Abbreviation: ComBCK

A \emph{commutative BCK-algebra} is a structure $\mathbf{A}=\langle A,\cdot ,0\rangle$ of type $\langle 2,0\rangle$ such that

(1): $((x\cdot y)\cdot (x\cdot z))\cdot (z\cdot y) = 0$

(2): $x\cdot 0 = x$

(3): $0\cdot x = 0$

(4): $x\cdot y=y\cdot x= 0 \Longrightarrow x=y$

(5): $x\cdot (x\cdot y) = y\cdot (y\cdot x)$

Remark: Note that the commutativity does not refer to the operation $\cdot$, but rather to the term operation $x\wedge y=x\cdot (x\cdot y)$, which turns out to be a meet with respect to the following partial order:

$x\le y \iff x\cdot y=0$, with $0$ as least element.

A \emph{commutative BCK-algebra} is a BCK-algebra $\mathbf{A}=\langle A,\cdot ,0\rangle$ such that

$x\cdot (x\cdot y) = y\cdot (y\cdot x)$

A \emph{commutative BCK-algebra} is a structure $\mathbf{A}=\langle A,\cdot ,0\rangle$ of type $\langle 2,0\rangle$ such that

(1): $(x\cdot y)\cdot z = (x\cdot z)\cdot y$

(2): $x\cdot (x\cdot y) = y\cdot (y\cdot x)$

(3): $x\cdot x = 0$

(4): $x\cdot 0 = x$

This definition shows that commutative BCK algebras form a variety.

Let $\mathbf{A}$ and $\mathbf{B}$ be commutative BCK-algebras. A morphism from $\mathbf{A}$ to $\mathbf{B}$ is a function $h:A\rightarrow B$ that is a homomorphism:

$h(x\cdot y)=h(x)\cdot h(y) \mbox{ and } h(0)=0$

Example 1:

$\begin{array}{lr} f(1)= &1 f(2)= &1 f(3)= &2 f(4)= &5 f(5)= &11 f(6)= &28 f(7)= &72 f(8)= &192 \end{array}$

Tarski algebras

MV-algebras

BCK-algebras