Is every finite orthomodular poset (which contains a zero element) atomistic?

I know that every finite poset which has a smallest element is atomic and that every atomic orthomodular lattice is atomistic. I also know that Greechie has given an example of an (infinite) atomic orthomodular poset which is not atomistic.

This leaves me wondering if there are also finite (atomic) orthomodular posets that are not atomistic?

EDIT: Here are some relevant definitions:

If $$P$$ is a partially ordered set (short: poset) with least element $$0$$, then an atom is defined as a minimal upper bound of zero, not equal to zero, i.e. there is no element in between zero and an atom.

A poset (with least element $$0$$) is called atomic if every element not equal to $$0$$ is either an atom itself or is greater than some atom.

A poset (with least element $$0$$) is called atomistic if every element is the supremum of the atoms which it is greater than (or equal to).

An orthocomplemented poset is a poset with least element $$0$$, greatest element $$1$$ and a unitary operation $$'$$ which is an orthocomplementation, i.e. for $$x,y \in P$$:

$$x \leq y \Rightarrow y' \leq x'$$

$$x'' = x$$

$$x\vee x' = 1, x \wedge x' = 0$$.

An orthomodular poset is an orthocomplemented poset for which the orthocomplementation fulfills the following:

$$x \perp y (:\Leftrightarrow x \leq y') \Rightarrow \exists x \vee y \in P$$

$$x \leq y \Rightarrow y = x \vee (x' \wedge y)$$

• A precise reference to Greechie's result would also be welcome. Commented Jul 12, 2021 at 2:30
• You have a false preliminary claim, so I'm not sure you asked the right question. The lattice $L=(\{1,2,4\};\mid)$ is a finite poset (so atomic) with anti-automorphism $x\mapsto4/x$ (so orthocomplemented) that is distributive (so modular). If every atomic orthomodular lattice were atomistic, then $L$ would be, but $4$ is not a supremum of atoms. If you did ask the right question, then I'll write this up as an answer. Commented Jul 12, 2021 at 4:26
• @JacobManaker: The orthocomplement must be more than just an anti-automorphism; it must also be a complement (i.e. it must satisfy $a\vee a^\perp=1$ and $a\wedge a^\perp=0$). Commented Jul 12, 2021 at 4:29
• @EricWofsey: Ooh! Thanks. Commented Jul 12, 2021 at 4:34
• @bof Of course, thanks for your comment. I'll give the definitions. Commented Jul 12, 2021 at 15:27

Suppose that a finite orthomodular poset $$P$$ is not atomistic. Then there exists a minimal element $$x\in P$$ which is not a join of atoms. We cannot have $$x=0$$ (since $$0$$ is the empty join of atoms), so there exists an atom $$a\leq x$$. By orthomodularity, we then have $$x=a\vee(a'\wedge x)$$. Note that $$a'\wedge x\neq x$$ since $$(a'\wedge x)\wedge a=0$$ but $$x\wedge a=a$$. Thus, $$a'\wedge x, so by minimality of $$x$$, $$a'\wedge x$$ is a join of atoms. Combining these atoms with $$a$$, we write $$x$$ as a join of atoms.
More generally, this argument applies to any well-founded poset with a least element and relative complements (i.e. whenever $$x\leq y$$, there exists $$z$$ such that $$x\wedge z=0$$ and $$x\vee z=y$$).