \{\eta \}$ and $\zeta \in K_{\eta }$. Then, $\eta \in G$ and $\eta *\zeta \in K\subseteq G$. Hence, $\zeta \in G$ since $G$ is GE-filter of $R$. Therefore, $K_{\eta } \subseteq G$. Hence, $K_{\eta }=\langle K\cup \{\eta \}\rangle$. $\square$

Proposition 3.6

Let $K$ and $G$ be two GE-filters of a transitive GE-algebra $R$. Then,

$$\begin{aligned} \langle K\cup G\rangle&=\{\beta \in R\mid \eta *(\zeta *\beta ) \\&= 1\text { for some }\eta \in K\text { and } \zeta \in G\} \end{aligned}$$

Proof

Let $J= \{\beta \in R\mid \eta *(\zeta *\beta ) = 1\text { for some }\eta \in K\text { and } \zeta \in G\}$ and $\beta \in J$. Then, $\eta *(\zeta *\beta ) = 1\text { for some }\eta \in K\text { and } \zeta \in G.$ Hence, $\beta \in \langle K\cup G\rangle$ by Theorem 3.3. Therefore, $J\subseteq \langle K\cup G\rangle .$ Conversely, assume that $\beta \in \langle K\cup G\rangle .$ Then, there are $\gamma _1,\gamma _2,....,\gamma _i,...,\gamma _n\in K\cup G$ such that $\gamma _n *(...*(\gamma _1*\beta )...) = 1.$ By (GE3), we can get that

$$\begin{aligned} \gamma _n *(...*(\gamma _{i+1}*(\gamma _i.... *(\gamma _1*\beta )...))...) = 1\in G\end{aligned}$$

such that $\gamma _1, \gamma _2,...,\gamma _i\in K$ and$\gamma _{i+1},...,\gamma _n\in G$. Since $\gamma _n\in G$ and $G$ is a GE-filter, we have

$$\begin{aligned} \gamma _{n-1}*(...*(\gamma _{i+1}*(\gamma _i...*(\gamma _1*\beta )...))...)\in G. \end{aligned}$$

By repeating this, we get $\gamma _i*(...*(\gamma _1*\beta )....)\in G.$ Take $c=\gamma _i*(...*(\gamma _1*\beta )....).$ By repeating (GE3), we get that

$$\begin{aligned} \gamma _i*(...*(\gamma _1*(c*\beta ))....)&= 1\in K, \end{aligned}$$

Since $\gamma _i\in K$ and $K$ is a GE-filter of $R$, we get $c*\beta \in K$. Put $\eta =\zeta *\beta$. Then, $\eta *(\zeta *\beta ) = (\zeta *\beta ) *(\zeta *\beta ) = 1$ and hence $\beta \in J$. Therefore, $J=\langle K\cup G\rangle$. $\square$

The intersection of two GE-filters $K$ and $G$ of a GE-algebra is a GE-filter of $R$, as can be seen. Also, $K\cap G$ is the infimum of both $K$ and $G$, as can be shown. The class of all GE-filters in a GE-algebra $R$ is designated by ${\mathfrak {O}}(R)$. The following theorem is now established.

Theorem 3.7

Let $R$ be a transitive GE-algebra. Then, ${\mathfrak {O}}(R)$ forms a complete distributive lattice.

Proof

For any two GE-filters $K_1$ and $K_2$ of a transitive GE-algebra, define

$$\begin{aligned} K_1\vee K_2= & {} \langle K_1 \cup K_2\rangle = \{\beta \in R\mid t*(r*\beta ) \\= & {} 1\text { for some } t\in K_1 \text { and } r\in K_2\}. \end{aligned}$$

Then, it is obvious that $({\mathfrak {O}}(R), \cap ,\vee )$ is a complete lattice with respect to set inclusion. Let $K_1,K_2,K_3\in {\mathfrak {O}}(R).$ Then, obviously, $K_1\cap (K_2 \vee K_3) \supseteq (K_1 \cap K_2) \vee (K_2 \cap K_3).$ Conversely, assume that $\beta \in K_1 \cap (K_2 \vee K_3).$ Then, $\beta \in K_1$ and $\beta \in K_2 \vee K_3.$ Then, there are $c\in K_2$ and $d\in K_3$ such that $c*(d*\beta ) = 1.$ Let $\alpha _1 = d*\beta$ and $\alpha _2 = \alpha _1 *\beta$. It can be observed that $\alpha _1\in K_1$ and $\alpha _2\in K_1.$ Now, $c*\alpha _1 = c*(d*\beta ) = 1\in K_2$ which implies that $\alpha _1=d*\beta \in K_2$. Hence, $\alpha _1\in K_1 \cap K_2$. Also, $d*\alpha _2 = d*(\alpha _1 *\beta ) = d*((d*\beta ) *\beta ) = d*((d*\beta ) *(d*\beta )) = d*1= 1\in K_3$. Then, $\alpha _2\in K_3$ and hence $\alpha _2\in K_1 \cap K_3.$ Now, by (GE3), $\alpha _1 *(\alpha _2 *\beta ) = (d*\beta ) *((\alpha _1 *\beta )*\beta ) = (d*\beta ) *(((d*\beta ) *\beta )*\beta ) = 1$. Hence, $\beta \in (K_1 \cap K_2) \vee (K_1 \cap K_3).$ Hence, $K_1 \cap (K_2 \vee K_3) \subseteq (K_1 \cap K_2) \vee (K_1 \cap K_3).$ Thus, $({\mathfrak {O}}(R), \cap ,\vee )$ is a complete distributive lattice. $\square$

Corollary 3.8

Let $R$ be a GE-algebra with transitivity. Then, with regard to the inclusion ordering $\subseteq$, the class ${\mathfrak {O}}(R)$ of all GE-filters of $R$ is a complete lattice in which for any class $\{K_{\zeta }\}_{\zeta \in \Delta }$ of GE-filters of $R$, inf$\{K_{\zeta }\}_{\zeta \in \Delta } =\cap _{\zeta \in \Delta } K_\alpha$ and sup$\{K_{\zeta }\}_{\zeta \in \Delta } =\langle \cup _{\zeta \in \Delta } K_\alpha \rangle .$

A GE-filter $K$ of a GE-algebra $R$ is said to be proper if $K\ne R$.

Definition 3.9

A proper GE-filter $N$ of a GE-algebra $R$ is said to be maximal if $\langle N\cup \{\beta \}\rangle = R$ for any $\beta \in R\setminus N$, where $\langle N\cup \{\beta \}\rangle$ is the GE-filter generated by $N\cup \{\beta \}$.

Example 3.10

Let $R=\{1, e, f, g, h, a, b\}$ be a set with the binary operation “$*$” in the following Cayley Table.

$$\begin{aligned}\begin{array}{c|ccccccc} *&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h&{}\quad a&{}\quad b\\ \hline 1&{} 1&{} e&{} f&{} g&{} h&{} a&{} b\\ e&{} 1&{} 1&{} 1&{} g&{} g&{} a&{} b\\ f&{} 1&{} e&{} 1&{} h&{} h&{} a&{} b\\ g&{} 1&{} e&{} 1&{} 1&{} 1&{} a&{} 1\\ h&{} 1&{} e&{} 1&{} 1&{} 1&{} a&{} b\\ a&{} 1&{} e&{} 1&{} g&{} g&{} 1&{} b\\ b&{} 1&{} 1&{} f&{} h&{} h&{} a&{} 1\end{array} \end{aligned}$$

Then, $(R, *, 1)$ is a GE-algebra and $H:=\{1,e,f,a,b\}$ is a proper GE-filter of $R$. Moreover, we have $\langle H\cup \{g\}\rangle =R=\langle H\cup \{h\}\rangle$, and so $H$ is a maximal GE-filter of $R$.

We now have a necessary and sufficient condition for any proper GE-filter to be maximal.

Theorem 3.11

Let $R$ be a transitive GE-algebra. Then, a proper GE-filter $K_1$ of $R$ is maximal if and only if $K_1 \subseteq K\subseteq R$ implies $K_1 =K$ or $K=R$ for any GE-filter $K$ of $R$.

Proof

Suppose $K_1$ is a maximal GE-filter of $R.$ Let $K$ be a GE-filter of $R$ such that $K_1\subseteq K\subseteq R.$ Suppose $K\ne R$. Then, we have to show that $K_1 =K$. Suppose $K_1 \ne K.$ Then, there exists $\beta \in K$ such that $\beta \notin K_1$. Since $K_1$ is a maximal GE-filter of $R$, we have $\langle K_1\cup \{\beta \}\rangle = R$. Let $\gamma \in R$. Then, $\gamma \in \langle K_1 \cup \{\beta \}\rangle .$ Then, $\beta *\gamma \in K_1 \subseteq K$ and hence, $\gamma \in K$. Therefore, $K=K_1$. Conversely, assume that the condition holds. Let $\beta \in R{\setminus }K_1$. Suppose $\langle K_1\cup \{\beta \}\rangle \ne R.$ Chose $\gamma \notin \langle K_1 \cup \{\beta \} \rangle$ and $\gamma \in R$. Hence, $K_1 \subseteq \langle K_1 \cup \{\beta \}\rangle \subset R.$ Then by assumption, we get $K_1 =\langle K_1 \cup \{\beta \}\rangle .$ Hence, $\beta \in K_1$ which is a contradiction. Thus, $K_1$ is a maximal GE-filter of $R.$ $\square$

Definition 3.12

A proper GE-filter $P$ of a GE-algebra $R$ is said to be prime if $Q\cap H\subseteq P$ implies $Q\subseteq P$ or $H\subseteq P$ for any two GE-filters $Q$ and $H$ of $R$.

Example 3.13

Consider the GE-algebra $R$ in Example 3.10. It is easy to verify that the set $K:=R\setminus \{a\}$ is a prime GE-filter of $R$.

Theorem 3.14

A proper GE-filter $P$ of a GE-algebra $R$ is prime if and only if $\langle \beta \rangle \cap \langle \gamma \rangle \subseteq P$ implies $\beta \in P$ or $\gamma \in P$ for all $\beta ,\gamma \in R.$

Proof

Assume that $P$ is a prime GE-filter of $R$. Let $\beta ,\gamma \in R$ be such that $\langle \beta \rangle \cap \langle \gamma \rangle \subseteq P$. Since $P$ is prime, it implies that $\beta \in \langle \beta \rangle \subseteq P$ or $\gamma \in \langle \gamma \rangle \subseteq P$. Conversely, assume that the condition holds. Let $K$ and $G$ be two GE-filters of $R$ such that $K\cap G\subseteq P.$ Let $\beta \in K$ and $\gamma \in G$. Then, $\langle \beta \rangle \subseteq K$ and $\langle \gamma \rangle \subseteq G.$ Hence, $\langle \beta \rangle \cap \langle \gamma \rangle \subseteq K\cap G\subseteq P.$ Then, $\beta \in P$ or $\gamma \in P$. Thus, $K\subseteq P$ or $G\subseteq P$. Therefore, $P$ is a prime GE-filter of $R$. $\square$

Theorem 3.15

Let $K$ be a GE-filter of a transitive GE-algebra $R$. Then, $P\cap L\subseteq K$ if and only if $\langle K\cup P\rangle \cap \langle K\cup L\rangle = K,$ for any GE-filters $P$ and $L$ of $R$.

Proof

Let $\langle K\cup P\rangle \cap \langle K\cup L\rangle =K$. Since $P\subseteq \langle K\cup P\rangle$ and $L\subseteq \langle K\cup L\rangle$, we get that $P\cap L\subseteq \langle K\cup P\rangle \cap \langle K\cup L\rangle =K.$ Therefore, $P\cap L\subseteq K.$ Conversely, assume that $P\cap L\subseteq K$. Clearly, $K\subseteq \langle K\cup P\rangle \cap \langle K\cup L\rangle .$ Let $t\in \langle K\cup P\rangle \cap \langle K\cup L\rangle .$ Since $K$ is a GE-filter of $R$, we get $(\beta _n*(...*(\beta _1*t)..))\in K,$ for some $n\in {\mathbb {N}}$ and $\beta _1,\beta _2,....\beta _n\in P$. It follows that, there exists $b_1\in K$ such that $\beta _n*(...*(\beta _1*t)..) = b_1$. By the similar argument, we have $\gamma _m*(...*(\gamma _1*t)..) = b_2$, for some $m\in {\mathbb {N}}, \gamma _1,\gamma _2,...,\gamma _m\in L$ and $b_2\in K.$ Hence, by repeating (GE3) and by (2.4), we get $\beta _n*(...*(\beta _1*(b_1*t))..)= 1$. Hence, $b_1*t\in P$. By the similar argument, we can show that $b_2*t\in L.$ Since $b_1,b_2\in K$, $b_1*t\le b_2 *(b_1 *t)$ and $b_2*t\le b_2 *(b_1 *t)$, we get $b_2*(b_1*t) \in P\cap L\subseteq K$. Hence, $t\in K$. Therefore, $\langle K\cup P\rangle \cap \langle K\cup L\rangle \subseteq K.$ Thus, $\langle K\cup P\rangle \cap \langle K\cup L\rangle =K$ $\square$

Corollary 3.16

Let $Q$ be a GE-filter of a transitive GE-algebra $R$. Then, for any $\eta ,\zeta \in R$,

$$\begin{aligned} \langle \eta \rangle \cap \langle \zeta \rangle \subseteq Q\text { if and only if } \langle Q\cup \{\eta \} \rangle \cap \langle Q\cup \{\zeta \} \rangle = Q. \end{aligned}$$

Theorem 3.17

In a transitive GE-algebra, every maximal GE-filter is a prime GE-filter.

Proof

Let $Q$ be a maximal GE-filter of a transitive GE-algebra $R$. Let $\langle \mu \rangle \cap \langle \nu \rangle \subseteq Q$ for some $\mu ,\nu \in R.$ Suppose $\mu \notin Q$ and $\nu \notin Q$. Then, $\langle Q\cup \{\mu \} \rangle = R$ and $\langle Q\cup \{\nu \} \rangle = R.$ Hence, $\langle Q\cup \{\mu \} \rangle \cap \langle Q\cup \{\nu \} \rangle = R.$ Hence, by Corollary 3.16, $\langle \mu \rangle \cap \langle \nu \rangle \nsubseteq Q$, which is a contradiction. Hence, $\mu \in Q$ or $\nu \in Q$. Therefore, $Q$ is a prime GE-filter of $R$. $\square$

Corollary 3.18

Let $R$ be a transitive GE-algebra. If $Q_1,Q_2,...Q_n$ and $Q$ are maximal GE-filters of $R$ such that $\bigcap \limits ^n_{j=1} Q_j\subseteq Q.$ There exists $i\in \{1,2,...,n\}$ such that $Q_i=Q$.

The following example shows that the converse of Theorem 3.17 is not valid.

Example 3.19

Let $R=\{1, e, f, g, h, a, b\}$ be a set with the binary operation “$*$” in the following Cayley Table.

$$\begin{aligned}\begin{array}{c|ccccccc} *&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h&{}\quad a&{}\quad b\\ \hline 1&{} 1&{} e&{} f&{} g&{} h&{} a&{} b\\ e&{} 1&{} 1&{} 1&{} h&{} h&{} b&{} b\\ f&{} 1&{} e&{} 1&{} g&{} g&{} a&{} a\\ g&{} 1&{} e&{} 1&{} 1&{} 1&{} a&{} b\\ h&{} 1&{} e&{} 1&{} 1&{} 1&{} a&{} b\\ a&{} 1&{} e&{} f&{} h&{} h&{} 1&{} 1\\ b&{} 1&{} e&{} f&{} g&{} h&{} 1&{} 1\end{array} \end{aligned}$$

Then, $(R,*, 1)$ is a transitive GE-algebra. It is routine to verify that $L:=\{1,e,f,g,h\}$ is a prime GE-filter of $R$. Note that $P:=\{1,f,g,h\}$ is a GE-filter of $R$ such that $P\subsetneq L\subsetneq R$. Hence, $L$ is not a maximal GE-filter of $R$.

Theorem 3.20

Let $R$ be a transitive GE-algebra and $G$ be a nonempty subset of $R$ such that $G$ is closed under $``\partial$”, where $\beta \partial \gamma := (\gamma *\beta )*\beta$, for any $\beta ,\gamma \in G$. If $K$ is a GE-filter of $R$ such that $K\cap G=\emptyset$, then there exist a prime GE-filter $M$ of $R$ such that $K\subseteq M$ and $M\cap G=\emptyset .$

Proof

Let $K$ be a GE-filter of $R$ such that $K\cap G=\emptyset .$ Consider

$$\begin{aligned} {\mathfrak {G}}=\{J\in {\mathfrak {G}}(R) \mid K\subseteq J\text { and } J\cap G=\emptyset \}. \end{aligned}$$

Clearly $K\in {\mathfrak {G}}$. Then, by Zorn’s lemma, ${\mathfrak {G}}$ has a maximal element, say $M$. Then, $K\subseteq M$ and $M\cap G=\emptyset .$ We prove that $M$ is a prime GE-filter of $R$. Suppose there are GE-filter $K,Q$ of $R$ such that $K\cap Q\subseteq M, K\nsubseteq M$ and $Q\nsubseteq M.$ By maximality of $M$, we have $\langle M\cup K\rangle \cap G\ne \emptyset$ and $\langle M\cup Q\rangle \cap G\ne \emptyset$. Let $t\in \langle M\cup K\rangle \cap G$ and $r\in \langle M\cup Q\rangle \cap G$. Since $t*(t\partial r) = t*((r*t) *t) = 1$ and $r*(t\partial r) = r*((r*t)*t) = 1$, we have $t\partial r\in \langle M\cup K\rangle \cap \langle M\cup Q\rangle .$ Also, $t,r\in G$ and $G$ is a GE-filter of $R$ implies that $t\partial r\in G$. Hence, $t\partial r\in (\langle M\cup K\rangle \cap \langle M\cup Q\rangle )\cap G.$ Therefore, $M\ne \langle M\cup K\rangle \cap \langle M\cup Q\rangle .$ Hence, by Theorem 3.15, $K\cap Q\nsubseteq M$ which is a contradiction. Therefore, $M$ is a prime GE-filter of $R$. $\square$

Corollary 3.21

Let $R$ be a transitive GE-algebra. Then, the following holds:

• (1) For any $\beta \in R\setminus \{1\},$ there exists a prime GE-filter $M$ such that $\beta \notin M$.

• (2) $\bigcap \{M\mid M\text { is a prime GE-filter of }$R$\} =\{1\}.$

• (3) Any proper GE-filter $K$ of $R$ can be expressed as the intersection of all prime GE-filters of $R$ containing $K$.

Theorem 3.22

If a GE-algebra $R$ is transitive, then ${\mathfrak {G}}(X)$ is a chain if and only if every proper GE-filter of $R$ is a prime GE-filter.

Proof

Suppose that ${\mathfrak {G}}(X)$ is a chain. Let $P$ be a proper GE-filter of $R$. Let $\mu ,\nu \in R$ be such that $\langle \mu \rangle \cap \langle \nu \rangle \subseteq P.$ Since $\langle \mu \rangle$ and $\langle \nu \rangle$ are GE-filters of $R,$ we get either $\langle \mu \rangle \subseteq \langle \nu \rangle$ or $\langle \nu \rangle \subseteq \langle \mu \rangle$. Hence, $\mu \in P$ or $\nu \in P$. Therefore, $P$ is a prime GE-filter of $R$.

Conversely, suppose that every proper GE-filter of $R$ is a prime GE-filter of $R$. Let $P$ and $L$ be two proper GE-filters of $R$. Since $P\cap L$ is a proper GE-filter of $R$, we get $P\subseteq P\cap L$ or $L\subseteq P\cap L.$ Hence, $P\subseteq L$ or $L\subseteq P$. Therefore, ${\mathfrak {G}}(X)$ is a chain. $\square$

4 Elitable GE-filters

In this section, the concept of elitable GE-filters is introduced and characterized. Some basic properties of elitable GE-filters are observed in terms of maximal GE-filters.

Definition 4.1

Let $K$ be a nonempty subset of a bordered GE-algebra $R$. Then, the elitable of $K$ is denoted by $K^{\otimes }$ and is defined as

Example 4.2

Let $R=\{0,1, e, f, g, h, a, b\}$ be a set with the binary operation “$*$” in the following Cayley Table.

$$\begin{aligned}\begin{array}{c|ccccccc} *&{}\quad 0&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h&{}\quad a\\ \hline 0&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1\\ 1&{} 0&{} 1&{} e&{} f&{} g&{} h&{} a\\ e&{} 0&{} 1&{} 1&{} 1&{} 0&{} 1&{} 1\\ f&{} g&{} 1&{} e&{} 1&{} g&{} e&{} e\\ g&{} 1&{} 1&{} e&{} 1&{} 1&{} e&{} e\\ h&{} 1&{} 1&{} 1&{} f&{} 1&{} 1&{} 1\\ a&{} 0&{} 1&{} 1&{} f&{} g&{} 1&{} 1\end{array} \end{aligned}$$

Then, $(R, *, 1)$ is a bordered GE-algebra. Given a nonempty subset $K$ of $R$, we have:

$$\begin{aligned} K^{\otimes }=\left\{ \begin{array}{ll} \{0, g, h\} &{}\mathrm{if}\;\, 0\in K, 1\notin K,\\ \{1, e, f, a\} &{}\mathrm{if}\;\, 0\notin K, 1\in K,\\ \emptyset &{}\mathrm{if}\;\, 0\notin K, 1\notin K,\\ R&{} \mathrm{if}\;\, 0\in K, 1\in K,\\ \end{array}\right. \end{aligned}$$

Given a nonempty subset $K$ in a bordered GE-algebra $R$, the elitable of $K$ may not be a GE-filter of $R$ as seen in the following example.

Example 4.3

In Example 4.2, if we take $K:=\{1,g,h\}$, then $K^{\otimes }=\{1, e, f, a\}$ and it is not a GE-filter of $R$ since $e*h=1$ and $e\in K^{\otimes }$ but $h\notin K^{\otimes }$.

Lemma 4.4

Let $R$ be a bordered GE-algebra and consider two elitable subsets $K$ and $G$ of $R$. Then, the following holds:

$$\begin{aligned}&K\subseteq G\Rightarrow K^{\otimes } \subseteq G^{\otimes }. \end{aligned}$$

(4.1)

$$\begin{aligned}&(K\cap G)^{\otimes }=K^{\otimes } \cap G^{\otimes }. \end{aligned}$$

(4.2)

Proof

Suppose $K\subseteq G$ and $\beta \in K^{\otimes }$. Then, and hence, $\beta \in G^{\otimes }$. Thus, (4.1) holds. Let $\beta \in (K\cap G)^{\otimes }$. Then, . Hence, and . Therefore, $\beta \in K^{\otimes }\cap G^{\otimes }$. Hence, $(K\cap G)^{\otimes }\subseteq K\cap G$. Suppose $\beta \in K^{\otimes }\cap G^{\otimes }$. Then, $\beta \in K^{\otimes }$ and $\beta \in G^{\otimes }$. Hence, and . Therefore, and hence, $\beta \in (K\cap G)^{\otimes }$. Therefore, $K^{\otimes }\cap G^{\otimes }\subseteq (K\cap G)^{\otimes }.$ Thus, (4.2) holds. $\square$

Lemma 4.5

Let $R$ be a bordered GE-algebra which is transitive. Then, for any $\beta , \gamma \in R$, we have

• (1) ,

• (2) ,

• (3) ,

• (4) ,

• (5) .

Proof

(1). Let $\beta \in R$. Then, by (GE1), (2.7) and (2.25),

Hence, , which gives .

(2). Let $\beta , \gamma \in R$. Then, by (2.21) and (2.25), .

(3). Let $\beta , \gamma \in R$. We can observe that . By (2.12), we get and so . Hence, by (GE1),(2.7), (2.20) and (2.21), we get

Thus, . Therefore, .

(4).  By (2.21), we have . Hence, by (2.25), (3) and (2.21), we get

(5). By (2.25), we get . Hence, . Also, by (4), we can observe that . Hence, (5) follows, since $R$ is transitive. $\square$

Theorem 4.6

Let $R$ be a bordered GE-algebra which is transitive. Then, for any GE-filter $K$ of $R$, we have the following:

• (1) $K^{\otimes }$ is a GE-filter of $R.$

• (2) $K\subseteq K^{\otimes }.$

• (3) $(K^{\otimes })^{\otimes } = K^{\otimes }.$

Proof

(1). Since , we have $1\in K^{\otimes }$. Let $\eta \in K^{\otimes }$ and $\eta *\zeta \in K^{\otimes }.$ Then, and . Since by Lemma 4.5(5) and , we have . Therefore, and hence, $\zeta \in K^{\otimes }$. Thus, $K^{\otimes }$ is a GE-filter of $R$.

(2). Let $\eta \in K$. Since and $\eta \in K$, we have . Hence, $\eta \in K^{\otimes }$. Thus, $K\subseteq K^{\otimes }$.

(3). Clearly $K^{\otimes } \subseteq (K^{\otimes })^{\otimes }$ by (2). Let $\eta \in (K^{\otimes })^{\otimes }$. Then, and hence, . Since implies that by (2.24), and $K$ is a GE-filter of $R$, we have . Therefore, $\eta \in K^{\otimes }$. Thus, $K^{\otimes }= (K^{\otimes })^{\otimes }.$ $\square$

The following example shows that if $R$ is a bordered GE-algebra which is not transitive, then the elitable of a GE-filter $K$ of $R$ may not be a GE-filter of $R$.

Example 4.7

Let $R=\{0,1,e,f,g,h\}$ be a set with a binary operation $*$ given in the following table:

$$\begin{aligned} \begin{array}{c|cccccc} *&{}\quad 0&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h\\ \hline 0&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1\\ 1&{} 0&{} 1&{} e&{} f&{} g&{} h\\ e&{} 0&{} 1&{} 1&{} 1&{} 0&{} 1\\ f&{} g&{} 1&{} e&{} 1&{} g&{} 1\\ g&{} 1&{} 1&{} e&{} 1&{} 1&{} 1\\ h&{} 1&{} 1&{} e&{} 1&{} 1&{} 1\end{array} \end{aligned}$$

Then, $R$ is a bordered GE-algebra which is not transitive. Let $K=\{1\}$. Then, $K$ is a GE-filter of $R$ and its elitable is $K^{\otimes }=\{1,e,f\}$. But $K^{\otimes }$ is not a GE-filter of $R$ since $e*h=1\in K^{\otimes }$ and $e\in K^{\otimes }$ but $h\notin K^{\otimes }$.

Definition 4.8

Let $R$ be a bordered GE-algebra. Then, a nonempty subset $K$ of $R$ is called an elitable GE-filter of $R$ if it is a GE-filter of $R$ and its elitable is $K$ itself.

Example 4.9

Consider a $4-$element Boolean algebra $R:= \{0, 1, e, e'\}$ with the partial order $\le$. If we define

$$\begin{aligned} x *y=\left\{ \begin{array}{ll} 1&{}\mathrm{if}\;\, x\le y\\ y &{}\mathrm{otherwise}\;\,\\ \end{array}\right. \end{aligned}$$

then $(R, *, 1)$ is a bordered GE-algebra. Let $K=\{1,e,e'\}$. Then, it can be easily verified that $K$ is an elitable GE-filter of $R$.

Example 4.10

In Example 4.2, we can observe that $G=\{1,e,f,a\}$ is an elitable GE-filter of $R$.

Theorem 4.11

Let $R$ be a bordered GE-algebra. Then, the intersection of two elitable GE-filters of $R$ is also an elitable GE-filter of $R$.

Proof

Let $K_1$ and $K_2$ be two elitable GE-filters of $R$. Then, $K_1^{\otimes }=K_1$ and $K_2^{\otimes }=K_2$ which imply from (4.2) that

$$\begin{aligned} K_1\cap K_2=K_1^{\otimes }\cap K_2^{\otimes } =(K_1\cap K_2)^{\otimes }. \end{aligned}$$

Hence, $K_1\cap K_2$ is an elitable GE-filter of $R$. $\square$

In the following example, we can find two elitable GE-filters of $R$ whose union is not an elitable GE-filter of $R$.

Example 4.12

Let $R=\{0,1,e,f,g,h\}$ be a set with a binary operation $*$ given in the following table:

$$\begin{aligned} \begin{array}{c|cccccc} *&{}\quad 0&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h\\ \hline 0&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1\\ 1&{} 0&{} 1&{} e&{} f&{} g&{} h\\ e&{} h&{} 1&{} 1&{} 1&{} h&{} h\\ f&{} g&{} 1&{} e&{} 1&{} g&{} g\\ g&{} f&{} 1&{} e&{} f&{} 1&{} 1\\ h&{} f&{} 1&{} e&{} f&{} 1&{} 1\end{array} \end{aligned}$$

Then, $R$ is a bordered GE-algebra. Let $K_1=\{1,e,f\}$ and $K_2=\{1,g,h\}$. Then, it is routine to verify that $K_1$ and $K_2$ are elitable GE-filters of $R$. But $K_1\cup K_2=\{1,e,f,g,h\}$ is not an elitable GE-filter of $R$ since $f*0=g\in K_1\cup K_2$ and $f\in K_1\cup K_2$ but $0\notin K_1\cup K_2$.

Theorem 4.13

For any transitive bordered GE-algebra $R$, the class ${\mathfrak {O}}^{\otimes }(R)$ of all elitable GE-filters of $R$ is a complete distributive lattice.

Proof

For any two elitable GE-filters $K$ and $G$ of $R$, define a relation $\le$ on ${\mathfrak {O}}^{\otimes }(X)$ by $K\le G\Leftrightarrow K\subseteq G$. Then, it can be observed that $({\mathfrak {O}}^{\otimes }(X),\le )$ is a partially ordered set. Consider $K\cap G=(K\cap G)^{\otimes }$ and $K\sqcup G= (K\vee G)^{\otimes },$ where

$$\begin{aligned}&K\vee G\\&\quad =\{\beta \in R\mid t*(r*\beta ) = 1 \text { for some } t\in K\text { and } r\in G\}. \end{aligned}$$

Obviously $(K\cap G)^{\otimes }$ is infimum of $K$ and $G$ in ${\mathfrak {O}}^{\otimes }(X)$. Also, $(K\vee G)^{\otimes }$ is the upper bound of $K^{\otimes }$ and $G^{\otimes }$. Let $M$ be any elitable GE-filter of $R$ such that $K\subseteq M$ and $G\subseteq M$. Let $\beta \in (K\vee G)^{\otimes }$. Then, , and hence, there exist $t\in K$ and $r\in G$ such that . Since $K\subseteq M$ and $G\subseteq M$, we get . Hence, $\beta \in M.$ Therefore, $(K\vee G)^{\otimes }$ is supremum of $K$ and $G$ in ${\mathfrak {O}}^{\otimes }(X)$. Hence, $({\mathfrak {O}}^{\otimes }(X),\cap , \sqcup )$ is a lattice. We can observe that the set theoretical intersection of an arbitrary set of elitable GE-filters of $R$ is an elitable GE-filter of $R$ again by (4.2). Hence, the set ${\mathfrak {O}}^{\otimes }(X)$ forms a complete lattice with respect to set inclusion. The least and greatest element of ${\mathfrak {O}}^{\otimes }(X)$ are $\{1\}$ and $R$, respectively. Now, for any $K, G, M\in {\mathfrak {O}}^{\otimes }(X)$, we obtain

$$\begin{aligned} (K\sqcup G) \cap (K\sqcup M)= & {} (K\vee G)^{\otimes } \cap (K\vee M)^{\otimes }\\= & {} ((K\vee G) \cap (K\vee M))^{\otimes }\\= & {} (K\vee (G\cap M))^{\otimes }\\= & {} K\sqcup (G\cap M). \end{aligned}$$

Therefore, $({\mathfrak {O}}^{\otimes }(R),\sqcup , \cap ,\{1\},R)$ is a distributive lattice. $\square$

Theorem 4.14

Given a GE-filter $K$ of a transitive bordered GE-algebra $R$, the following are equivalent:

• (1) $K$ is an elitable GE-filter.

• (2) implies $\eta \in K,$ for all $\eta \in R$.

• (3) implies $\zeta \in K$, for all $\eta ,\zeta \in R.$

Proof

(1) $\Rightarrow$ (2) Assume (1) and . Then, $\eta \in K^{\otimes }=K$. Thus, (2) follows.

(2) $\Rightarrow$ (3) Assume (2) and $\eta ,\zeta \in R$ such that . We can observe that and hence, . Since $K$ is an GE-filter of $R$ and , we get . Therefore, $\zeta \in K$ by (2). Thus, (3) follows.

(3) $\Rightarrow$ (1) Assume (3). Let $\eta \in K^{\otimes }$. Then, . Therefore, and hence, $\eta \in K$ by (3). Thus, $K^{\otimes }\subseteq K$, and hence, $K$ is an elitable GE-filter of $R$. $\square$

Theorem 4.15

If $K$ is an elitable GE-filter of a transitive bordered GE-algebra $R$, then for and $\beta \in K$ imply that $\gamma \in K$.

Proof

Let $K$ be an elitable GE-filter of $R$ and $\beta ,\gamma \in R$ be such that and $\beta \in K$. Then, and hence, since $K$ is GE-filter of $R$. Therefore, $\gamma \in K$ by Theorem 4.14. $\square$

Theorem 4.16

Every maximal GE-filter of a transitive bordered GE-algebra $R$ is an elitable GE-filter of $R$.

Proof

Let $H$ be a maximal GE-filter of $R$. Clearly, $H\subseteq H^{\otimes }$ by Theorem 4.6(2). Now, we prove that $H^{\otimes }\subseteq H$. Suppose $H^{\otimes }\nsubseteq H$. Then, there exists $\beta \in H^{\otimes }$ such that $\beta \notin H$. Hence, and $\langle H\cup \{\beta \}\rangle =R$. Since $0\in R$, we have Since $0\le \beta$, we have $(\beta *0)*0\le (\beta *0) *\beta$ by (2.12). That is . Since and $H$ is a GE-filter of $R$, we get $\beta \in H$ which is a contradiction. Hence, $H^{\otimes }\subseteq H$. Thus, $H$ is an elitable GE-filter of $R$. $\square$

The following example shows that the converse of Theorem 4.16 is not valid.

Example 4.17

Let $R=\{0,1, e, f, g, h, a\}$ be a set with the binary operation “$*$” in the following Cayley Table.

$$\begin{aligned}\begin{array}{c|ccccccc} *&{}\quad 0&{}\quad 1&{}\quad e&{}\quad f&{}\quad g&{}\quad h&{}\quad a\\ \hline 0&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1\\ 1&{} 0&{} 1&{} e&{} f&{} g&{} h&{} a\\ e&{} 0&{} 1&{} 1&{} f&{} a&{} h&{} a\\ f&{} 0&{} 1&{} 1&{} 1&{} g&{} h&{} g\\ g&{} h&{} 1&{} 1&{} 1&{} 1&{} h&{} 1\\ h&{} a&{} 1&{} 1&{} f&{} a&{} 1&{} a\\ a&{} h&{} 1&{} 1&{} 1&{} 1&{} h&{} 1\end{array} \end{aligned}$$

Then, $(R, *, 1)$ is a transitive bordered GE-algebra. Consider two GE-filters $K=\{1,e,f\}$ and $G=\{1,e,f,g,a\}$ of $R$. Then, we can observe that $K$ is an elitable GE-filter of $R$, but $K$ is not maximal GE-filter of $R$ since $K\subseteq G\subseteq R$ but $K\ne G$ and $G\ne R$.

We provide conditions for a subset of a transitive bordered GE-algebra to be an elitable GE-filter.

Theorem 4.18

Let $K$ be a nonempty subset of a transitive bordered GE-algebra $R$ such that its elitable is $K$ itself. If $K$ satisfies (2.18) and

(4.3)

then $K$ is an elitable GE-filter of $R$.

Proof

Suppose $K^{\otimes }=K$ and satisfies (2.18) and (4.3). Since $K\ne \emptyset$, there exists $\eta \in K$. As $\eta \le 1$, we have $1\in K$ by (2.18). Let $\eta , \zeta \in R$ be such that $\eta \in K$ and $\eta *\zeta \in K$. Then, and by (2.10) and (2.18), which induces by (4.3). Since and , we have by (2.18). Now, by (2.21). Hence, it follows from (2.24) and (2.18) that . Since by (2.2), we have by (2.24) and hence by (2.18). Thus, $\zeta \in K$, and so $K$ is a GE-filter of $R$. Therefore, $K$ is an elitable GE-filter of $R$. $\square$

Corollary 4.19

If $R$ is antisymmetric, then, for all $\eta ,\zeta \in R$, the following are equivalent.

• (1) $R$ is duplex.

• (2) implies $\eta =\zeta .$

• (3)

Corollary 4.20

If $R$ is antisymmetric, then the conditions below are equivalent.

• (1) $R$ is duplex.

• (2) Every GE-filter is an elitable GE-filter.

• (3) Every principal GE-filter is an elitable GE-filter.

Open Problem. We were unable to identify an example of an elitable GE-filter for an infinite GE-algebra. Is it possible to construct an elitable GE-filter for an infinite GE-algebra?

5 Conclusion

In this paper, we have introduced the notions of maximal GE-filter and prime GE-filter in a GE-algebra and studied the relation between them. We have characterized prime GE-filter in terms of the GE-filter generated by a subset of a transitive GE-algebra. We generalized Stone’s theorem to transitive GE-algebras. We have introduced the notion of elitable GE-filter of a bordered GE-algebra and investigated its properties. We have observed that the class of all elitable GE-filters of a transitive bordered GE-algebra is a complete distributive lattice. We have given equivalent conditions for a GE-filter of a transitive bordered GE-algebra to be elitable GE-filter. We have provided conditions for a subset of a transitive bordered GE-algebra to be elitable GE-filter.