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:

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

Proof
Follows directly from the definition of divisor:

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