# Boolean Algebra - Simplifying a Function

Simplify the function $$f(x_2, x_1, x_0) = f(x) = \overline{\overline{x_0} \cdot x_1 + x_2 + x_1}$$ such that the final minimized version only contains literals and the operators $$"+" \text{and} "\cdot"$$.

$$f(x) = \overline{\overline{x_0} \cdot x_1 + x_2 + x_1}\stackrel{\text{Order of Operations}}{=} \overline{(\overline{x_0} \cdot x_1) + (x_2 + x_1)} \stackrel{De Morgan}= \overline{\overline{x_0} + x_1} \cdot \overline{x_2+x_1} \stackrel{De Morgan} = x_0 \cdot \overline{x_1} \cdot \overline{x_2} \cdot \overline{x_1} \stackrel{Idempotence}= x_0 \cdot \overline{x_1} \cdot \overline{x_2}.$$

However, my textbook gives $$f(x) = \overline{x_1} \cdot \overline{x_2}$$ as the solution. Where's my mistake?

It should be $$\overline{\overline{x_{0}}\cdot{x_{1}}}\cdot\overline{x_{2}+x_{1}}$$ instead of $$\overline{\overline{x_{0}}+{x_{1}}}\cdot\overline{x_{2}+x_{1}}$$
$$\overline{\overline{x_{0}}\cdot x_{1}+x_{2}+x_{1}}=\overline{x_{1}(\overline{x_{0}}+1)+x_{2}}=\overline{x_1{}+x_{2}}=\overline{x_{1}}\cdot\overline{x_{2}}$$