Order Sum of Ordered Sets is Ordered

Theorem
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Then the order sum $\struct {S_1, \preccurlyeq_1} \oplus \struct {S_2, \preccurlyeq_2}$ of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ is also an ordered set.

Proof
Let $\struct {S, \preccurlyeq} := \struct {S_1, \preccurlyeq_1} \oplus \struct {S_2, \preccurlyeq_2}$.

By definition:


 * $\forall \tuple {a, b}, \tuple {c, d} \in S: \tuple {a, b} \preccurlyeq \tuple {c, d} \iff \begin {cases} b = 0 \text { and } d = 1 \\ b = d = 0 \text { and } a \preccurlyeq_1 c \\ b = d = 1 \text { and } a \preccurlyeq_2 c \end {cases}$

We check in turn each of the criteria for an ordering:

Reflexivity
We have by definition of set union:
 * $\forall \tuple {a, n} \in S_1 \sqcup S_2: a \in S_1 \lor a \in S_1$

and so either $a \preccurlyeq_1 a$ or $a \preccurlyeq_2 a$.

Hence by definition of $\oplus$:
 * $\forall \tuple {a, n} \in S: \tuple {a, n} \preccurlyeq \tuple {a, n}$

So $\preccurlyeq$ is reflexive.

Transitivity
We proceed on a case-by-case basis.

Let $\tuple {a, n_a}, \tuple {b, n_b}, \tuple {c, n_c} \in S$.

Let $\tuple {a, n_a} \preccurlyeq \tuple {b, n_b}$ and $\tuple {b, n_b} \preccurlyeq \tuple {c, n_c}$.

Suppose $a, b, c \in S_1$.

Then:

Suppose $a, b, c \in S_2$.

Then:

Suppose $a, b \in S_1$ and $c \in S_2$.

Then:

Then we have that:

Suppose $a \in S_1$ and $b, c \in S_2$.

Then:

Then we have that:

There are four other cases to consider:
 * $a, c \in S_1, b \in S_2$
 * $b, c \in S_1, a \in S_2$
 * $c \in S_1, a, b \in S_2$
 * $b \in S_1, a, c \in S_2$.

None of these can happen as otherwise one of $\lnot \paren {\tuple {a, n_a} \preccurlyeq \tuple {b, n_b} }$ or $\lnot \paren {\tuple {b, n_b} \preccurlyeq \tuple {c, n_c} }$ would be the case.

Thus in all cases we have that:


 * $\tuple {a, n_a} \preccurlyeq \tuple {b, n_b} \text { and } \tuple {b, n_b} \preccurlyeq \tuple {c, n_c} \implies \tuple {a, n_a} \preccurlyeq \tuple {c, n_c}$

So we have shown that $\preccurlyeq$ is transitive.

Antisymmetry
Suppose $\tuple {a, n_a} \preccurlyeq \tuple {b, n_b}$ and $\tuple {b, n_b} \preccurlyeq \tuple {a, n_a}$.

It cannot be the case that $a \in S_1$ and $b \in S_2$ because then $\lnot \paren {\tuple {b, n_b} \preccurlyeq \tuple {a, n_a} }$.

It cannot be the case that $a \in S_2$ and $b \in S_1$ because then $\lnot \paren {\tuple {a, n_a} \preccurlyeq \tuple {b, n_b} }$

So either $a, b \in S_1$ or $a, b \in S_2$.

If $a, b \in S_1$ then $a = b$ by antisymmetry of $\preccurlyeq_1$.

If $a, b \in S_2$ then $a = b$ by antisymmetry of $\preccurlyeq_2$.

Thus in all cases it can be seen that:
 * $\tuple {a, n_a} \preccurlyeq \tuple {b, n_b} \text { and } \tuple {b, n_b} \preccurlyeq \tuple {a, n_a} \implies \tuple {a, n_a} = \tuple {b, n_b}$

So $\preccurlyeq$ is antisymmetric.

So we have shown that $\preccurlyeq$ is reflexive, transitive and antisymmetric.

Thus by definition, $\preccurlyeq$ is an ordering and so $\struct {S_1, \preccurlyeq_1} \oplus \struct {S_2, \preccurlyeq_2}$ is an ordered set.