Trivial Ring is Commutative Ring

Theorem
Let $\struct {R, +, \circ}$ be a trivial ring.

Then $\struct {R, +, \circ}$ is a commutative ring.

Proof
First we need to show that a trivial ring is actually a ring in the first place.

Taking the ring axioms in turn:

$(\text A)$: Ring Addition forms Group
$\struct {R, +}$ is a group:

This follows from the definition.

$(\text M 0)$: Closure of Ring Product
$\struct {R, \circ}$ is closed:

From Ring Product with Zero, we have $x \circ y = 0_R \in R$.

$(\text M 1)$: Associativity of Ring Product
$\circ$ is associative on $\struct {R, +, \circ}$:


 * $x \circ \paren {y \circ z} = 0_R = \paren {x \circ y} \circ z$

$(\text D)$: Distributivity of Ring Product over Addition
$\circ$ distributes over $+$ in $\struct {R, +, \circ}$:


 * $x \circ \paren {y + z} = 0_R$ by definition.

Then:

and the same for $\paren {y + z} \circ x$.

Commutative
From the definition of trivial ring:
 * $\forall x, y \in R: x \circ y = 0_R = y \circ x$

Hence its commutativity.