﻿ 广义优势多粒度直觉模糊粗糙集及规则获取
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 智能系统学报  2017, Vol. 12 Issue (6): 883-888  DOI: 10.11992/tis.201706034 0

### 引用本文

LIANG Meishe, MI Jusheng, ZHAO Tianna. Generalized dominance-based multi-granularity intuitionistic fuzzy rough set and acquisition of decision rules[J]. CAAI Transactions on Intelligent Systems, 2017, 12(6), 883-888. DOI: 10.11992/tis.201706034.

### 文章历史

1. 河北师范大学 数学与信息科学学院，河北 石家庄 050024;
2. 石家庄职业技术学院 科技发展与校企合作部，河北 石家庄 050081

Generalized dominance-based multi-granularity intuitionistic fuzzy rough set and acquisition of decision rules
LIANG Meishe1,2, MI Jusheng1, ZHAO Tianna1
1. College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang 050024, China;
2. Office of Science & Technology Administration, Shijiazhuang University of Applied Technology, Shijiazhuang 050081, China
Abstract: The dominance relation rough set model is the main method of data mining when researching order information systems. In this paper, we attempt to enrich the present model and make it more effective for practical problems using the following methods: Firstly, defining three types of dominance relations using triangular norms and co-norms in an intuitionistic fuzzy decision information system; here, three types of dominance class were obtained; secondly, establishing a generalized dominance-based multi-granularity intuitionistic fuzzy rough set model and discussing its properties; thirdly, establishing the decision rules for obtaining the logic connective " OR” in the intuitionistic fuzzy decision information system; and finally, using an example to illustrate the effectiveness of the model.
Key words: multi-granularity    rough set    triangular norm    dominance relation    intuitionistic fuzzy information system    decision rules

1 预备知识 1.1 直觉模糊信息系统

1) ${{A}} \subseteq {{B}} \Leftrightarrow \forall x \in U$ ${\mu _{{A}}}\left( x \right) \leqslant {\mu _{{B}}}\left( x \right)$ ${\nu _{{A}}}\left( x \right) \geqslant {\nu _{{B}}}\left( x \right)$

2) ${{A}} \!=\! {{B}} \Leftrightarrow \forall x \in U$ ${\mu _{{A}}}\left( x \right) = {\mu _{{B}}}\left( x \right)$ ，且 ${\nu _{{A}}}\left( x \right) \! \geqslant \! {\nu _{{B}}}\left( x \right)$

3) $\sim {{A}} = \left\{ {\left\langle {{\nu _{{A}}}\left( x \right),{\mu _{{A}}}\left( x \right)} \right\rangle \left| {x \in U} \right.} \right\}$

4) ${{A}} \cap {{B}} \Leftrightarrow \left\{ {\left\langle {{\mu _{{A}}}\left( x \right) \wedge {\mu _{{B}}}\left( x \right),{\nu _{{A}}}\left( x \right) \vee {\nu _{{B}}}\left( x \right)} \right\rangle \left| {x \in U} \right.} \right\}$

5) ${{A}} \cup { B} \Leftrightarrow \left\{ {\left\langle {{\mu _{{A}}}\left( x \right) \vee {\mu _{{{B}}}}\left( x \right),{\nu _{{A}}}\left( x \right) \wedge {\nu _{{{B}}}}\left( x \right)} \right\rangle \left| {x \in U} \right.} \right\}$

$\alpha = \left( {{u_\alpha },{v_\alpha }} \right)$ ，其中 ${\mu _\alpha }$ ${\nu _\beta } \in \left[ {0,1} \right]$ ，且 $0 \leqslant {\mu _\alpha } + {\nu _\alpha } \leqslant 1$ ，则称α为一个直觉模糊数。全体直觉模糊数集合记为IFN。它的得分函数 $s\left( \alpha \right) = {u_\alpha } - {v_\alpha }$ ，精确函数为 $h\left( \alpha \right) = {u_\alpha } + {v_\alpha }$ ，利用得分函数和精确函数就可以给出比较两个直觉模糊数大小的方法。

1.2 三角模算子

1) $N\left( 0 \right) = 1$ $N\left( 1 \right) = 0$ (边界性)；

2) $a \leqslant b$ ，则 $N\left( a \right) \geqslant N(b)$ (单调性)；

$\forall a \in \left[ {0,1} \right]$ 均有 $N\left( a \right) = 1 - a$ 成立，称N为标准模糊补算子，记为Ns

1) $T\left( {a,1} \right) = a$ (边界性)；

2) 若 $b \leqslant c$ ，则 $T\left( {a,b} \right) \leqslant T\left( {a,c} \right)$ (单调性)；

3) $T\left( {a,b} \right){\rm{ = }}T\left( {b,a} \right)$ (交换性)；

4) $T\left( {a,T\left( {b,c} \right)} \right){\rm{ = }}T\left( {T\left( {a,b} \right),c} \right)$ (结合性)；

1) $S\left( {a,0} \right) = a$ (边界性)；

2) 若 $b \leqslant c$ ，则 $S\left( {a,b} \right) \leqslant S\left( {a,c} \right)$ (单调性)；

3) $S\left( {a,b} \right){\rm{ = }}S\left( {b,a} \right)$ (交换性)；

4) $S\left( {a,S\left( {b,c} \right)} \right){\rm{ = }}S\left( {S\left( {a,b} \right),c} \right)$ (结合性)；

S为三角模余模(t-余模)。

TST关于模糊补算子N满足是对偶的当且仅当 $\forall a,b \in \left[ {0,1} \right]$ $N\left( {T\left( {a,b} \right)} \right) = {S_T}\left( {N\left( a \right),N\left( b \right)} \right)$ $N\left( {{S_T}\left( {a,b} \right)} \right) = T\left( {N\left( a \right),N\left( b \right)} \right)$ ，称 $\left( {T,{S_T},N} \right)$ 为对偶三元组。常见的对偶三元组有：

$\min \text{-} \max :\left( {\min \left( {a,b} \right),\max \left( {a,b} \right),{N_S}} \right)$

${\rm{product \text{-} sum}}:\left( {ab,a + b - 1,{N_S}} \right)$

${\rm{Lukasiewicz}}:\left( {\max (0,a + b - 1),\min (1,a + b),{N_S}} \right)$

2 多粒度直觉模糊粗糙集 2.1 直觉模糊信息系统中的优势关系

 $\begin{array}{c}AV_a^1\left( x \right) = \displaystyle\frac{1}{2}\left( {\min \left( {{u_a}\left( x \right),N\left( {{v_a}\left( x \right)} \right)} \right) + } \right.\\\left. {\max \left( {{u_a}\left( x \right),N\left( {{v_a}\left( x \right)} \right)} \right)} \right) = \\{\left( {{u_a}\left( x \right) + N\left( {{v_a}\left( x \right)} \right)} \right)/2}\end{array}$
 $\begin{array}{c}AV_a^2\left( x \right) = \displaystyle\frac{1}{2}\left( {{u_a}\left( x \right) \cdot N\left( {{v_a}\left( x \right)} \right) + {u_a}\left( x \right) + } \right.\\\left. {N\left( {{v_a}\left( x \right)} \right) - {u_a}\left( x \right) \cdot N\left( {{v_a}\left( x \right)} \right)} \right) = \\\left( {{u_a}\left( x \right) + N\left( {{v_a}\left( x \right)} \right)} \right)/2\end{array}$
 $\begin{array}{c}AV_a^3\left( x \right) = \displaystyle\frac{1}{2}\left( {\max \left( {0,{u_a}\left( x \right) + N\left( {{v_a}\left( x \right)} \right) - 1} \right) + } \right.\\\left. {\min \left( {1,{u_a}\left( x \right) + N\left( {{v_a}\left( x \right)} \right)} \right)} \right) = \\\displaystyle\frac{1}{2}\left( {\max \left( {0,{u_a}\left( x \right) - {v_a}\left( x \right)} \right) + } \right.\\\left. {\min \left( {1,{u_a}\left( x \right) + 1 - {v_a}\left( x \right)} \right)} \right)\end{array}$

1) 若 ${u_a}\left( x \right) - {v_a}\left( x \right) \geqslant 0$ $\min \left( {1,{u_a}\left( x \right) + 1 - {v_a}\left( x \right)} \right) = 1$ $\max \left( {0,{u_a}\left( x \right) - {v_a}\left( x \right)} \right) = {u_a}\left( x \right) - {v_a}\left( x \right)$ ，则有 $AV_a^3 = {u_a}\left( x \right) +$ $N\left( {{v_a}\left( x \right)} \right)$

2) 若 ${u_a}\left( x \right) - {v_a}\left( x \right) < 0$ ，有 $\min \left( {1,{u_a}\left( x \right) + 1 - {v_a}\left( x \right)} \right) =$ ${u_a}\left( x \right) \!+\! 1 \!-\! {v_a}\left( x \right)$ $\max \left( {0,{u_a}\left( x \right) \!-\! {v_a}\left( x \right)} \right) \!=\! 0$ ，则 $AV_a^3 \!=\! {u_a}\left( x \right) \!+$ $N\left( {{v_a}\left( x \right)} \right)$ ，证毕。

1) ${x_4} \leqslant {x_2} \leqslant {x_3} \leqslant {x_5} \leqslant {x_6} \leqslant {x_1}$

2) ${x_2} = {x_6} \leqslant {x_4} = {x_3} \leqslant {x_5} \leqslant {x_1}$

3) ${x_4} = {x_5} \leqslant {x_3} \leqslant {x_2} \leqslant {x_6} = {x_1}$

4) ${x_4} \leqslant {x_2} = {x_3} = {x_5} \leqslant {x_6} \leqslant {x_1}$

1) $R_{ \cdot ,B}^{\leqslant}$ 满足自反性和传递性；

2) ${x_j} \in \left[ {{x_k}} \right]_{ \cdot ,B}^{\leqslant} \Leftrightarrow \left[ {{x_j}} \right]_{ \cdot ,B}^{\leqslant} \subseteq \left[ {{x_k}} \right]_{ \cdot ,B}^{\leqslant}$

3) $\left[ {{x_i}} \right]_{ \cdot ,B}^{\leqslant} = \cup \left\{ {\left[ {{x_j}} \right]_{ \cdot ,B}^{\leqslant} :{x_j} \in \left[ {{x_i}} \right]_{ \cdot ,B}^{\leqslant} } \right\}$

4) $U = \bigcup\nolimits_{i = 1}^m {\left[ {{x_i}} \right]_{ \cdot ,B}^{\leqslant} }$

5) $\left[ {{x_i}} \right]_{ \cdot ,B}^{\leqslant} = \left[ {{x_j}} \right]_{ \cdot ,B}^{\leqslant} \Leftrightarrow {T_{{a_i}}}\left( {{x_i}} \right) = {T_{{a_i}}}\left( {{x_j}} \right)$ $\forall {a_i} \in B$

2.2 多粒度优势直觉模糊粗糙集

 $\begin{array}{c}\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}O}} } \left( X \right) = \left\{ {x \in U:\left[ x \right]_{T,{A_1}}^{\leqslant} \subseteq X \vee \left[ x \right]_{T,{A_2}}^{\leqslant} } \subseteq \right.\\\left. { X \vee \cdots \vee \left[ x \right]_{T,{A_n}}^{\leqslant} \subseteq X} \right\}\\[5pt]\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( X \right) = \sim \underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}O}} } \left( { \sim X} \right)\end{array}$

 $\begin{array}{c}\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( X \right) = \left\{ {x \in U:\left[ x \right]_{T,{A_1}}^{\leqslant} \subseteq X \wedge \left[ x \right]_{T,{A_2}}^{\leqslant} } \subseteq \right.\\\left. { X \wedge \cdots \wedge \left[ x \right]_{T,{A_n}}^{\leqslant} \subseteq X} \right\},\\\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( X \right) = \sim \underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( { \sim X} \right),\end{array}$

3 多粒度优势粗糙直觉模糊集及决策规则获取

3.1 多粒度优势粗糙直觉模糊集

 $\begin{array}{l}\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}O}} } \left( d \right)\left( x \right) = \vee _{i = 1}^n\left\{ { \wedge \left\{ {d\left( y \right):y \in \left[ x \right]_{T,{A_i}}^{\leqslant}} \right\}} \right\}\\[5pt]\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( d \right)\left( x \right) = \wedge _{i = 1}^n\left\{ { \vee \left\{ {d\left( y \right):y \in \left[ x \right]_{T,{A_i}}^{\geqslant} } \right\}} \right\}\end{array}$

 $\begin{array}{c}\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( {{f_d}} \right)\left( x \right) = \wedge _{i = 1}^n\left\{ { \wedge \left\{ {{f_d}\left( y \right):y \in \left[ x \right]_{T,{A_i}}^{\leqslant}} \right\}} \right\}\\[5pt]\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( {{f_d}} \right)\left( x \right) = \vee _{i = 1}^n\left\{ { \vee \left\{ {{f_d}\left( y \right):y \in \left[ x \right]_{T,{A_i}}^ {\geqslant} } \right\}} \right\}\end{array}$

 $\begin{array}{c}\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} O}} } \left( {{f_d}} \right) = \left\{ {\left\langle {0.6,0.3} \right\rangle ,\left\langle {0.6,0.3} \right\rangle ,\left\langle {0.0,1.0} \right\rangle ,} \right.\\[4.5pt]\left\langle {0.5,0.4} \right\rangle ,\left\langle {0.4,0.6} \right\rangle \left. {,\left\langle {0.3,0.6} \right\rangle ,\left\langle {0.0,0.9} \right\rangle ,\left\langle {0.6,0.3} \right\rangle } \right\}\\[4.5pt]\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}O}} } \left( {{f_d}} \right) = \left\{ {\left\langle {0.7,0.2} \right\rangle ,\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.4,0.6} \right\rangle ,} \right.\\[4.5pt]\left\langle {0.5,0.4} \right\rangle ,\left. {\left\langle {0.4,0.6} \right\rangle ,\left\langle {0.4,0.6} \right\rangle ,\left\langle {0.4,0.6} \right\rangle ,\left\langle {0.6,0.3} \right\rangle } \right\}\\[4.5pt]\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( {{f_d}} \right) = \left\{ {\left\langle {0.0,1.0} \right\rangle ,\left\langle {0.0,1.0} \right\rangle ,\left\langle {0.0,1.0} \right\rangle ,} \right.\\[4.5pt]\left. {\left\langle {0.0,1.0} \right\rangle ,\left\langle {0.0,1.0} \right\rangle ,\left\langle {0.0,1.0} \right\rangle ,\left\langle {0.0,1.0} \right\rangle ,\left\langle {0.0,1.0} \right\rangle } \right\}\\[4.5pt]\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}P}} } \left( {{f_d}} \right) = \left\{ {\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.8,0.1} \right\rangle } \right.,\\[4.5pt]\left. {\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.8,0.1} \right\rangle ,\left\langle {0.8,0.1} \right\rangle ,} \right\}\end{array}$

1) $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right) \subseteq {f_d} \subseteq \overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)$ $\underline {{\mkern 1mu} \sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}P}} } \left( {{f_d}} \right) \subseteq$ ${f_d} \subseteq \overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}P}} } \left( {{f_d}} \right)$

2) $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right) = \cup _{i = 1}^n\underline {R_{T,{A_i}}^{{\leqslant} O}} \left( {{f_d}} \right)$ $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} O}} } \left( {{f_d}} \right) =\cap _{i = 1}^n$ $\overline {R_{T,{A_i}}^{ {\leqslant} O}} \left( {{f_d}} \right)$

3) $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( {{f_d}} \right) = \cap _{i = 1}^n\underline {R_{T,{A_i}}^{ {\leqslant} P}} \left( {{f_d}} \right)$ $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( {{f_d}} \right) =\cup _{i = 1}^n$ $\overline {R_{T,{A_i}}^{ {\leqslant} P}} \left( {{f_d}} \right)$

4) $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}O}} } \left( {{f_d}} \right)} \right) = \underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant}O}} } \left( {{f_d}} \right)$ $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} }$ $\left( {\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}O}} } \left( {{f_d}} \right)} \right) = \overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)$

5) $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( {\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( {{f_d}} \right)} \right) = \underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( {{f_d}} \right)$ $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} }$ $\left( {\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( {{f_d}} \right)} \right) = \overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} P}} } \left( {{f_d}} \right)$

1) $\forall x \in U$ ，根据定义15有 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}O}} } \left( d \right)\left( x \right) =$ $\vee _{i = 1}^n\left\{ { \wedge \left\{ {d\left( y \right):y \in \left[ x \right]_{T,{A_i}}^{\leqslant} } \right\}} \right\}$ 。由近似关系的自反性知 $\wedge \left\{ {d\left( y \right):y \in \left[ x \right]_{T,{A_i}}^{\leqslant} } \right\}{\leqslant} {f_d}\left( x \right)$ ，从而 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( d \right)\left( x \right){\leqslant} \vee _{i = 1}^n$ $\left( {{f_d}\left( x \right)} \right) = {f_d}\left( x \right)$ ，即 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right) \subseteq {f_d}$ 成立。同理 ${f_d} \subseteq$ $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} O}} } \left( {{f_d}} \right)$

2) $\forall x \in U$ ，由定义15有 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)\left( x \right) = \vee _{i = 1}^n$ $\left\{ { \wedge \left\{ {d\left( y \right):y \in \left[ x \right]_{T,{A_i}}^{\leqslant}} \right\}} \right\} = \vee _{i = 1}^n\underline {R_{T,{A_i}}^{{\leqslant} O}} \left( {{f_d}} \right)\left( x \right)$ ，即 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right) =$ $\cup _{i = 1}^n\underline {R_{T,{A_i}}^{{\leqslant} O}} \left( {{f_d}} \right)$

3) 同理 $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{ {\leqslant} O}} } \left( {{f_d}} \right) = \cap _{i = 1}^n\overline {R_{T,{A_i}}^{ {\leqslant} O}} \left( {{f_d}} \right)$

4) 由1)可知 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}O}} } \left( {\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)} \right) \subseteq$ $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}O}} }$ (fd)，因此只要证明 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right) \! \subseteq \! \underline {\sum\nolimits_{i = 1}^n \! {R_{T,{A_i}}^{{\leqslant} O}} } \! \left( {\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)} \right)$ 即可。由2)可知 $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant}O}} } \left( {\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)} \right) = \cup _{i = 1}^n\underline {R_{T,{A_i}}^{{\leqslant} O}}$ $\left( {\underline {\sum\nolimits_{i = 1}^n \, {R_{T,{A_i}}^{{\leqslant} O}} } \, \left( {{f_d}} \right)} \right) = \cup _{i = 1}^n \, \underline {R_{T,{A_i}}^{{\leqslant} O}} \, \left( { \cup _{i = 1}^n \, \underline {R_{T,{A_i}}^{{\leqslant} O}} \, \left( {{f_d}} \right)} \right) \supseteq \cup _{i = 1}^n \cup _{i = 1}^n\underline {R_{T,{A_i}}^{{\leqslant} O}}$ $\left( {\underline {R_{T,{A_i}}^{{\leqslant}O}} \left( {{f_d}} \right)} \right) = \cup _{i = 1}^n\underline {R_{T,{A_i}}^{ {\leqslant} O}} \left( {{f_d}} \right) = \underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)$ 。类似的，易证 $\left( {\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)} \right) = \overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)$

1) $\underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( {{f_d}} \right) \subseteq \underline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right)$

2) $\overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} O}} } \left( {{f_d}} \right) \subseteq \overline {\sum\nolimits_{i = 1}^n {R_{T,{A_i}}^{{\leqslant} P}} } \left( {{f_d}} \right)$

3.2 决策规则获取

“at least”决策规则：

 $\begin{array}{c}{T_{{a_1}}}\left( y \right) \geqslant {T_{{a_1}}}\left( x \right) \vee {T_{{a_2}}}\left( x \right) \vee \cdots \vee {T_{{a_n}}}\left( y \right) \geqslant {T_{{a_n}}}\left( x \right) \to \\[5pt]{f_d}\left( y \right) \geqslant \underline {\sum\nolimits_{i = 1}^n {R_{T,{a_i}}^{ \leqslant O}} } \left( {{f_d}} \right)\left( x \right);\end{array}$

“at most”决策规则：

 $\begin{array}{c}{T_{{a_1}}}\left( y \right) \leqslant {T_{{a_1}}}\left( x \right) \vee {T_{{a_2}}}\left( x \right) \vee \cdots \vee {T_{{a_n}}}\left( y \right) \leqslant {T_{{a_n}}}\left( x \right) \to \\[5pt]{f_d}\left( y \right) \leqslant \underline {\sum\nolimits_{i = 1}^n {R_{T,{a_i}}^{ \leqslant P}} } \left( {{f_d}} \right)\left( x \right).\end{array}$

1) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.86 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.56 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.04 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.56 \to {f_d}\left( y \right) \geqslant \left( {0.6,0.3} \right)$

2) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.86 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.08 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.10 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.02 \to {f_d}\left( y \right) \geqslant \left( {0.6,0.3} \right)$

3) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.03 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.02 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.02 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.86 \to {f_d}\left( y \right) \geqslant \left( {0.0,1.0} \right)$

4) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.00 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.81 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.81 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.72 \to {f_d}\left( y \right) \geqslant \left( {0.5,0.4} \right)$

5) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.02 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.01 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 1.0 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.76 \to {f_d}\left( y \right) \geqslant \left( {0.4,0.6} \right)$

6) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.12 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.06 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.90 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.02 \to {f_d}\left( y \right) \geqslant \left( {0.3,0.6} \right)$

7) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.00 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.02 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.81 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 0.08 \to {f_d}\left( y \right) \geqslant \left( {0.0,0.9} \right)$

8) ${T_{{a_1}}}\left( y \right) \,\geqslant\, 0.86 \vee {T_{{a_2}}}\left( y \right) \,\geqslant\, 0.81 \vee {T_{{a_3}}}\left( y \right) \,\geqslant\, 0.02 \,\, \vee$ ${T_{{a_4}}}\left( y \right) \geqslant 1.0 \to {f_d}\left( y \right) \geqslant \left( {0.6,0.3} \right)$

4 结束语

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