Does a BCL algebra define a partial order? A BCL algebra is a universal algebra with a binary operation denoted as "$*$" and a $0$-ary operation (constant) denoted as "$0$", satisying the following axioms:
(1) $x * x = 0$;
(2) if $x * y = 0$ and $y * x = 0$, then $x=y$;
(3) $(((x * y) * z) * ((x * z) * y)) * ((z * y) * x) = 0$ 
This definition is from the article "A New Branch of the Pure Algebra: BCL-Algebras" by
Yonghong Liu published in Advances in Pure Mathematics, 2011, 1, 297-299.
The author also defines a relationship $x\leqslant y$ iff $x*y = 0$. Without any proof, this relationship is said to be partial order. While it is obviously reflexive and antisymmetric, I find it impossible to proof that it is also transitive. Can anybody help?
 A: The paper you cite says
in theorem 2.1 3) that

Any a BCH-algebra is a BCL-algebra

The paper http://emis.library.cornell.edu/journals/NSJOM/Papers/25_1/NSJOM_25_1_075_082.pdf
gives in example 1 a four element BCH-algebra where the "order" is not transitive.
Note: I easily found that with no experience about such algebras, and only a little experience about google scholar.
Edit: in the introduction of the cited paper, cases where the "order" is transitive are considered.
A: Proof: If (0*x)0=0, 0(0*x)=0, by axiom(2) we have 0*x=0; If（x*0)x=0, x(x*0)=0, by axiom (2), we have x*0=x. Note (x*0)x=x(x*0).
Yonghong Liu
x*z=(((x*z)0)(0*0))*((z*x)*0) =(((x*z)*(y*z))*((x*y))*(z*y))*((z*x)*(y*x)) =(((x*y)z)((x*z)y))((z*y)*x)=0.
A: BCL algebra define a partial order. Answer 
Proof: (i) Reflexivity: If xx=0, then x⩽x. (ii) Anti-symmety: If x⩽y and y⩽x, then xy=0 and yx=0, by axiom (2),we have x=y. (iii) Transitivity: If x⩽y and y⩽z, then xy=0 and y*z=0, since x*0=x, by axiom (3), we have xz=(((xz)0)(0*0))((zx)0) =(((xz)(yz))((xy))(zy))((zx)(yx)) =(((xy)z)((xz)y))((z*y)*x)=0. we see that x⩽z. The proof is now complete.
Yonghong Liu
See "Filtrations and Deductive Systems in BCL+ Algebras". 
DOI:10.9734/BJMCS/2015/15901
Theorem 3.1
