The Boolean expression given is:

$f(a, b, c) = \left( \overline{\overline{a} \cdot (\overline{a} + b + \overline{c}) + a} \right) \cdot \overline{\overline{a} \cdot \overline{b} \cdot \overline{c}}$

Let's break this expression into parts and simplify it step by step:

First Term:

$\overline{\overline{a} \cdot (\overline{a} + b + \overline{c}) + a}$

1. Inside the parentheses, first look at $\overline{a} \cdot (\overline{a} + b + \overline{c})$:

   • This is a product term where $\overline{a}$ is ANDed with a sum term.
   • By the Absorption Law in Boolean algebra, $\overline{a} \cdot (\overline{a} + b + \overline{c}) = \overline{a}$.

   So, this simplifies to: $\overline{a} + a$

2. $\overline{a} + a$ is always equal to 1 (Complement Law), so: $\overline{\overline{a} \cdot (\overline{a} + b + \overline{c}) + a} = \overline{1} = 0$

Second Term:

$\overline{\overline{a} \cdot \overline{b} \cdot \overline{c}}$

• The term $\overline{a} \cdot \overline{b} \cdot \overline{c}$ is a simple AND of three negated variables.
• Applying double negation, we get: $\overline{\overline{a} \cdot \overline{b} \cdot \overline{c}} = a \cdot b \cdot c$

Final Expression:

Now substitute the results of the two terms back into the original function: $f(a, b, c) = 0 \cdot (a \cdot b \cdot c) = 0$

Thus, the Boolean function simplifies to 0.

This means the output of $f(a, b, c)$ is always 0, regardless of the values of $a$, $b$, or $c$.

Would you like a further explanation or have any questions about this simplification? Here are some related questions that could deepen your understanding:

1. What is the Absorption Law in Boolean algebra?
2. How does the Complement Law simplify Boolean expressions?
3. What is the significance of double negation in Boolean functions?
4. Can the output of this function ever be different from zero?
5. What are the most common Boolean algebra rules used in simplifications?

Tip: When simplifying Boolean expressions, always start by applying basic laws like Complement and Identity to reduce complexity step by step.