西南大学学报 (自然科学版)  2020, Vol. 42 Issue (2): 41-47.  DOI: 10.13718/j.cnki.xdzk.2020.02.007
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  • 基于模糊否定与生成子的模糊蕴涵及性质    [PDF全文]
    彭祖明    
    长江师范学院 数学与统计学院, 重庆 涪陵 408100
    摘要:首先,提出了一种利用模糊否定和连续单调增函数构建模糊蕴涵的新方法.利用该方法,可以只利用模糊否定构造模糊蕴涵.同时,对所得模糊蕴涵的特性,如左单位元性质、交换原则、序性质等进行了探讨,并获得了一些成果.
    关键词模糊蕴涵    模糊否定    聚合函数    特性    

    模糊蕴涵作为经典布尔蕴涵的推广,是模糊逻辑系统、模糊控制、模糊决策的基础.模糊蕴涵的构建方法主要有两类:其一是由模糊逻辑连接词(如t模、t余模、copulas函数)及模糊否定构造而成,亦称为(AN)-蕴涵[1-2],如(SN)-蕴涵[3-4]R-蕴涵[4-5]QL-蕴涵[4-6]、(UN)-蕴涵[7]、(RU)-蕴涵[8]、概率蕴涵[9]等;其二是由单调函数或已知模糊蕴涵的组合来构造,如f-蕴涵、g-蕴涵[10]h-蕴涵[11-12]、(g,min)-蕴涵[13]、推广h-蕴涵[14]R-序和蕴涵[15]e-生成蕴涵[16]等.受文献[17]中构建semicopulas的方法以及文献[1]中构建模糊蕴涵的方法的启示,本文提出了通过模糊否定和单调连续函数来构造一类新的(AN)-蕴涵的方法.在该方法中,如果所给的连续单调函数为恒等函数,则只需模糊否定就可构造模糊蕴涵.同时,本文对所得的模糊蕴涵的基本特性,如左单位元性质、序性质、交换原则、恒等原则及逆否对称性进行了探讨.

    定义1 [4]   对任意的xx1x2yy1y2∈[0, 1],如果函数I:[0, 1]2→[0, 1]满足下列条件:

    (I1) x1x2,则I(x1y)≥I(x2y);

    (I2) y1y2,则I(xy1)≤I(xy2);

    (I3) I(0,0)=1,I(1,1)=1,I(1,0)=1.

    则称I为模糊蕴涵,所有模糊蕴涵构成的集合记成FI.

    定义2 [4]   模糊蕴涵I满足:

    (ⅰ)I(1,y)=yy∈[0, 1],则称I具备左单位元性质;

    (ⅱ) I(xI(yz))= I(yI(xz)),xyz∈[0, 1],则称I满足交换原则;

    (ⅲ) I(xx)=1,x∈[0, 1],则称I满足恒等原则;

    (ⅳ) I(xy)=1$ \Leftrightarrow $xyxy∈[0, 1],则称I满足序性质;

    (ⅴ) I(xy)=I(N(y),N(x)),xy∈[0, 1],则称I关于N满足逆否对称性,其中N为模糊否定;

    (ⅵ) I(xN(y))=I(yN(x)),xy∈[0, 1],其中N为模糊否定,则称I关于N满足右逆否对称性;

    (ⅶ)如果存在tT,使得I(T(xy),z)=I(xI(yz)),xyz∈[0, 1],则称I满足输入律.

    定义3 [4]  如果函数N:[0, 1]→[0, 1]满足N(0)=1,N(1)=0,且N单调递减,则称N为模糊否定.

    定义4 [4]  已知模糊否定N,如果对任意的x∈[0, 1],有N(N(x))=x,则称N为强的.如果N(x)在区间[0, 1]连续且严格单调递减,则称N为严格的.称N(x)=1-x为标准的模糊否定,记作NC(x).

    定理1 [4]  模糊否定N为强的当且当存在严格增的连续函数f:[0, 1]→[0,∞),使得f(0)=0,且N(x)=f-1(f(1)-f(x)),x∈[0, 1].称函数f为模糊否定N的生成子.

    引理1 [4]   已知函数I:[0, 1]2→[0, 1]及模糊否定N,如果I具备左单位元性质及关于N的逆否对称性,则N=NI为强的,其中NII的自然否定,定义为NI(x)=I(x,0).

    定义5 [4]   如果函数φ:[0, 1]→[0, 1]连续,严格增并满足边界条件φ(0)=0,φ(1)=1,则称φ为序自同构.记Φ为所有从[0, 1]到[0, 1]的序自同构构成的集合.

    定义6 [4]   已知φΦ,如果函数fg:[0, 1]n→[0, 1]满足g=fφ,即g(x1x2,…,xn)=φ-1(f(φ(x1),φ(x2),…,φ(x3))),则称函数fgφ结合的.

    NI的定义可知,给定模糊蕴涵I,可得NI.反过来,给定N,如何得到一个模糊蕴涵I?在现有的文献中还未有研究.文献[1]提出了利用模糊否定N构建simicopulas的方法,受此启发,本文构造了一个双变量函数IN:[0, 1]2→[0, 1],定义为

    $ {I_N}\left( {x,y} \right) = \min \left\{ {1,N\left( x \right) \wedge y - N\left( {N\left( x \right) \vee y} \right) + 1} \right\}\;\;\;\;x,y \in \left[ {0,1} \right] $

    易证IN为模糊蕴涵.如果考虑严格增的连续函数g:[0, 1]→[0,∞),则可得双变量函数Ig,N:[0, 1]2→[0, 1],定义如下:

    定义7   已知连续函数g:[0, 1]→[0,∞)严格增,且g(0)=0,N为模糊否定.定义双变量函数Ig,N:[0, 1]2→[0, 1]为

    $ {I_{g,N}}\left( {x,y} \right) = {g^{ - 1}}\min \left\{ {g\left( 1 \right),g\left( {N\left( x \right) \wedge y} \right) - g\left( {N\left( {N\left( x \right) \vee y} \right)} \right) + g\left( 1 \right)} \right\}\;\;\;\;\;x,y \in \left[ {0,1} \right] $

    称函数g为生成子.

    定理2  已知生成子g及模糊否定N,则双变量函数

    $ {I_{g,N}}\left( {x,y} \right) = {g^{ - 1}}\min \left\{ {g\left( 1 \right),g\left( {N\left( x \right) \wedge y} \right) - g\left( {N\left( {N\left( x \right) \vee y} \right)} \right) + g\left( 1 \right)} \right\}\;\;\;\;x,y \in \left[ {0,1} \right] $

    为模糊蕴涵.

      直接利用定义1可以证明.

    注1    (ⅰ)如果函数g=x,则模糊蕴涵Ig,N可只由模糊否定N构成,即

    $ {I_{g,N}}\left( {x,y} \right) = \min \left\{ {1,N\left( x \right) \wedge y - N\left( {N\left( x \right) \vee y} \right) + 1} \right\}\;\;\;\;x,y \in \left[ {0,1} \right] $

    显然,定义7中所得的模糊蕴涵Ig,N属于(AN)-模糊蕴涵.事实上,设

    $ A\left( {x,y} \right) = \min \left\{ {g\left( 1 \right),g\left( {x \wedge y} \right) - g\left( {N\left( {x \vee y} \right)} \right) + g\left( 1 \right)} \right\}\;\;\;\;x,y \in \left[ {0,1} \right] $

    显然,A(xy)为聚合函数,且Ig,N(xy)=A(N(x),y).

    (ⅱ)如果

    $ N\left( x \right) = {N_{{D_2}}}\left( x \right) = \left\{ {\begin{array}{*{20}{c}} 1&{x < 1}\\ 0&{x = 1} \end{array}} \right. $

    $ {I_{g,N}}\left( {x,y} \right) = \left\{ \begin{array}{l} 0\;\;\;\;\;x = 1,y < 1\\ 1\;\;\;\;\;\;否则 \end{array} \right. $

    其中xy∈[0, 1].易见Ig,N满足交换原则和恒等原则,但是不满足左单位元性质及序性质.

    (ⅲ)如果

    $ N\left( x \right) = {N_{{D_1}}}\left( x \right) = \left\{ {\begin{array}{*{20}{c}} 1&{x = 0}\\ 0&{x > 1} \end{array}} \right. $

    $ {I_{g,N}}\left( {x,y} \right) = \left\{ \begin{array}{l} 0\;\;\;\;\;x > 0,y = 0\\ 1\;\;\;\;\;\;否则 \end{array} \right. $

    其中xy∈[0, 1].易见Ig,N交换原则和恒等原则,但是不满足左单位元性质及序性质.

    下面主要研究模糊蕴涵Ig,N的基本性质,这些性质主要在模糊否定N为连续(强、严格)的情形下所得.

    性质1  模糊蕴涵Ig,N满足左单位元性质当且当g为模糊否定N的生成子,即

    $ N\left( x \right) = {g^{ - 1}}\left( {g\left( 1 \right) - g\left( x \right)} \right)\;\;\;\;x \in \left[ {0,1} \right] $

      设x∈[0, 1],则Ig,N满足左单位元性质$ \Leftrightarrow $Ig,N(1,x)=x$ \Leftrightarrow $g-1(g(1)-g(N(x)))=x$ \Leftrightarrow $N(x)=g-1(g(1)-g(x)).

    注2   由性质1可知,如果g为模糊否定N的加法生成子,则模糊蕴涵Ig,N的自然否定为N,而且Ig,N可表示为Ig,N(xy)= min{1,g-1(g(1)+g(y)-g(x))},xy∈[0, 1].

    性质2  模糊蕴涵Ig,N满足恒等原则当且当

    $ \left\{ {x \in \left[ {0,1} \right]:x < N\left( x \right)} \right\} \cap \left\{ {x \in \left[ {0,1} \right]:x < N\left( {N\left( x \right)} \right)} \right\} = \emptyset $

      必要性  设Ig,N满足序性质,即对任意的x∈[0, 1],有Ig,N(xx)=1.如果

    $ \left\{ {x \in \left[ {0,1} \right]:x < N\left( x \right)} \right\} \cap \left\{ {x \in \left[ {0,1} \right]:x < N\left( {N\left( x \right)} \right)} \right\} = \emptyset $

    则存在x0∈(0,1),使得x0N(x0)和x0N(N(x0)),从而

    $ \begin{array}{l} {I_{g.N}}\left( {{x_0},{x_0}} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right) \wedge {x_0}} \right) - g\left( {N\left( {N\left( {{x_0}} \right) \vee {x_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {{x_0}} \right) - g\left( {N\left( {N\left( {{x_0}} \right)} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {g\left( {{x_0}} \right) - g\left( {N\left( {N\left( {{x_0}} \right)} \right)} \right) + g\left( 1 \right)} \right) < \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {g\left( 1 \right)} \right) = 1 \end{array} $

    这与Ig,N(xx)=1矛盾.

    充分性  因为{x∈[0, 1]:xN(x)}∩{x∈[0, 1]:xN(N(x))}=$\emptyset $,则xN(x)可推出xN(N(x)).如果x满足xN(x),可得

    $ \begin{array}{l} {I_{g.N}}\left( {x,x} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( x \right) \wedge x} \right) - g\left( {N\left( {N\left( x \right) \vee x} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( x \right) - g\left( {N\left( {N\left( x \right)} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {g\left( 1 \right)} \right) = 1 \end{array} $

    如果x满足xN(x),可得

    $ \begin{array}{l} {I_{g.N}}\left( {x,x} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( x \right) \wedge x} \right) - g\left( {N\left( {N\left( x \right) \vee x} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( x \right)} \right) - g\left( {N\left( x \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {g\left( 1 \right)} \right) = 1 \end{array} $

    从而Ig,N满足恒等原则.

    性质3  模糊蕴涵Ig,N满足序性质当且当N为强的.

      必要性  由Ig,N满足序性质,则Ig,N满足恒等原则.由性质2,xN(x)可推出xN(N(x)),x∈[0, 1].设N不强,则存在x0∈(0,1),使得x0N(N(x0)).

    如果x0N(x0),则x0N(N(x0)),且存在y0∈[0, 1],使得N(N(x0))<y0x0N(x0),从而

    $ \begin{array}{l} {I_{g,N}}\left( {{x_0},{y_0}} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right) \wedge {y_0}} \right) - g\left( {N\left( {N\left( {{x_0}} \right) \vee {y_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {{y_0}} \right) - g\left( {N\left( {N\left( {{x_0}} \right)} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {g\left( 1 \right)} \right) = 1 \end{array} $

    这与Ig,N满足序性质矛盾.

    如果x0=N(x0),则N(x0)=N(N(x0))=x0,矛盾.

    如果x0N(x0),则N(N(x0))≥N(x0),考虑下面两种情形:

    情形1   N(N(x0))=N(x0),则

    $ \begin{array}{l} {I_{g,N}}\left( {{x_0},N\left( {{x_0}} \right)} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right) \wedge N\left( {{x_0}} \right)} \right) - g\left( {N\left( {N\left( {{x_0}} \right) \vee N\left( {{x_0}} \right)} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right)} \right) - g\left( {N\left( {{x_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {g\left( 1 \right)} \right) = 1 \end{array} $

    矛盾.

    情形2   N(N(x0))>N(x0),由恒等原则,N(x0)≥N(N(N(x0))),从上面的证明知N(x0)>N(N(N(x0)))不成立,则N(x0)=N(N(N(x0))). x0N(N(x0))比较,只有两种可能,即x0N(N(x0))或x0N(N(x0)).

    如果x0N(N(x0)),则存在y0∈[0, 1],使得N(N(x0))<y0x0,所以y0N(x0).因为N(x0)=N(N(N(x0))),则N(y0)=N(x0),但是

    $ \begin{array}{l} {I_{g,N}}\left( {{x_0},{y_0}} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right) \wedge {y_0}} \right) - g\left( {N\left( {N\left( {{x_0}} \right) \vee {y_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right)} \right) - g\left( {N\left( {{x_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( 1 \right)} \right\}} \right) = 1 \end{array} $

    矛盾.

    如果x0N(N(x0)),则存在y0∈[0, 1],使得x0y0N(N(x0)),所以y0N(x0).因为N(x0)=N(N(N(x0))),则N(y0)=N(x0),但是

    $ \begin{array}{l} {I_{g,N}}\left( {{y_0},{x_0}} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{y_0}} \right) \wedge {x_0}} \right) - g\left( {N\left( {N\left( {{y_0}} \right) \vee {x_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( {{x_0}} \right)} \right) - g\left( {N\left( {{x_0}} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( 1 \right)} \right\}} \right) = 1 \end{array} $

    矛盾.

    综上所述,对任意的x∈[0, 1],有x=N(N(x)),即N为强的.

    充分性  设xy,因为N为强的,则

    $ \begin{array}{l} {I_{g,N}}\left( {x,y} \right) = {g^{ - 1}}\left( {\min \left\{ {g\left( 1 \right),g\left( {N\left( x \right) \wedge y} \right) - g\left( {N\left( {N\left( x \right) \vee y} \right)} \right) + g\left( 1 \right)} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ \begin{array}{l} {g^{ - 1}}(\min \{ g(1),g(N(x)) - g(N(y)) + g(1)\} )\;\;\;\;\;y \ge N\left( x \right)\\ {g^{ - 1}}(\min \{ g(1),g(y) - g(N(N(x))) + g(1)\} )\;\;\;\;\;y < N\left( x \right) \end{array} \right. = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ \begin{array}{l} 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y \ge N\left( x \right)\\ {g^{ - 1}}(\min \{ g(1),g(y) - g(x) + g(1)\} )\;\;\;\;\;\;\;\;y < N\left( x \right) \end{array} \right. = 1 \end{array} $

    反之,当Ig,N(xy)=1时,如果yN(x),则

    $ {g^{ - 1}}(\min \{ g(1),g(N(x)) - g(N(y)) + g(1)\} ) = 1 $

    所以

    $ \min \{ g(1),g(N(x)) - g(N(y)) + g(1)\} = g(1) $

    从而g(N(x))≥g(N(y)),即xy.如果yN(x),同理xy.从而Ig,N满足序性质.

    推论1  模糊蕴涵(Ig,N)φ满足序性质当且当N为强的.

    性质4   模糊蕴涵Ig,N的自然否定NIg,N为强的,且g为其生成子当且当N为强的.

       NIg,N(x)=Ig,N(x,0)=g-1(g(1)-g(N(N(x)))),如果N为强的,则

    $ {N_{{I_{g,N}}}}(x) = {g^{ - 1}}(g(1) - g(x)) $

    因为g:[0, 1]→[0,∞)严格增且连续,则NIg,N为强的,g为其生成子.反之,如果NIg,N为强的,且g为其生成子,则NIg,N(x)=g-1(g(1)-g(x)),从而

    $ N(N(x)) = {g^{ - 1}}\left( {g(1) - g\left( {{N_{{I_{g,N}}}}(x)} \right)} \right) = {g^{ - 1}}\left( {g(1) - g\left( {{g^{ - 1}}(g(1) - g(x))} \right)} \right) = x $

    N为强.

    性质5   已知N为连续模糊否定,如果模糊蕴涵Ig,N关于模糊否定N'满足逆否对称性,则N=N',且N为强的.

      设xy∈[0, 1],因为Ig,N关于N′满足逆否对称性,则

    $ {I_{g.N}}(x,y) = {I_{g,N}}(N'(y),N'(x)) $ (1)

    在等式(1)中令y=0,得Ig,N(x,0)=Ig,N(1,N'(x)),即

    $ {g^{ - 1}}(g(1) - g(N(N(x)))) = {g^{ - 1}}\left( {g(1) - g\left( {N\left( {N'(x)} \right)} \right)} \right) $

    所以对任意的x∈[0, 1],有

    $ N(N(x)) = N\left( {N'(x)} \right) $ (2)

    在等式(1)中令x=1,得Ig,N(1,y)=Ig,N(N'(y),0),即

    $ {g^{ - 1}}(g(1) - g(N(y))) = {g^{ - 1}}\left( {g(1) - g\left( {N\left( {N\left( {{N^\prime }(y)} \right)} \right)} \right)} \right) $

    从而得

    $ N(y) = N\left( {N\left( {N'(y)} \right)} \right)\;\;\;\;y \in [0,1] $ (3)

    由(2),(3)式,可得N(x)=N(N(N(x))),即N为强的.又由(2)式可得N=N'.

    性质6   若模糊否定N为强的,则模糊蕴涵Ig,N只关于N满足逆否对称性.

      易证Ig,N(N(y),N(x))=Ig,N(xy),再由性质5,可得模糊蕴涵Ig,N只关于N满足逆否对称性.

    性质7  若N为连续的模糊否定,则模糊蕴涵Ig,N满足交换原则当且当N为强的,且g为其生成子.

      必要性  由Ig,N满足交换原则,则

    $ {I_{g,N}}\left( {x,{I_{g,N}}(y,z)} \right) = {I_{g,N}}\left( {y,{I_{g,N}}(x,z)} \right)\;\;\;\;x,y,z \in [0,1] $

    z=0,y=1,可得

    $ {I_{g,N}}(x,0) = {I_{g,N}}\left( {1,{N_{{I_{g,N}}}}(x)} \right)\;\;\;\;x \in [0,1] $

    所以有

    $ N(N(x)) = N\left( {{N_{{I_{g,N}}}}(x)} \right)\;\;\;\;\;x \in [0,1] $

    从而

    $ N(N(x)) = N\left( {{g^{ - 1}}(g(1) - g(N(N(x))))} \right)\;\;\;x \in [0,1] $

    $ \tilde N(x) = {g^{ - 1}}(g(1) - g(x))\;\;\;x \in [0,1] $

    显然,${\tilde N}$为强的,则有

    $ N(N(x)) = N\left( {{{\tilde N}^ \circ }(N(x))} \right)\;\;\;\;x \in [0,1] $

    u=N(N(x)),因为N为连续的模糊否定,则u=N(N (u)),u∈[0, 1].注意到${\tilde N}$ (${\tilde N}$(u))=u,则N=${\tilde N}$,即N为强的,且g为其生成子.

    充分性  由g为强模糊否定N的生成子知N(x)=g-1(g(1)-g(x)),x∈[0, 1],从而

    $ {I_{g,N}}\left( {x,{I_{g,N}}(y,z)} \right) = \left\{ {\begin{array}{*{20}{l}} {{g^{ - 1}}(g(z) - g(y) - g(x) + 2g(1))}&{y \ge z,g(x) \ge g(z) - g(y) + g(1)}\\ 1&{否则} \end{array}} \right. $
    $ {I_{g,N}}\left( {y,{I_{g,N}}(x,z)} \right) = \left\{ {\begin{array}{*{20}{l}} {{g^{ - 1}}(g(z) - g(x) - g(y) + 2g(1))}&{x \ge z,g(y) \ge g(z) - g(x) + g(1)}\\ 1&{否则} \end{array}} \right. $

    考虑下面4种情形:

    情形1   如果xzyz,则Ig,N(yIg,N(xz))=1=Ig,N(xIg,N(yz)).

    情形2   如果xzyzIg,N(yIg,N(xz))=1,因为g(y)-g(z)≥g(1)-g(x)不成立,则Ig,N(xIg,N(yz))=1,所以Ig,N(yIg,N(xz))=Ig,N(xIg,N(yz)).

    情形3   如果xzyz,同情形2,Ig,N(yIg,N(xz))=Ig,N(xIg,N(yz)).

    情形4   如果xzyz,显然,Ig,N(yIg,N(xz))=Ig,N(xIg,N(yz)).

    由以上讨论,可得Ig,N满足交换原则.

    推论2   如果N为强的,且g为其加法生成子,则Ig,N关于其自然否定满足逆否对称性.

    推论3   如果N为连续模糊否定,则下列结论等价:

    (ⅰ) Ig,N满足交换原则;

    (ⅱ) Ig,N为(SN)-蕴涵,且S为阿基米德t余模S(xy)=g-1(min{g(1),g(x)+g(y)});

    (ⅲ) Ig,N关于tT满足输入律.

    性质8   设g1g2:[0, 1]→[0,∞)为严格增且连续的函数,满足g1(0)=0,g2(0)=0,模糊否定N1N2连续,则Ig1N1=Ig2N2当且当N1=N2且存在常数k>0,使得g2=kg1.

      类似于文献[18]中命题2.2的证明.

    本文提出了利用模糊否定及单调函数构建模糊蕴涵的方法.当单调函数为恒等函数时,则可直接利用模糊否定构造模糊蕴涵.对所给模糊蕴涵的基本特性进行了探讨,并得到了一些结果.对于如何将所得的模糊蕴涵拓展到二值模糊集[19]或凸vague集环境[20]是一个值得研究的问题.

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    Fuzzy Implication from Fuzzy Negation and Generator and Its Properties
    PENG Zu-ming    
    College of Mathematics and Statistics, Yangtze Normal University, Fuling Chongqing 408100, China
    Abstract: In this paper, a new way of generating fuzzy implications is introduced, which produces a fuzzy implication from a fuzzy negation and a continuous monotone function. The proposed method can generate a fuzzy implication from a fuzzy negation only. Also, some properties such as the left neutrality property, the exchange principle and the ordering property are studied, and some results are obtained.
    Key words: fuzzy implication    fuzzy negation    aggregation function    property    
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