8.1.3Cut set of dynamic fuzzy sets

In the transformation of DF set and common set, an important concept is the horizontal section set $\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right).$

Definition 8.4 Assume$\left(\overleftarrow{A},\overrightarrow{A}\right)\in DF\left(U\right),\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)\in \left[0,1\right]\times \left[\leftarrow ,\to \right],$ we note the following:

$\begin{array}{l}\left(1\right){\overleftarrow{A}}_{\overleftarrow{A}}=\left\{\left|\overleftarrow{u}\right|\overleftarrow{u}\in U,\overleftarrow{A}\left(\overleftarrow{u}\right)\ge \overleftarrow{\lambda }\right\}\text{or}{\overleftarrow{A}}_{\overleftarrow{\lambda }}=\left\{\overrightarrow{u}|\overrightarrow{u}\in U,\overrightarrow{A}\left(\overrightarrow{u}\right)\ge \overrightarrow{\lambda }\right\},\left(0\le \left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)<1\right),\text{we}\\ \text{note}\text{that}\left({\overleftarrow{A}}_{{}^{\overleftarrow{\lambda }}},{\overrightarrow{A}}_{{}^{\overrightarrow{\lambda }}}\right)\text{is}\text{the}\text{horizontal}\text{section}\text{set}\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)\text{of}\left(\overleftarrow{A},\overrightarrow{A}\right).\\ \left(2\right){\overleftarrow{A}}_{\overleftarrow{A}}=\left\{\left|\overleftarrow{u}\right|\overleftarrow{u}\in U,\overleftarrow{A}\left(\overleftarrow{u}\right)\ge \overleftarrow{\lambda }\right\}\text{or}{\overleftarrow{A}}_{\overleftarrow{\lambda }}=\left\{\overrightarrow{u}|\overrightarrow{u}\in U,\overrightarrow{A}\left(\overrightarrow{u}\right)\ge \overrightarrow{\lambda }\right\},\left(0\le \left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)<1\right),\text{we}\\ \text{note}\text{that}\left({\overleftarrow{A}}_{{}^{\overleftarrow{\lambda }}},{\overrightarrow{A}}_{{}^{\overrightarrow{\lambda }}}\right)\text{is}\text{the}\text{weak}\text{section}\text{set}\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)\text{of}\left(\overleftarrow{A},\overrightarrow{A}\right).\end{array}$

According to definition, for $\forall \left(\overleftarrow{u},\overrightarrow{u}\right)\in U,\text{if}\left(\overleftarrow{A}\left(\overleftarrow{u}\right),\overleftarrow{A}\left(\overleftarrow{u}\right)\right)\ge \left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right),\text{then}\left(\overleftarrow{u},\overrightarrow{u}\right)\in$$({\overleftarrow{A}}_{\overleftarrow{\lambda }},{\overrightarrow{A}}_{\overrightarrow{\lambda }}).$ That means, in the horizontal $(\overleftarrow{\lambda },\overrightarrow{\lambda }),(\overleftarrow{u},\overrightarrow{u})$ belongs to the DF set $(\overleftarrow{A},\overrightarrow{A}).$ If $(\overleftarrow{A}(\overleftarrow{u}),\overrightarrow{A}(\overrightarrow{u}))<(\overleftarrow{\lambda },\overrightarrow{\lambda }),\text{then}\text{(}\overleftarrow{u},\overrightarrow{u})\notin ({\overleftarrow{A}}_{\overleftarrow{\lambda }},{\overrightarrow{A}}_{\overrightarrow{\lambda }}).$ That means, in the horizontal $(\overleftarrow{\lambda },\overrightarrow{\lambda }),\text{(}\overleftarrow{u},\overrightarrow{u})$ does not belong to the DF set $(\overleftarrow{A},\overrightarrow{A}).$ Therefore, a DF set can be regarded as an unclear dynamic set with only moving boundary.

Property 8.1 Assume $(\overleftarrow{A},\overrightarrow{A}),(\overleftarrow{B},\overrightarrow{B})\in D\text{F}(U),$

then

$\begin{array}{l}{\left(\left(\overleftarrow{A},\overrightarrow{A}\right)\cup \left(\overleftarrow{B},\overrightarrow{B}\right)\right)}_{\left(\overleftarrow{A},\overrightarrow{A}\right)}={\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overleftarrow{A},\overrightarrow{A}\right)}\cup {\left(\overleftarrow{B},\overrightarrow{B}\right)}_{\left(\overleftarrow{A},\overrightarrow{A}\right)}\\ {\left(\left(\overleftarrow{A},\overrightarrow{A}\right)\cap \left(\overleftarrow{B},\overrightarrow{B}\right)\right)}_{\left(\overleftarrow{A},\overrightarrow{A}\right)}={\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overleftarrow{A},\overrightarrow{A}\right)}\cap {\left(\overleftarrow{B},\overrightarrow{B}\right)}_{\left(\overleftarrow{A},\overrightarrow{A}\right)}.\end{array}$

Obviously, for the finite number of DF set in DF(U), these conclusions still hold.

Namely:

$\begin{array}{l}{\left({\displaystyle \underset{i=1}{\overset{N}{\cup }}\left({\overleftarrow{A}}_j,{\overrightarrow{A}}_i\right)}\right)}_{\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)}={\displaystyle \underset{i=1}{\overset{n}{\cup }}\left({\overleftarrow{A}}_{i_{\overleftarrow{\lambda }}},{\overleftarrow{A}}_{i_{\overrightarrow{\lambda }}}\right)}\\ {\left({\displaystyle \underset{i=1}{\overset{N}{\cap }}\left({\overleftarrow{A}}_j,{\overrightarrow{A}}_i\right)}\right)}_{\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)}={\displaystyle \underset{i=1}{\overset{n}{\cap }}\left({\overleftarrow{A}}_{i_{\overleftarrow{\lambda }}},{\overleftarrow{A}}_{i_{\overrightarrow{\lambda }}}\right)}.\end{array}$

Property 8.2 If ${({\overleftarrow{A}}_{\overleftarrow{t}},{\overrightarrow{A}}_{\overrightarrow{t}})}_{(\overleftarrow{t},\overrightarrow{t})\in T}\subseteq D\text{F}(U),$ then

$\begin{array}{l}{\displaystyle \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cup }\left({\overleftarrow{A}}_{\overleftarrow{t}},{\overrightarrow{A}}_{\overrightarrow{t}}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)}=\left({\displaystyle \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cup }\left({\overleftarrow{A}}_{\overleftarrow{t}},{\overrightarrow{A}}_{\overrightarrow{t}}\right)}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\\ {\displaystyle \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cap }\left({\overleftarrow{A}}_{\overleftarrow{t}},{\overrightarrow{A}}_{\overrightarrow{t}}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)}=\left({\displaystyle \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cap }\left({\overleftarrow{A}}_{\overleftarrow{t}},{\overrightarrow{A}}_{\overrightarrow{t}}\right)}\right)\left(\overleftarrow{x},\overrightarrow{x}\right).\end{array}$

Property 8.3 Assume $({\overleftarrow{\lambda }}_1,{\overrightarrow{\lambda }}_1),({\overleftarrow{\lambda }}_2,{\overrightarrow{\lambda }}_2)\in [0,1]\times [\leftarrow ,\to ];(\overleftarrow{A},\overrightarrow{A})\in D\text{F}(U).\text{If}({\overleftarrow{\lambda }}_2,{\overrightarrow{\lambda }}_2)\le$$({\overleftarrow{\lambda }}_1,{\overrightarrow{\lambda }}_1),$ then

${\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left({\overleftarrow{\lambda }}_2,{\overrightarrow{\lambda }}_2\right)}\supset {\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left({\overleftarrow{\lambda }}_1,{\overrightarrow{\lambda }}_1\right)}.$

Property 8.4 Assume

$\left(\overleftarrow{A},\overrightarrow{A}\right)\left(\underset{t\in T}{\vee }\left({\overleftarrow{\lambda }}_t,{\overrightarrow{\lambda }}_t\right)\right)={\displaystyle \underset{t\in T}{\cup }\left({\overleftarrow{A}}_{{\overleftarrow{\lambda }}_t},{\overrightarrow{A}}_{{\overrightarrow{\lambda }}_t}\right).}$

8.1.4Dynamic fuzzy sets decomposition theorem

According to the concept of horizontal section set the on the last section, we can get the following theorem:

Theorem 8.2 The horizontal section set $\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)$ and the weak horizontal section set have the following properties:

Theorem 8.3 If $\left\{{\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overleftarrow{t},\overrightarrow{t}\right)};\left(\overleftarrow{t},\overrightarrow{t}\right)\in T\right\}\subset D\text{F}\left(\overleftarrow{X},\overrightarrow{X}\right),$ then we have the following properties:

Proof: If then it exits $\left({\overleftarrow{t}}_0,{\overrightarrow{t}}_0\right)\in T.$ Let satisfied $\left(\overleftarrow{x},\overrightarrow{x}\right)\in {\left(\left({\overleftarrow{A}}_{{\overleftarrow{t}}_0},{\overrightarrow{A}}_{{\overrightarrow{t}}_0}\right)\right)}_{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)},\text{Therefore,}\left({\overleftarrow{A}}_{{\overleftarrow{t}}_0},{\overrightarrow{A}}_{{\overrightarrow{t}}_0}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)>\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right),$ namely, $\underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{Sup}\left({\overleftarrow{A}}_{{\overleftarrow{t}}_0},{\overrightarrow{A}}_{{\overrightarrow{t}}_0}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)>\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right);\text{thus,}\left(\overleftarrow{x},\overrightarrow{x}\right)\in \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cup }{\left(\left({\overleftarrow{A}}_{\overleftarrow{t}},{\overrightarrow{A}}_{\overrightarrow{t}}\right)\right)}_{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)}.$ So we prove property

(1). The rest of the situation is similar to the certificate. [Prove up].

Theorem 8.4 Assume $\left(\overleftarrow{A},\overrightarrow{A}\right)\in DF\left(\overleftarrow{X},\overrightarrow{X}\right),\left\{\left({\overleftarrow{a}}_{\overleftarrow{t}},{\overrightarrow{a}}_{\overrightarrow{t}}\right);\left(\overleftarrow{t},\overrightarrow{t}\right)\in T\right\}\subset \left[\left(\overleftarrow{0},\overrightarrow{0}\right),\left(\overleftarrow{1},\overrightarrow{1}\right)\right].$ Then we have

where

$\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)=\underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\vee }\left({\overleftarrow{\alpha }}_{\overleftarrow{t}},{\overrightarrow{\alpha }}_{\overrightarrow{t}}\right),\left(\overleftarrow{\beta },\overrightarrow{\beta }\right)=\underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\vee }\left({\overleftarrow{\alpha }}_{\overleftarrow{t}},{\overrightarrow{\alpha }}_{\overrightarrow{t}}\right).$

Proof: According to

$\begin{array}{l}\left({\overleftarrow{A}}_{\overrightarrow{\alpha }},{\overrightarrow{A}}_{\overrightarrow{\alpha }}\right)=\left\{\left(\overleftarrow{x},\overrightarrow{x}\right);\left(\overrightarrow{A},\overleftarrow{A}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\ge \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\vee }\left({\overleftarrow{\alpha }}_{\overleftarrow{t}},{\overrightarrow{\alpha }}_{\overrightarrow{t}}\right)\right\}\\ ={\displaystyle \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cap }\left(\overleftarrow{x},\overrightarrow{x}\right);\left(\overleftarrow{A},\overrightarrow{A}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)}\ge \left({\overleftarrow{\alpha }}_{\overleftarrow{t}},{\overrightarrow{\alpha }}_{\overrightarrow{t}}\right)\\ ={\displaystyle \underset{\left(\overleftarrow{t},\overrightarrow{t}\right)\in T}{\cap }\left({\overleftarrow{A}}_{{\overleftarrow{\alpha }}_{\overleftarrow{t}}},{\overrightarrow{A}}_{{\overrightarrow{\alpha }}_{\overrightarrow{t}}}\right)}.\end{array}$

Then we know

[Prove up].

(2) It is Similar to permit, so we omitted.

Theorem 8.5 For any $\left(\overleftarrow{A},\overrightarrow{A}\right)\in D\text{F}\left(\overleftarrow{X},\overrightarrow{X}\right),$ we have

$\begin{array}{l}\left(1\right)\left({\overleftarrow{A}}_{\overrightarrow{\alpha }},{\overrightarrow{A}}_{\overrightarrow{\alpha }}\right)={\displaystyle \underset{\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)<\left(\overrightarrow{\alpha },\overleftarrow{\alpha }\right)}{\cap }\left({\overleftarrow{A}}_{\overleftarrow{\lambda }},{\overrightarrow{A}}_{\overrightarrow{\lambda }}\right)}\\ \left(2\right)\left({\overleftarrow{A}}_{\overrightarrow{\alpha }},{\overrightarrow{A}}_{\overrightarrow{\alpha }}\right)={\displaystyle \underset{\left(\overleftarrow{\lambda },\overrightarrow{\lambda }\right)>\left(\overrightarrow{\alpha },\overleftarrow{\alpha }\right)}{\cap }\left({\overleftarrow{A}}_{\overleftarrow{\lambda }},{\overrightarrow{A}}_{\overrightarrow{\lambda }}\right)}\end{array}$

Definition 8.5 Assume $\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\in \left[\left(\overleftarrow{0},\overrightarrow{0}\right),\left(\overleftarrow{1},\overrightarrow{1}\right)\right],\left(\overleftarrow{A},\overrightarrow{A}\right)\in D\text{F}\left(\overleftarrow{X},\overrightarrow{X}\right),$ then the number product of $\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\mathrm{with}\left(\overleftarrow{A},\overrightarrow{A}\right)\text{is}\left(\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\left(\overleftarrow{A},\overrightarrow{A}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)=\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\wedge \left(\overleftarrow{A},\overrightarrow{A}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\right).$

Theorem 8.6 (DF set decomposition theorem) For any we have

$\begin{array}{l}\left(\overrightarrow{A},\overrightarrow{A}\right)={\displaystyle \underset{\left(\overrightarrow{\alpha },\overleftarrow{\alpha }\right)\in \left(\overleftarrow{0},\overrightarrow{0}\right),\left(\dot{1},\overrightarrow{1}\right)}{\cup }\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right){\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)}}\\ \left(\overrightarrow{A},\overrightarrow{A}\right)={\displaystyle \underset{\left(\overrightarrow{\alpha },\overleftarrow{\alpha }\right)\in \left(\overleftarrow{0},\overrightarrow{0}\right),\left(\dot{1},\overrightarrow{1}\right)}{\cup }\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right){\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)}}.\end{array}$

Proof: Since

${\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overrightarrow{\alpha },\overleftarrow{\alpha }\right)}\left(\overleftarrow{x},\overrightarrow{x}\right)=\{\begin{array}{l}\left(\overleftarrow{1},\overrightarrow{1}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\in \left({\overleftarrow{A}}_{\overleftarrow{\alpha }},{\overrightarrow{A}}_{\overrightarrow{\alpha }}\right)\\ \left(\overleftarrow{0},\overrightarrow{0}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\in \left({\overleftarrow{A}}_{\overleftarrow{\alpha }},{\overrightarrow{A}}_{\overrightarrow{\alpha }}\right)\end{array},$

then we have

$\begin{array}{l}\left({\displaystyle \underset{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\in \left[\left(\overleftarrow{0},\overrightarrow{0}\right),\left(\overrightarrow{1},\overleftarrow{1}\right)\right]\right)}{\cup }\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right){\left(\overleftarrow{A},\overrightarrow{A}\right)}_{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)}}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\\ =\sup \left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\cdot \left(\left({\overleftarrow{A}}_{\overleftarrow{\alpha }},{\overrightarrow{A}}_{\overrightarrow{\alpha }}\right)\left(\overleftarrow{x},\overrightarrow{x}\right)\right)\\ =\underset{\left(\overrightarrow{x},\overrightarrow{x}\right)\in \left(\overleftarrow{A},\overrightarrow{A}\right)\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)}{\sup }\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\\ =\underset{\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\in \left({\overleftarrow{A}}_{\overrightarrow{x}},{\overrightarrow{A}}_{\overrightarrow{x}}\right)}{\sup }\left(\overleftarrow{\alpha },\overrightarrow{\alpha }\right)\\ =\left(\overleftarrow{A},\overrightarrow{A}\right)\left(\overrightarrow{x},\overleftarrow{x}\right)).\end{array}$