Exponent Combination Laws/Product of Powers/Proof 2/Lemma

Theorem
Let $x_1, x_2, y_1, y_2 \in \R_{>0}$ be strictly positive real numbers.

Let $\epsilon \in \left({0, \,.\,.\, \min \left \{{y_1, y_2, 1}\right \} }\right)$.

Then:
 * $\left\vert{x_1 - y_1}\right\vert < \epsilon \land \left\vert{x_2 - y_2}\right\vert < \epsilon \implies \left \vert{x_1 x_2 - y_1 y_2}\right \vert < \epsilon \left({y_1 + y_2 + 1}\right)$

Proof
First:

The same logic, mutatis mutandis, shows that $0 < y_2 - \epsilon$.

From Negative of Absolute Value: Corollary 3:

Hence:

Subtracting $y_1 y_2$ from all sections of the inequality:
 * $- \epsilon \left({y_1 + y_2}\right) - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \left({y_1 + y_2}\right) + \epsilon^2$

If follows that:

Hence the result.