# Divisor Relation in Integral Domain is Transitive

## Theorem

Let $\struct {D, +, \circ}$ be an integral domain.

Let $x, y, z \in D$.

Then:

$x \divides y \land y \divides z \implies x \divides z$

### Corollary

The divisibility relation is a transitive relation on $\Z$, the set of integers.

That is:

$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$

## Proof

Let $x \divides y \land y \divides z$.

Then from the definition of divisor, we have:

$x \divides y \iff \exists s \in D: y = s \circ x$
$y \divides z \iff \exists t \in D: z = t \circ y$

Then:

$z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$

Thus:

$\exists \paren {t \circ s} \in D: z = \paren {t \circ s} \circ x$

and the result follows.

$\blacksquare$