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

Proof
First:

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

From Negative of Absolute Value/Corollary 3:

Hence:

Subtracting $y_1y_2$ 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.