Dual of Total Ordering is Total Ordering

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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.

$\blacksquare$


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