Before proving our main results, we first list two lemmas which will be used in the following section.

#### Lemma 2.1

([17]). *Suppose that the function* *ϕ*: [0, ∞) → [0, ∞) *is upper semicontinuous with* *ϕ*(0) = 0 *and* *ϕ*(*x*) < *x* *for all* *x* > 0. *Then there exists a strictly increasing continuous function α* : [0, ∞) → [0, ∞) *such that* *α*(0) = 0 *and* *ϕ*(*x*) ≤ *α*(*x*) < *x* *for all* *x* > 0. *The function* *α* *is invertible and for any* *x* > 0, $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*α*^{–n} = ∞, *where* *α*^{–n} *denotes the* *n*-*th iterates of* *α*^{–1} *and* *α*^{–1} *denotes the inverse of* *α*.

#### Lemma 2.2

([17]). *Suppose that* (*X*, *F*) *is a PM*-*space and* *α*: [0, ∞) → [0, ∞) *is a strictly increasing function satisfying* *α*(0) = 0 *and* *α*(*x*) < *x* *for all* *x* > 0. *If* *x*, *y* *are two members in* *X* *such that*

$$\begin{array}{}{\displaystyle {F}_{x,y}(\alpha (\u03f5))\ge {F}_{x,y}(\u03f5),}\end{array}$$

*for all* *ϵ* > 0, *then* *x* = *y*.

Now, we state and prove our main result.

#### Theorem 2.3

*Let* *A*, *B*, *S*, *T*, *L* *and* *M* *be self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1]. *Suppose that the inequality* (1) *of Lemma 1.7 holds. If the pairs* (*L*, *AB*) *and* (*M*, *ST*) *share the* (*CLR*_{(AB)(ST)}) *property*, *then* (*L*, *AB*) *and* (*M*, *ST*) *have a coincidence point each*.

*Moreover*, *if*

*both the pairs* (*L*, *AB*) *and* (*M*, *ST*) *are weakly compatible*.

*AB* = *BA*, *LA* = *AL*, *MS* = *SM* *and* *ST* = *TS*.

*Then* *A*, *B*, *S*, *T*, *L* *and* *M* *have a unique common fixed point*.

#### Proof

Since the pairs (*L*, *AB*) and (*M*, *ST*) share the (*CLR*_{(AB)(ST)}) property, there exist two sequences {*x*_{n}}, {*y*_{n}} in *X* such that

$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}L{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}AB{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}ST{y}_{n}=\underset{n\to \mathrm{\infty}}{lim}M{y}_{n}=z,\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}z\in AB(X)\cap ST(X).}\end{array}$$

Since *z* ∈ *ST*(*X*), there exists a point *u* ∈ *X* such that *STu* = *z*. Putting *p* = *x*_{n} and *q* = *u* in inequality (1), it yields that

$$\begin{array}{}{\displaystyle {F}_{L{x}_{n},Mu}(\varphi (x))\ge min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{STu,Mu}(x),{F}_{STu,L{x}_{n}}(\beta x),{F}_{AB{x}_{n},Mu}((1+\beta )x),{F}_{AB{x}_{n},STu}(x)\}}\end{array}$$

Letting *n* → ∞, we obtain that

$$\begin{array}{}{\displaystyle {F}_{z,Mu}(\varphi (x))\ge min\{{F}_{z,z}(x),{F}_{z,Mu}(x),{F}_{z,z}(\beta x),{F}_{z,Mu}((1+\beta )x),{F}_{z,z}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}=min\{1,{F}_{z,Mu}(x),1,{F}_{z,Mu}((1+\beta )x),1\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}={F}_{z,Mu}(x).}\end{array}$$

From Lemma 2.1, there exists a strictly increasing continuous function *α* : [0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{z, Mu}(*α*(*x*)) ≥ *F*_{z, Mu}(*ϕ*(*x*)) ≥ *F*_{z, Mu}(*x*), for all *x* > 0. By Lemma 2.2, we obtain that *z* = *Mu*. Hence, *z* = *Mu* = *STu*, which shows *u* is a coincidence point of the pair (*M*, *ST*).

As *z* ∈ *AB*(*X*), there exists a point *v* ∈ *X* such that *ABv* = *z*, putting *p* = *v*, *q* = *y*_{n} in inequality (1), we have

$$\begin{array}{}{\displaystyle {F}_{Lv,M{y}_{n}}(\varphi (x))\ge min\{{F}_{ABv,Lv}(x),{F}_{ST{y}_{n},M{y}_{n}}(x),{F}_{ST{y}_{n},Lv}(\beta x),{F}_{ABv,M{y}_{n}}((1+\beta )x),{F}_{ABv,ST{y}_{n}}(x)\}.}\end{array}$$

Letting *n* → ∞, we obtain that

$$\begin{array}{}{\displaystyle {F}_{Lv,z}(\varphi (x))\ge min\{{F}_{z,Lv}(x),{F}_{z,z}(x),{F}_{z,Lv}(\beta x),{F}_{z,z}((1+\beta )x),{F}_{z,z}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=min\{{F}_{z,Lv}(x),1,{F}_{z,Lv}(\beta x),1,1\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}={F}_{z,Lv}(x).}\end{array}$$

From Lemma 2.1, there exists a strictly increasing continuous function *α* : [0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{Lv, z}(*α*(*x*)) ≥ *F*_{Lv, z}(*ϕ*(*x*)) ≥ *F*_{Lv, z}(*x*), for all *x* > 0. By Lemma 2.2, we obtain that *z* = *Lv*. Hence, *z* = *ABv* = *Lv*, which shows *v* is a coincidence point of the pair (*L*, *AB*).

Since the pair (*M*, *ST*) is weakly compatible, and by the previous proof, *z* = *Mu* = *STu*, then *MSTu* = *STMu*, it yields that *Mz* = *STz*. And since the pair (*L*, *AB*) is weakly compatible, and by the previous proof, *z* = *ABv* = *Lv*, then *LABv* = *ABLv*, it yields that *Lz* = *ABz*. Letting *p* = *z*, *q* = *u* in inequality (1), we obtain:

$$\begin{array}{}{\displaystyle {F}_{Lz,Mu}(\varphi (x))\ge min\{{F}_{Mu,Lz}(x),{F}_{STu,Mu}(x),{F}_{STu,Lz}(\beta x),{F}_{ABz,Mu}((1+\beta )x),{F}_{ABz,STu}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{z,Lz}(x),{F}_{z,z}(x),{F}_{z,Lz}(\beta x),{F}_{Lz,z}((1+\beta )x),{F}_{Lz,z}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{z,Lz}(x),1,{F}_{z,Lz}(\beta x),{F}_{Lz,z}((1+\beta )x),{F}_{Lz,z}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}={F}_{z,Lz}(x).}\end{array}$$

From Lemma 2.1, there exists a strictly increasing continuous function *α* :[0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{Lz, Mu}(*α*(*x*)) ≥ *F*_{Lz, Mu}(*ϕ*(*x*)) ≥ *F*_{Lz, Mu}(*x*), for all *x* > 0. By Lemma 2.2, we obtain that *Lz* = *Mu*. Therefore, *Lz* = *Mu* = *z*. Thus, *z* = *Lz* = *ABz*.

Sequentially, letting *p* = *z*, *q* = *z* in inequality (1), we obtain:

$$\begin{array}{}{\displaystyle {F}_{Lz,Mz}(\varphi (x))\ge min\{{F}_{Mz,Lz}(x),{F}_{STz,Mz}(x),{F}_{STz,Lz}(\beta x),{F}_{ABz,Mz}((1+\beta )x),{F}_{ABz,STz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{Mz,Lz}(x),{F}_{Mz,Mz}(x),{F}_{Mz,Lz}(\beta x),{F}_{Lz,Mz}((1+\beta )x),{F}_{Lz,Mz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{Mz,Lz}(x),1,{F}_{Mz,Lz}(\beta x),{F}_{Lz,Mz}((1+\beta )x),{F}_{Lz,Mz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}={F}_{Lz,Mz}(x).}\end{array}$$

From Lemma 2.1, there exists a strictly increasing continuous function *α* : [0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{Lz, Mz}(*α*(*x*)) ≥ *F*_{Lz, Mz}(*ϕ*(*x*)) ≥ *F*_{Lz, Mz}(*x*), for all *x* > 0. By Lemma 2.2, we obtain *Lz* = *Mz* = *z* = *STz* = *ABz*. Hence, *AB*, *ST*, *L* and *M* have a common fixed point *z*.

Letting *p* = *z*, *q* = *Sz* in inequality (1), we obtain:

$$\begin{array}{}{\displaystyle {F}_{z,Sz}(\varphi (x))\ge min\{{F}_{ABz,Lz}(x),{F}_{STSz,MSz}(x),{F}_{STSz,Lz}(\beta x),{F}_{ABz,MSz}((1+\beta )x),{F}_{ABz,STSz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=min\{{F}_{z,z}(x),{F}_{Sz,Sz}(x),{F}_{Sz,z}(\beta x),{F}_{z,Sz}((1+\beta )x),{F}_{z,Sz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=min\{1,1,{F}_{Sz,z}(\beta x),{F}_{z,Sz}((1+\beta )x),{F}_{z,Sz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}={F}_{z,Sz}(x).}\end{array}$$

From Lemma 2.1, there exists a strictly increasing continuous function *α* :[0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{z, Sz}(*α*(*x*)) ≥ *F*_{z, Sz}(*ϕ*(*x*)) ≥ *F*_{z, Sz}(*x*), for all *x* > 0. By Lemma 2.2, we obtain *z* = *Sz*. Thus, *z* = *Sz* = *STz* = *TSz* = *Tz*.

Letting *p* = *Az*, *q* = *z* in inequality (1), we obtain:

$$\begin{array}{}{\displaystyle {F}_{Az,z}(\varphi (x))\ge min\{{F}_{ABAz,LAz}(x),{F}_{STz,Mz}(x),{F}_{STz,Lz}(\beta x),{F}_{ABAz,Mz}((1+\beta )x),{F}_{ABAz,STz}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=min\{{F}_{Az,Az}(x),{F}_{z,z}(x),{F}_{z,z}(\beta x),{F}_{Az,z}((1+\beta )x),{F}_{Az,z}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=min\{1,1,1,{F}_{Az,z}((1+\beta )x),{F}_{Az,z}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}={F}_{Az,z}(x).}\end{array}$$

From Lemma 2.1, there exists a strictly increasing continuous function *α* :[0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{Az, z}(*α*(*x*)) ≥ *F*_{Az, z}(*ϕ*(*x*)) ≥ *F*_{Az, z}(*x*), for all *x* > 0. By Lemma 2.2, we obtain *z* = *Az*. Hence, *z* = *Az* = *ABz* = *BAz* = *Bz*. Thus, combing with the above proof, we have *z* = *Az* = *Bz* = *Lz* = *Mz* = *Sz* = *Tz*.

Then, *A*, *B*, *S*, *T*, *L* and *M* have a common fixed point *z*.

(Uniqueness). Assume that *t* is another common fixed point of *A*, *B*, *S*, *T*, *L* and *M*. It follows that *t* = *At* = *Bt* = *Lt* = *Mt* = *St* = *Tt*. Letting *p* = *z*, *q* = *t* in inequality (1), we obtain:

$$\begin{array}{}{\displaystyle {F}_{Lz,Mt}(\varphi (x))\ge min\{{F}_{ABz,Lz}(x),{F}_{STt,Mt}(x),{F}_{STt,Lz}(\beta x),{F}_{ABz,Mt}((1+\beta )x),{F}_{ABz,STt}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{z,z}(x),{F}_{t,t}(x),{F}_{t,z}(\beta x),{F}_{z,t}((1+\beta )x),{F}_{z,t}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{1,1,{F}_{t,z}(\beta x),{F}_{z,t}((1+\beta )x),{F}_{z,t}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}={F}_{z,t}(x).}\end{array}$$

It yields that *F*_{z, t}(*ϕ*(*x*)) ≥ *F*_{z, t}(*x*). From Lemma 2.1, there exists a strictly increasing continuous function *α* :[0, ∞) → [0, ∞) such that *α*(0) = 0 and *ϕ*(*x*) ≤ *α*(*x*) < *x* for all *x* > 0. Therefore, *F*_{z, t}(*α*(*x*)) ≥ *F*_{z, t}(*ϕ*(*x*)) ≥ *F*_{z, t}(*x*), for all *x* > 0. By Lemma 2.2, we obtain *z* = *t*. Thus, *A*, *B*, *S*, *T*, *L* and *M* have a unique common fixed point *z*.□

If we take *B* = *T* = *I*(I ≡ the identity mapping on *X*), we have:

#### Corollary 2.4

*Let* *A*, *S*, *L* *and* *M* *be self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1]. *Suppose that the inequality*

$$\begin{array}{}{\displaystyle {F}_{Lp,Mq}(\varphi (x))\ge min\{{F}_{Ap,Lp}(x),{F}_{Sq,Mq}(x),{F}_{Sq,Lp}(\beta x),{F}_{Ap,Mq}((1+\beta )x),{F}_{Ap,Sq}(x)\}}\end{array}$$

*holds. If the pairs* (*L*, *A*) *and* (*M*, *S*) *share the* (*CLR*_{(AS)}) *property*, *then* (*L*, *A*) *and* (*M*, *S*) *have a coincidence point each. Moreover*, *if both the pairs* (*L*, *A*) *and* (*M*, *S*) *are weakly compatible*, *LA* = *AL*, *MS* = *SM*, *then* *A*, *S*, *L* *and* *M* *have a unique common fixed point*.

Now, we illustrate an example to show that our main result of Theorem 2.3 is valid, and at the same time, the existing literature does not hold.

#### Example 2.6

*Let* *X* = [0, 3), *with the metric* *d* *defined by* *d*(*x*, *y*) = | *x* – *y* | *and define* *F*_{x, y}(*u*) = *H*( *u* – *d*(*x*, *y*)) *for all* *x*, *y* ∈ *X*, *u* > 0*(refer to [15, Example 3.2]). It is obviously that the space* *X* *is not complete*, *since it is not a closed interval in real numbers* ℝ. *We define* *t*(*a*, *b*) = min {*a*, *b*} *for all* *a*, *b* ∈ [0, 1]. *Let* *A*, *B*, *S*, *T*, *L* *and* *M* *be self mappings on* *X* *defined as*

$$\begin{array}{}{\displaystyle A(x)=\left\{\begin{array}{ll}& 2-x,\phantom{\rule{1em}{0ex}}0\le x<1,\\ & 2,\phantom{\rule{2em}{0ex}}1\le x<3.\end{array}\right.\phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}B(x)=\left\{\begin{array}{ll}& 2,\phantom{\rule{2em}{0ex}}x=0,\\ & 1/x,\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0<x<1,\\ & 2,\phantom{\rule{2em}{0ex}}1\le x<3.\end{array}\right.}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}S(x)=\frac{1}{2}x+1,\phantom{\rule{1em}{0ex}}0\le x<3,\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}T(x)=\frac{1}{3}(x+4),\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\le x<3.}\end{array}$$

*And* *L*(*x*) = *M*(*x*) = 2. *By a simple calculation*, *we can check the conditions in Theorem 2.3 hold true*.

*Consider two sequences* $\begin{array}{}{\displaystyle \{{x}_{n}\}=\{1+\frac{1}{n}\}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\{{y}_{n}\}=\{2-\frac{1}{n}\}.}\end{array}$ *Then* *Lx*_{n} = 2, *ABx*_{n} = *A*(2) = 2, *My*_{n} = 2, $\begin{array}{}{\displaystyle ST{y}_{n}=S(\frac{1}{3}({y}_{n}+4))=\frac{1}{6}(2-\frac{1}{n})+\frac{5}{3}=2-\frac{1}{6n}}\end{array}$ *which consequently it yields that*

$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}L{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}AB{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}ST{y}_{n}=\underset{n\to \mathrm{\infty}}{lim}M{y}_{n}=2,\text{\hspace{0.17em}}where\text{\hspace{0.17em}}2\in AB(X)\cap ST(X).}\end{array}$$

*Therefore*, *the pairs* (*L*, *AB*) *and* (*M*, *ST*) *have the* (*CLR*_{(AB)(ST)}) *property. It is obvious that* $\begin{array}{}{\displaystyle ST(X)=[\frac{5}{3},\frac{13}{6})}\end{array}$ *is not closed in* *X*.

*Check the inequality* (1). *Let* *ϕ* : [0, ∞] → [0, ∞] *defined by* *ϕ*(*t*) = *kt*, *k* ∈ (0, 1) *be an upper semicontinuous function with* *ϕ*(0) = 0 *and* *ϕ*(*t*) < *t* *for all* *t* > 0. *For any* *p*, *q* ∈ ℝ *and* *x* > 0, *we have* *F*_{Lp, Mq}(*kx*) = 1 *and*

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}min\{{F}_{ABp,Lp}(x),{F}_{STq,Mq}(x),{F}_{STq,Lp}(\beta x),{F}_{ABp,Mq}((1+\beta )x),{F}_{ABp,STq}(x)\}}\\ {\displaystyle =min\{1,{F}_{STq,Mq}(x),{F}_{STq,Lp}(\beta x),1,{F}_{ABp,STq}(x)\}}\\ {\displaystyle =min\{{F}_{STq,Mq}(x),{F}_{STq,Lp}(\beta x),{F}_{ABp,STq}(x)\}}\\ {\displaystyle =min\{{F}_{STq,2}(x),{F}_{STq,2}(\beta x),{F}_{2,STq}(x)\}}\\ {\displaystyle ={F}_{STq,2}(x)}\\ {\displaystyle \le 1.}\end{array}$$

*Then* *F*_{Lp, Mq}(*kx*) ≥ min{*F*_{ABp, Lp}(*x*), *F*_{STq, Mq}(*x*), *F*_{STq, Lp}(*βx*), *F*_{ABp, Mq}((1+*β*)*x*), *F*_{ABp, STq}(*x*)} *holds for* *x*, *y* ∈ *X*, *β* ≥ 1 *and* *x* > 0.

*It is obviously that* *L*(*x*) = *AB*(*x*) = {2} *for* 1 ≤ *x* < 3, *and* *L*(*AB*)(*x*) = (*AB*)*L*(*x*) = {2}. *Then the weakly compatibility of the pair* (*L*, *AB*) *is satisfied. And* *M*(*x*) = *ST*(*x*) = {2} *for* *x* = 1, *and* *M*(*ST*)(*x*) = 2 = *ST*(2) = (*ST*)*M*(*x*) *for* *x* = 1. *Then the weakly compatibility of the pair* (*M*, *ST*) *is also satisfied*.

*AB* = *BA* = {2}, *ST* = *TS* = $\begin{array}{}{\displaystyle \frac{1}{6}x+\frac{5}{3},}\end{array}$ *LA* = *LA* = {2}, *and* *SM* = *MS* = {2}.

*Thus*, *all the conditions of Theorem 2.3 are satisfied*, *but 2 is a unique common fixed point of* *A*, *B*, *S*, *T*, *L* *and* *M*.

#### Theorem 2.7

*Let* *A*, *B*, *S*, *T*, *L* *and* *M* *be self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1]. *Suppose that the conditions* *(i)*-*(iv) of Lemma 1.7 hold. Then* (*L*, *AB*) *and* (*M*, *ST*) *have a coincidence point each*.

*Moreover*, *if*

*both the pairs* (*L*, *AB*) *and* (*M*, *ST*) *are weakly compatible*.

*AB* = *BA*, *LA* = *AL*, *MS* = *SM* *and* *ST* = *TS*.

*Then* *A*, *B*, *S*, *T*, *L* *and* *M* *have a unique common fixed point*.

#### Proof

Since the conditions (i)-(iv) of Lemma 1.7 hold, thus the pairs (*L*, *AB*) and (*M*, *ST*) have the (*CLR*_{(AB)(ST)}) property. The rest of proof can be completed along the routine of the proof of Theorem 2.3. In order to avoid tedious presentation, we omit the rest of proof.□

It can be noted that the conclusion in Example 2.6 does not hold if we utilize Theorem 2.7. Indeed, conditions (3) of Lemma 1.7 are not satisfied, i.e., the closure of *ST*(*X*). So we give another example, and obtain the corresponding uniqueness of common fixed point which was proposed in Theorem 2.7.

#### Example 2.8

*Assume the same conditions of Example 2.6*, *except that*

$$\begin{array}{}{\displaystyle S(x)=T(x)=\left\{\begin{array}{ll}& \frac{8}{3},\phantom{\rule{1em}{0ex}}x=0,\\ & \frac{4}{3},\phantom{\rule{1em}{0ex}}x\in (0,1],\\ & \frac{2x+2}{3},\phantom{\rule{1em}{0ex}}x\in (1,3).\end{array}\right.}\end{array}$$

*And* *L*(*x*) = *M*(*x*) = 2. *First*, *we can check the conditions in Lemma 1.7*.

$\begin{array}{}{\displaystyle L(X)=2,ST(X)=[\frac{4}{3},\frac{22}{9}].\text{\hspace{0.17em}}Thus,L(X)\subseteq ST(X).}\end{array}$

*Take* *x*_{n} = 1 – 1/*n* ∈ *X*. *Then* $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*AB*(*x*_{n}) = $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*AB*(1 – 1/*n*) = {2} *and* $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*L*(*x*_{n}) = $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*L*(1 – 1/*n*) = {2}. *Therefore*, $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*AB*(*x*_{n}) = $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*L*(*x*_{n}). *It yields that the pair* (*L*, *AB*) *satisfies the property* (*E*, *A*).

*ST*(*X*) = $\begin{array}{}{\displaystyle [\frac{4}{3},\frac{22}{9}].}\end{array}$ *It is a closed interval in* ℝ, *of course*, *it is closed subset of* *X*.

*Check the inequality* (1). *Let* *ϕ*: [0, ∞] → [0, ∞] *defined by* *ϕ*(*t*) = *kt*, *k* ∈ (0, 1) *be an upper semicontinuous function with* *ϕ*(0) = 0 *and* *ϕ*(*t*) < *t* *for all* *t* > 0. *For any* *p*, *q* ∈ ℝ *and* *x* > 0, *we have* *F*_{Lp, Mq}(*kx*) = 1 *and*

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}min\{{F}_{ABp,Lp}(x),{F}_{STq,Mq}(x),{F}_{STq,Lp}(\beta x),{F}_{ABp,Mq}((1+\beta )x),{F}_{ABp,STq}(x)\}}\\ {\displaystyle =min\{1,{F}_{STq,Mq}(x),{F}_{STq,Lp}(\beta x),1,{F}_{ABp,STq}(x)\}}\\ {\displaystyle =min\{{F}_{STq,Mq}(x),{F}_{STq,Lp}(\beta x),{F}_{ABp,STq}(x)\}}\\ {\displaystyle =min\{{F}_{STq,1}(x),{F}_{STq,1}(\beta x),{F}_{1,STq}(x)\}}\\ {\displaystyle ={F}_{STq,1}(x)}\\ {\displaystyle \le 1.}\end{array}$$

*Then* *F*_{Lp, Mq}(*kx*) ≥ min{*F*_{ABp, Lp}(*x*), *F*_{STq, Mq}(*x*), *F*_{STq, Lp}(*βx*), *F*_{ABp, Mq}((1 + *β*)*x*), *F*_{ABp, STq}(*x*)} *holds for* *x*, *y* ∈ *X*, *β* ≥ 1 *and* *x* > 0.

*Besides*, *we should check weak compatibility of* (*M*, *ST*). *M*(*x*) = *ST*(*x*) = {2} *for* *x* = 2, *and* *M*(*ST*)(*x*) = *M*(2) = 2 = *ST*(2) = (*ST*)*M*(*x*) for *x* = 2. *Then the weakly compatibility of the pair* (*M*, *ST*) *is also satisfied*.

At the last, $\begin{array}{}{\displaystyle ST=TS=\left\{\begin{array}{ll}& \frac{4}{3},\phantom{\rule{1em}{0ex}}x=0,\\ & \frac{22}{9},\phantom{\rule{1em}{0ex}}x\in (0,1],\\ & \frac{4x+10}{9},\phantom{\rule{1em}{0ex}}x\in (1,3).\end{array}\right.and\text{\hspace{0.17em}}SM=MS=\{\frac{4}{3}\}.}\end{array}$

*Thus all the conditions of Theorem 2.7 are satisfied. From Theorem 2.7*, *A*, *B*, *S*, *T*, *L* *and* *M* *have a unique common fixed point in* *X*. *In fact*, *by the definition of* *A*, *B*, *S*, *T*, *L* *and* *M*, 2 *is the unique common fixed point of* *A*, *B*, *S*, *T*, *L* *and* *M* *in* *X*.

Instead of the (*CLR*_{(AB)(ST)}) property of (*L*, *AB*) and (*M*, *ST*) in Theorem 2.3, we utilize the common property (E.A.) to obtain fixed point theorems.

#### Theorem 2.9

*Let* *A*, *B*, *S*, *T*, *L* *and* *M* *be self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1]. *Suppose that the inequality* (1) *of Lemma 1.7 and the following hypotheses hold*:

*the pairs* (*L*, *AB*) *and* (*M*, *ST*) *have the common property (E.A.);*

*ST*(*X*) *and* *AB*(*X*) *is closed subset of* *X*.

*Then* (*L*, *AB*) *and* (*M*, *ST*) *have a coincidence point each*.

*Moreover*, *if*

*both the pairs* (*L*, *AB*) *and* (*M*, *ST*) *are weakly compatible*.

*AB* = *BA*, *LA* = *AL*, *MS* = *SM* *and* *ST* = *TS*.

*Then* *A*, *B*, *S*, *T*, *L* *and* *M* *have a unique common fixed point*.

#### Proof

If the pairs (*L*, *AB*) and (*M*, *ST*) have the common property (E.A.), then there exist two sequences {*x*_{n}} and {*y*_{n}} in *X* such that

$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}L{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}AB{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}ST{y}_{n}=\underset{n\to \mathrm{\infty}}{lim}M{y}_{n}=z,\text{\hspace{0.17em}}\text{for some}\text{\hspace{0.17em}}z\in X.}\end{array}$$

Since *ST*(*X*) is closed, then $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*STy*_{n} = *z* = *STu* for some *u* ∈ *X*. And *AB*(*X*) is closed, then $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*ABx*_{n} = *z* = *ABv* for some *v* ∈ *X*. The rest of the proof can runs on the lines of Theorem 2.3.□

#### Corollary 2.10

*The result of Theorem 2.9 holds if condition (b’) is substituted for condition (**b*):

*(b’)* *L*(*X*) ⊆ *ST*(*X*) *and* *M*(*X*) ⊆ *AB*(*X*) *where* ⋅ *denoted the closure*.

#### Corollary 2.11

*The result of Theorem 2.9 holds if condition (b”) is substituted for condition (b)*:

*(b”)* *L*(*X*) *and* *M*(*X*) *is closed subset of* *X*, *and* *L*(*X*)⊆ *ST*(*X*) *and* *M*(*X*) ⊆ *AB*(*X*).

#### Example 2.12

*Assume the same conditions of* Example 2.6 *hold*, *except that*

$$\begin{array}{}{\displaystyle S(x)=T(x)=\left\{\begin{array}{ll}& \frac{8}{3},\phantom{\rule{1em}{0ex}}x=0,\\ & \frac{4}{3},\phantom{\rule{1em}{0ex}}x\in (0,1],\\ & \frac{6x+2}{5},\phantom{\rule{1em}{0ex}}x\in (1,3).\end{array}\right.}\end{array}$$

$\begin{array}{}{\displaystyle ST(X)=\{\frac{8}{3}\}\cup \{\frac{4}{3}\}\cup (\frac{2}{5},4)}\end{array}$ *is not closed subset of* *X*, *but conditions (b’) and (b”) of Corollary 2.10 and Corollary 2.11 are satisfied*, *2 is a unique common fixed point of* *A*, *B*, *S*, *T*, *L* *and* *M*.

In order to show that the common property (*E.A*) of two pairs (*L*, *AB*) and (*M*, *ST*) can be deduced from containment of *L*(*X*) ⊆ *ST*(*X*) and property (*E.A*) of (*L*, *AB*), we propose the following theorem.

#### Theorem 2.14

*Let* *A*, *B*, *S*, *T*, *L* *and* *M* *be self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1]. *Suppose that the inequality* (1) *of Lemma 1.7 and the following hypotheses hold*:

*L*(*X*) ⊆ *ST*(*X*)*;*

*ST*(*X*) *is closed in* *X* *and* (*L*, *AB*) *satisfies the property* (*E.A*).

*Then* (*L*, *AB*) *and* (*M*, *ST*) *share the common property* (*E.A*).

#### Proof

Since (*L*, *AB*) satisfies the property (*E.A*), there exists a sequence {*x*_{n}} ⊂ *X* such that

$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}L{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}AB{x}_{n}=z,\text{\hspace{0.17em}}\text{for some}\text{\hspace{0.17em}}z\in X.}\end{array}$$

Since *L*(*X*) ⊆ *ST*(*X*), for each *x*_{n}, there exists a corresponding *y*_{n} ∈ *X* such that *Lx*_{n} = *STy*_{n}. Therefore, we have

$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}L{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}AB{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}ST{y}_{n}=z,\text{\hspace{0.17em}}\text{for some}\text{\hspace{0.17em}}z\in X.}\end{array}$$

It suffices to show that $\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}}\end{array}$*My*_{n} = *z*. Substituting *p* = *x*_{n}, *q* = *y*_{n} in inequality (1), we obtain

$$\begin{array}{}{\displaystyle {F}_{L{x}_{n},M{y}_{n}}(\varphi (x))\ge min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{ST{y}_{n},M{y}_{n}}(x),{F}_{ST{y}_{n},L{x}_{n}}(\beta x),{F}_{AB{x}_{n},M{y}_{n}}((1+\beta )x),{F}_{AB{x}_{n},ST{y}_{n}}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{L{x}_{n},M{y}_{n}}(x),{F}_{L{x}_{n},L{x}_{n}}(\beta x),{F}_{AB{x}_{n},M{y}_{n}}((1+\beta )x),{F}_{AB{x}_{n},L{x}_{n}}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{L{x}_{n},M{y}_{n}}(x),1,{F}_{AB{x}_{n},M{y}_{n}}((1+\beta )x),{F}_{AB{x}_{n},L{x}_{n}}(x)\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\ge min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{L{x}_{n},M{y}_{n}}(x),t({F}_{AB{x}_{n},L{x}_{n}}(\beta x),{F}_{L{x}_{n},M{y}_{n}}(x))\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{L{x}_{n},M{y}_{n}}(x),min\{({F}_{AB{x}_{n},L{x}_{n}}(\beta x),{F}_{L{x}_{n},M{y}_{n}}(x))\}\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=\ge min\{{F}_{AB{x}_{n},L{x}_{n}}(x),{F}_{L{x}_{n},M{y}_{n}}(x)\}.}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}={F}_{AB{x}_{n},L{x}_{n}}(x).}\end{array}$$

Letting *n* → ∞, we obtain that *F*_{ABxn, Lxn}(*x*) → *F*_{z, z}(*x*) = 1. So,

$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}L{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}AB{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}ST{y}_{n}=\underset{n\to \mathrm{\infty}}{lim}M{y}_{n}=z,\text{\hspace{0.17em}}\text{for some}\text{\hspace{0.17em}}z\in X.}\end{array}$$

Thus, (*L*, *AB*) and (*M*, *ST*) share the common property (*E.A*).□

Next, we extend common fixed point theorem of six self-mappings to six finite families of self mappings in Menger spaces.

#### Theorem 2.16

*Let* $\begin{array}{}{\displaystyle \{{A}_{i}{\}}_{i=1}^{m},\{{B}_{r}{\}}_{r=1}^{n},\{{S}_{k}{\}}_{k=1}^{e},\{{T}_{h}{\}}_{h=1}^{f},\{{L}_{j}{\}}_{j=1}^{c}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\{{M}_{v}{\}}_{v=1}^{d}}\end{array}$ *be six finite families of self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1] *where* *A* = *A*_{1}*A*_{2} ⋯ *A*_{m}, *B* = *B*_{1}*B*_{2} ⋯ *B*_{n}, *S* = *S*_{1}*S*_{2} ⋯ *S*_{e}, *T* = *T*_{1}*T*_{2} ⋯ *T*_{f}, *L* = *L*_{1}*L*_{2} ⋯ *L*_{c} and M = *M*_{1}*M*_{2} ⋯ *M*_{d}. *Suppose that the inequality* (1) *of Lemma 1.7 holds. If the pairs* (*L*, *AB*) *and* (*M*, *ST*) *share the* (*CLR*_{(AB)(ST)}) *property*, *then* (*L*, *AB*) *and* (*M*, *ST*) *have a coincidence point each*.

*Moreover*, *if*

*both the pairs* (*L*, *AB*) *and* (*M*, *ST*) *are weakly compatible*.

*AB* = *BA*, *LA* = *AL*, *MS* = *SM* *and* *ST* = *TS*.

*Then* *A*, *B*, *S*, *T*, *L* *and* *M* *have a unique common fixed point*.

#### Proof

The proof can be completed on the lines of Theorem 4.2 in Imdad et al. [12].□

When *A*_{1} = *A*_{2} = ⋯ = *A*_{m} = *A*, *B*_{1} = *B*_{2} = ⋯ = *B*_{n} = *B*, *S*_{1} = *S*_{2} = ⋯ = *S*_{e} = *S*, *T*_{1} = *T*_{2} = ⋯ = *T*_{f} = *T*, *L*_{1} = *L*_{2} = ⋯ = *L*_{c} = *L* and *M*_{1} = *M*_{2} = ⋯ = *M*_{d} = *M*, then we have the following corollary:

#### Corollary 2.17

*Let* *A*, *B*, *S*, *T*, *L* *and* *M* *of self mappings of a Menger space* (*X*, *F*, *t*), *with continuous* *t*-*norm with* *t*(*x*, *x*) ≥ *x* *for all* *x* ∈ [0, 1]. *Suppose that*

*the pairs* (*L*^{c}, *A*^{m}B^{n}) *and* (*M*^{d}, *S*^{e}T^{f}) *share the* (*CLR*_{(AmBn)(SeTf)}) *property*,

*there exists an upper semicontinuous function* *ϕ* : [0, ∞) → [0, ∞) *with* *ϕ*(0) = 0 *and* *ϕ*(*x*) < *x* *for all* *x* > 0 *such that*

$$\begin{array}{}{\displaystyle {F}_{{L}^{c}p,{M}^{d}q}(\varphi (x))\ge min\{{F}_{{A}^{m}{B}^{n}p,{L}^{c}p}(x),{F}_{{S}^{e}{T}^{f}q,{M}^{d}q}(x),{F}_{{S}^{e}{T}^{f}q,{L}^{c}p}(\beta x),{F}_{{A}^{m}{B}^{n}p,{M}^{d}q}((1+\beta )x),{F}_{{A}^{m}{B}^{n}p,{S}^{e}{T}^{f}q}(x)\},}\end{array}$$

*for all* *p*, *q* ∈ *X*, *β*≥ 1 *and* *x* > 0. *Then* (*L*^{c}, *A*^{m}B^{n}) *and* (*M*^{d}, *S*^{p}T^{q}) *have a coincidence point each*.

*Moreover*, *if* *AB* = *BA*, *LA* = *AL*, *MS* = *SM* *and* *ST* = *TS*. *Then* *A*, *B*, *S*, *T*, *L* *and* *M* *have a unique common fixed point*.

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