- #1

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*distributive*if for all a,b,c, [tex]\begin{align}

a\wedge(b\vee c) &=(a\wedge b)\vee(a\wedge c)\\

a\vee(b\wedge c) &=(a\vee b)\wedge(a\vee c).

\end{align}

[/tex] A

*complementation*is a unary operation (on a bounded lattice) [itex]a\mapsto a^\circ[/itex] such that for all a,b, [tex]

\begin{align}

\text{(1)} & a^\circ{}^\circ=a\\

\text{(2)} & a\wedge a^\circ=0\text{ and }a\vee a^\circ =1

\end{align}

[/tex] An

*orthocomplementation*is a complementation that satisfies the additional requirement [tex]\text{(3)}\ a\leq b\ \Rightarrow\ b^\circ\leq a^\circ.[/tex] I want to know if the following two statements (found in two different books) are equivalent:

1. A complemented distributive lattice is called a Boolean algebra.

2. An orthocomplemented distributive lattice is called a Boolean algebra.

Every orthocomplementation is obviously a complementation. Is it possible to prove that every complementation (on a distributive lattice) is an orthocomplementation?

I haven't made much progress. I just noticed the equivalences [tex]

\begin{align}

a\leq b\ &\Leftrightarrow\ a\wedge b=a,\ a\vee b=b\\

b^\circ\leq a^\circ\ &\Leftrightarrow a^\circ\wedge b^\circ =b^\circ,\ a^\circ\vee b^\circ=a^\circ,

\end{align}[/tex] which make me think that the condition that defines an orthocomplementation is equivalent to de Morgan's laws.