In this relationship matrix, we know that the relation value between positive membership and positive membership is 1. Similarly, the relation value between neutral memberships, negative memberships, refusal memberships are 1. Since the three component functions of picture fuzzy set are independent of each other, the relation value among positive membership, neutral membership and negative membership are 0. We assume that the values of the relationship between the refusal membership and positive membership, neutral membership, negative membership are
, respectively. It’s clear that
are not greater than one, and they add up to one.
Proof.
Let $A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ and $B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ be two picture fuzzy sets on X.
Taking a close explanation of the similarity measure
$S(A,B)$:
(S1) For
$A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle \mid {x_{i}}\in X\}$,
$B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle \mid {x_{i}}\in X\}$, we have
$-1\leqslant {\mu _{A}}({x_{i}})-{\mu _{B}}({x_{i}})\leqslant 1$,
$-1\leqslant {\eta _{A}}({x_{i}})-{\eta _{B}}({x_{i}})\leqslant 1$ and
$-1\leqslant {\nu _{A}}({x_{i}})-{\nu _{A}}({x_{i}})\leqslant 1$. Hence,
Besides,
Thus, we can get
From the above analysis, we get
$0\leqslant S(A,B)\leqslant 1$.
(S2) Obviously, $S(A,B)=S(B,A)$.
(S3)
Then from the above analysis, we get
$S(A,B)=1$ if and only if
$A=B$.
(S4) Let
$A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle \mid {x_{i}}\in X\}$,
$B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ and
$C=\{\langle {\mu _{C}}({x_{i}})$,
${\eta _{C}}({x_{i}}),{\nu _{C}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ be three PFSs that satisfy the relation
$A\subseteq B\subseteq C$, we have:
For
$a,b,c,\in [0,1]$,
$a+b+c\leqslant 1$,
$x,y,z\in [0,1]$,
$x+y+z\in [0,1]$.
$1-g,1-h,1-l,1-g-h,1-g-l,1-h-l\in [0,1]$, we define a function
f as:
We have
and
Therefore,
$\frac{\partial f}{\partial x}\leqslant 0$, for
$0\leqslant x\leqslant a$. It means that
f is a decreasing function of
x, when
$y=b$,
$z=c$,
$x\leqslant a$. Similarly,
$\frac{\partial f}{\partial x}\geqslant 0$, for
$a<x\leqslant 1$. It means that
f is an increasing function of
x, when
$y=b$,
$z=c$,
$x>a$.
In the same way, we can get $\frac{\partial f}{\partial y}\leqslant 0$ for $0\leqslant y\leqslant b$. It means that f is a decreasing function of y, when $x=a$, $z=c$, $y\leqslant b$. $\frac{\partial f}{\partial y}\geqslant 0$ for $b<y\leqslant 1$. It means that f is an increasing function of y, when $x=a$, $z=c$, $y>b$. $\frac{\partial f}{\partial z}\leqslant 0$ for $0\leqslant z\leqslant c$. It means that f is a decreasing function of z, when $x=a$, $y=b$, $z\leqslant c$. $\frac{\partial f}{\partial z}\geqslant 0$ for $c<z\leqslant 1$. It means that f is an increasing function of z, when $x=a$, $y=b$, $z>c$.
(1) Suppose that
$A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle |{x_{i}}\in X\}$,
$B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle |{x_{i}}\in X\}$ and
$C=\{\langle {\mu _{C}}({x_{i}}),{\eta _{C}}({x_{i}}),{\nu _{C}}({x_{i}})|{x_{i}}\in X\rangle \}$ satisfying
we can obtain:
then:
i.e.
Thus, we have:
$S(A,B)\geqslant S(A,C)$ is founded. Similarly, if we suppose
$a={\mu _{C}}$,
$b={\eta _{C}}$,
$c={\nu _{C}}$, then we have
$S(B,C)\geqslant S(A,C)$.
(2) Suppose that
$A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle \mid {x_{i}}\in X\}$,
$B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ and
$C=\{\langle {\mu _{C}}({x_{i}}),{\eta _{C}}({x_{i}}),{\nu _{C}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ satisfying
we can obtain
Thus, we have:
$S(A,B)\geqslant S(A,C)$ is founded. Similarly, if we suppose
$a={\mu _{C}}$,
$b={\eta _{C}}$,
$c={\nu _{C}}$, then we have
$S(B,C)\geqslant S(A,C)$.
(3) Suppose that
$A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle \mid {x_{i}}\in X\}$,
$B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ and
$C=\{\langle {\mu _{C}}({x_{i}}),{\eta _{C}}({x_{i}}),{\nu _{C}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ satisfying
we can obtain
Thus we have
$S(A,B)\geqslant S(A,C)$ is founded. Similarly, if we suppose
$a={\mu _{C}}$,
$b={\eta _{C}}$,
$c={\nu _{C}}$, then we can get
$S(B,C)\geqslant S(A,C)$.
Furthermore, picture fuzzy sets $A=\{\langle {\mu _{A}}({x_{i}}),{\eta _{A}}({x_{i}}),{\nu _{A}}({x_{i}})\rangle \mid {x_{i}}\in X\}$, $B=\{\langle {\mu _{B}}({x_{i}}),{\eta _{B}}({x_{i}}),{\nu _{B}}({x_{i}})\rangle \mid {x_{i}}\in X\}$ and $C=\{\langle {\mu _{C}}({x_{i}}),{\eta _{C}}({x_{i}}),{\nu _{C}}({x_{i}})\rangle \mid {x_{i}}\in X\}$, $A\subseteq B\subseteq C$ also satisfying
(4) $({\mu _{A}}({x_{i}})<{\mu _{B}}({x_{i}})={\mu _{C}}({x_{i}}))\wedge ({\nu _{A}}({x_{i}})\geqslant {\nu _{B}}({x_{i}})>{\nu _{C}}({x_{i}}))$.
(5) $({\mu _{A}}({x_{i}})<{\mu _{B}}({x_{i}})={\mu _{C}}({x_{i}}))\wedge ({\nu _{A}}({x_{i}})\geqslant {\nu _{B}}({x_{i}})={\nu _{C}}({x_{i}}))\wedge ({\eta _{B}}({x_{i}})\leqslant {\eta _{C}}({x_{i}}))$.
(6) $({\mu _{A}}({x_{i}})={\mu _{B}}({x_{i}})<{\mu _{C}}({x_{i}}))\wedge ({\nu _{A}}({x_{i}})>{\nu _{B}}({x_{i}})\geqslant {\nu _{C}}({x_{i}}))$.
(7) $({\mu _{A}}({x_{i}})={\mu _{B}}({x_{i}})={\mu _{C}}({x_{i}}))\wedge ({\nu _{A}}({x_{i}})>{\nu _{B}}({x_{i}})={\nu _{C}}({x_{i}}))\wedge ({\eta _{B}}({x_{i}})\leqslant {\eta _{C}}({x_{i}}))$.
(8) $({\mu _{A}}({x_{i}})={\mu _{B}}({x_{i}})<{\mu _{C}}({x_{i}}))\wedge ({\nu _{A}}({x_{i}})={\nu _{B}}({x_{i}})\geqslant {\nu _{C}}({x_{i}}))\wedge ({\eta _{A}}({x_{i}})\leqslant {\eta _{B}}({x_{i}}))$.
(9) $({\mu _{A}}({x_{i}})={\mu _{B}}({x_{i}})={\mu _{C}}({x_{i}}))\wedge ({\nu _{B}}({x_{i}})>{\nu _{C}}({x_{i}}))\wedge ({\eta _{A}}({x_{i}})\leqslant {\eta _{B}}({x_{i}}))$.
Similarly, these cases can be proved.
Summing up the above, Eq. (
13) satisfies Definition
2.4. □