Given that R is an integral domain. I'm trying to prove that divisibility on this set constitutes a partial ordering. In particular, I have defined the relation $y \leq_{\,d} x$ on R by $y|x$. Reflexivity is, of course, very easy. Let $x \in R$ be arbitrary. Then, $x|x$ and as a result $x \leq_{\,d} x$. Anti-symmetry is where I'm running into a little difficulty. I can very easily prove for arbitrary nonzero $x,y \in R$, that x and y are associates by the following.
Suppose that $x \leq_{\,d} y$ and $y \leq_{\,d} x$. Then, $x|y$ and $y|x$. Moreover, $y=lx$ and $x=py$ for some $l,p \in R$. Hence, $y=lpy$, and since R is an integral domain, $lp=1$ or $y=0$. Since $y \neq 0$ by construction, $lp=1$ and it follows that x and y are associates.
Any ideas on how to prove instead that $x=y$?
It is only a preorder. If $u$ is any non-trivial unit, and $x$ is any non-zero element, then $x|ux|x$, but $x \neq ux$.