Equivalence of Definitions of Irreducible Element of Ring

Proof
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.

$(1)$ implies $(2)$
Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 1.

By definition:
 * $x$ has no non-trivial factorization in $D$.

Let $x = y \circ z$ for some $y, z \in D$.

By definition, it cannot be the case that neither $y$ nor $z$ are units of $D$.

So either $y$ or $z$ is a unit of $D$.

, suppose $y$ is a unit of $D$.

Then by definition $z$ is an associate of $x$.

Contrariwise, suppose $z$ is a unit of $D$.

Then by definition $y$ is an associate of $x$.

Thus both $y$ and $z$ are either a unit of $D$ or an associate of $x$.

$x = y \circ z$ is an arbitrary factorization of $x$ in $D$.

It follows that the only divisors of $x$ are its associates and the units of $D$.

Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 2.

$(2)$ implies $(1)$
Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 2.

Then by definition:
 * the only divisors of $x$ are its associates and the units of $D$.

Let $x = y \circ z$.

Then either:
 * $y$ is an associate of $x$ and $z$ is a unit of $D$

or:
 * $z$ is an associate of $x$ and $y$ is a unit of $D$.

In either case, $y \circ z$ is a trivial factorization of $x$.

Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 1.