General Positivity Rule in Ordered Integral Domain

Theorem
Let $\left({D, +, \times}\right)$ be an ordered integral domain, whose positivity property is denoted $P$.

Let $S \subset D$ be a subset of $D$ such that:
 * $\forall s \in S: P \left({x}\right)$

Then the following are true:


 * $\displaystyle \forall n \in \N_{>0}: \forall s_i \in S: P \left({\sum_{i \mathop = 1}^n s_i}\right)$


 * $\displaystyle \forall n \in \N_{>0}: \forall s_i \in S: P \left({\prod_{i \mathop = 1}^n s_i}\right)$

where:
 * $\displaystyle \sum_{i \mathop = 1}^n s_i = s_1 + s_2 + \cdots + s_n$


 * $\displaystyle \prod_{i \mathop = 1}^n s_i = s_1 \times s_2 \times \cdots \times s_n$

That is, the ring sum and ring product of any number of elements in $D$ which have the positivity property also have the positivity property.

Corollary
Let $P \left({x}\right)$ where $x \in D$.

Then:
 * $P \left({n \cdot x}\right)$ and $P \left({x^n}\right)$

Proof
Trivially true for $n=1$, and true by definition for $n=2$.

Suppose it is true for all $n$ up to $n=k$ for some $k \in \N$.

Then:


 * $\displaystyle \forall s_i \in S: P \left({\sum_{i \mathop = 1}^k s_i}\right)$


 * $\displaystyle \forall s_i \in S: P \left({\prod_{i \mathop = 1}^k s_i}\right)$

Take any $\displaystyle \sum_{i \mathop = 1}^{k+1} s_i = \sum_{i \mathop = 1}^k s_i + s_{k+1}$.

Then $\displaystyle P \left({\sum_{i \mathop = 1}^k s_i}\right)$ and $P \left({s_{k+1}}\right)$ so $\displaystyle P \left({\sum_{i \mathop = 1}^{k+1} s_i}\right)$ because the positivity property holds for all $s_i \in S$.

Similarly $\displaystyle P \left({\prod_{i \mathop = 1}^k s_i}\right)$ and $P \left({s_{k+1}}\right)$ so $\displaystyle P \left({\prod_{i \mathop = 1}^{k+1} s_i}\right)$.

The result follows by the Principle of Strong Induction.

Proof of Corollary
From the definition of power of an element:
 * $\displaystyle n \cdot x = \sum_{i \mathop = 1}^n x$


 * $\displaystyle x^n = \prod_{i \mathop = 1}^n x$

The result then follows directly from the main proof.