﻿ 模糊f代数的基本性质研究 The Study of Elementary Property of Fuzzy f Algebra

Pure Mathematics
Vol. 14  No. 05 ( 2024 ), Article ID: 88724 , 11 pages
10.12677/pm.2024.145207

The Study of Elementary Property of Fuzzy f Algebra

Hengyuan Zhou

School of Science, Xihua University, Chengdu Sichuan

Received: Mar. 8th, 2024; accepted: Apr. 14th, 2024; published: May 31st, 2024

ABSTRACT

This article first discussed that any disjoint complement in fuzzy f algebra is a fuzzy l-ideal, and studied some relationships in fuzzy f algebra when the associative law holds. Then, the definition of fuzzy orthogonal operators was given, and the conditions for two fuzzy orthogonal operators to be equal were studied. Finally, the fuzzy Archimedean f algebra was introduced, and the conditions under which f2 and fg equal zero were discussed.

Keywords:Fuzzy f Algebra, Fuzzy Archimedean f Algebra

1. 引言

1994年，Beg I，在Zadeh教授提出的模糊概念和Venugopalan提出的模糊序集定义的基础上，首次提出了模糊Riesz空间的概念，并在 [1] 中讨论了模糊Riesz空间的部分性质。1997年，Beg I讨论了模糊Archimedean空间的基本性质。然后，Hong L在 [2] 中研究了在模糊Riesz子空间、模糊理想、模糊带以及模糊投影带等的基本性质。2021年，Gui R等人在 [3] 中讨论了模糊Riesz空间中的模糊正算子、模糊序连续算子以及模糊序有界线性算子等的性质。2022年，Cheng在 [4] 中讨论了模糊正线性算子的基本性质。

2. 预备知识

1) 假设 $x=y$ ，则 $\mu \left(x,x\right)=1$ (自反性)

2) 假设 $x,y\in X$ ，如果 $\mu \left(x,y\right)>0$$\mu \left(y,x\right)>0$ ，则 $x=y$ (反对称性)

3) 假设 $x,z\in X$ ，则 $\mu \left(x,z\right)\ge \underset{y\in X}{\vee }\left[\mu \left(x,y\right)\wedge \mu \left(y,z\right)\right]$ (传递性)

$U\left(A\right)\left(y\right)=\left\{\begin{array}{l}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(↑x\right)\left(y\right)\le \frac{1}{2},\text{\hspace{0.17em}}x\in A\\ \left({\cap }_{x\in A}↑x\right)\left(y\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}其他\end{array}$

$L\left(A\right)\left(y\right)=\left\{\begin{array}{l}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(↓x\right)\left(y\right)\le \frac{1}{2},\text{\hspace{0.17em}}x\in A\\ \left({\cap }_{x\in A}↓x\right)\left(y\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}其他\end{array}$

1) $z\in U\left(A\right)$

2) 如果 $y\in U\left(A\right)$ ，则 $y\in U\left(z\right)$

1) $z\in L\left(A\right)$

2) 如果 $y\in L\left(A\right)$ ，则 $y\in L\left(z\right)$

1) 对任意 $x\in E$ ，假设 ${x}_{1},{x}_{2}\in E$ ，使得 $\mu \left({x}_{1},{x}_{2}\right)>\frac{1}{2}$ ，则

$\mu \left({x}_{1},{x}_{2}\right)\le \mu \left({x}_{1}+x,{x}_{2}+x\right)$

2) 对任意非负实数 $\alpha$ ，假设 ${x}_{1},{x}_{2}\in E$ ，使得 $\mu \left({x}_{1},{x}_{2}\right)>\frac{1}{2}$ ，则

$\mu \left({x}_{1},{x}_{2}\right)\le \mu \left(\alpha {x}_{1},\alpha {x}_{2}\right)$

1) 如果 $\mu \left(\underset{_}{0},x\right)>\frac{1}{2}$$\mu \left(\underset{_}{0},y\right)>\frac{1}{2}$ ，则 $\mu \left(\underset{_}{0},x+y\right)>\frac{1}{2}$

2) 如果 $\mu \left(\underset{_}{0},x\right)>\frac{1}{2}$$\alpha \ge 0$ ，则 $\mu \left(\underset{_}{0},\alpha x\right)>\frac{1}{2}$

3) 如果 $\mu \left({x}_{1},{x}_{2}\right)>\frac{1}{2}$$\alpha \le 0$ ，则 $\mu \left(\alpha {x}_{2},\alpha {x}_{1}\right)>\frac{1}{2}$

$|{x}_{1}|\wedge |{x}_{2}|=\underset{_}{0}$

$g\in D$ ，那么称D是模糊solid的。E的模糊solid子向量空间I是模糊序理想。假设B是E中的模糊序理想。如果 $D\subset B$$x=\mathrm{sup}D$ ，则 $x\in B$ ，那么称B是模糊带。

${A}^{d}=\left\{x\in E|x\perp y,\forall y\in A\right\}$

$\mu \left(x\wedge \left({x}_{1}+{x}_{2}\right),x\wedge {x}_{1}+x\wedge {x}_{2}\right)>\frac{1}{2}$

$\mu \left(|{x}_{n}-x|,{r}_{n}\right)>\frac{1}{2}$${r}_{n}↓0$ ，则称 ${\left\{{x}_{n}\right\}}_{n\in N}$ 模糊序收敛于x，且 $x\in X$ ，记做 ${x}_{n}\stackrel{{f}_{o}}{\to }x$ ，同时也称x是序列

${\left\{{x}_{n}\right\}}_{n\in N}$ 的模糊序极限。

1)如果 $P\left(k\right)\subset H$ 是模糊序有界的当且仅当 $k\subset K$ 是模糊序有界的，则称算子P是模糊序有界的；

2)如果在K中有 ${k}_{\lambda }\stackrel{{f}_{o}}{\to }0$ 能推出在H中有 $P\left({k}_{\lambda }\right)\stackrel{{f}_{o}}{\to }0$ ，则称算子P是模糊序连续的。

$f,g\in E$ ，并且 $\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$\mu \left(\underset{_}{0},g\right)>\frac{1}{2}$ ，有 $\mu \left(\underset{_}{0},fg\right)>\frac{1}{2}$ ，则称E为模糊Riesz代数(又可以称模糊格序代数)。在模糊Riesz代数E中，如果 $f\wedge g=\underset{_}{0}$ ，对于任意 $w\in E$$\mu \left(\underset{_}{0},w\right)>\frac{1}{2}$ ，能得到 $\left(fw\right)\wedge g=\left(wf\right)\wedge g=\underset{_}{0}$

3. 模糊f代数

$\mu \left(|f|,|g|\right)>\frac{1}{2}$ 。若 $h\in D$ ，则 $\mu \left(|f|\wedge |h|,|g|\wedge |h|\right)>\frac{1}{2}$ ，即

$\mu \left(|f|\wedge |h|,0\right)>\frac{1}{2}$

$fg\wedge h=\underset{_}{0},\text{\hspace{0.17em}}gf\wedge h=\underset{_}{0}$

$fg,gf\in {D}^{d}$ 。E中的任意不交补是模糊双边理想，得证。

1) 如果 $f,g\in E$$\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$\mu \left(\underset{_}{0},g\right)>\frac{1}{2}$ ，则有

${\left(f\vee g\right)}^{2}={f}^{2}\vee {g}^{2}$${\left(f\wedge g\right)}^{2}={f}^{2}\wedge {g}^{2}$

2) 如果 $f,g\in E$$\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$\mu \left(\underset{_}{0},g\right)>\frac{1}{2}$ ，且k是任意非负整数，则有

$\mu \left[{\left(fg-gf\right)}^{+},{f}^{2}\vee f{\left(g-kf\right)}^{+}\right]>\frac{1}{2}$

$\mu \left[{\left(fg-gf\right)}^{+},{g}^{2}\vee {\left(f-kg\right)}^{+}g\right]>\frac{1}{2}$

3) 如果 $f,g\in E$$\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$\mu \left(\underset{_}{0},g\right)>\frac{1}{2}$ ，且n是任意正整数，则有

$\mu \left(n|fg-gf|,{f}^{2}\vee {g}^{2}\right)>\frac{1}{2}$

$\begin{array}{c}{\left(f\vee g\right)}^{2}=\left(f\vee g\right)\left(f\vee g\right)\\ =\left\{f\left(f\vee g\right)\right\}\vee \left\{g\left(f\vee g\right)\right\}\\ =\left({f}^{2}\vee fg\right)\vee \left(gf\vee {f}^{2}\right)\\ =\left({f}^{2}\vee {g}^{2}\right)\vee \left(fg\vee gf\right)\end{array}$

2) 假设 $fg-gf=f\left(g-kf\right)-\left(g-kf\right)f$

$\mu \left\{{\left[f\left(g-kf\right)-\left(g-kf\right)f\right]}^{+},{\left[f\left(g-kf\right)\right]}^{+}+{\left[\left(kf-g\right)f\right]}^{+}\right\}>\frac{1}{2}$ ，故有

$\mu \left\{{\left(fg-gf\right)}^{+},{\left[f\left(g-kf\right)\right]}^{+}+{\left[\left(kf-g\right)f\right]}^{+}\right\}>\frac{1}{2}$ $\mu \left[{\left(fg-gf\right)}^{+},f{\left(g-kf\right)}^{+}+{\left(kf-g\right)}^{+}f\right]>\frac{1}{2}$

$f{\left(g-kf\right)}^{+}+{\left(g-kf\right)}^{-}f=\left[f{\left(g-kf\right)}^{+}\vee {\left(g-kf\right)}^{-}f\right]+\left[f{\left(g-kf\right)}^{+}\wedge {\left(g-kf\right)}^{-}f\right]$

$f{\left(g-kf\right)}^{+}+{\left(g-kf\right)}^{-}f=\left[f{\left(g-kf\right)}^{+}\vee {\left(g-kf\right)}^{-}f\right]$

$\mu \left[{\left(fg-gf\right)}^{+},f{\left(g-kf\right)}^{+}\vee {\left(g-kf\right)}^{-}f\right]>\frac{1}{2}$

$\mu \left[{\left(fg-gf\right)}^{+},{z}_{k}\right]>\frac{1}{2}$

$\mu \left\{{z}_{k+1},\left[{f}^{2}\vee f{\left(g-kf\right)}^{+}\right]\wedge {y}_{k+1}\right\}>\frac{1}{2}$

$\mu \left\{{z}_{k+1},\left[{f}^{2}\vee f{\left(g-kf\right)}^{+}\right]\wedge \left[f{\left[g-\left(k+1\right)f\right]}^{+}\vee {\left[g-\left(k+1\right)f\right]}^{-}f\right]\right\}>\frac{1}{2}$

$\mu \left({z}_{k+1},{f}^{2}\vee \left(f{\left(g-kf\right)}^{+}\wedge \left(f{\left(g-\left(k+1\right)f\right)}^{+}\vee {\left(g-\left(k+1\right)f\right)}^{-}f\right)\right)\right)>\frac{1}{2}$

$\mu \left({z}_{k+1},{f}^{2}\vee f{\left(g-\left(k+1\right)f\right)}^{+}\vee \left(f{\left(g-kf\right)}^{+}\wedge {\left(g-\left(k+1\right)f\right)}^{-}f\right)\right)>\frac{1}{2}$

$\mu \left(f{\left(g-kf\right)}^{+}\wedge {\left(g-\left(k+1\right)f\right)}^{-}f,f{\left(g-kf\right)}^{+}\wedge {\left(g-kf\right)}^{-}+f{\left(g-kf\right)}^{+}\wedge {f}^{2}\right)>\frac{1}{2}$

$\mu \left(f{\left(g-kf\right)}^{+}{\left(g-\left(k+1\right)f\right)}^{-}f,\underset{_}{0}+{f}^{2}\right)>\frac{1}{2}$ 。所以 $\mu \left({z}_{k+1},{f}^{2}\vee f{\left(g-\left(k+1\right)f\right)}^{+}\right)>\frac{1}{2}$

$\mu \left[{z}_{k},{f}^{2}\vee f{\left(g-kf\right)}^{+}\right]>\frac{1}{2}$

3) 由(2)可知，令 $k=n$$n{\left(fg-gf\right)}^{+}={\left[f\left(ng\right)-\left(ng\right)f\right]}^{+}$ ，则有

$\mu \left({\left(f\left(ng\right)-\left(ng\right)f\right)}^{+},{f}^{2}\vee f{\left(ng-nf\right)}^{+}\right)>\frac{1}{2}$

$\mu \left({f}^{2}+nf{\left(g-f\right)}^{+}-{f}^{2}\vee {g}^{2},nf{\left(g-f\right)}^{+}\right)>\frac{1}{2}$$\mu \left({g}^{2}+n{\left(g-f\right)}^{-}g-{f}^{2}\vee {g}^{2},n{\left(g-f\right)}^{-}g\right)>\frac{1}{2}$

$\mu \left(n{\left(fg-gf\right)}^{+}-{f}^{2}\vee {g}^{2},nf{\left(g-f\right)}^{+}\right)>\frac{1}{2}$

$\mu \left(n{\left(fg-gf\right)}^{+}-{f}^{2}\vee {g}^{2},n{\left(g-f\right)}^{-}g\right)>\frac{1}{2}$

$n{\left(fg-gf\right)}^{+}-{f}^{2}\vee {g}^{2}\in L\left\{nf{\left(g-f\right)}^{+}\wedge n{\left(g-f\right)}^{-}g\right\}$

$\mu \left(n{\left(fg-gf\right)}^{+}-{f}^{2}\vee {g}^{2},nf{\left(g-f\right)}^{+}\wedge n{\left(g-f\right)}^{-}g\right)>\frac{1}{2}$

$\mu \left(n{\left(gf-fg\right)}^{+},{f}^{2}\vee {g}^{2}\right)>\frac{1}{2}$

${f}^{2}\vee {g}^{2}\in U\left\{n{\left(fg-gf\right)}^{+},n{\left(gf-fg\right)}^{+}\right\}$

${f}^{2}\vee {g}^{2}\in U\left\{n{\left(fg-gf\right)}^{+}\vee n{\left(fg-gf\right)}^{+}\right\}$

$\mu \left(n{\left(fg-gf\right)}^{+}\vee n{\left(gf-fg\right)}^{+},{f}^{2}\vee {g}^{2}\right)>\frac{1}{2}$ 。又利用

$n|fg-gf|=n{\left(fg-gf\right)}^{+}\vee n{\left(fg-gf\right)}^{-}=n{\left(fg-gf\right)}^{+}\vee n{\left(gf-fg\right)}^{+}$

1) $\mu \left(\underset{_}{0},e\right)>\frac{1}{2}$

2) 如果 $\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$f\in E$${f}^{-1}$ 存在，则 $\mu \left(\underset{_}{0},{f}^{-1}\right)>\frac{1}{2}$

3) 如果 $\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$f\in E$${f}^{-1}$ 存在，则 $\mu \left(f\wedge {f}^{-1},e\right)>\frac{1}{2}$

4) 如果 $f,g\in E$$\mu \left(\underset{_}{0},f\right)>\frac{1}{2}$$\mu \left(\underset{_}{0},g\right)>\frac{1}{2}$ ，使得 ${f}^{-1},{g}^{-1}$ 存在，则 ${\left(f\vee g\right)}^{-1}$${\left(f\wedge g\right)}^{-1}$ 存在且满足 ${\left(f\vee g\right)}^{-1}={f}^{-1}\wedge {g}^{-1}$${\left(f\wedge g\right)}^{-1}={f}^{-1}\vee {g}^{-1}$

2) 令 $u={f}^{-1}$ ，则得

$e=fu=f\left({u}^{+}-{u}^{-}\right)=f{u}^{+}-f{u}^{-}$$e=uf=\left({u}^{+}-{u}^{-}\right)f={u}^{+}f-{u}^{-}f$

$\left(f{u}^{+}\right)\wedge \left(f{u}^{-}\right)=\underset{_}{0}$$\left({u}^{+}f\right)\wedge \left({u}^{-}f\right)=\underset{_}{0}$

3) 根据定理3.3(1)，可知 ${f}^{2}\wedge e={\left(f\wedge e\right)}^{2}$ 。根据 $\mu \left({\left(f\wedge e\right)}^{2},fe\right)>\frac{1}{2}$ ，即

$\mu \left({\left(f\wedge e\right)}^{2},f\right)>\frac{1}{2}$

$\mu \left({f}^{-1}\left({f}^{2}\wedge e\right),{f}^{-1}f\right)>\frac{1}{2}$

4) 由3)可知 $\mu \left(\left({f}^{-1}g\right)\wedge \left({g}^{-1}f\right),e\right)>\frac{1}{2}$$\mu \left(\left(g{f}^{-1}\right)\wedge \left(f{g}^{-1}\right),e\right)>\frac{1}{2}$

$\begin{array}{l}\left({f}^{-1}\wedge {g}^{-1}\right)\left(f\vee g\right)\\ =\left[{f}^{-1}\left(f\vee g\right)\right]\wedge \left[{g}^{-1}\left(f\vee g\right)\right]\\ =\left[e\vee \left({f}^{-1}g\right)\right]\wedge \left[\left({g}^{-1}f\right)\vee e\right]\\ =e\vee \left[\left({f}^{-1}g\right)\wedge \left({g}^{-1}f\right)\right]\\ =e\end{array}$

$fg=\left(f\wedge g\right)\left(f\vee g\right)$

$\begin{array}{l}fg-\left(f\wedge g\right)\left(f\vee g\right)\\ =fg-\left(f\wedge g\right)\left(f+g-f\wedge g\right)\\ =\left(f-\left(f\wedge g\right)\right)\left(g-\left(f\wedge g\right)\right)\\ =\underset{_}{0}\end{array}$

4. 模糊正交算子

$F{L}_{o}\left(E\right)$ 。显然 $F{L}_{o}\left(E\right)$$F{L}_{b}\left(E\right)$ 模糊线性子空间，即 $\mu \left({T}_{1},{T}_{2}\right)>\frac{1}{2}⇔\mu \left({T}_{1}f,{T}_{2}f\right)>\frac{1}{2}$$\forall f\in E$

1) 对于 $f,g\in E$ ，有 $f\perp g⇒Tf\perp g$

2) $Tf\in {\left\{f\right\}}^{dd}$$\forall f\in E$

3) 对于E中任意模糊带B， $T\left(B\right)\subset B$

4) 对于E中任意非空子集D， $T\left({D}^{d}\right)\subset {D}^{d}$

2) $⇒$ 3)

3) $⇒$ 4)

4) $⇒$ 1)

$\left\{f\in E:Tf=0\right\}$ )。因为 $|Tf|=|T|f||=|T||f|$$\forall f\in E$ ，任意 $\mu \left(g,f\right)>\frac{1}{2}$$g\in E$$f\in {K}_{T}$ ，有

$\mu \left(Tg,Tf\right)>\frac{1}{2}$ ，可得 $Tg=0$ ，所以 ${K}_{T}$ 是模糊序理想。如果 $B\subset {K}_{T}$$f=\mathrm{sup}B$ ，由任意模糊正交算子T

1) ${K}_{T}={K}_{|T|}={K}_{{T}^{+}}\cap {K}_{{T}^{-}}$

2) ${K}_{T}={R}_{T}^{d}$

2) 因为 ${K}_{T}={K}_{|T|}$${R}_{T}^{d}={R}_{|T|}^{d}$ 。假设 $\mu \left(0,T\right)>\frac{1}{2}$ ，如果有 $\mu \left(0,a\right)>\frac{1}{2}$$a\in {R}_{T}^{d}$ ，可得 $a\wedge Ta=\underset{_}{0}$ 。所以

$Ta=Ta\wedge Ta=\underset{_}{0}$ 。即 $a\in {K}_{T}$

$\left(na-na\wedge b\right)\wedge \left\{Tb-T\left(na\wedge b\right)\right\}=\underset{_}{0}$ 。又由 $T\left(na\wedge b\right)=\underset{_}{0}$ ，可得

$\left(a-a\vee {n}^{-1}b\right)\wedge Tb=\underset{_}{0},\forall n=1,2,3,\cdots$

$Tf=\left({T}_{1}-{T}_{2}\right)f={T}_{1}f-{T}_{2}f=\underset{_}{0}$

5. 模糊Archimedean-f代数

1) $f\in N$ 当且仅当 ${f}^{2}=\underset{_}{0}$

2) N是E中的模糊带；

3) 如果 $f\in N$ ，则 $fg=\underset{_}{0}$$\forall g\in E$

4) N是模糊零f代数(即 $fg=\underset{_}{0}$$\forall f,g\in N$ )， ${N}^{d}$ 是模糊半素f代数；

5) 如果 $f\in E$ ，则 ${f}^{2}\in {N}^{d}$ 。(因此如果 $f,g\in E$ ，则 $fg\in {N}^{d}$ )

${\left(n{f}^{k-1}-{f}^{k-2}\right)}^{+}\wedge {\left({f}^{k-2}-n{f}^{k-1}\right)}^{+}=\underset{_}{0}$ ，可得 ${\left(n{f}^{k-1}-{f}^{k-2}\right)}^{+}\wedge \left(n{f}^{k-1}\right)=0$$\forall n\in Ν$

Archimedean空间定义，得 ${f}^{k-1}=\underset{_}{0}$

2) 假设 $\left\{{a}_{\tau }\right\}$ 是非负集合， ${a}_{\tau }↑a$ 且对于任意的 $\tau$${a}_{\tau }^{2}=\underset{_}{0}$ 。如果 $\tau \ge {\tau }_{0}$ ，有 $\mu \left({a}_{\tau }{a}_{{\tau }_{0}},{a}_{\tau }^{2}\right)>\frac{1}{2}$ 。由模糊

3) 假设 $f\in E$${f}^{2}=\underset{_}{0}$ 。E中的模糊正交算子 ${T}_{f}$ 满足 ${T}_{f}f=\underset{_}{0}$${T}_{f}g=0$$\forall g\in {\left\{f\right\}}^{d}$ 。因为 $\left\{f\right\}\cup {\left\{f\right\}}^{d}$ 是E中的模糊序稠密子集，所以由定理4.4可得 ${T}_{f}=0$ ，即 $fg=\underset{_}{0}$$\forall g\in E$

4) 由3)可知 $fg=\underset{_}{0}$$\forall f,g\in N$ 。因为E是模糊f代数， ${N}^{d}$ 是模糊双边理想。如果 $f\in {N}^{d}$${f}^{2}=\underset{_}{0}$ ，则 $f\in N$ 。因此可得 $f=\underset{_}{0}$ 。所以 ${N}^{d}$ 是半素的。

5) 假设非负元素 $f\in E$ 。因为存在 ${w}_{\tau }\in N\oplus {N}^{d}$ ，使得 $\mu \left(0,{w}_{\tau }\right)>\frac{1}{2}$${w}_{\tau }↑f$ ，所以 ${\left(N\oplus {N}^{d}\right)}^{dd}=E$ ，则 ${w}_{\tau }={a}_{\tau }+{b}_{\tau }$$\mu \left(0,{a}_{\tau }\right)>\frac{1}{2}$$\mu \left(0,{b}_{\tau }\right)>\frac{1}{2}$${a}_{\tau }\in N$${b}_{\tau }\in {N}^{d}$ 。由 ${w}_{\tau }^{2}={b}_{\tau }^{2}\in {N}^{d}$${w}_{\tau }^{2}↑{f}^{2}$ 可得 ${f}^{2}\in {N}^{d}$

6. 结论

The Study of Elementary Property of Fuzzy f Algebra[J]. 理论数学, 2024, 14(05): 537-547. https://doi.org/10.12677/pm.2024.145207

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