Strict Weak Ordering Induces Partition

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

Then $$S$$ can be partitioned into equivalence classes whose equivalence relation is "is non-comparable".

That is, each of the partitions $$A$$ of $$S$$ is a relational structure $$\left({\mathbb S, <}\right)$$ such that:


 * $$\mathbb S$$ is the set of these partitions of $$S$$;


 * $$<$$ is the strict total ordering on $$\mathbb S$$ induced by $$\prec$$.

Proof
From the definition of strict weak ordering, we define the symbol $$\Bumpeq$$ as:
 * $$a \Bumpeq b \ \stackrel {\mathbf {def}} {=\!=} \ \neg a \prec b \and \neg b \prec a$$

that is, $$a \Bumpeq b$$ means "$$a$$ and $$b$$ are non-comparable".

Checking in turn each of the criteria for equivalence:

Reflexive
As $$\prec$$ is antireflexive, by definition $$\forall a \in S: \neg a \prec a$$.

Hence by the Rule of Idempotence $$\neg a \prec a \and \neg a \prec a$$ and so $$\forall a \in S: a \Bumpeq a$$.

Symmetric
We have that $$a \Bumpeq b$$ is defined as being $$\neg a \prec b \and \neg b \prec a$$.

It follows from the Rule of Commutation that $$\neg b \prec a \and \neg a \prec b$$, and so $$b \Bumpeq a$$.

Transitive
The relation $$a \Bumpeq b$$ is defined as being transitive in a strict weak ordering.