Intuitionistic $(\lambda ,\mu )$-fuzzy sets in Γ-semigroups

Abstract

We first introduce $(\lambda ,\mu )$-fuzzy ideals and $(\lambda ,\mu )$-fuzzy interior ideals of an ordered Γ-semigroup. Then we prove that in regular and in intra-regular ordered semigroups the $(\lambda ,\mu )$-fuzzy ideals and the $(\lambda ,\mu )$-fuzzy interior ideals coincide. Lastly, we introduce $(\lambda ,\mu )$-fuzzy simple ordered Γ-semigroup and characterize the simple ordered Γ-semigroups in terms of $(\lambda ,\mu )$-fuzzy interior ideals.

1 Introduction and preliminaries

The formal study of semigroups began in the early twentieth century. Semigroups are important in many areas of mathematics, for example, coding and language theory, automata theory, combinatorics and mathematical analysis.

Γ-semigroups were first defined by Sen and Saha [1] as a generalization of semigroups and studied by many researchers [2–13].

The concept of fuzzy sets was first introduced by Zadeh [14] in 1965, and then the fuzzy sets have been used in the reconsideration of classical mathematics. Recently, Yuan [15] introduced the concept of a fuzzy subfield with thresholds. A fuzzy subfield with thresholds λ and μ is also called a $(\lambda ,\mu )$-fuzzy subfield. Yao continued to research $(\lambda ,\mu )$-fuzzy normal subfields, $(\lambda ,\mu )$-fuzzy quotient subfields, $(\lambda ,\mu )$-fuzzy subrings and $(\lambda ,\mu )$-fuzzy ideals in [16–19].

In this paper, we study $(\lambda ,\mu )$-fuzzy ideals in ordered Γ-semigroups. This can be seen as an application of [19] and as a generalization of [20, 21].

Let $S=\{x,y,z,\dots \}$ and $\Gamma =\{\alpha ,\beta ,\gamma ,\dots \}$ be two non-empty sets. An ordered Γ-semigroup $S_{\Gamma }=(S,\Gamma ,\le )$ is a poset $(S,\le )$, and there is a mapping $S\times \Gamma \times S\to S$ (images to be denoted by $a\alpha b$) such that, for all $x,y,z\in S$, $\alpha ,\beta ,\gamma \in \Gamma$, we have

• (1)
  $$(x\beta y)\gamma z=x\beta (y\gamma z)$$

  ;

• (2)
  $$x\le y\Rightarrow \{\begin{array}{c}x\alpha z\le y\alpha z,\hfill \\ z\alpha x\le z\alpha y.\hfill \end{array}$$

If $(S,\Gamma ,\le )$ is an ordered Γ-semigroup and A is a subset of S, we denote by $(A]$ the subset of S defined as follows:

$$(A]=\{t\in S|t\le a\text{ for some }a\in A\}.$$

Given an ordered Γ-semigroup S, a fuzzy subset of S (or a fuzzy set in S) is an arbitrary mapping $f:S\to [0,1]$, where $[0,1]$ is the usual closed interval of real numbers. For any $t\in [0,1]$, $f_t$ is defined by $f_t=\{x\in S|f(x)\ge t\}$.

For each subset A of S, the characteristic function $f_A$ is a fuzzy subset of S defined by

$$f_A(x)=\{\begin{array}{cc}1\hfill & \text{if }x\in A,\hfill \\ 0\hfill & \text{if }x\notin A.\hfill \end{array}$$

In the following, we will use S, $S_{\Gamma }$ or $(S,\Gamma ,\le )$ to denote an ordered Γ-semigroup. In the rest of this paper, we will always assume that $0\le \lambda <\mu \le 1$.

2 Intuitionistic $(\lambda ,\mu )$-fuzzy Γ-ideals

In what follows, we will use S to denote a Γ-semigroup unless otherwise specified.

Definition 1 For an IFS $A=(f_A,g_A)$ in S, consider the following axioms:

($\Gamma S_1$) $f_A(x\gamma y)\vee \lambda \ge f_A(x)\wedge f_A(y)\wedge \mu$,

($\Gamma S_2$) $g_A(x\gamma y)\wedge \mu \le f_A(x)\vee f_A(y)\vee \lambda$

for all $x,y\in S$ and $\gamma \in \Gamma$. Then $A=(f_A,g_A)$ is called a first (resp. second) intuitionistic $(\lambda ,\mu )$-fuzzy Γ-subsemigroup (briefly $(\lambda ,\mu )$-$IF\Gamma S_1$ (resp. $(\lambda ,\mu )$-$IF\Gamma S_2$)) of S if it satisfies ($\Gamma S_1$) (resp. $\Gamma S_2$).

$$A=(f_A,g_A)$$

is called an intuitionistic $(\lambda ,\mu )$-fuzzy Γ-subsemigroup (briefly $(\lambda ,\mu )$-$IF\Gamma S$) of S if it is both a first and a second intuitionistic fuzzy Γ-subsemigroup.

Theorem 1 If U is a Γ-subsemigroup of S, then $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IF\Gamma S$ of S.

Proof Let $x,y\in S$ and $\gamma \in \Gamma$.

• (1) If $x,y\in U$, then $x\gamma y\in U$ from the hypothesis. Thus

  $${\chi }_U(x\gamma y)\vee \lambda =1\vee \lambda =1\ge {\chi }_U(x)\wedge {\chi }_U(y)\wedge \mu$$

and

$$\widetilde{{\chi }_U}(x\gamma y)\wedge \mu =(1-{\chi }_U(x\gamma y))\wedge \mu =(1-1)\wedge \mu =0\le \widetilde{{\chi }_U}(x)\vee \widetilde{{\chi }_U}(y)\vee \lambda .$$

• (2) If $x\notin U$ or $y\notin U$, then ${\chi }_U(x)=0$ or ${\chi }_U(y)=0$. Thus

  $${\chi }_U(x\gamma y)\vee \lambda \ge 0={\chi }_U(x)\wedge {\chi }_U(y)\wedge \mu$$

and

$$\widetilde{{\chi }_U}(x\gamma y)\wedge \mu \le 1=\widetilde{{\chi }_U}(x)\vee \widetilde{{\chi }_U}(y)\vee \lambda .$$

And we complete the proof. □

Theorem 2 Let U be a non-empty subset of S. If $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IF\Gamma S_1$ or $(\lambda ,\mu )$-$IF\Gamma S_2$ of S, then U is a Γ-subsemigroup of S.

Proof (1) Suppose that $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IF\Gamma S_1$ of S. For any $u,v\in U$ and $\gamma \in \Gamma$, we need to show that $u\gamma v\in U$. From ($\Gamma S_1$), we know that

$${\chi }_U(u\gamma v)\vee \lambda \ge {\chi }_U(u)\wedge {\chi }_U(v)\wedge \mu =1\wedge 1\wedge \mu =\mu .$$

Notice that $\lambda <\mu$, thus ${\chi }_U(u\gamma v)\ge \mu >0$.

And also because U is a crisp set of S, then we conclude that ${\chi }_U(u\gamma v)=1$; that is, $u\gamma v\in U$. Thus U is a Γ-subsemigroup of S.

• (2) Now assume that $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IF\Gamma S_2$ of S. For any $u,v\in U$ and $\gamma \in \Gamma$, we also need to show that $u\gamma v\in U$. It follows from ($\Gamma S_2$) that

  $$\widetilde{{\chi }_U}(x\gamma y)\wedge \mu \le \widetilde{{\chi }_U}(x)\vee \widetilde{{\chi }_U}(y)\vee \lambda =0\vee 0\vee \lambda =\lambda .$$

Notice that $\lambda <\mu$, thus $\widetilde{{\chi }_U}(x\gamma y)\le \lambda$.

And also because U is a crisp set of S, then we conclude that $\widetilde{{\chi }_U}(x\gamma y)=0$, i.e., ${\chi }_U(u\gamma v)=1$. That is, $u\gamma v\in U$. Thus U is a Γ-subsemigroup of S. □

Definition 2 For an IFS $A=(f_A,g_A)$ in S, consider the following axioms:

($L\Gamma I_1$) $f_A(x\gamma y)\vee \lambda \ge f_A(y)\wedge \mu$,

($L\Gamma I_2$) $g_A(x\gamma y)\wedge \mu \le g_A(y)\vee \lambda$

for all $x,y\in S$ and $\gamma \in \Gamma$. Then $A=(f_A,g_A)$ is called a first (resp. second) intuitionistic $(\lambda ,\mu )$-fuzzy left Γ-ideal (briefly $(\lambda ,\mu )$-$IFL\Gamma I_1$ (resp. $(\lambda ,\mu )$-$IFL\Gamma I_2$)) of S if it satisfies ($L\Gamma I_1$) (resp. ($L\Gamma I_2$)).

$$A=(f_A,g_A)$$

is called an intuitionistic $(\lambda ,\mu )$-fuzzy left Γ-ideal (briefly $(\lambda ,\mu )$-$IFL\Gamma I$) of S if it is both a first and a second intuitionistic $(\lambda ,\mu )$-fuzzy left Γ-ideal.

Definition 3 For an IFS $A=(f_A,g_A)$ in S, consider the following axioms:

($R\Gamma I_1$) $f_A(x\gamma y)\vee \lambda \ge f_A(x)\wedge \mu$,

($R\Gamma I_2$) $g_A(x\gamma y)\wedge \mu \le g_A(x)\vee \lambda$

for all $x,y\in S$ and $\gamma \in \Gamma$. Then $A=(f_A,g_A)$ is called a first (resp. second) intuitionistic $(\lambda ,\mu )$-fuzzy right Γ-ideal (briefly $(\lambda ,\mu )$-$IFR\Gamma I_1$ (resp. $(\lambda ,\mu )$-$IFR\Gamma I_2$)) of S if it satisfies ($R\Gamma I_1$) (resp. ($R\Gamma I_2$)).

$$A=(f_A,g_A)$$

is called an intuitionistic $(\lambda ,\mu )$-fuzzy right Γ-ideal (briefly $(\lambda ,\mu )$-$IFR\Gamma I$) of S if it is both a first and a second intuitionistic $(\lambda ,\mu )$-fuzzy right Γ-ideal.

Definition 4 For an IFS $A=(f_A,g_A)$ in S, it is called an intuitionistic $(\lambda ,\mu )$-fuzzy Γ-ideal (briefly $(\lambda ,\mu )$-$IF\Gamma I$) of S if it is both an intuitionistic fuzzy left and an intuitionistic fuzzy right Γ-ideal.

Theorem 3 If $A=(f_A,g_A)$ is a $(\lambda ,\mu )$-$IFL\Gamma I_1$ of S. U is a left-zero Γ-subsemigroup of S. For any $x,y\in U$, one of the following must hold:

• (1)
  $$f_A(x)=f_A(y)$$

  ;

• (2)
  $$f_A(x)\ne f_A(y)\Rightarrow$$

  ($f_A(x)\vee f_A(y)\le \lambda$, or $f_A(x)\wedge f_A(y)\ge \mu$).

Proof Let $x,y\in U$. Since U is left-zero, we have that $x\gamma y=x$ and $y\gamma x=y$ for all $\gamma \in \Gamma$.

From the hypothesis, we have that

$$f_A(x)\vee \lambda =f_A(x\gamma y)\vee \lambda \ge f_A(y)\wedge \mu$$

and

$$f_A(y)\vee \lambda =f_A(y\gamma x)\vee \lambda \ge f_A(x)\wedge \mu .$$

Obviously, if $f_A(x)=f_A(y)$, then the previous two inequalities hold.

Suppose $f_A(x)<f_A(y)$. If $f_A(x)\vee f_A(y)>\lambda$ and $f_A(x)\wedge f_A(y)<\mu$, four cases are possible:

• (1) If $f_A(x)>\lambda$ and $f_A(x)<\mu$, then $f_A(x)\vee \lambda =f_A(x)<f_A(y)$. Note that $f_A(x)<\mu$, we obtain that $f_A(x)<f_A(y)\wedge \mu$; that is, $f_A(x)\vee \lambda <f_A(y)\wedge \mu$. This is a contradiction to the previous proposition.

• (2) If $f_A(x)>\lambda$ and $f_A(y)<\mu$, then $f_A(x)\vee \lambda =f_A(x)<f_A(y)=f_A(y)\wedge \mu$. This is a contradiction to the previous proposition.

• (3) If $f_A(y)>\lambda$ and $f_A(x)<\mu$, then from $f_A(x)<\mu$ and $f_A(x)<f_A(y)$, we obtain that $f_A(x)<f_A(y)\wedge \mu$. From $\lambda <f_A(y)$ and $\lambda <\mu$, we conclude that $\lambda <f_A(y)\wedge \mu$. So, $f_A(x)\vee \lambda <f_A(y)\wedge \mu$. This is a contradiction to the previous proposition.

• (4) If $f_A(y)>\lambda$ and $f_A(y)<\mu$, then from $F_A(X)<f_A(y)$ and $\lambda <f_A(y)$, we obtain $F_A(X)\vee \lambda <f_A(y)=f_A(y)\wedge \mu$. This is a contradiction to the previous proposition.

If $f_A(y)<f_A(x)$, we can prove the results dually.

Thus if $f_A(y)\ne f_A(x)$, then $f_A(x)\vee f_A(y)\le \lambda$ or $f_A(x)\wedge f_A(y)\ge \mu$. □

Similarly, we can prove the following three theorems.

Theorem 4 If $A=(f_A,g_A)$ is a $(\lambda ,\mu )$-$IFR\Gamma I_1$ of S. U is a right-zero Γ-subsemigroup of S. For any $x,y\in U$, one of the following must hold:

• (1)
  $$f_A(x)=f_A(y)$$

  ;

• (2)
  $$f_A(x)\ne f_A(y)\Rightarrow (f_A(x)\vee f_A(y)\le \lambda ,\mathit{\text{or}}{f}_A(x)\wedge f_A(y)\ge \mu )$$

  .

Theorem 5 If $A=(f_A,g_A)$ is a $(\lambda ,\mu )$-$IFL\Gamma I_2$ of S. U is a left-zero Γ-subsemigroup of S. For any $x,y\in U$, one of the following must hold:

• (1)
  $$g_A(x)=g_A(y)$$

  ;

• (2)
  $$g_A(x)\ne g_A(y)\Rightarrow (g_A(x)\vee g_A(y)\le \lambda ,\mathit{\text{or}}{g}_A(x)\wedge g_A(y)\ge \mu )$$

  .

Theorem 6 If $A=(f_A,g_A)$ is a $(\lambda ,\mu )$-$IFR\Gamma I_2$ of S. U is a right-zero Γ-subsemigroup of S. For any $x,y\in U$, one of the following must hold:

• (1)
  $$g_A(x)=g_A(y)$$

  ;

• (2)
  $$g_A(x)\ne g_A(y)\Rightarrow (g_A(x)\vee g_A(y)\le \lambda ,\mathit{\text{or}}{g}_A(x)\wedge g_A(y)\ge \mu )$$

  .

Lemma 1 If U is a left Γ-ideal of S, then $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFL\Gamma I$ of S.

Proof Let $x,y\in S$ and $\gamma \in \Gamma$.

• (1) If $y\in U$, then $x\gamma y\in U$ since U is a left Γ-ideal of S. It follows that

  $${\chi }_U(x\gamma y)\vee \lambda =1\vee \lambda =1\ge {\chi }_U(y)\wedge \mu$$

and

$$\widetilde{{\chi }_U}(x\gamma y)\wedge \mu =0\wedge \mu =0\le \widetilde{{\chi }_U}(y)\vee \lambda .$$

• (2) If $y\notin U$, then ${\chi }_U(y)=0$. It follows that

  $${\chi }_U(x\gamma y)\vee \lambda \ge 0={\chi }_U(y)\wedge \mu$$

and

$$\widetilde{{\chi }_U}(x\gamma y)\wedge \mu \le 1=\widetilde{{\chi }_U}(y)\vee \lambda .$$

Consequently, $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFL\Gamma I$ of S. □

Theorem 7 Let S be regular. If $E_S$ is a left-zero Γ-subsemigroup of S, then, for all $e,e^{\prime }\in E_S$, we have

$${\chi }_{L[e]}\left(e^{\prime }\right)={\chi }_{L[e]}(e)$$

or

$${\chi }_{L[e]}\left(e^{\prime }\right)\vee {\chi }_{L[e]}(e)\le \lambda ,$$

or

$${\chi }_{L[e]}\left(e^{\prime }\right)\wedge {\chi }_{L[e]}(e)\ge \mu .$$

Proof Since S is regular, $E_S$ is non-empty. Let $e=e\gamma e,e^{\prime }=e^{\prime }{\gamma }^{\prime }{e}^{\prime }\in E_S$, where $\gamma ,{\gamma }^{\prime }\in \Gamma$. Because S is regular, $L[e]=S\Gamma e$. From the fact that $L[e]$ is a left Γ-ideal of S, we obtain that $\widetilde{L[e]}=({\chi }_{L[e]},\widetilde{{\chi }_{L[e]}})$ is a $(\lambda ,\mu )$-$IFL\Gamma I_1$ of S by the previous lemma.

Applying Theorem 3, we obtain the results. □

The following theorem can be proved in a similar way.

Theorem 8 Let S be regular. If $E_S$ is a left-zero Γ-subsemigroup of S, then, for all $e,e^{\prime }\in E_S$, we have

$$\widetilde{{\chi }_{L[e]}}\left(e^{\prime }\right)=\widetilde{{\chi }_{L[e]}}(e)$$

or

$$\widetilde{{\chi }_{L[e]}}\left(e^{\prime }\right)\vee \widetilde{{\chi }_{L[e]}}(e)\le \lambda ,$$

or

$$\widetilde{{\chi }_{L[e]}}\left(e^{\prime }\right)\wedge \widetilde{{\chi }_{L[e]}}(e)\ge \mu .$$

Theorem 9 Let S be regular. If for all $e,e^{\prime }\in E_S$ we have

$${\chi }_{L[e]}\left(e^{\prime }\right)={\chi }_{L[e]}(e)$$

or

$${\chi }_{L[e]}\left(e^{\prime }\right)\vee {\chi }_{L[e]}(e)\le \lambda ,$$

or

$${\chi }_{L[e]}\left(e^{\prime }\right)\wedge {\chi }_{L[e]}(e)\ge \mu ,$$

then $E_S$ is a left-zero Γ-subsemigroup of S.

Proof Since S is regular, $E_S$ is non-empty. Let $e=e\gamma e,e^{\prime }=e^{\prime }{\gamma }^{\prime }{e}^{\prime }\in E_S$, where $\gamma ,{\gamma }^{\prime }\in \Gamma$. Because S is regular, $L[e]=S\Gamma e$. From the fact that $L[e]$ is a left Γ-ideal of S, we obtain that $\widetilde{L[e]}=({\chi }_{L[e]},\widetilde{{\chi }_{L[e]}})$ is a $(\lambda ,\mu )$-$IFL\Gamma I_1$ of S by the previous lemma.

• (1) If ${\chi }_{L[e]}(e^{\prime })={\chi }_{L[e]}(e)=1$, then $e^{\prime }\in L[e]=S\Gamma e$. Thus

  $$e^{\prime }=x\beta e=x\beta (e\gamma e)=(x\beta e)\gamma e=e^{\prime }\gamma e$$

for some $x\in S$ and $\beta \in \Gamma$.

• (2)
  $${\chi }_{L[e]}(e^{\prime })\vee {\chi }_{L[e]}(e)\le \lambda$$

  will never happen since ${\chi }_{L[e]}(e^{\prime })\vee {\chi }_{L[e]}(e)=1$.

• (3) If ${\chi }_{L[e]}(e^{\prime })\wedge {\chi }_{L[e]}(e)\ge \mu$, that is, ${\chi }_{L[e]}(e^{\prime })\ge \mu$, then ${\chi }_{L[e]}(e^{\prime })=1$. And so $e^{\prime }\in L[e]=S\Gamma e$. The following proof will be the same as in case (1).

Consequently, $E_S$ is a left-zero Γ-subsemigroup of S. □

The following theorem can be proved similarly.

Theorem 10 Let S be regular. If for all $e,e^{\prime }\in E_S$ we have

$$\widetilde{{\chi }_{L[e]}}\left(e^{\prime }\right)=\widetilde{{\chi }_{L[e]}}(e)$$

or

$$\widetilde{{\chi }_{L[e]}}\left(e^{\prime }\right)\vee \widetilde{{\chi }_{L[e]}}(e)\le \lambda ,$$

or

$$\widetilde{{\chi }_{L[e]}}\left(e^{\prime }\right)\wedge \widetilde{{\chi }_{L[e]}}(e)\ge \mu ,$$

then $E_S$ is a left-zero Γ-subsemigroup of S.

3 Intuitionistic $(\lambda ,\mu )$-fuzzy interior Γ-ideals

Definition 5 For an IFS $A=(f_A,g_A)$ in S, consider the following axioms:

($I\Gamma I_1$) $f_A(x\beta s\gamma y)\vee \lambda \ge f_A(s)\wedge \mu$,

($I\Gamma I_2$) $g_A(x\beta s\gamma y)\wedge \mu \le g_A(s)\vee \lambda$

for all $s,x,y\in S$ and $\beta ,\gamma \in \Gamma$. Then $A=(f_A,g_A)$ is called a first (resp. second) intuitionistic $(\lambda ,\mu )$-fuzzy interior Γ-ideal (briefly $(\lambda ,\mu )$-$IFI\Gamma I_1$ (resp. $(\lambda ,\mu )$-$IFI\Gamma I_2$)) of S if it satisfies ($I\Gamma I_1$) (resp. ($I\Gamma I_2$)).

$$A=(f_A,g_A)$$

is called an intuitionistic $(\lambda ,\mu )$-fuzzy interior Γ-ideal (briefly $(\lambda ,\mu )$-$IFI\Gamma I$) of S if it is both a first and a second intuitionistic $(\lambda ,\mu )$-fuzzy interior Γ-ideal.

Theorem 11 Every $(\lambda ,\mu )$-$IF\Gamma I$ of S is a $(\lambda ,\mu )$-$IFI\Gamma I$ of S.

Proof Let $A=(f_A,g_A)$ be a $(\lambda ,\mu )$-$IF\Gamma I$ of S. For all $s,x,y\in S$ and $\beta ,\gamma \in \Gamma$, we have

$$\begin{array}{rcl}{f}_A(x\beta s\gamma y)\vee \lambda & =& f_A(x\beta s\gamma y)\vee \lambda \vee \lambda \ge (f_A(x\beta s)\wedge \mu )\vee \lambda \\ =& (f_A(x\beta s)\vee \lambda )\wedge (\mu \vee \lambda )\ge f(s)\wedge \mu .\end{array}$$

Similarly, we have $g_A(x\beta s\gamma y)\wedge \mu \le g_A(s)\vee \lambda$.

So, $A=(f_A,g_A)$ is a $(\lambda ,\mu )$-$IFI\Gamma I$ of S. □

Theorem 12 If S is regular, then every $(\lambda ,\mu )$-$IFI\Gamma I$ of S is a $(\lambda ,\mu )$-$IF\Gamma I$ of S.

Proof Let $A=(f_A,g_A)$ be a $(\lambda ,\mu )$-$IFI\Gamma I$ of S and $x,y\in S$. Since S is regular, there exist $s,s^{\prime }\in S$ and $\beta ,{\beta }^{\prime },\gamma ,{\gamma }^{\prime }\in \Gamma$ such that $x=x\beta s\gamma x$ and $y=y{\beta }^{\prime }{s}^{\prime }{\gamma }^{\prime }y$. Thus

$$f_A(x\alpha y)\vee \lambda =f_A\left(x\alpha \left(y{\beta }^{\prime }{s}^{\prime }{\gamma }^{\prime }y\right)\right)\vee \lambda =f_A\left(x\alpha y{\beta }^{\prime }\left(s^{\prime }{\gamma }^{\prime }y\right)\right)\vee \lambda \ge f_A(y)\wedge \mu$$

and

$$g_A(x\alpha y)\wedge \mu =g_A\left(x\alpha \left(y{\beta }^{\prime }{s}^{\prime }{\gamma }^{\prime }y\right)\right)\wedge \mu =g_A\left(x\alpha y{\beta }^{\prime }\left(s^{\prime }{\gamma }^{\prime }y\right)\right)\wedge \mu \le g_A(y)\vee \lambda$$

for all $\alpha \in \Gamma$. It follows that $A=(f_A,g_A)$ is a $(\lambda ,\mu )$-$IFL\Gamma I$ of S. Similarly, we can prove that A is a $(\lambda ,\mu )$-$IFR\Gamma I$ of S. This completes the proof. □

Theorem 13 If U is an interior Γ-ideal of S, then $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFI\Gamma I$ of S.

Proof Let $s,x,y\in S$ and $\beta ,\gamma \in \Gamma$.

• (1) If $s\in U$, then $x\beta s\gamma y\in U$ since U is an interior Γ-ideal of S. So,

  $${\chi }_U(x\beta s\gamma y)\vee \lambda =1\vee \lambda =1\ge {\chi }_U(s)\wedge \mu$$

and

$$\widetilde{{\chi }_U}(x\beta s\gamma y)\wedge \mu =0\wedge \mu =0\le \widetilde{{\chi }_U}(s)\vee \lambda .$$

• (2) If $s\notin U$, then ${\chi }_U(s)=0$. Thus

  $${\chi }_U(x\beta s\gamma y)\vee \lambda \ge 0={\chi }_U(s)\wedge \mu$$

and

$$\widetilde{{\chi }_U}(x\beta s\gamma y)\wedge \mu \le 1=\widetilde{{\chi }_U}(s)\vee \lambda .$$

Consequently, we obtain that $\tilde{U}$ is a $(\lambda ,\mu )$-$IFI\Gamma I$ of S. □

Theorem 14 Let S be regular and U be a non-empty subset of S. If $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFI\Gamma I_1$ or $(\lambda ,\mu )$-$IFI\Gamma I_2$ of S, then U is an interior Γ-ideal of S.

Proof It is obvious that U is a Γ-subsemigroup of S by Theorem 2.

Case 1. Suppose that $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFI\Gamma I_1$ of S and $x\in S\Gamma U\Gamma S$. Thus $x=s\beta u\gamma t$ for some $s,t\in S$, $u\in U$ and $\beta ,\gamma \in \Gamma$. It follows from ($I\Gamma I_1$) that

$${\chi }_U(x)\vee \lambda ={\chi }_U(s\beta u\gamma t)\vee \lambda \ge {\chi }_U(u)\wedge \mu =1\wedge \mu =\mu .$$

Notice that $\lambda <\mu$, we obtain that ${\chi }_U(x)\ge \mu$, that is, ${\chi }_U(x)=1$. So, $x\in U$. Thus U is an interior Γ-ideal of S.

Case 2. Suppose that $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFI\Gamma I_2$ of S and $x\in S\Gamma U\Gamma S$. Then $x=s\beta u\gamma t$ for some $s,t\in S$, $u\in U$ and $\beta ,\gamma \in \Gamma$. Using ($I\Gamma I_2$), we conclude that

$$\widetilde{{\chi }_U}(x)\wedge \mu =\widetilde{{\chi }_U}(s\beta u\gamma t)\wedge \mu \le \widetilde{{\chi }_U}(u)\vee \lambda =0\vee \lambda =\lambda .$$

Notice that $\lambda <\mu$, we obtain that $\widetilde{{\chi }_U}(x)\le \lambda$, that is, $\widetilde{{\chi }_U}(x)=0$. So, $x\in U$. Thus U is an interior Γ-ideal of S. □

4 Intuitionistic $(\lambda ,\mu )$-fuzzy simple Γ-semigroups

Definition 6 S is called first (resp. second) intuitionistic $(\lambda ,\mu )$-fuzzy left simple if for any $(\lambda ,\mu )$-$IFL\Gamma I_1$ (resp. $(\lambda ,\mu )$-$IFL\Gamma I_2$) $A=(f_A,g_A)$ of S, we have $f_A(a)\vee \lambda \ge f_A(b)\wedge \mu$ (resp. $g_A(a)\vee \lambda \ge g_A(b)\wedge \mu$) for all $a,b\in S$.

S is said to be intuitionistic $(\lambda ,\mu )$-fuzzy left simple if it is both first and second intuitionistic $(\lambda ,\mu )$-fuzzy left simple.

Theorem 15 If S is left simple, then S is intuitionistic $(\lambda ,\mu )$-fuzzy left simple.

Proof Let $A=(f_A,g_A)$ be a $(\lambda ,\mu )$-$IFL\Gamma I$ of S and $x,x^{\prime }\in S$. Because S is left simple, there exist $s,s^{\prime }\in S$ and $\gamma ,{\gamma }^{\prime }\in \Gamma$ such that $x=s\gamma x^{\prime }$ and $x^{\prime }=s^{\prime }{\gamma }^{\prime }x$. Thus, since A is a $(\lambda ,\mu )$-$IFL\Gamma I$ of S, we have that

and

Consequently, S is intuitionistic $(\lambda ,\mu )$-fuzzy left simple. □

Theorem 16 If S is first or second intuitionistic $(\lambda ,\mu )$-fuzzy left simple, then S is left simple.

Proof Let U be a left Γ-ideal of S. Suppose that S is first (or second) intuitionistic $(\lambda ,\mu )$-fuzzy left simple. Because $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFL\Gamma I$ of S, $\tilde{U}=({\chi }_U,\widetilde{{\chi }_U})$ is a $(\lambda ,\mu )$-$IFL\Gamma I_1$ (and $(\lambda ,\mu )$-$IFL\Gamma I_2$) of S. □

5 Conclusion and further research

In this paper, we generalized the results of [20, 21]. We introduced $(\lambda ,\mu )$-fuzzy ideals and $(\lambda ,\mu )$-fuzzy interior ideals of an ordered Γ-semigroup and we got some interesting results. When $\lambda =0$ and $\mu =1$, we meet ordinary fuzzy ideals and fuzzy interior ideals. From this point of view, $(\lambda ,\mu )$-fuzzy ideals and $(\lambda ,\mu )$-fuzzy interior ideals are more general concepts than fuzzy ones.

In [19], Yao gave the definition of $(\lambda ,\mu )$-fuzzy bi-ideals in semigroups. One can study $(\lambda ,\mu )$-fuzzy bi-ideals in ordered Γ-semigroups. We would like to explore this in next papers.