生成关联函数的单调函数的特征

Chen Meng, Yun-Mao Zhang, Xue-ping Wang
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引用次数: 0

摘要

双位函数 $T: [0,1]^2\rightarrow [0,1]$ 的关联性由 $T(x,y)=f^{(-1)}(F(f(x),f(y)))$ 定义,其中 $F:[0,\infty]^2\rightarrow[0,\infty]$ 是一个关联函数,$f:[0,1]\rightarrow[0、\infty]$是一个单调函数,当$f(x^{+})满足$f(x^{+})\in/mbox{Ran}(f)$时,或者当$f(x^{+})\notin/mbox{Ran}(f)$满足$f(x)\neq f(y)$时,对于所有$x\in[0,1]$和$f^{(-1)}:[0,\infty]\rightarrow[0,1]$是$f$的伪逆,这只取决于$f$范围的性质。通过应用单调函数 $f$ 的性质,提出了 $T$ 是关联函数的必要条件和充分条件。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
The characterizations of monotone functions which generate associative functions
Associativity of a two-place function $T: [0,1]^2\rightarrow [0,1]$ defined by $T(x,y)=f^{(-1)}(F(f(x),f(y)))$ where $F:[0,\infty]^2\rightarrow[0,\infty]$ is an associative function, $f: [0,1]\rightarrow [0,\infty]$ is a monotone function which satisfies either $f(x)=f(x^{+})$ when $f(x^{+})\in \mbox{Ran}(f)$ or $f(x)\neq f(y)$ for any $y\neq x$ when $f(x^{+})\notin \mbox{Ran}(f)$ for all $x\in[0,1]$ and $f^{(-1)}:[0,\infty]\rightarrow[0,1]$ is a pseudo-inverse of $f$ depends only on properties of the range of $f$. The necessary and sufficient conditions for the $T$ to be associative are presented by applying the properties of the monotone function $f$.
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