For integers $a$ and $b$ the following are equivalent: $a\mid b$ and $b\mid a$; $b=ua$ for a unit $u$; $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$. Being associates is an equivalence relation whose class of $a$ is $\{a,-a\}$

Every integer $x$ satisfies $x=\mathrm{\mid }x\mathrm{\mid }$ or $x=-\mathrm{\mid }x\mathrm{\mid }$: by totality either $x\ge 0$, when $\mathrm{\mid }x\mathrm{\mid }=x$, or $x\le 0$, when $\mathrm{\mid }x\mathrm{\mid }=-x$ and so $x=-(-x)=-\mathrm{\mid }x\mathrm{\mid }$.

$1\mathrm{\ne }0$, $0<1$, $\mathrm{\mid }1\mathrm{\mid }=1$ and $\mathrm{\mid }-1\mathrm{\mid }=1$. Indeed $1=0$ would give $-1=-0=0=1$, contradicting $1\mathrm{\ne }-1$ in [L6]. By totality either $0\le 1$ or $1\le 0$; in the second case adding $-1$ gives $0\le -1$, and $-1\mathrm{\ne }0$ (else $1=0$), so $0<-1$, hence $0<(-1)(-1)=1$, which with $1\le 0$ contradicts antisymmetry. So $0\le 1$, and $0<1$ since $1\mathrm{\ne }0$; then $\mathrm{\mid }1\mathrm{\mid }=1$, while $0\le 1$ gives $-1\le 0$ by compatibility with addition, so $\mathrm{\mid }-1\mathrm{\mid }=-(-1)=1$.

Claim 1 implies claim 3. Suppose $a\mid b$ and $b\mid a$. If $a=0$ then $0\mid b$ forces $b=0$, so $\mathrm{\mid }a\mathrm{\mid }=0=\mathrm{\mid }b\mathrm{\mid }$. If $a\mathrm{\ne }0$, then $b\mid a$ with $a\mathrm{\ne }0$ gives $b\mathrm{\ne }0$ and $\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$, and $a\mid b$ with $b\mathrm{\ne }0$ gives $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }b\mathrm{\mid }$; antisymmetry gives $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$.

$\sim$ is reflexive, since $a\mid a$; symmetric, since its defining condition is unchanged when $a$ and $b$ are interchanged; and transitive, since $a\mid b$, $b\mid c$ give $a\mid c$ and $c\mid b$, $b\mid a$ give $c\mid a$. So it is an equivalence relation.

Claim 3 implies claim 1. Suppose $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$. By step 1.1, $b$ is $\mathrm{\mid }b\mathrm{\mid }$ or $-\mathrm{\mid }b\mathrm{\mid }$, that is $\mathrm{\mid }a\mathrm{\mid }$ or $-\mathrm{\mid }a\mathrm{\mid }$; and, again by step 1.1, $\mathrm{\mid }a\mathrm{\mid }$ is $a$ or $-a$. Hence $b=a$ or $b=-a$. If $b=a$ then $a\mid b$ and $b\mid a$ by reflexivity. If $b=-a$ then $a\mid -a=b$ by [L4]; and $b=-a$ gives $a=-(-a)=-b$, so $b\mid -b=a$, again by [L4].

Claim 2 implies claim 3. If $b=ua$ with $u$ a unit then $u=1$ or $u=-1$, so $\mathrm{\mid }u\mathrm{\mid }=1$ by step 1.2, and $\mathrm{\mid }b\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }$.

Claim 3 implies claim 2. By step 2.1, $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$ gives $b=a=1\cdot a$ or $b=-a=(-1)a$, and $1$ and $-1$ are units.

The three claims are equivalent: claim 3 implies claim 1 by step 2.1 and claim 1 implies claim 3 by step 1.3, while claim 3 implies claim 2 by step 3.1 and claim 2 implies claim 3 by step 2.2.

The class of $a$ is $[a]=\{b:a\sim b\}=\{b:\mathrm{\mid }b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\}=\{a,-a\}$: the middle equality is the equivalence of claims 1 and 3, and the last holds because $\mathrm{\mid }-a\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }$ by [L7], while conversely $\mathrm{\mid }b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }$ gives $b=a$ or $b=-a$ by step 2.1. At $a=0$ this set is $\{0\}$, since $-0=0$; and for $a\mathrm{\ne }0$ it has exactly two elements, since by totality either $0<a$, when adding $-a$ gives $-a<0<a$, or $a<0$, when adding $-a$ gives $a<0<-a$; in both cases $a\mathrm{\ne }-a$.

By [L9] the classes $\{a,-a\}$ are nonempty, cover $\mathbb{Z}$, and any two are equal or disjoint; together with steps 4.1, 1.4 and 5.1 this is the full statement.