Integers form Ordered Integral Domain/Proof 1

Proof
From Integers form Integral Domain we have that $\struct {\Z, +, \times}$ forms an integral domain.

From Natural Numbers are Non-Negative Integers we have that the set $\N$ can be considered as a subset of the integers.

Then we have that $\struct {\N_{>0}, +, \times}$ is a (commutative) semiring.

So it follows by definition of semiring, in particular the fact that $+$ and $\times$ on $\N_{> 0}$ are closed, that:
 * $\forall a, b \in \N_{> 0}: a + b \in \N_{>0}$


 * $\forall a, b \in \N_{> 0}: a \times b \in \N_{>0}$

It follows that we can define a property $P$ on $\Z$ such that:
 * $\forall x \in \Z: \map P x \iff x \in \N_{>0}$

Checking that $P$ fulfils the conditions for it to be the (strict) positivity property:


 * $(1): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a + b}$

Follows directly from the fact that $\map P x \iff x \in \N_{> 0}$.


 * $(2): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a \times b}$

Follows directly from the fact that $\map P x \iff x \in \N_{> 0}$.


 * $(3): \quad \forall a \in \Z: \map P a \lor \map P {-a} \lor a = 0$

This follows from the definition of the integers as the Inverse Completion of Natural Numbers.

If $a \in \Z - \N$ then $\exists b \in \N: a + b = 0$ and so $a = -b$.

Hence the conditions are fulfilled and $\struct {\Z, +, \times}$ forms an ordered integral domain.