A study of bi-bases of ternary semigroups

Abstract

In this paper, we introduce the bi-bases of a ternary semigroup. The results of this paper are based on the bi-ideals generated by a non-empty subset of a ternary semigroup. Moreover, we define the quasi-order relation of a ternary semigroup and study some of their interesting properties.

1 Introduction and preliminaries

The idea of investigation of n-ary algebras, i.e. the sets with one n-ary operation, was given by Kasner [3]. In particular, n-ary semigroups are known as ternary semigroups for n = 3 with one associative operation [6]. Kerner [4] expressed many applications of ternary structures in physics. The concept of ideal in ternary semigroup was given by Sioson [7]. He also defined regular ternary semigroups. The properties of quasi-ideals and bi-ideals in ternary semigroups were studied by Dixit and Dewan [1].

Tamura [8] introduced the notion of a (right)left base of semigroup. Later, Fabrici described a semigroup structure containing one-sided bases [2]. Thongkam and Changphas [9] introduced the notion of left bases and right bases of a ternary semigroup. Kumoon and Changphas [5] introduced the concept of bi-bases in the semigroups and discussed some interesting results.

To start with, we need the following.

Definition 1.1

[6] A non-empty set S is called a ternary semigroup if there exists a ternary operation $S \times S \times S \rightarrow S$, written as$(x_{1}, x_{2}, x_{3}) \rightarrow [x_{1}x_{2}x_{3}]$, satisfying the following identity for any $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ $\in S,$

$$\begin{aligned} {[[}x_{1}x_{2}x_{3}]x_{4}x_{5}] = [x_{1}[x_{2}x_{3}x_{4}]x_{5}] = [x_{1}x_{2}[x_{3}x_{4}x_{5}]].\ \end{aligned}$$

For non-empty subsets A, B and C of a ternary semigroup S,

$$\begin{aligned}{}[ABC]:= \{[abc]: a \in A, b \in B \,{\textrm{and }}\, c \in C \}. \end{aligned}$$

If $A =\{a\}$, then we write $[\{a\}BC]$ as [aBC] and similarly if $B =\{b\}$ or $C =\{c\}$, we write [AbC] and [ABc], respectively. Throughout the paper, we denote $[x_{1}x_{2}x_{3}]$ by $x_{1}x_{2}x_{3}$ and [ABC] as ABC.

Definition 1.2

[7] A non-empty subset B of a ternary semigroup S is called a ternary sub-semigroup of S, if $BBB \subseteq B$.

Definition 1.3

[1] A non-empty subset B of a ternary semigroup S is called a bi-ideal of S if BSBSB $\subseteq$ B.

Proposition 1.1

[1] Let B be a non-empty subset of a ternary semigroup S without identity. Then $B \cup [BBB] \cup [BSBSB]$ is the smallest bi-ideal of S containing B.

Remark 1.1

In this paper, smallest bi-ideal of S containing B is denoted by $(B)_{b}$.

2 Main results

Definition 2.1

Let S be a ternary semigroup. A non-empty subset B of S is called a bi-base of S if it satisfies the following conditions:

• (1) $S=(B)_{b}$ (i.e. $S=B \cup BBB \cup BSBSB$);

• (2) If A is a subset of B such that $S =(A)_{b}$, then $A=B$.

Example 2.1

Let S = $\{a, b, c, d\}$ with xyz = $(x \circ y) \circ z$, for all x, y, z $\in$ S and a ternary operation, $'\circ '$ given by the following table:

Then S is a ternary semigroup. Let B = $\{b, c, d\}$, then clearly $(B)_{b}$ = $B\cup BBB\cup BSBSB$ = S and there is no proper subset A of B such that $S=(A)_{b}$. This shows that B is a bi-base of S.

Theorem 2.1

If B is a bi-base of a ternary semigroup S and $a, b \in B$ such that $a\in bbb \cup bSbSb$, then $a = b$.

Proof

Let B be a bi-base of a ternary semigroup S and $a, b \in B$ such that $a \in bbb\cup bSbSb$, and suppose that $a \ne b$. Consider A = ${B}{\setminus } \{a\}$, then $A\subseteq B$. Since $a\ne b$, therefore $b\in A$. Clearly, $(A)_{b}\subseteq S$. Let $x\in S$, then by $S=(B)_{b}$, we have $x \in B \cup BBB \cup BSBSB$. Now, three cases arise:

Case 1: For an element $x\in B$, we have two subcases

Subcase 1.1: If $x \ne a$, then $x \in B{\setminus } \{a\}=A\subseteq (A)_{b}$.

Subcase 1.2: If $x=a$, then by assumption, we have $x = a \in bbb\cup bSbSb\subseteq AAA\cup ASASA \subseteq (A)_{b}$.

Case 2: If $x \in BBB$. Then for $x=b_{1}b_{2}b_{3}$, for some $b_{1},b_{2},b_{3}\in B$, we have eight subcases

Subcase 2.1: If $b_{1} = a = b_{2} = b_{3}$. Then, by assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1} b_{2} b_{3}\\ &{}\in &{} (bbb \cup bSbSb)(bbb \cup bSbSb)(bbb \cup bSbSb) \\ &{}\subseteq &{} (AAA\cup ASASA)(AAA\cup ASASA)(AAA\cup ASASA )\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Subcase 2.2: If $b_{1} \ne a$, $b_{2} = a$, $b_{3} = a$. Then, by assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1} b_{2} b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})(bbb\cup bSbSb)(bbb\cup bSbSb)\\ &{}=&{} ({B}{\setminus } \{a\}bbb\cup {B}{\setminus } \{a\}bSbSb)(bbb\cup bSbSb)\\ &{}=&{} {B}{\setminus } \{a\}bbbbbb\cup {B}{\setminus } \{a\}bbbbSbSb\cup {B}{\setminus } \{a\}bSbSbbbb\cup {B}{\setminus } \{a\}bSbSbbSbSb\\ &{}\subseteq &{} AAAAAAA\cup AAAASASA\cup AASASAAAA\cup AASASAASASA\\ &{}\subseteq &{} ASASA\subseteq (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 2.3 for $b_{2}\ne a$, $b_{1}=a$, $b_{3}=a$ and subcase 2.4 for $b_{3}\ne a$, $b_{1}=a$, $b_{2}=a$.

Subcase 2.5: If $b_{1}\ne a$, $b_{2}\ne a$, $b_{3}=a$. Then, by assumption and $A={B}{\setminus }$ $\{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}b_{2}b_{3} \\ &{}\in &{} ({B}{\setminus } \{a\})({B}{\setminus } \{a\})(bbb\cup bSbSb)\\ &{}=&{} {B}{\setminus } \{a\}{B}{\setminus } \{a\}bbb\cup {B}{\setminus } \{a\}{B}{\setminus } \{a\}bSbSb\\ &{}\subseteq &{} AAAAA\cup AAASASA\\ &{}\subseteq &{} ASASA\\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 2.6 for $b_{2}\ne a$, $b_{3}\ne a$, $b_{1}=a$ and subcase 2.7 for $b_{1}\ne a$, $b_{3}\ne a$, $b_{2}=a$.

Subcase 2.8: If $b_{1} \ne a$, $b_{2}\ne a$ and $b_{3}\ne a$. By assumption and for $A={B} {\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}b_{2}b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})({B}{\setminus } \{a\})({B}{\setminus } \{a\})\\ &{}\subseteq &{} AAA\\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Case 3: If $x\in BSBSB$. Then $x=b_{1}s_{1}b_{2}s_{2}b_{3}$, for some $b_{1},b_{2},b_{3}\in B$ and $s_{1},s_{2}\in S$. Again, we have eight subcases.

Subcase 3.1: If $b_{1} = a = b_{2} = b_{3}$. By assumption

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\\ &{}\in &{} (bbb\cup bSbSb)S(bbb\cup bSbSb)S(bbb\cup bSbSb)\\ &{}\subseteq &{} (AAA\cup ASASA)S(AAA\cup ASASA)S(AAA\cup ASASA )\\ &{}\subseteq &{} ASASA\\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Subcase 3.2: If $b_{1} \ne a$, $b_{2}=a$, $b_{3}=a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x&{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})S(bbb\cup bSbSb)S(bbb\cup bSbSb)\\ &{}=&{} ({B}{\setminus } \{a\}Sbbb\cup {B}{\setminus } \{a\}SbSbSb)S(bbb\cup bSbSb)\\ &{}=&{} {B}{\setminus } \{a\}SbbbSbbb\cup {B}{\setminus } \{a\}SbbbSbSbSb\cup {B}{\setminus } \{a\}SbSbSbSbbb \cup {B}{\setminus } \{a\}SbSbSbSbSbSb\\ &{}\subseteq &{} ASAAASAAA\cup AAAASASA\cup ASASASAAAA\cup ASASASASASASA\\ &{}\subseteq &{} ASASA\\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 3.3 for $b_{2}\ne a$, $b_{1}=a$, $b_{3}=a$ and the subcase 3.4 for $b_{3}\ne a$,$b_{1}=a$, $b_{2}=a$.

Subcase 3.5: If $b_{1}\ne a$,$b_{2}\ne a$,$b_{3}=a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})S({B}{\setminus } \{a\})S(bbb\cup bSbSb)\\ &{}=&{} {B}{\setminus } \{a\}S{B}{\setminus } \{a\}Sbbb\cup {B}{\setminus } \{a\}S{B}{\setminus } \{a\}SbSbSb\\ &{}\subseteq &{} ASASAAA\cup ASASASASA\\ &{}\subseteq &{} ASASA\\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 3.6 for $b_{2}\ne a$, $b_{3}\ne a$, $b_{1}=a$ and subcase 3.7 for $b_{1}\ne a$, $b_{3}\ne a$, $b_{2}=a$.

Subcase 3.8: If $b_{1}\ne a$, $b_{2}\ne a$ and $b_{3}\ne a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})S({B}{\setminus } \{a\})S({B}{\setminus } \{a\})\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Thus in all cases $x \in (A)_{b}$, it implies $(A)_{b}=S$, which is a contradiction, as B is a bi-base of S. Hence, we have $a = b$ as required. $\square$

Theorem 2.2

Let B be a bi-base of a ternary semigroup S and $a, b, c, d \in B$ such that $a\in bcd\cup bScSd$, then $a=b$ or $a=c$ or $a=d$.

Proof

Assume that $a\in bcd\cup bScSd$ and if possible $a\ne b$, $a\ne c$ and $a\ne d$. Consider A = ${B} {\setminus } \{a\}$, then $A\subseteq B$. Since $a\ne b$, $a\ne c$ and $a\ne d$, we have $b, c, d \in A$. Therefore $(A)_{b}\subseteq S$. Let $x\in S$. Then, by $S=(B)_{b}$, we have $x\in B\cup BBB\cup BSBSB$. Now, three cases arise:

Case 1: For $x \in B$, we have two subcases

Subcase 1.1: If $x \ne a$, then by assumption $x\in {B} {\setminus } \{a\}=A \subseteq (A)_{b}$.

Subcase 1.2: If $x = a$, then by hypothesis $x = a \in bcd\cup bScSd$ $\subseteq$ AAA $\cup$ ASASA $\subseteq$ $(A)_{b}$.

Case 2: If $x \in BBB$. Then $x = b_{1}b_{2}b_{3}$, for some $b_{1}, b_{2}, b_{3} \in B$. Now, we have eight subcases

Subcase 2.1: If $b_{1} = a = b_{2} = b_{3}$. By assumption and $A=B {\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}b_{2}b_{3}\\ &{}\in &{} (bcd\cup bScSd)(bcd\cup bScSd)(bcd\cup bScSd)\\ &{}\subseteq &{} (AAA\cup ASASA)(AAA\cup ASASA)(AAA\cup ASASA )\\ &{}\subseteq &{} ASASA \\ &{} \subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Subcase 2.2: If $b_{1}\ne a$, $b_{2}=a$, $b_{3}=a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}b_{2}b_{3}\in ({B}{\setminus } \{a\})(bcd\cup bScSd)(bcd\cup bScSd)\\ &{}=&{} ({B}{\setminus } \{a\}bcd\cup {B}{\setminus } \{a\}bScSd)(bcd\cup bScSd)\\ &{}=&{} {B}{\setminus } \{a\}bcdbcd \cup {B}{\setminus } \{a\}bcdbScSd\cup {B}{\setminus } \{a\}bScSdbcd\cup {B}{\setminus } \{a\}bScSdbScSd\\ &{}\subseteq &{} AAAAAAA\cup AAAASASA\cup AASASAAAA\cup AASASAASASA\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 2.3 for $b_{2}\ne a$, $b_{1}=a$, $b_{3}=a$ and subcase 2.4 for $b_{3}\ne a$, $b_{1}=a$, $b_{2}=a$.

Subcase 2.5: If $b_{1}\ne a$, $b_{2}\ne a$, $b_{3}=a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}b_{2}b_{3}\in ({B}{\setminus } \{a\})({B}{\setminus } \{a\})(bcd\cup bScSd)\\ &{}=&{} {B}{\setminus } \{a\}{B}{\setminus } \{a\}bcd\cup {B}{\setminus } \{a\}{B}{\setminus } \{a\}bScSd\\ &{}\subseteq &{} AAAAA \cup AAASASA\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 2.6 for $b_{2}\ne a$, $b_{3}\ne a$, $b_{1}=a$ and the subcase 2.7 for $b_{1}\ne a$, $b_{3}\ne a$, $b_{2}=a$.

Subcase 2.8: If $b_{1}\ne a$, $b_{2}\ne a$ and $b_{3}\ne a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x&{}=&{} b_{1}b_{2}b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})({B}{\setminus } \{a\})({B}{\setminus } \{a\})\\ &{}\subseteq &{} AAA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Case 3: If $x\in BSBSB$. Then $x = b_{1}s_{1}b_{2}s_{2}b_{3}$, for some $b_{1},b_{2},b_{3}\in B$ and $s_{1},s_{2}\in S$. Again, we have eight subcases

Subcase 3.1: If $b_{1} = a = b_{2} = b_{3}$. By assumption, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\\ &{}\in &{} (bcd\cup bScSd)S(bcd\cup bScSd)S(bcd\cup bScSd)\\ &{}\subseteq &{} (AAA\cup ASASA)S(AAA\cup ASASA)S(AAA\cup ASASA )\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Subcase 3.2: If $b_{1}\ne a$, $b_{2}=a$, $b_{3}=a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3} \\ &{}\in &{}({B}{\setminus } \{a\})S(bcd\cup bScSd)S(bcd\cup bScSd)\\ &{}=&{} ({B}{\setminus } \{a\}Sbcd\cup {B}{\setminus } \{a\}SbScSd)S(bcd\cup bScSd)\\ &{}=&{} {B}{\setminus } \{a\}SbcdSbcd\cup {B}{\setminus } \{a\}SbcdSbScSd\cup {B}{\setminus } \{a\}SbScSdSbcd\cup {B}{\setminus } \{a\}SbScSdSbScSd\\ &{}\subseteq &{} ASAAASAAA\cup AAAASASA\cup ASASASAAAA\cup ASASASASASASA\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 3.3 for $b_{2}\ne a$, $b_{1}=a$, $b_{3}=a$ and subcase 3.4 for $b_{3}\ne a$,$b_{1}=a$, $b_{2}=a$.

Subcase 3.5: If $b_{1}\ne a$, $b_{2}\ne a$, $b_{3}=a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x &{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\in ({B}{\setminus } \{a\})S({B}{\setminus } \{a\})S(bcd\cup bScSd)\\ &{}=&{} {B}{\setminus } \{a\}S{B}{\setminus } \{a\}Sbcd\cup {B}{\setminus } \{a\}S{B}{\setminus } \{a\}SbScSd\\ &{}\subseteq &{} ASASAAA\cup ASASASASA\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Similarly, we can prove the subcase 3.6 for $b_{2}\ne a$, $b_{3}\ne a$, $b_{1}=a$ and subcase 3.7 for $b_{1}\ne a$, $b_{3}\ne a$, $b_{2}=a$.

Subcase 3.8: If $b_{1}\ne a$, $b_{2}\ne a$ and $b_{3}\ne a$. By assumption and $A={B}{\setminus } \{a\}$, we have

$$\begin{aligned} \begin{array}{lll} x&{}=&{} b_{1}s_{1}b_{2}s_{2}b_{3}\\ &{}\in &{} ({B}{\setminus } \{a\})S({B}{\setminus } \{a\})S({B}{\setminus } \{a\})\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Thus in all cases x $\in$ $(A)_{b}$. It follows that $(A)_{b}=S$, which is a contradiction as B is a bi-base S. Hence, $a=b$ or $a=c$ or $a=d$. $\square$

Definition 2.2

Let S be a ternary semigroup.Then a quasi-order on S is defined as $a\leqslant _{b} b$ $\Leftrightarrow (a)_{b}\subseteq (b)_{b}$ for any $a, b\in S$.

Theorem 2.3

Let B be a bi-base of a ternary semigroup S. Then $a\leqslant _{b} b$, $a\leqslant _{b} c$ and $b\leqslant _{b} c$ if and only if $a=b=c$ for any $a, b, c \in S$.

Proof

It is straightforward. $\square$

Theorem 2.4

Let B be a bi-base of a ternary semigroup S such that $a, b, c, d \in B$ and $s \in S$, then following statements are true:

• (1) If $a \in bcd \cup bcdbcdbcd \cup bcdSbcdSbcd$, then $a=b$ or $a=c$ or $a=d$.

• (2) If $a\in bscsd\cup bscsdsbscsdsbscsd \cup bscsdSbscsdSbscd$, then $a=b$ or $a=c$ or $a=d$.

Proof

(1) Assume that $a \in bcd\cup bcdbcdbcd \cup bcdSbcdSbcd$ and suppose that $a\ne b$, $a\ne c$ and $a\ne d$. Let $A={B}{\setminus } \{a\}$. Then $A\subseteq B$. Since $a\ne b$, $a\ne c$ and $a\ne d$, we have $b, c, d \in A$. We show that $B \subseteq (A)_{b}$. Let $x\in B$ and if $x\ne a$, then $x\in A$, and so $x\in (A)_{b}$. If $x=a$ then by assumption, we have

$$\begin{aligned} \begin{array}{lll} x &{}\in &{} bcd\cup bcdbcdbcd \cup bcdSbcdSbcd\\ &{}\subseteq &{} AAA\cup AAAAAAAAA\cup AAASAAASAAA\\ &{}\subseteq &{} ASASA \\ &{}\subseteq &{} (A)_{b}. \end{array} \end{aligned}$$

Thus $B\subseteq (A)_{b}$. This implies that $(B)_{b}\subseteq (A)_{b}$. Since B is a bi-base of S, therefore $S=(B)_{b}\subseteq (A)_{b}\subseteq S$. It implies $S=(A)_{b}$, which is a contradiction. Hence $a=b$ or $a=c$ or $a=d$.

(2) The proof is similar to (1). $\square$

Theorem 2.5

Let B be a bi-base of a ternary semigroup S. Then the following statements are true:

• (1) For any $a, b, c, d \in B$ if $a\ne b$, $a\ne c$ and $a\ne d$ then $a\nleq _{b}bcd$.

• (2) For any $a, b, c, d \in B$ and $s\in S$, if $a\ne b$, $a\ne c$ and $a\ne d$, then $a\nleq _{b}bscsd$.

Proof

(1) For any $a, b, c, d \in B$, let $a\ne b$, $a\ne c$ and $a\ne d$. Suppose that $a\leqslant _{b}bcd$, we have

$$\begin{aligned} \begin{array}{lll} a &{}\subseteq &{} (a)_{b} \\ &{}\subseteq &{} (bcd)_{b} \\ &{}=&{} bcd\cup bcdbcdbcd \cup bcdSbcdSbcd. \end{array} \end{aligned}$$

By Theorem 2.4(1), it follows that $a=b$ or $a=c$ or $a=d$. This contradict the assumption. Hence $a\nleq _{b}bcd$.

(2) The proof is similar to (1). $\square$

Theorem 2.6

A non-empty subset B of a ternary semigroup S is a bi-base of S if and only if it satisfies the following conditions

• (1) For any $x\in S$,

  • (a) there exists $b\in B$ such that $x\leqslant _{b} b$,

  • (b) there exists $b_{1},b_{2},b_{3}\in B$ such that $x\leqslant _{b} b_{1}b_{2}b_{3}$,

  • (c) there exists $b_{1},b_{2},b_{3}\in B$, $s\in S$ such that $x\leqslant _{b} b_{1}sb_{2}sb_{3}$.

• (2) For any $a, b, c, d \in B$, let $a\ne b$, $a\ne c$ and $a\ne d$, then $a\nleq _{b}bcd$.

• (3) For any $a, b, c, d \in B$ and $s\in S$, let $a\ne b$, $a\ne c$ and $a\ne d$, then $a\nleq _{b}bscsd$.

Proof

Suppose that B is a bi-base of S, then $S=(B)_{b}$. To prove (1), let $x\in S$, it implies $x\in B\cup BBB\cup BSBSB$. Now, three cases arise:

Case 1: If $x\in B$. Then, $x=b$, for some $b\in B$. This implies $(x)_{b}\subseteq (b)_{b}$. Hence $x\leqslant _{b} b$.

Case 2: If $x\in BBB$, then $x = b_{1}b_{2}b_{3}$, for some $b_{1}, b_{2}, b_{3} \in B$. It implies $(x)_{b}\subseteq (b_{1}b_{2}b_{3})_{b}$. Hence, $x\leqslant _{b} b_{1}b_{2}b_{3}$.

Case 3: If $x \in BSBSB$, then $x=b_{1}sb_{2}sb_{3}$ for some $b_{1}, b_{2}, b_{3} \in B$ and $s\in S$. It implies $(x)_{b}\subseteq (b_{1}sb_{2}sb_{3})_{b}$. Hence $x\leqslant _{b} b_{1}sb_{2}sb_{3}$. Proofs of (2) and (3) are similar to the Theorem 2.5.

Conversely, suppose that (1), (2) and (3) holds. Then, we have to prove that B is a bi-base of S. Clearly $(B)_{b}\subseteq S$ and by (1) $S\subseteq (B)_{b}$ and so $S=(B)_{b}$. Now, it remains to show that B is minimal subset of S. Suppose that $S=(A)_{b}$ for some $A\subseteq B$. Since $A\subseteq B$, there exists $b\in {B}{\setminus } A$. Since $b\in B\subseteq S=(A)_{b}$ and $b\notin A$, it implies $b\in AAA\cup ASASA$. Now, two cases arise:

Case 1: If $b\in AAA$, then $b = a_{1}a_{2}a_{3}$, for some $a_{1}, a_{2}, a_{3}\in A$. As $b\notin A$ so, $b\ne a_{1}$, $b\ne a_{2}$ and $b\ne a_{3}$. It implies $a_{1}, a_{2}, a_{3}\in A$. Since $b=a_{1}a_{2}a_{3}$, so $(b)_{b}\subseteq (a_{1}a_{2}a_{3})_{b}$. It follows that $b\leqslant _{b} a_{1}a_{2}a_{3}$ which contradict (2). Hence B is minimal subset of S such that $(B)_{b}=S$.

Case 2: If $b\in ASASA$, then $b = a_{1}sa_{2}sa_{3}$, for some $a_{1},a_{2},a_{3}\in A$ and $s\in S$. As $b\notin A$ so, $b\ne a_{1}$, $b\ne a_{2}$ and $b\ne a_{3}$. It implies $a_{1},a_{2},a_{3}\in A$. Since $b = a_{1}sa_{2}sa_{3}$, therefore $(b)_{b}\subseteq (a_{1}sa_{2}sa_{3})_{b}$.

Thus, $b\leqslant _{b} a_{1}sa_{2}sa_{3}$ which contradict (3). Thus there exists no proper subset A of B such that $(A)_{b}=S$. Hence, B is bi-base of S. $\square$