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If we remove an edge from the Hasse diagram of a finite lattice, as long as any vertex maintain at least one upward edge and at least one downward edge, do we still always have a lattice from the resulting Hasse diagram?

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Here is a counter-example:

enter image description here

This is a lattice but deleting the four highlighted edges produces a non-lattice. Of course, it is possible to get a somewhat smaller counter-example, but in this one the symmetry makes the lattice/non-lattice property clear.

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