Product of Positive Strictly Increasing Mappings is Strictly Increasing

Theorem
Let $A$ be an ordered set.

Let $B$ be an ordered field.

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

Let $h \colon A \to B$ be defined by $h(x) = f(x)g(x)$.

Then $h$ is strictly increasing.

Proof
Suppose that $x,y \in A$ and $x \prec y$.

If $h(x) = 0$, then $f(x)=0$ or $g(x) = 0$.

Then $f(y)>0$ and $g(y)>0$, so $h(y) > h(x)$.

Otherwise, $h(x) > 0$, so $f(x)>0$ and $g(x)>0$

Then $f(x) < f(y)$ and $g(x) < g(y)$ by the definition of strictly increasing.

$f(x)g(x) < f(y)g(x)$ because $g(x)>0$.

$f(y)g(x) < f(y)g(y)$ because $f(x)>0$.

By transitivity,

$h(x) = f(x)g(x) < f(y)g(y) = h(y)$.