Supremum does not Precede Infimum

Theorem
Let $$\left({S, \preceq}\right)$$ be a partially ordered set.

Let $$T \subseteq S$$ admit both a supremum $$M$$ and an infimum $$m$$.

Then $$m \preceq M$$.

Proof
By definition of supremum:


 * $$\forall a \in T: a \preceq M$$

By definition of infimum:


 * $$\forall a \in T: m \preceq a$$

The result follows from transitivity of ordering.