(¬(¬A ∧ ¬B) ∧ (¬A ∨ B)) ∨ (C ∨ (C ∧ B))

Let's simplify the logical expression:

\[
(¬(¬A ∧ ¬B) ∧ (¬A ∨ B)) ∨ (C ∨ (C ∧ B))
\]

### Step 1: Simplify the inner part \( ¬(¬A ∧ ¬B) \)
This can be simplified using De Morgan's law:

\[
¬(¬A ∧ ¬B) = A ∨ B
\]

Now, the expression becomes:

\[
((A ∨ B) ∧ (¬A ∨ B)) ∨ (C ∨ (C ∧ B))
\]

### Step 2: Simplify \( (C ∨ (C ∧ B)) \)
Using the absorption law \( C ∨ (C ∧ B) = C \), the second part simplifies to:

\[
(C ∨ (C ∧ B)) = C
\]

Now the expression becomes:

\[
((A ∨ B) ∧ (¬A ∨ B)) ∨ C
\]

### Step 3: Simplify \( (A ∨ B) ∧ (¬A ∨ B) \)
We now apply the distributive property:

\[
(A ∨ B) ∧ (¬A ∨ B) = (A ∧ ¬A) ∨ (A ∧ B) ∨ (B ∧ ¬A) ∨ (B ∧ B)
\]

Simplifying the terms:

- \( A ∧ ¬A = \text{False} \)
- \( B ∧ B = B \)

Thus, the expression reduces to:

\[
(A ∧ B) ∨ (B ∧ ¬A) ∨ B
\]

Now, using the absorption law \( (A ∧ B) ∨ B = B \) and \( (B ∧ ¬A) ∨ B = B \), the expression simplifies to:

\[
B
\]

### Step 4: Combine with the remaining part
The expression becomes:

\[
B ∨ C
\]

This is the fully simplified form of the given logical expression.

### Final Answer:
\[
B ∨ C
\]

Do you want more details or have any questions?

Here are some related questions:
1. How does De Morgan's law apply in logical expressions?
2. What is the absorption law in logic?
3. How does distributive property work in boolean algebra?
4. Can we simplify \( C ∧ B \) if we know the truth values of C and B?
5. How does Boolean logic apply to digital circuits?

**Tip**: Always look for patterns like contradictions (\( A ∧ ¬A \)) and identities to simplify logic expressions efficiently.

Discover how to simplify the Boolean expression (¬(¬A ∧ ¬B) ∧ (¬A ∨ B)) ∨ (C ∨ (C ∧ B)) step by step. This problem explores core concepts like De Morgan's Law, Absorption Law, and Distributive Property, suitable for students in Grades 10-12.

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