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We know that a finite lattice $L$ is modular if and only if for all $s,t,u \in L$ such that $s \le u$, we have $$ s \vee (t \wedge u)=(s \vee t) \wedge u. $$

If we remove the condition $s \le u$, that is, if
for all $s,t,u \in L$ a finite lattice $L$ satisfies $s \vee (t \wedge u)=(s \vee t) \wedge u$ for all $s,t,u \in L,$ what is such an $L$ would be?

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up vote 4 down vote accepted

Suppose that $L$ satisfies $$s\lor(t\land u)=(s\lor t)\land u\tag{1}$$ for some $s,t,u\in L$. Then $$s\le s\lor(t\land u)=(s\lor t)\land u\le u\;,$$ so $(1)$ already implies that $s\le u$. Thus, if $L$ satisfies $(1)$ for all $s,t,u\in L$, $s\le u$ for all $s,u\in L$, and $|L|=1$. $L$ is not a very interesting lattice!

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