Fuzzy rank functions in the set of all binary systems

Abstract

In this paper, we introduce fuzzy rank functions for groupoids, and we investigate their roles in the semigroup of binary systems by using the notions of right parallelisms and $\rho$-shrinking groupoids.

Background

The notion of a fuzzy subset of a set was introduced by Zadeh (1965). His seminal paper has opened up new insights and applications in a wide range of scientific fields. Rosenfeld (1971) used the notion of a fuzzy subset to set down corner stone papers in several areas of mathematics. Mordeson and Malik (1998) published a remarkable book, Fuzzy commutative algebra, presented a fuzzy ideal theory of commutative rings and applied the results to the solution of fuzzy intersection equations. The book included all the important work that has been done on L-subspaces of a vector space and on L-subfields of a field.

In the study of groupoids $(X,*)$ defined on set X, it has also proven useful to investigate the semigroups $(Bin(X), \Box )$ where Bin(X) is the set of all binary systems (groupoids) $(X,*)$ along with an associative product operation $(X,*)\,\Box \,(X, \bullet ) = (X, \Box )$ such that $x\,\Box \,y = (x*y)\,\bullet \,(y*x)$ for all $x, y\in X$. Thus, e.g., it becomes possible to recognize that the left-zero-semigroup $(X,*)$ with $x*y = x$ for all $x, y\in X$ acts as the identity of this semigroup [see Kim and Neggers (2008)]. Fayoumi (2011) introduced the notion of the center ZBin(X) in the semigroup Bin(X) of all binary systems on a set X, and showed that a groupoid $(X,\bullet )\in ZBin(X)$ if and only if it is a locally-zero groupoid. Han et al. (2012) introduced the notion of hypergroupoids $(HBin(X),\Box )$, and showed that $(HBin(X), \Box )$ is a supersemigroup of the semigroup $(Bin(X), \Box )$ via the identification $x \longleftrightarrow \{x\}$. They proved that $(HBin^*(X), \ominus , [\emptyset ])$ is a BCK-algebra.

In this paper, we introduce fuzzy rank functions for groupoids, and we investigate their roles in the semigroup of binary systems by using the notions of right parallelisms and $\rho$-shrinking groupoids.

Preliminaries

Given a non-empty set X, we let $Bin(X)$ denote the collection of all groupoids $(X,*)$, where $*: X\times X\rightarrow X$ is a map and where $*(x, y)$ is written in the usual product form. Given elements $(X,*)$ and $(X,\bullet )$ of $Bin(X)$, define a product “$\Box$” on these groupoids as follows:

$$(X,*) \,\Box \,(X,\bullet ) =(X,\Box )$$

where

$$x\,\Box \,y = (x*y)\,\bullet \, (y*x)$$

for any $x, y\in X$. Using that notion, H. S. Kim and J. Neggers proved the following theorem.

Theorem 1

(Kim and Neggers 2008) $(Bin(X), \Box )$ is a semigroup, i.e., the operation “$\Box$” as defined in general is associative. Furthermore, the left- zero-semigroup is the identity for this operation.

Fuzzy rank functions for groupoids

Given a groupoid $(X,*)$ in Bin(X), a map $\rho : X\rightarrow [0, \infty )$ (or $\rho : X\rightarrow [0, 1]$) is said to be:

• (i) a (fuzzy) rank-subalgebra: $\rho (x*y)\ge \min \{\rho (x), \rho (y)\}$,

• (ii) a (fuzzy) rank-co-subalgebra: $\rho (x*y) \le \max \{\rho (x), \rho (y)\}$,

• (iii) a (fuzzy) rank-d-function: $\rho (x*y) =\max \{ \rho (x) -\rho (y), 0\}$,

• (iv) a (fuzzy) symmetric-rank-function: $\rho (x*y)= \max \{\rho (x)-\rho (y), \rho (y) -\rho (x)\}\,\hbox{for\,all}\,x,y \in X.$

Note that if $\rho : X\rightarrow [0, \infty )$ is a fuzzy subset of X, then it is a fuzzy rank-subalgebra as well. Thus these algebraic structures are special cases of the classes of fuzzy rank-subalgebras. Each of the types listed above serve as some idea of measure of size when the binary operation “$*$” is considered as corresponding very roughly to a (group) sum (product), a left-zero-semigroup, a d-algebra (BCK-algebra) or an absolute (value) difference. For the sake of being able to make comparisons in the behavior of interactions of rank-type and groupoid it seems a better idea to consider these simultaneously rather than study only isolated cases without considering common aspects as well as distinguishing ones.

Example 2

• (a) Let $(X,*)$ be a left-zero-semigroup, i.e., $x*y = x$ for all $x, y\in X$. If $\rho : X\rightarrow [0, \infty )$ is any map, then $\rho (x*y) = \rho (x)$ for all $x, y\in X$. It follows that $\rho (x*y) \ge \min \{\rho (x), \rho (y)\}$ and $\rho (x*y) \le \max \{\rho (x), \rho (y)\}$ for all $x, y\in X$, which shows that every function $\rho : X\rightarrow [0, \infty )$ is both a (fuzzy) rank-subalgebra and a (fuzzy) rank-co-subalgebra.

• (b) Let $(X,*)$ be a left-zero-semigroup and let $\rho : X \rightarrow [0, \infty )$ be a (fuzzy) rank-d-function. Then it is a zero function. In fact, if we assume that there exists an $x_0\in X$ such that $\rho (x_0)> 0$, then $0< \rho (x_0) = \rho (x_0* y) =\max \{\rho (x_0)-\rho (y), 0\}$ for all $y\in X$. If we take $y:=x_0$, then it leads to a contradiction.

• (c) Let $(X,*)$ be a left-zero-semigroup and let $\rho : X \rightarrow [0, \infty )$ be a (fuzzy) symmetric-rank-function. Assume that there exists $x_0\in X$ such that $0< \rho (x_0)$. Since $\rho$ is a (fuzzy) symmetric-rank-function, we have $\rho (x) = \rho (x*y) = \max \{\rho (x) -\rho (y), \rho (y)-\rho (x)\}$ for all $x, y\in X$. If we take $x:= x_0, y:= y_0$, then $0< \rho (x_0)= \rho (x_0* x_0) =\max \{ \rho (x_0)-\rho (x_0), \rho (x_0)-\rho (x_0)\}=0$, a contradiction. This shows that $\rho$ is a zero function.

If $(X,*)$ is a right-zero-semigroup, i.e., $x*y = y$ for all $x, y\in X$, then any function $\rho : X\rightarrow [0, \infty )$ is both a (fuzzy) rank-subalgebra and a (fuzzy) rank-co-subalgebra of $(X,*)$, while if $\rho$ is either a (fuzzy) rank-d-function or a (fuzzy) symmetric-rank-function, then it is a zero function.

A groupoid $(X,*)$ is said to be selective (Neggers 1976; Neggers and Kim 1996) if $x*y\in \{x, y\}$ for all $x, y\in X$. For example, every left-(right-)zero-semigroup is selective. Given a selective groupoid $(X,*)$, we may construct a digraph via $x\rightarrow y \Leftrightarrow x*y =y$ for all $x, y\in X$. Hence selective groupoids are interpretable as digraphs on the (vertex) set X.

Proposition 3

Let $(X,*)$ be a selective groupoid and let $\rho : X\rightarrow [0, \infty )$ be a (fuzzy) rank-d-function of X. If $x\in X$ such that $\rho (x)>0$, then any vertex $y (\not = x)$ in X with $x\nrightarrow y$ has $\rho (y)=0$.

Proof

Let $y (\not = x)$ in X with $x\nrightarrow y$. Since $(X,*)$ is selective and $x\nrightarrow y$, we have $x*y = x$. It follows from the fact that $\rho$ is a rank-d-function that $\rho (x) = \rho (x*y) = \max \{\rho (x)-\rho (y), 0\}$ and hence that $\rho (x) = \rho (x) -\rho (y)$, proving that $\rho (y)= 0$. $\square$

Theorem 4

Let $(X,*)$ be a selective groupoid and let $\rho : X\rightarrow [0, \infty )$ be a (fuzzy) rank-d-function of X. If $x\in X$ such that $\rho (x)>0$, then there exists at most one vertex $x\in X$ such that $\rho (x)>0$.

Proof

Assume that there are two vertices x and y in X such that $\rho (x)>0, \rho (y)>0$. It follows that $x\rightarrow y, y\rightarrow x$ by Proposition 3. Now $x\rightarrow y$ implies that $x*y= y$. Hence $0 < \rho (y) =\rho (x*y) =\max \{\rho (x)-\rho (y), 0\}= \rho (x)-\rho (y)$. This shows that $2\rho (y) = \rho (x)$. Similarly, $y\rightarrow x$ implies $2\rho (x) = \rho (y)$. Thus we obtain $\rho (x) = 2\rho (y) = 4\rho (x)$, which implies $\rho (x)=0$, a contradiction. $\square$

Proposition 5

If $(X,*)$ is a selective groupoid, then every (fuzzy) rank-d-function of $(X, *)$ is a zero function.

Proof

If $(X,*)$ is a selective groupoid, then $x*x = x$ for all $x\in X$. Since $\rho$ is a (fuzzy) rank-d-function, we obtain $\rho (x) = \rho (x*x) = \max \{\rho (x)-\rho (x), 0\} = 0$, proving the proposition. $\square$

Given a map $\rho : X\rightarrow [0, \infty )$, we define

$${\sum _i(\rho )}:= \left\{ (X,*)\in Bin(X)| \text {the\,condition}\, (i) \text {holds\,for}\, (X,*)\right\} \quad (i=1, 2, 3, 4).$$

Proposition 6

$(\sum _i(\rho ), \Box )$ is a subsemigroup of $(Bin(X), \Box )$ where $i= 1, 2, 3$.

Proof

Let $(X,*), (X,\bullet )\in \sum _1(\rho )$. Then $\rho (x*y)\ge \min \{\rho (x), \rho (y)\}$ and $\rho (x\bullet y)\ge \min \{\rho (x), \rho (y)\}$ for all $x, y\in X$. If we let $(X,\Box ):= (X,*)\,\Box \,(X,\bullet )$, then for any $x, y\in X$, we have $x\,\Box \,y = (x*y)\bullet (y*x)$. It follows that

$$\begin{aligned} \rho (x\,\Box \,y)&= {} \rho ((x*y)\,\bullet \, (y*x))\\&\ge {} \min \{\rho (x*y), \rho (y*x)\}\\&\ge {} \min \{ \min \{\rho (x), \rho (y)\}, \min \{\rho (y), \rho (x)\}\}\\&= {} \min \{ \rho (x), \rho (y)\} \end{aligned}$$

This shows that $(X,*)\,\Box \,(X,\bullet )= (X, \Box )\in \sum \nolimits _1(\rho )$. Hence $(\sum \nolimits _1(\rho ), \Box )$ is a subsemigroup of $(Bin(X), \Box )$. Similarly, we may obtain that $(\sum \nolimits _i(\rho ), \Box )$ is also a subsemigroup of $(Bin(X), \Box )$ where $i= 2, 3$. $\square$

Note that $(\sum \nolimits _4(\rho ), \Box )$ need not be a subsemigroup of $(Bin(X), \Box )$. Let $(X,*), (X,\bullet )\in \sum \nolimits _1(\rho )$ and let $(X,\Box ):= (X,*)\,\Box \,(X,\bullet )$. If we take $x, y\in X$ such that $\rho (x) > \rho (y)$, then we have

$$\begin{aligned} \rho (x\,\Box \,y)&= {} \rho ((x*y)\,\bullet \, (y*x))\\&= {} \max \{\rho (x*y)-\rho (y*x), \rho (y*x)-\rho (y*x)\}\\&= {} \max \{ \max \{\rho (x)-\rho (y), \rho (y)-\rho (x)\}\\&\quad- \max \{\rho (y)-\rho (x), \rho (x)-\rho (y)\},\\&\quad \max \{\rho (y)-\rho (x), \rho (x)-\rho (y)\} \\&\quad-\max \{\rho (x)-\rho (y), \rho (y)-\rho (x)\} \}\\&= {} \min \left\{ [\rho (x)- \rho (y)] - [\rho (x) -\rho (y)],\right. \\&\left. \quad [\rho (x)- \rho (y)] - [\rho (x) -\rho (y)]\right\} \\&= {} 0 < \rho (x) -\rho (y) \\&= {} \max \{ \rho (x) -\rho (y), \rho (y) -\rho (x)\}, \end{aligned}$$

which shows that $(\sum \nolimits _4(\rho ), \Box )$ is not a subsemigroup of $(Bin(X), \Box )$.

Let $\rho : X\rightarrow [0, \infty )$ be a map. A groupoid $(X,*)$ is said to have a $\rho$-chain n if there exist $x_1, \ldots , x_n\in X$ such that $\rho (x_1)< \rho (x_2)< \cdots < \rho (x_n)$. We denote the $\rho$-chain by $\langle x_1, \ldots , x_n \rangle$. A groupoid $(X,*)$ is said to have the $\rho$-height n if $\langle x_1, \ldots , x_n \rangle$ is the largest maximal $\rho$-chain in $(X,*)$.

Proposition 7

Let $\rho : X\rightarrow [0, \infty )$ be a map and let $Bin(X) = \sum _1(\rho )$. Then the $\rho$-height of $(X,*)$ is ≥2 for any $(X,*)\in Bin(X)$.

Proof

Assume there exist $x, y, z\in X$ such that $\rho (x)< \rho (y) < \rho (z)$. Let $(X,*)$ be a groupoid such that $x= y*z$. Then $(X,*) \in Bin(X) = \sum \nolimits _1(\rho )$. It follows that $\rho (y*z) =\rho (x) < \rho (y) = \min \{\rho (y), \rho (z)\}$. This shows that $(X,*)\not \in \sum \nolimits _1(\rho ) = Bin(X)$, a contradiction. $\square$

Proposition 8

Let $\rho : X\rightarrow [0, \infty )$ be a map and let $Bin(X) = \sum _1(\rho )$. If $\rho$ has two values $a, b\in [0, \infty )$ with $a< b$, then there exists uniquely an $\widehat{x} \in X$ such that $\rho (\widehat{x})= b$ and $\rho (y) = a$ for all $y\not = \widehat{x}$ in X.

Proof

Let $\widehat{x} \in X$ such that $\rho (\widehat{x}) = b > a$. If $y\in X$ such that $\widehat{x}\not = y$, then $\rho (y) = a$. In fact, if $\rho (y)=b$, then we may take a groupoid $(X,*)$ in Bin(X) such that $\rho (\widehat{x} * y) = a$, since $Bin(X) = \sum\nolimits _1(\rho )$. It follows that $a = \rho (\widehat{x} * y) \ge \min \{ \rho (\widehat{x}), \rho (y)\}= b$, a contradiction. We claim that such an $\widehat{x}$ is unique. Assume that there are two elements $x^{\prime }, \widehat{x}$ in X such that $\rho (\widehat{x}) = b = \rho (x^{\prime })$. A groupoid $(X,\bullet )$ satisfying $\rho (\widehat{x}\,\bullet \, x^{\prime }) = a$ may then be selected. It follows that $a= \rho (\widehat{x}\,\bullet \, x^{\prime }) \ge \min \{ \rho (\widehat{x}), \rho (x^{\prime })\}= b$, a contradiction. $\square$

Proposition 9

Let $(X,*)\in Bin(X)$ and let $\rho : X\rightarrow [0, \infty )$ be a (fuzzy) rank-d-function of X. Then

• (1) $Ker(\rho ) \not = \emptyset$,

• (2) if $\rho (x) \le \rho (y)$ then $x*y \in Ker(\rho )$,

• (3) $Ker(\rho )$ is a right ideal of $(X,*)$.

Proof

(1) Given $x\in X$, $\rho (x*x) = \max \{\rho (x)-\rho (x), 0\} = 0$. It follows that $x*x\in Ker(\rho )$. (2) If $\rho (x) \le \rho (y)$ then $\rho (x*y) = \max \{\rho (x)-\rho (y), 0\}= 0$ and hence $x*y\in Ker(\rho )$. (3) Given $x\in Ker(\rho ), y\in X$, we have $\rho (x*y) = \max \{\rho (x)-\rho (y), 0\} = \max \{0-\rho (y), 0\} = 0$, which shows $x*y\in Ker(\rho )$. $\square$

Right parallelisms

Let $\rho _1, \rho _2$ be mappings from X to $[0, \infty )$. The map $\rho _1$ is said to be right parallel to $\rho _2$ if $\rho _1(a)\le \rho _1(b)$ implies $\rho _2(a)\le \rho _2(b)$, and we denote it by $\rho _1\, ||\, \rho _2$.

Proposition 10

If $\rho _1$ is a (fuzzy) rank-subalgebra of $(X,*)$ and $\rho _1 || \rho _2$, then $\rho _2$ is also a (fuzzy) rank-subalgebra of $(X,*)$.

Proof

If $\rho _1$ is a rank-subalgebra of $(X,*)$, then $\rho _1(x*y)\ge \min \{\rho _1(x),$ $\rho _1(y)\}$ for all $x, y\in X$. Without loss of generality, we let $\rho _1(x*y) \ge \rho _1(x)$. Since $\rho _1|| \rho _2$, we obtain $\rho _2(x*y) \ge \rho _2(x)$. It follows that $\rho _2(x*y)\ge \min \{\rho _2(x), \rho _2(y)\}$. $\square$

Given maps $\rho _i: X\rightarrow [0, \infty )$, we define

$$\begin{aligned} (\rho _1 + \rho _2)(x):&= {} \rho _1(x) + \rho _2(x)\\ (\rho _1 \,\bullet \, \rho _2)(x):&= {} \rho _1(x)\rho _2(x) \end{aligned}$$

for all $x\in X$.

Note that $\rho _1 + \rho _2$ need not be a (fuzzy) rank-subalgebra of $(X,*)$ even though $\rho _1$ and $\rho _2$ are (fuzzy) rank-subalgebras of $(X,*)$. In fact,

$$\begin{aligned} (\rho _1 + \rho _2)(x*y)&= {} \rho _1(x*y) + \rho _2(x*y)\\&\ge {} \min \{\rho _1(x), \rho _1(y)\} + \min \{\rho _2(x), \rho _2(y)\} \end{aligned}$$

and

$$\min \{ (\rho _1 + \rho _2)(x), (\rho _1 + \rho _2)(y)\} = \min \{\rho _1(x) + \rho _2(x), \rho _1(y) + \rho _2(y)\}$$

In the real numbers, it is not always true that $\min \{a, b\} + \min \{c, d\} \ge \min \{a+c, b+d\}$, which shows that $\rho _1 + \rho _2$ need not be a (fuzzy) rank-subalgebra of $(X,*)$.

Proposition 11

Let $\rho _1, \rho _2$ be mappings from X to $[0, \infty )$. If $\rho _1$ is a constant for all $x\in X$ and if $\rho _1|| \rho _2$, then $\rho _2$ is also a constant function on X.

Proof

Straightforward. $\square$

Note that any map $\rho _1$ is right parallel to $\rho _2$ if $\rho _2$ is a constant function. The ‘right parallel’ relation “||” is reflexive and transitive, but it is not an anti-symmetric, i.e., || is a quasi order, but not a partial order on $\{\rho | \rho : X\rightarrow [0, \infty )$ : a map $\}$.

For example, we let $X:=[0, \infty )$ and let $\rho _1: X\rightarrow [0, \infty )$ be the identity map, and let $\rho _2: X\rightarrow [0, \infty )$ be a map defined by $\rho _2(x):= e^ x$. Then $\rho _1(a)\le \rho _1(b)$ if and only if $\rho _2(a)\le \rho _2(b)$ for all $a, b\in X$, i.e., $\rho _1|| \rho _2, \rho _2|| \rho _1$, but $\rho _1 \not = \rho _2$.

Proposition 12

If $\rho _1$ is a (fuzzy) rank-subalgebra of $(X,*)$ and $\rho _1|| \rho _2$, then $\rho _1\,\bullet \,\rho _2$ is also a (fuzzy) rank-subalgebra of $(X,*)$.

Proof

If $\rho _1$ is a rank-subalgebra of $(X,*)$, then $\rho _1(x*y)\ge \min \{\rho _1(x),$ $\rho _1(y)\}$ for all $x, y\in X$. If we assume $\rho _1(x*y)\ge \rho _1(x)$, then

$$\begin{aligned} (\rho _1\,\bullet \,\rho _2)(x*y)&= {} \rho _1(x*y) \rho _2(x*y)\\&\ge {} \rho _1(x) \rho _1(y)\\&= {} (\rho _1\,\bullet \,\rho _2)(x)\\&\ge {} \min \{(\rho _1\,\bullet \,\rho _2)(x), (\rho _1\,\bullet \,\rho _2)(x)\}, \end{aligned}$$

which proves the proposition. $\square$

Corollary 13

If $\rho _1$ is a (fuzzy) rank-co-subalgebra of $(X,*)$ and $\rho _1|| \rho _2$, then $\rho _1\,\bullet \,\rho _2$ and $\rho _2$ are also (fuzzy) rank-co-subalgebras of $(X,*)$.

Proof

The proofs are similar to Propositions 10 and 12. $\square$

In the above note, we mentioned that $\rho _1 + \rho _2$ need not be a rank-subalgebra of $(X,*)$. Using the notion of the right parallelism, we prove the following.

Proposition 14

If $\rho _i$ is a rank-subalgebra of $(X,*)$ $(i= 1,2)$ and if $\rho _1|| \rho _2$, then $\rho _1 + \rho _2$ is also a rank-subalgebra of $(X,*)$.

Proof

Since $\rho _1$ is a rank-subalgebra of $(X,*)$, we have $\rho _1(x*y)\ge \min \{\rho _1(x), \rho _1(y)\}$ for all $x, y\in X$. Without loss of generality, we let $\rho _1(x*y)\ge \rho _1(x)$. Since $\rho _1|| \rho _2$, we obtain $\rho _2(x*y)\ge \rho _2(x)$. It follows that $\rho _1(x*y) + \rho _2(x*y) \ge \rho _1(x) + \rho _2(x)$ and hence $(\rho _1 + \rho _2)(x*y) \ge \min \{ (\rho _1 + \rho _2)(x), (\rho _1 + \rho _2)(y)\}$, proving the proposition. $\square$

Corollary 15

If $\rho _i$ is a rank-co-subalgebra of $(X,*)$ $(i= 1,2)$ and if $\rho _1|| \rho _2$, then $\rho _1 + \rho _2$ is also a rank-co-subalgebra of $(X,*)$.

Proof

Similar to Proposition 14. $\square$

Proposition 16

Let $X:={\mathbf{R}}$ be the set of all real numbers. Let $\rho _1, \rho _2$ be mappings from X to $[0, \infty )$ and let $\rho _2(x):=x$ for all $x\in X$. If $\rho _1|| \rho _2$, then $\rho _1$ is strictly increasing.

Proof

Assume that there are $x, y\in X$ such that $x<y, \rho _1(x) \ge \rho _1(y)$. Since $\rho _1|| \rho _2$, we obtain $x=\rho _2(x)\ge \rho _2(y) = y$, a contradiction. $\square$

Theorem 17

If $\rho _i$ is a rank-d-function of $(X,*)$ $(i= 1,2)$ and if $\rho _1|| \rho _2$, then $\rho _1 + \rho _2$ is also a rank-d-function of $(X,*)$.

Proof

Given $x, y\in X$, we have four cases: (i) $\rho _1(x) \ge \rho _1(y), \rho _2(x) \ge \rho _2(y)$; (ii) $\rho _1(x) \ge \rho _1(y), \rho _2(x) < \rho _2(y)$; (iii) $\rho _1(x) < \rho _1(y), \rho _2(x) \ge \rho _2(y)$; (iv) $\rho _1(x)< \rho _1(y), \rho _2(x) < \rho _2(y)$. Since $\rho _1|| \rho _2$, the cases (ii) and (iii) are removed. For the case (i), we have $\rho _1(x*y) =\rho _1(x) -\rho _1(y)$ and $\rho _2(x*y)=\rho _2(x) -\rho _2(y)$. It follows that

$$\begin{aligned} (\rho _1+ \rho _2)(x*y)&= {} \rho _1(x*y) + \rho _2(x*y)\\&= {} \rho _1(x) - \rho _1(y) + \rho _2(x) - \rho _2(y)\\&= {} (\rho _1 +\rho _2)(x)- (\rho _1 +\rho _2)(y)\\&\le {} \max \{(\rho _1+\rho _2)(x)-(\rho _1+\rho _2)(y), 0\}, \end{aligned}$$

For the case (iv), we have $\rho _1(x*y) = 0 = \rho _2(x*y)$. It follows that $(\rho _1+ \rho _2)(x*y) = \rho _1(x*y) + \rho _2(x*y)=0 =\max \{(\rho _1+\rho _2)(x)-(\rho _1+\rho _2)(y), 0\}$, proving the theorem. $\square$

Note that if $\rho _1|| \rho _2, \rho _2 || \rho _3$, then $\rho _1 || \rho _2 + \rho _3$, and if $\rho _1 || \rho _2, \rho _1 || \rho _3$, then $\rho _1 || \rho _2 + \rho _3$.

$\rho$-Shrinking groupoids

A groupoid $(X,*)$ is said to be $\rho$-shrinking if $\rho : X\rightarrow [0, \infty )$ is a map satisfying the condition:

$$\rho (x*y) \le \min \{\rho (x), \rho (y)\}$$

for all $x, y\in X$.

Example 18

Let $X:=[0, \infty )$ and let $x*y:= x +y$ for all $x, y\in X$. If we define $\rho (x):=x$ for all $x\in X$, then $\rho (x*y)\ge \max \{\rho (x), \rho (y)\}$. It follows that $(e^{-\rho })(x*y) = e^{-x}e^{-y}\le \min \{e^{-x}, e^{-y}\}= \min \{ (e^{-\rho })(x),$ $(e^{-\rho })(y)\}$, which shows that $(X,*)$ is $e^{-\rho }$-shrinking.

Theorem 19

If $(X,*)$ is both $\rho _1$-shrinking and $\rho _2$-shrinking and if $\rho _1|| \rho _2$, then $(X,*)$ is both $(\rho _1 + \rho _2)$-shrinking and $\rho _1\,\bullet \,\rho _2$-shrinking.

Proof

Since $(X,*)$ is $\rho _1$-shrinking, we have $\rho _1(x*y) \le \min \{\rho _1(x),$ $\rho _1(y)\}$ for all $x, y\in X$. The condition $\rho _1 || \rho _2$ implies that $\rho _2(x*y) \le \min \{\rho _2(x), \rho _2(y)\}$. Hence

$$\begin{aligned} (\rho _1+ \rho _2)(x*y)&= {} \rho _1(x*y) + \rho _2(x*y)\\&\le {} \min \{\rho _1(x), \rho _1(y)\} + \min \{\rho _2(x), \rho _2(y)\} \end{aligned}$$

Without loss of generality, we may assume $\rho _1(x)\le \rho _1(y)$. Then $\rho _2(x)\le \rho _2(y)$, since $\rho _1 || \rho _2$. It follows that $(\rho _1+ \rho _2)(x*y) \le \rho _1(x) + \rho _2(y) = (\rho _1+ \rho _2)(x)$. Hence $(X,*)$ is $(\rho _1 + \rho _2)$-shrinking. Similarly, if we assume $\rho _1(x)\le \rho _1(y)$, then

$$\begin{aligned} (\rho _1\,\bullet \, \rho _2)(x*y)&= {} \rho _1(x*y)\rho _2(x*y)\\&\le {} \rho _1(x)\rho _1(x)\\&= {} (\rho _1\,\bullet \,\rho _2)(x)\\&= {} \min \{(\rho _1\,\bullet \,\rho _2)(x), (\rho _1\,\bullet \,\rho _2)(y)\}, \end{aligned}$$

which shows that $(X,*)$ is $\rho _1\,\bullet \, \rho _2$-shrinking. $\square$

Proposition 20

If $(X,*)$ and $(X,\bullet )$ are $\rho$-shrinking and if $(X,\Box ):=(X,*)\,\Box \,(X,\bullet )$, then $(X,\Box )$ is also $\rho$-shrinking.

Proof

If $(X,\Box ):=(X,*)\,\Box \,(X,\bullet )$, then for all $x, y\in X$, we have

$$\begin{aligned} \rho (x\,\Box \,y)&= {} \rho ((x*y)\,\bullet \,(y*x))\\&\le {} \min \{\rho (x*y), \rho (y*x)\}\\&\le {} \min \{\min \{\rho (x), \rho (y)\}, \min \{\rho (y), \rho (x)\} \\&= {} \min \{\rho (x), \rho (y)\}, \end{aligned}$$

showing that $(X,\Box )$ is $\rho$-shrinking. $\square$

Proposition 20 shows that the collection of all $\rho$-shrinking groupoids forms a subsemigroup of $(Bin(X), \Box )$.

Given maps $\rho : X\rightarrow [0, \infty )$ and $\sigma : Y\rightarrow [0, \infty )$, we define a map $[\rho , \sigma ]: X\times Y\rightarrow [0, \infty )$ by $[\rho , \sigma ](x):= \rho (x) + \sigma (y)$ as a sort of “inner product” ranking. Given groupoids $(X,*)$ and $(Y,\bullet )$, we define a Cartesian product $(X\times Y, \nabla )$ where $(x, y)\nabla (x^{\prime }, y^{\prime }):= (x*x^{\prime }, y\,\bullet \, y^{\prime })$ for all $(x, y), (x^{\prime }, y^{\prime })\in X\times Y$.

Proposition 21

If $(X,*)$ is $\rho$-shrinking and $(Y,\bullet )$ is $\sigma$-shrinking, then $(X\times Y, \nabla )$ is $[\rho , \sigma ]$-shrinking.

Proof

Since $(X,*)$ is $\rho$-shrinking and $(Y,\bullet )$ is $\sigma$-shrinking, we have $\rho (x*y)\le \min \{\rho (x), \rho (y)\}$ and $\sigma (x\,\bullet \, y) \le \min \{\sigma (x), \sigma (y)\}$ for all $x, y\in X$. It follows that

$$\begin{aligned} \rho (x*y) + \sigma (x\,\bullet \, y)&\le {} \min \{\rho (x), \rho (y)\} + \min \{\sigma (x), \sigma (y)\}\\&\le {} \min \{\rho (x) + \sigma (x), \rho (y)+ \sigma (y)\}\\&= {} \min \{[\rho , \sigma ](x), [\rho ,\sigma ](y)\}, \end{aligned}$$

which proves the proposition. $\square$

Conclusions

Above, we introduced four (fuzzy) rank functions in the semigroup of all binary systems (i.e., groupoids), and we investigated their roles related to selective groupoids and the notion of Bin(X). Using the notion of “right parallelism”, we showed that if $\rho _i$ is a (fuzzy) rank-subalgebra (resp., (fuzzy) rank-d-function) of $(X,*)$ ($i= 1,2$) and if $\rho _1|| \rho _2$, then $\rho _1 + \rho _2$ is also a (fuzzy) rank-subalgebra (resp., (fuzzy) rank-d-function) of $(X,*)$. By introducing the notion of $\rho$-shrinking to groupoids, we found that if $(X,*)$ is both $\rho _1$-shrinking and $\rho _2$-shrinking and if $\rho _1|| \rho _2$, then it is both $(\rho _1 + \rho _2)$-shrinking and $\rho _1\,\bullet \,\rho _2$-shrinking. This research may provide hyper-fuzzy rank functions in the set of all binary systems naturally, and thus several well-developed theorems/propositions in the areas of soft fuzzy theory and intuitionistic fuzzy set theory can then possibly be applied in future research also.