**Solution:** Suppose there exist some element a in a poset P, such that a<a.

Then by definition a≤a and a≠a

By anti-symmetry a≤a, a≤a ⟹a = a

Thus we get a contradiction.

*Again, a < b, b < c ⟹ a ≤ b, a ≠ b*

### b ≤ c, b ≠ c

So a ≤ b, b ≤ c ⟹ a ≤ c [by transitivity]

If a = c, then c ≤ b, b ≤ c ⟹ b =c, a contradiction.

Hence a < c. ** (Proved)**