= Two cube edges in every direction force many vertices
{title2=$|\mathcal F|\ge\lceil3n/2\rceil+1$}
For a family in the $n$-dimensional <Boolean lattice>, the loss of trace cardinality on deleting coordinate $i$ equals the number of pairs $S,S\cup\{i\}$ both in the family. If every direction has at least two such edges, <downward coordinate compression> preserves that property. The resulting <down-set> contains the empty set and all $n$ singletons. Each coordinate must also occur in a two-element member. Those pairs cover all $n$ coordinates and number at least $\lceil n/2\rceil$, giving the bound. For $n\ge2$, the empty set, all singletons and a minimum collection of pairs covering every coordinate attain it.
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