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 $\epsilon \in \openint 0 {\min \set {y_1, y_2, 1} }$.

Then:
 * $\size {x_1 - y_1} < \epsilon \land \size {x_2 - y_2} < \epsilon \implies \size {x_1 x_2 - y_1 y_2} < \epsilon \paren {y_1 + y_2 + 1}$

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:
 * $-\epsilon \paren {y_1 + y_2} - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \paren {y_1 + y_2} + \epsilon^2$

If follows that:

Hence the result.