A BE-algebra $(B,*,1)$ is a type $<2,0>$ algebra satisfying the identities:
- $x*x=1$
2.$x*1=1$
3.$1*x=x$
4.$x*(y*z)=y*(x*z)$.
Define a relation $R(x,y)$ that holds iff $x*y=1$. It can be proven that $R$ is a partial order if the identity $(x*y)*y=(y*x)*x$ holds.
My question is whether the converse is true.
That is to say, if $R$ is a partial order, does the identity $(x*y)*y=(y*x)*x$ hold?
The converse is not true in general. Please see Example 3.8 (ii)[1]. [1] S. S. Ahn, Y. H. Kim and J. M. Ko, Filters in commutative BE-algebras, Commun. Korean Math. Soc. 27(2012), No. 2, pp. 233-242.
Best regards Akbar Rezaei Department of Mathematics, Payame Noor University, Tehran, Iran. Email: [email protected]
Here is that example:
\begin{array}{c|cccc} \ast & 1 & a & b & 0 \\ \hline 1 & 1 & a & b & 0 \\ a & 1 & 1 & b & 0 \\ b & 1 & a & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ \end{array}
We have $(a\ast0)\ast0=0\ast0=1\neq a=1\ast a=(0\ast a)\ast a$, so that the identity is not satisfied. However, the order defined by $R$ is a diamond lattice with top $1$, bottom $0$, and atoms $a$ and $b$.