Product of Absolute Values on Ordered Integral Domain

Theorem
Let $\struct {D, +, \times, le}$ be an ordered integral domain whose zero is denoted by $0_D$.

For all $a \in D$, let $\size a$ denote the absolute value of $a$.

Then:
 * $\size a \times \size b = \size {a \times b}$

Proof
Let $P$ be the (strict) positivity property on $D$.

Let $<$ be the (strict) total ordering defined on $D$ as:
 * $a < b \iff a \le b \land a \ne b$

Let $N$ be the strict negativity property on $D$.

We consider all possibilities in turn.

$(1): \quad a = 0_D$ or $b = 0_D$

In this case, both the $\size a \times \size b$ and the  are equal to zero.

So:


 * $\size a \times \size b = \size {a \times b}$

$(2): \quad \map P a, \map P b$

First:

Then:

So:
 * $\size a \times \size b = \size {a \times b}$

$(3): \quad \map P a, \map N b$

First:

Then:

So:
 * $\size a \times \size b = \size {a \times b}$

Similarly $\map N a, \map P b$.

$(4): \quad \map N a, \map N b$

First:

Then:

So:
 * $\size a \times \size b = \size {a \times b}$

In all cases the result holds.