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.