Proof: Suppose a is greatest element of P
Then x ≤ a ∀ x ∈ P ——————————— (i)
If a is not maximal then there exist some y ∈ P such that a < y
i.e., a ≤ y, a ≠ y.
But by (i) y ≤ a and so y = a, a contradiction.
Hence a is maximal.
Again, suppose b is another maximal element of P. Since a is greatest and b ∈ P, b ≤ a
But b ≠ a and so b < a, a contradiction as b is maximal.
Hence greatest element is unique maximal element of a poset.