Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive

Theorem
Let $\left({D, +, \times}\right)$ be an ordered integral domain.

Then:
 * $\forall x \in D: x \ne 0 \iff P \left({x \times x}\right)$

where $P \left({x \times x}\right)$ denotes that $x \times x$ has the positivity property.

That is, the square of any element of an ordered integral domain is positive if and only if that element is non-zero.

Proof
Suppose $x = 0$. Then $x \times x = 0 \times 0 = 0$ by the properties of the ring zero.

Now suppose $x \ne 0$.

One of two cases applies:
 * $P \left({x}\right)$


 * $\neg P \left({x}\right)$

Let $P \left({x}\right)$.

Then by definition, $P \left({x \times x}\right)$.

Now suppose $\neg P \left({x}\right)$.

Then by the trichotomy law of ordered integral domains, we have that $P \left({- x}\right)$.

Then again by definition, $P \left({\left({-x}\right) \times \left({-x}\right)}\right)$.

But by Product of Ring Negatives, $\left({-x}\right) \times \left({-x}\right) = x \times x$.

So again $P \left({x \times x}\right)$.

Hence the result.