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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?

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  • $\begingroup$ You seem to have an error in (2); do you mean $y*x=0$? $\endgroup$ Apr 11, 2014 at 19:09

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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.

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  • $\begingroup$ This answers my question. Even more useful was to learn that there is such a great tool like scholar.google.com. But then, here is my next questions: is there a narrower class of algebras where this "order" is transitive $\endgroup$ Apr 11, 2014 at 21:59
  • $\begingroup$ Theorem 2.1 is in fact bogus: the BCI algebra $\langle\mathbb Z,-,0\rangle$ clearly violates axiom (3). $\endgroup$ Jul 30, 2014 at 21:13
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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.

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    $\begingroup$ Did you mean this as an update to your previous answer? It is better to edit your existing answer (there is an "edit" button under your post) rather than adding a new one. $\endgroup$ Jul 31, 2014 at 14:51
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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

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  • $\begingroup$ How do we get x*0 = x? Is there as an axiom missing from the OP? $\endgroup$ Jul 30, 2014 at 19:53
  • $\begingroup$ And how do we get any of the subsequent equations, for that matter? The only I can understand is the last one (which is just axiom (3)). $\endgroup$ Jul 30, 2014 at 19:58
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    $\begingroup$ You can use $\LaTeX$ syntax here, thanks to MathJax. That would make your answer much more readable. More details. $\endgroup$ Jul 31, 2014 at 14:51

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