Dual of Total Ordering is Total Ordering

Theorem
Let $\preccurlyeq$ be a total ordering.

Then its dual ordering $\succcurlyeq$ is also a total ordering.

Proof
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

From Dual Ordering is Ordering we have that $\succcurlyeq$ is an ordering.

Let $x, y \in S$.

Then $x \preccurlyeq y$ or $y \preccurlyeq x$ since $\preccurlyeq$ is total.

By definition of dual ordering, also $y \succcurlyeq x$ or $x \succcurlyeq y$.

Hence the result, by definition of total ordering.