Anticommutativity of External Direct Product

Theorem
Let $\left(S, \circ_1 \right)$ be a commutative structure and $\left(T, \circ_2 \right)$ be a anticommutative structure.

Then their external direct product $\left(S \times T, \circ \right)$ is an anticommutative structure.

Sufficient Condition
Let $\left(a, b \right), \left(c, d \right) \in S \times T$ and $\left(a, b \right) \neq \left(c, d \right)$.

We have that:


 * $\left(a, b \right) \circ \left(c, d \right) = \left(a \circ_1 c, b \circ_2 d \right)$

And


 * $\left(c, d \right) \circ \left(a, b \right) = \left(c \circ_1 a, d \circ_2 b \right)$

By definition of $\circ_2$:


 * $b \circ_2 d \neq d \circ_2 b$

Hence:


 * $\left(a, b \right) \circ \left(c, d \right) \neq \left(c, d \right) \circ \left(a, b \right)$

Necessary Condition
Let $\left(a, b \right), \left(c, d \right) \in S \times T$ and $\left(a, b \right) \circ \left(c, d \right) \neq \left(c, d \right) \circ \left(a, b \right)$.

Suppose $\left(a, b \right) = \left(c, d \right)$.

Then clearly:


 * $\left(a, b \right) \circ \left(c, d \right) = \left(c, d \right) \circ \left(a, b \right)$

Which contradicts our assumption.

So it must be the case that:


 * $\left(a, b \right) \neq \left(c, d \right)$