1. Introduction

In 1991, O. G. Xi [1] applied the concept of fuzzy sets to BCK-algebras. After that, concept of fuzzy sets to BCK-algebras has been considered by a number of authors, amongst them, Y. B. Jun [2]. E. Y. Deeba (1980) [3] introduced the concept of lattice filters. Y. B. Jun, S. M. Hong and J. Meng (1998) [4] introduced the concept of fuzzy BCK-Filter. For the general development of BCK-algebras, the filter theory plays an important role as well as ideal theory. The purpose of this paper is to give the fuzzification of BCK-algebras and we study relationships between BCK-filters and lattice filters in BCK-algebras.

2. Preliminaries

(For more details of BCK/BCI-algebras we refer to [5] [6] and [7].)

An algebra $\left(X;\ast ,0\right)$ of type (2, 0) is said to be a BCK-algebra if it satisfies the following

(I) $\left(\left(x\ast y\right)\ast \left(x\ast z\right)\right)\ast \left(z\ast y\right)=0$

(II) $\left(x\ast \left(x\ast y\right)\right)\ast y=0$

(III) $x\ast x=0$

(IV) $0\ast x=0$

(V) $x\ast y=0$ and $y\ast x=0$ imply $x=y$

A BCK-algebras can be (partially) ordered by $x\le y$ if and only if $x\ast y=0$. The following statements are true in any BCK-algebras: for all $x,y,z$,

(p₁) $\left(x;\le \right)$ is a partially ordered set.

(p₂) $x\ast 0=x$.

(p₃) $\left(x\ast y\right)\ast z=\left(x\ast z\right)\ast y.$

(p₄) $x\ast \left(x\ast \left(x\ast y\right)\right)=x\ast y.$

(p₅) $x\le y$ implies $x\ast z\le y\ast z$ and $z\ast y\le z\ast x.$

BCK-algebra X satisfying the identity $x\wedge y=y\wedge x$ where $x\wedge y=y\ast \left(y\ast x\right)$ for all $x,y\in X$ is said to be commutative. If there is an element $\mathcal{l}$ of BCK-algebra X satisfying $x\le \mathcal{l}$ for all x in X, the element $\mathcal{l}$ is called the unit of X. A BCK-algebra with the unit is called to be bounded. In a bounded BCK-algebra, we denote $\mathcal{l}\ast x$ by $N_x$ for every $x\in X$. In bounded commutative BCK-algebra denote $x\vee y=N_{\left(N_x\wedge N_y\right)}$.

(p₆) In bounded BCK-algebra we have $N_x\ast N_y\le y\ast x$ ) [8].

3. Fuzzy BCK-Algebra

Definition 3.1. Zadeh [9] A fuzzy subset of a BCK-algebra X is a function $\mu :X\to \left[0,1\right]$.

Definition 3.2. [10] Let X be a BCK-algebra. A fuzzy set $\mu$ in X is called a fuzzy subalgebra of X if, for all $x,y\in X$,

$\mu \left(x\ast y\right)\ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}$

Definition 3.3. [11] Let X be a BCK-algebra and let I be a non empty subset of X. Then I it is called to be an ideal of X if, for all $x,y$ in X.

(a) $0\in I$

(b) $x\ast y\in I$ and $y\in I$ imply $x\in I$.

Definition 3.4. [12]. Let $\mu$ be a fuzzy set of X. For $t\in \left[0,1\right]$, the set ${\mu }_t=\left\{x\in X:\mu \left(x\right)\ge t\right\}$ is called a level subset of $\mu$.

Theorem 3.5. Let X be a BCK-algebra and let $\mu$ be an arbitrary fuzzy subalgebra of X, then for every $t\in \left[0,1\right]$, ${\mu }_t$ is an ideal of X, when ${\mu }_t\ne \varphi$.

Proof. For any element $x\in X$ we have

$\mu \left(0\right)=\mu \left(x\ast x\right)\ge \min \left\{\mu \left(x\right),\mu \left(x\right)\right\}=\mu \left(x\right)=t$, where $t\in \left[0,1\right]$,

so that $0\in {\mu }_t$. Let $x\ast y\in {\mu }_t$ and $y\in {\mu }_t$, and for any $x,y$ in X, denote

$t=\min \left\{\mu \left(x\right),\mu \left(y\right)\right\}.$

Then by the hypothesis $\mu$ is a fuzzy subalgebra of X, we have

$\mu \left(x\ast y\right)\ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=t$.

If $\min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=\mu \left(y\right)=t$, then $\mu \left(x\right)\ge t$ so that $x\in {\mu }_t$. Or $\min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=\mu \left(x\right)=t$ which implies $x\in {\mu }_t$. Hence ${\mu }_t$ is an ideal of X.

Definition 3.6. [13] A partially ordered set $\langle X,\le \rangle$ is said to be a lower semilattice if every pair of elements in X has a greatest lower bound (meet ( $\wedge$ )); it is called to be an upper semilattice if every pair of elements in X has least upper bound (Join ( $\vee$ )). If $\langle X,\le \rangle$ is both an upper and a lower semilattice, then it is called a lattice.

Definition 3.7. Let X be a bounded BCK-algebra, $\mu$ be a fuzzy set in X. Then $\mu$ is called a fuzzy BCK-filter of X if, for all $x,y\in X$

(FF₁) $\mu \left(\mathcal{l}\right)\ge \mu (\; x\; )$

(FF₂) $\mu \left(x\right)\ge \min \left\{\mu \left(N_{\left(N_x\ast N_y\right)}\right),\mu \left(y\right)\right\}$

Theorem 3.8. A BCK-algebra $\left(X;\ast ,0\right)$ is commutative if and only if it is lower semilattice with respect to BCK-order $\le$.

Theorem 3.9. Let $\mu$ be a fuzzy BCK-filter of bounded commutative BCK-algebra. Then for every $t\in \left[0,1\right]$, ${\mu }_t$ is an upper semilattice with respect to BCK-order $\le$, when ${\mu }_t\ne \varphi .$

Proof. For bounded commutative BCK-algebra $x=N_{N_x}$ for all $x\in X$ (by Y.B.Jun) and $N_x\wedge N_y\le N_x,N_y$ (by Theorem 3.7), then we have $x=N_{N_x}\le N_{\left(N_x\wedge N_y\right)}=x\vee y$ and $y=N_{N_y}\le N_{\left(N_x\wedge N_y\right)}=x\vee y$ for all $x,y\in {\mu }_t.$

This shows that $x\vee y$ is a common upper-bounded of x and y. Next if $x,y\le z$, then $N_z\le N_x,N_y$. (by Theorem 3.7) we obtain $N_z\le N_x\wedge N_y,N_{\left(N_x\wedge N_y\right)}\le N_{N_z}$ thus $x\vee y\le z$. Hence $x\vee y$ is a least upper bound of x and y, so it remains to prove $x\vee y\in {\mu }_t$, from (p₆) we have $N_{\left(x\vee y\right)}\ast N_y\le y\ast \left(x\vee y\right)=0$, and so $N_{\left(x\vee y\right)}\ast N_y=0$.

Hence $\mu \left(x\vee y\right)\ge \min \left\{\mu \left(N_{\left(N_{\left(x\vee y\right)}\right)}\ast N_y\right),\mu \left(y\right)\right\}=\min \left\{\mu \left(N_0\right),\mu \left(y\right)\right\}=\mu \left(y\right)$, so that $x\vee y\in {\mu }_t$. Hence $\left({\mu }_t,\le \right)$ is upper semilattice. This completes the proof. □

Theorem 3.10. Let $\mu$ be a fuzzy BCK-filter of a bounded commutative BCK-algebra X. If $y\le x$ for all $x,y\in X$, then for every $t\in \left[0,1\right]$, ${\mu }_t$ is a lower semilattice with respect to BCK-order $\le$, when ${\mu }_t\ne \varphi .$

Proof. Let $x,y\in {\mu }_t$. By (II) we know that $\left(x\wedge y\right)\le x,y$. Let z be any element of X such that $z\le x,y$. Then $z\ast x=z\ast y=0$, so

$z=z\ast 0=z\ast \left(z\ast x\right)=x\ast \left(x\ast z\right)$.

By the same reason, we have $z=y\ast \left(y\ast z\right)$, hence

$z=x\ast \left(x\ast z\right)=x\ast \left(x\ast \left(y\ast \left(y\ast z\right)\right)\right)\le x\ast \left(x\ast y\right)=x\wedge y$.

This says that for all $x,y$ in ${\mu }_t$, $x\wedge y$ it is the greatest lower bound, so it is remain to prove $x\wedge y\in {\mu }_t$. Then from (p₆) we have $N_{\left(x\wedge y\right)}\ast N_y\le y\ast \left(x\wedge y\right)=y\ast \left(y\ast \left(y\ast x\right)\right)=y\ast \left(y\ast 0\right)=y\ast y=0$, and so $N_{\left(x\wedge y\right)}\ast N_y=0$.

Hence $\begin{array}{c}\mu \left(x\wedge y\right)\ge \min \left\{\mu \left(N_{N_{\left(x\wedge y\right)}}\ast N_y\right),\mu \left(y\right)\right\}\\ =\min \left\{\mu \left(N_0\right),\mu \left(y\right)\right\}\\ =\min \left\{\mu \left(\mathcal{l}\right),\mu \left(y\right)\right\}=\mu ,\end{array}$

so that $x\wedge y\in {\mu }_t\left(y\right)$. Hence $\left(\mu ,\le \right)$ is lower semilattice. This completes the proof. □

Theorem 3.11. Let $\mu$ be a fuzzy BCK-filter of a bounded commutative BCK-algebra X. If $y\le x$ for all $x,y\in X$, then for every $t\in \left[0,1\right]$, ${\mu }_t$ a lattice with respect to BCK-order $\le$, when ${\mu }_t\ne \varphi .$

Proof. Since $\mu$ fuzzy BCK-filter of a bounded commutative BCK-algebra X. and $y\le x$ for all $x,y\in X$, then for every $t\in \left[0,1\right]$, ${\mu }_t$ is lower semilattice with respect to BCK-order $\le$, (from Theorem 3.10). also ${\mu }_t$ is upper semilattice with respect to BCK-order $\le$, (from Theorem 3.9) combining with Definition 3.6, thus ${\mu }_t$ is a lattice with respect to BCK-order $\le$.

Definition 3.12. [14] A non-empty subset F of a BCK-algebra X is said to be a lattice filter if it satisfies that (D₁) $x\in F$ and $x\le y$ implies $y\in F$ ; (D₂) $x,y\in F$ g.L.b. $\left\{x,y\right\}\in F$.

Theorem 3.13. Let X be a bounded commutative BCK-algebra, $\mu$ be a fuzzy BCK-filter of X, then for every $t\in \left[0,1\right]$, ${\mu }_t$ is a lattice filter of X, when ${\mu }_t\ne \varphi$ and $y\le x$ for all $x,y$ in X.

Proof. Assume $\mu$ is a fuzzy BCK-filter of X, then for all $t\in \left[0,1\right]$, let $x,y\in X$ be such that $y\in {\mu }_t$ and $y\le x$. Then $N_x\ast N_y\le y\ast x=0$ and so $N_x\ast N_y=0$. Hence

$\begin{array}{c}\mu \left(x\right)\ge \min \left\{\mu \left(N_{\left(N_x\ast N_y\right)}\right),\mu \left(y\right)\right\}\\ =\min \left\{\mu \left(N_0\right),\mu \left(y\right)\right\}\\ =\min \left\{\mu \left(\mathcal{l}\right),\mu \left(y\right)\right\}\\ =\mu (\; y\; )\end{array}$

So that $x\in {\mu }_t$. This shows that ${\mu }_t$ satisfies (D₁). The proof of (D’₂) is similar to the proof of Theorem 3.10 and omitted. Hence ${\mu }_t$ is a lattice filter of X. This completes the proof.

Theorem 3.14. Let X be a BCK-algebra and $\mu$ be an arbitrary fuzzy subalgebra of X. Then for every a,x and y in X and for every $t\in \left[0,1\right]$, the following hold when ${\mu }_t\ne \varphi$ :

(a) $a\ast \left(a\ast \left(\left(\cdots \left(\left(a\ast x\right)\ast 0\right)\ast \cdots \right)\ast 0\right)\right)\in {\mu }_t$

(b) if X it is positive implicative, then $\left(x\ast {}^{n}y\right)\in {\mu }_t$ where $x\ast {}^{n}y$ is recursively defined as follows $x\ast {}^{1}y=x\ast y,x\ast {}^{n+1}y=\left(x\ast {}^{n}y\right)\ast y$ for any $n\in \mathbb{N}$ (where $\mathbb{N}$ is the set of all the natural numbers).

(c) $\left(x\ast y\right)\ast x\in {\mu }_t$

(d) if X is bounded commutative, then $N_x\ast N_y\in {\mu }_t$

(e) if X is positive implicative, then $\left(x\ast y\right)\ast y\in {\mu }_t$

(f) if X is implicative, then $\left(x\ast y\right)\ast \left(y\ast \left(x\ast y\right)\right)\in {\mu }_t$

(g) if X is commutative, them $x\ast \left(y\ast \left(y\ast x\right)\right)\in {\mu }_t$

Proof. For any $x,y$ in X denote $t=\min \left\{\mu \left(x\right),\mu \left(y\right)\right\}$

$\begin{array}{l}\mu \left(a\ast \left(a\ast \left(\left(\cdots \left(\left(a\ast x\right)\ast 0\right)\ast \cdots \right)\ast 0\right)\right)\right)\\ =\mu \left(a\ast x\right)\ge \min \left\{\mu \left(a\right),\mu \left(x\right)\right\}=t\end{array}$

So $a\ast \left(a\ast \left(\left(\cdots \left(\left(a\ast x\right)\ast 0\right)\ast \cdots \right)\ast 0\right)\right)\in {\mu }_t$. Hence (a) holds. In any positive implicative BCK-algebra, the following identity hold $x\ast {}^{n}y=x\ast y$ for any $n\in \mathbb{N}$ thus $\mu \left(x\ast {}^{n}y\right)=\mu \left(x\ast y\right)\ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=t$ so $x\ast {}^{n}y\in {\mu }_t$. Hence (b) holds.

$\begin{array}{c}\mu \left(\left(x\ast y\right)\ast x\right)=\mu \left(\left(x\ast x\right)\ast y\right)=\mu \left(0\ast x\right)=\mu \left(0\right)\\ =\mu \left(x\ast x\right)\ge \min \left\{\mu \left(x\right),\mu \left(x\right)\right\}=t,\end{array}$

so $\left(x\ast y\right)\ast x\in {\mu }_t$ and (c) holds.

$\begin{array}{c}\mu \left(N_x\ast N_y\right)=\mu \left(\left(\mathcal{l}\ast x\right)\ast \left(\mathcal{l}\ast y\right)\right)=\mu \left(\left(\mathcal{l}\ast \left(\mathcal{l}\ast y\right)\right)\ast x\right)\\ =\mu \left(y\ast \left(y\ast \mathcal{l}\right)\ast x\right)=\mu \left(\left(y\ast 0\right)\ast x\right)\\ =\mu \left(y\ast x\right)\ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=t,\end{array}$

thus $\mu \left(N_x\ast N_y\right)\ge t.$

Which implies $N_x\ast N_y\in {\mu }_t$, proving (d).

(e) Since X is a positive implicative BCK-algebra, it follows that

$\begin{array}{c}\mu \left(\left(x\ast y\right)\ast y\right)=\mu \left(\left(x\ast y\right)\ast \left(y\ast y\right)\right)=\mu \left(\left(x\ast y\right)\ast 0\right)\\ =\mu \left(x\ast y\right)\ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=t,\end{array}$

thus $\left(x\ast y\right)\ast y\in {\mu }_t$. Hence proving (e).

(f) Since X is an implicative BCK-algebra, it follows that

$\mu \left(\left(x\ast y\right)\ast \left(y\ast \left(x\ast y\right)\right)\right)=\mu \left(x\ast y\right)\ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=t,$

thus $\mu \left(\left(x\ast y\right)\ast \left(y\ast \left(x\ast y\right)\right)\right)\ge t$, so $\left(x\ast y\right)\ast \left(y\ast \left(x\ast y\right)\right)\in {\mu }_t.$

Hence (f) holds.

(g) Form commutative of X and it follows that

$\begin{array}{c}\mu \left(x\ast \left(y\ast \left(y\ast x\right)\right)\right)=\mu \left(x\ast \left(x\ast \left(x\ast y\right)\right)\right)=\mu \left(x\ast y\right)\\ \ge \min \left\{\mu \left(x\right),\mu \left(y\right)\right\}=t.\end{array}$

Thus $\mu \left(x\ast \left(y\ast \left(y\ast x\right)\right)\right)\ge t$ so $x\ast \left(y\ast \left(y\ast x\right)\right)\in {\mu }_t.$

Hence (g) holds. This finishes the proof.

4. Conclusions

1) A level subset of a fuzzy sub-algebra of BCK-algebras X is an ideal X.

2) A level subset of the fuzzy BCK-filter of bounded commutative BCK-algebra X is an upper semilattice with respect to BCK-order ≤.

3) A level subset of a fuzzy BCK-filter of abounds commutative BCK algebras X is a lower semilattice with respect to BCK-order ≤ and If $y\le x$ for all $x,y\in X$.

4) A level subset of a fuzzy BCK-filter of abounds commutative BCK algebras X is a lattice with respect to BCK-order ≤.

5) A level subset of a fuzzy BCK-filter of a bounded commutative BCK-algebra X is a lattice filter with respect to BCK-order ≤ and if $y\le x$ for all $x,y\in X$.