Element of Integral Domain is Divisor of Itself

Theorem
Let $\left({D, +, \circ}\right)$ be an integral domain whose unity is $1_D$.

Then every element of $D$ is a divisor of itself:


 * $\forall x \in D: x \mathrel \backslash x$

Proof
Follows directly from the definition of divisor:


 * $\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$