Consider the following lattice Find the number of complements for the element a

Concept: In a bounded lattice with bottom c and top g, an element x is a complement of a if it satisfies BOTH conditions at once: a ∨ x = g (their least upper bound is the top) and a ∧ x = c (their greatest lower bound is the bottom). Missing either condition disqualifies x — a candidate that is comparable to a (lies on a chain above or below it) usually fails immediately, because then one of the two operations just returns a or x itself, never the top or bottom.

Application — check every element of the lattice: reading the Hasse diagram, c is the bottom, g is the top, a is covered by both h and j, b is covered by both i and j, and d, e, f are each covered only by j. Using these cover relations to compute the join and meet of a with every other element:

Why h fails (this is the case flagged in the report): since a lies directly below h in the diagram, h is itself a common upper bound of {a, h} — being the larger of the two — and it is the least such upper bound, so a ∨ h = h — i.e. LUB{a, h} = h, confirming the report's observation. Because h ≠ g, the join condition already fails, so h cannot be counted. (h also fails the meet condition: since a ≤ h, a ∧ h = a ≠ c — a double failure, not a single one.)

Cross-check: i is the only row in the table where BOTH columns land on the required values simultaneously (a ∨ i = g and a ∧ i = c). Every other element of the 10-element lattice — b, c, d, e, f, g, h, j — fails at least one condition, so no other candidate qualifies.

Conclusion: a has exactly 1 complement — the element i.