Product of Positive Strictly Increasing Mappings is Strictly Increasing

Theorem
Let $A$ be an ordered set.

Let $B$ be an ordered field.

Let $f, g: A \to B$ be strictly increasing mappings with positive values.

Let $h: A \to B$ be defined by $\map h x = \map f x \map g x$.

Then $h$ is strictly increasing.

Proof
Let $x, y \in A$ such that $x < y$.

If $\map h x = 0$.

By the definition of strictly increasing:
 * $\map f y > \map f x \ge 0$

and:
 * $\map g y > \map g x \ge 0$

So:
 * $\map h y > 0 = \map h x$

If $\map h x \ne 0$, then $\map h x > 0$ so:
 * $\map f x > 0$

and:
 * $\map g x > 0$

Also:
 * $\map f x < \map f y$

and:
 * $\map g x < \map g y$

by the definition of strictly increasing.

Because $\map g x > 0$:
 * $\map f x \map g x < \map f y \map g x$

Because $\map f y > 0$:
 * $\map f y \map g x < \map f y \map g y$

By transitivity:


 * $\map h x = \map f x \map g x < \map f y \map g y = \map h y$