Answer
\[
\boxed{A \cap ((B \cup A^c) \cap B^c) = \emptyset}
\]
---
## 💡 Cited Properties (from Theorem 6.2.2):
- **(2b)**: Associative Law for intersection
- **(3a)**: Distributive Law: \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)
- **(5b)**: Complement Law: \(A \cap A^c = \emptyset\)
- **(4a)**: Identity Law: \(A \cup \emptyset = A\)
- **(5b)** again: \(B \cap B^c = \emptyset\)
- **(4b)**: Identity Law: \(A \cap \emptyset = \emptyset\)
Work Step by Step
We are asked to **simplify** the expression:
\[
A \cap ((B \cup A^c) \cap B^c)
\]
Citing a property from **Theorem 6.2.2** at every step.
---
## ✅ Step-by-step Simplification:
Start with:
\[
A \cap \left( (B \cup A^c) \cap B^c \right)
\]
---
### **Step 1: Apply Associative Law (2b)**
Group all intersections together:
\[
= (A \cap (B \cup A^c)) \cap B^c
\]
---
### **Step 2: Distribute \(A\) over \((B \cup A^c)\)** using **Distributive Law (3a)**
\[
= [(A \cap B) \cup (A \cap A^c)] \cap B^c
\]
---
### **Step 3: Simplify \(A \cap A^c\)** using **Complement Law (5b)**
\[
A \cap A^c = \emptyset
\Rightarrow
= [(A \cap B) \cup \emptyset] \cap B^c
\]
---
### **Step 4: Use Identity Law (4a)**
\[
(A \cap B) \cup \emptyset = A \cap B
\]
So now:
\[
= (A \cap B) \cap B^c
\]
---
### **Step 5: Distribute Intersection** (though not needed here — look carefully)
We now have:
\[
A \cap B \cap B^c
\]
Group as:
\[
A \cap (B \cap B^c)
\]
Now apply **Complement Law (5b)**:
\[
B \cap B^c = \emptyset
\Rightarrow A \cap \emptyset = \emptyset
\]
