Complement of Strict Total Ordering

Theorem
Let $$\left({S; \prec}\right)$$ be a relational structure such that $$\prec$$ is a strict total ordering.

Then the complement of $$\prec$$ is a weak total ordering.

Proof
We need to show that $$\not \prec$$ is a weak total ordering.

First we check in turn each of the criteria for an ordering:

Reflexivity
By Complement of Reflexive Relation, it follows directly that $$\not \prec$$ is reflexive.

Transitivity
Let $$a \not \prec b$$ and $$b \not \prec c$$.

Now, suppose that $$a \prec c$$.

As $$b \not \prec c$$, from the Trichotomy Law, one of the following is the case:


 * $$b = c$$ and so $$a \prec b$$;


 * $$c \prec b$$ and $$a \prec b$$ from transitivity of $$\prec$$.

Both are an immediate contradiction.

So $$a \not \prec c$$ and so $$\not \prec$$ is transitive.

Antisymmetry
Suppose $$a \not \prec b$$.

Then from the Trichotomy Law, either $$a = b$$ or $$b \prec a$$.

Similarly, if $$b \not \prec a$$ we have that either $$a = b$$ or $$a \prec b$$.

As $$\prec$$ is asymmetric, $$b \prec a \iff \neg a \prec b$$.

So if $$a \not \prec b$$ and $$b \not \prec a$$, it follows that $$a = b$$.

So $$\not \prec$$ is antisymmetric.

The fact that $$\not \prec$$ is connected follows directly from the Trichotomy Law.