Ring is Ideal of Itself

Theorem
Let $\left({R, +, \circ}\right)$ be a ring.

Then $R$ is an ideal of $R$.

Proof
From Null Ring and Ring Itself Subrings, $\left({R, +, \circ}\right)$ is a subring of $\left({R, +, \circ}\right)$.

Also, $\forall x, y \in \left({R, +, \circ}\right): x \circ y \in R$, thus fulfilling the condition for $R$ to be an ideal of $R$.