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 $$R$$.