Lattice with property weaker than distributivity

60 Views Asked by At

Consider a (finite?) lattice $L$ with bottom $\perp$.

Suppose that I want the following property to hold:

For every $x_1,\dots,x_n,y\in L$, $$ \bigwedge_i x_i = \perp $$ implies $$ \bigwedge_i (y \vee x_i) = y\;. $$

Does it have a name? Clearly this property is true if $L$ is distributive. But is it actually weaker? Is it satisfied by other "famous" type of lattices? Has it been studied?

1

There are 1 best solutions below

0
On

Clearly this property is true if L is distributive. But is it actually weaker?

Yes, it is weaker. Your condition, namely

$$ (\bigwedge x_i=\bot)\Rightarrow (\bigwedge (y\vee x_i) = y) $$ is a quasiidentity in the language with symbols $\vee, \wedge, \bot$. It fails in $M_3$ and $N_5$, so a lattice satisfying this condition will not have any minimal distributivity failures involving $\bot$. But any lattice with a new least element $\bot$ adjoined will satisfy this quasiidentity. That is, the ordinal sum $\{\bot\}+L$ satisfies this condition for any $L$.

If you alter your condition so that it says $$ (\bigwedge x_i=z)\Rightarrow (\bigwedge (y\vee x_i) = (y\vee z)), $$ then you get distributivity, but you should not expect to get distributivity with a condition that depends in an essential way on a reference to the bottom element.