**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.*