Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 3

Proof
By the definition of divisor:
 * $x \divides y$ and $y \divides x$

Let $x \divides y$ and $y \divides x$.

Then $\exists s, t \in D$ such that:
 * $(1): \quad y = t \circ x$

and:
 * $(2): \quad x = s \circ y$

If either $x = 0_D$ or $y = 0_D$, then so must be the other (as an integral domain has no zero divisors by definition).

So $x = 1_D \circ y$ and $y = 1_D \circ x$, and the result holds.

Otherwise:

So:
 * $s \circ t = 1_D$

and both $s \in U_D$ and $t \in U_D$.

The result follows.