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A lattice is called adiamond if it admits no sublattice equivalent to the diamond lattice $M_3$ below:
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The top interval of a lattice is the interval between the meet of all the maximal elements and the greatest element. A lattice is called selftop is it is equal to its top interval.

Let $\mathcal{L}$ be a selftop finite lattice, $M$ the set of its maximal elements, and $e$ the smallest element.

Question: Is $\mathcal{L}$ adiamond iff $\forall S \subsetneq M$, $\bigwedge_{b \in S}b >e$?

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No, neither implication holds.

If you take $\mathcal L$ to be the 8-element Boolean lattice, then remove one atom, the resulting poset is a selftop adiamond lattice which fails your condition. That is, the three coatoms meet to $0$, but one subset of two coatoms also meets to $0$.

On the other hand, start with any nonempty finite lattice $\mathcal K$ and form the poset that is the parallel sum of $\mathcal K$ and a singleton $z$. Now adjoin a top element $1$ and a bottom element $0$ to this poset. This is a selftop lattice with two maximal elements ($z$ and the top of $\mathcal K$), and no one of the coatoms is $0$. Hence this lattice satisfies your condition that no proper subset of coatoms meets to $0$. But there is no reason for this lattice to be adiamond, since $\mathcal K$ is an interval of arbitrary structure.

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