First, it is to be established that $\struct {A, \oplus, \circ}$ is a ring in the first place. This we do by verifying the ring axioms. === Axiom $(\text A 0)$: Closure for $\oplus$ === Let $x, y \in A$. It is to be shown that $x \oplus y \in A$, i.e. that $x \oplus y$ is an idempotent element of $R$. Compute as follows: {{begin-eqn}} {{eqn | l = \paren {x \oplus y} \circ \paren {x \oplus y}       | r = \paren {x + y - 2 x \circ y} \circ \paren {x + y - 2 x \circ y}       | c = Definition of $\oplus$ }} {{eqn | r = \paren {x + y - 2 x \circ y} \circ x + \paren {x + y - 2 x \circ y} \circ y - \paren {x + y - 2 x \circ y} \circ \paren {2 x \circ y}       | c = {{Ring-axiom|D}} }} {{eqn | r = x \circ x + y \circ x - \paren {2 x \circ y} \circ x + x \circ y + y \circ y - \paren {2 x \circ y} \circ y }} {{eqn | o =        | ro= -       | r = x \circ \paren {2 x \circ y} - y \circ \paren {2 x \circ y} + \paren {2 x \circ y} \circ \paren {2 x \circ y}       | c = {{Ring-axiom|D}} }} {{eqn | r = x \circ x + y \circ x - 2 x \circ y \circ x + x \circ y + y \circ y - 2 x \circ y \circ y }} {{eqn | o =        | ro= -       | r = 2 x \circ x \circ y - 2 y \circ x \circ y + 4 x \circ y \circ x \circ y       | c = Product of Integral Multiples }} {{eqn | r = x + x \circ y - 2 x \circ y + x \circ y + y - 2 x \circ y - 2 x \circ y - 2 x \circ y + 4 x \circ y       | c = $\circ$ is commutative, $x, y$ are idempotent }} {{eqn | r = x + y - 2 x \circ y       | c = Integral Multiple Distributes over Ring Addition }} {{eqn | r = x \oplus y       | c = Definition of $\oplus$ }} {{end-eqn}} Hence $x \oplus y \in A$, as desired. {{qed|lemma}} === Axiom $(\text A 1)$: Associativity of $\oplus$ === Let $x, y, z \in A$. It is to be shown that $\oplus$ is associative, that is: :$\paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z}$ This is shown by the following computation: {{begin-eqn}} {{eqn | l = \paren {x \oplus y} \oplus z       | r = \paren {x \oplus y} + z - 2 \paren {x \oplus y} \circ z }} {{eqn | r = x + y - 2 x \circ y + z - 2 \paren {x + y - 2 x \circ y} \circ z }} {{eqn | r = x + y + z - 2 x \circ y - 2 x \circ z - 2 y \circ z + 4 x \circ y \circ z }} {{eqn | r = x + \paren {y + z - 2 y \circ z} - 2 x \circ \paren {y + z - 2 y \circ z} }} {{eqn | r = x + \paren {y \oplus z} - 2 x \circ \paren {y \oplus z} }} {{eqn | r = x \oplus \paren {y \oplus z} }} {{end-eqn}} {{finish|justify equalities}} {{qed|lemma}} === Axiom $(\text A 2)$: Commutativity of $\oplus$ === Let $x, y \in A$. It is to be shown that $\oplus$ is commutative, that is: :$x \oplus y = y \oplus x$ This is shown by the following computation: {{begin-eqn}} {{eqn | l = x \oplus y       | r = x + y - 2 x \circ y }} {{eqn | r = y + x - 2 y \circ x       | c = $+, \circ$ are commutative }} {{eqn | r = y \oplus x }} {{end-eqn}} {{qed|lemma}} === Axiom $(\text A 3)$: Identity for $\oplus$ === Let $x \in A$, and let $0_R$ be the zero of $R$. By Ring Zero is Idempotent, $0_R \in A$. It suffices to show that: :$x \oplus 0_R = x$ since {{Ring-axiom|A2}} was already verified above. Now: {{begin-eqn}} {{eqn | l = x \oplus 0_R       | r = x + 0_R - 2 x \circ 0_R }} {{eqn | r = x       | c = Ring Product with Zero }} {{end-eqn}} as desired. {{qed|lemma}} === Axiom $(\text A 4)$: Inverses for $\oplus$ === Let $x \in A$. It is to be shown that there exists $y \in A$ such that: :$x \oplus y = y \oplus x = 0_R$ by $(A3)$ above. In fact, one has: {{begin-eqn}} {{eqn | l = x \oplus x       | r = x + x - 2 x \circ x }} {{eqn | r = 2 x - 2 x       | c = $x$ is an idempotent element }} {{eqn | r = 0_R }} {{end-eqn}} so that each $x \in A$ is its own inverse for $\oplus$. {{qed|lemma}} === Axiom $(\text M 0)$: Closure for $\circ$ === Let $x, y \in A$. It is to be shown that: :$x \circ y \in A$ that is, that $x \circ y$ is idempotent. We have the following computation: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}       | r = x \circ x \circ y \circ y       | c = $\circ$ is associative and commutative }} {{eqn | r = x \circ y       | c = $x, y$ are idempotent elements }} {{end-eqn}} {{qed|lemma}} === Axiom $(\text M 1)$: Associativity of $\circ$ === Immediate from Restriction of Associative Operation is Associative. {{qed|lemma}} === Axiom $(\text D)$: Distributivity === By Restriction of Commutative Operation is Commutative, $\circ$ is commutative on $A$. Thus to establish distributivity, it suffices to verify, for $x, y, z \in A$: :$x \circ \paren {y \oplus z} = \paren {x \circ y} \oplus \paren {x \circ z}$ To this end, we compute as follows: {{begin-eqn}} {{eqn | l = x \circ \paren {y \oplus z}       | r = x \circ \paren {y + z - 2 y \circ z} }} {{eqn | r = x \circ y + x \circ z - 2 x \circ y \circ z }} {{eqn | r = x \circ y + x \circ z - 2 x \circ x \circ y \circ z       | c = $x$ is an idempotent element }} {{eqn | r = x \circ y + x \circ z - 2 x \circ y \circ x \circ z       | c = $\circ$ is commutative }} {{eqn | r = \paren {x \circ y} \oplus \paren {x \circ z} }} {{end-eqn}} {{qed|lemma}} Therefore, having verified all ring axioms, we conclude $\struct {A, \oplus, \circ}$ is a ring. By assumption all $x \in A$ are idempotent elements for $\circ$. Thus $\circ$ is an idempotent operation on $A$. Consequently, $\struct {A, \oplus, \circ}$ is an idempotent ring. {{qed}}