Integers under Addition form Totally Ordered Group

Theorem
Let $\left({\Z, +}\right)$ denote the additive group of integers.

Let $\le$ be the usual ordering on $\Z$.

Then the ordered structure $\left({\Z, +, \le}\right)$ is a totally ordered group.

$(1)$
By Integer Addition is Closed, $\left({\Z, +}\right)$ is an algebraic structure.

$(2)$
$\le$ is an ordering on $\Z$.

Thus, $\left({\Z, \le}\right)$ is an ordered set.

$(3)$
By Ordering Preserved on Integers by Addition and Integer Addition is Commutative, $\le$ is compatible with $+$.

Thus, $\left({\Z, +, \le}\right)$ is an ordered structure.

$\left({\Z, +, \le}\right)$ is a Totally Ordered Structure
By definition, the ordered structure $\left({\Z, +, \le}\right)$ is a totally ordered structure iff $\le$ is a total ordering.

This follows from Ordering on Integers is Total Ordering.

$\left({\Z, +, \le}\right)$ is a Totally Ordered Group
By definition, the totally ordered structure $\left({\Z, +, \le}\right)$ is a totally ordered group iff $\left({\Z, +}\right)$ is a group.

This follows from Integers under Addition form Abelian Group.