Upper Bound is Lower Bound for Inverse Ordering

Definition
Let $$\left({S, \preceq}\right)$$ be a poset.

Let $$T \subseteq S$$.

Let $$M$$ be an upper bound for $$\left({T, \preceq}\right)$$.

Let $$\succeq$$ be the inverse of $$\preceq$$.

Then $$M$$ is a lower bound for $$\left({T, \succeq}\right)$$.

Corollary
Let $$m$$ be a lower bound for $$\left({T, \preceq}\right)$$.

Then $$m$$ is an upper bound for $$\left({T, \succeq}\right)$$.

Proof
Let $$M$$ be an upper bound for $$\left({T, \preceq}\right)$$.

We have from Inverse Ordering that $$\succeq$$ is also an ordering.

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

By definition of inverse relation, it follows that:
 * $$\forall a \in T: M \succeq a$$

That is, $$M$$ is a lower bound for $$\left({T, \succeq}\right)$$.

Proof of Corollary
We have that $$\succeq$$ is an ordering whose inverse is $$\preceq$$.

The result follows by reversing the argument of the main proof.