Answer
\[
\boxed{
(A - B) - (B - C) = A - B
}
\]
**Justified using:**
- Set Difference Law (12)
- De Morgan’s Law (9a)
- Distributive Law (3b)
- Idempotent and associative laws (7, 2)
Work Step by Step
We are asked to verify the identity:
\[
(A - B) - (B - C) = A - B
\]
for all sets \(A, B, C\), using **algebraic set identities** from Theorem 6.2.2.
---
## ✅ Step-by-step Proof:
We will simplify the **left-hand side** and try to show it equals \(A - B\).
---
### Step 1: Apply the **Set Difference Law (12)**
\[
A - B = A \cap B^c
\quad \text{and} \quad
B - C = B \cap C^c
\]
So:
\[
(A - B) - (B - C)
= (A \cap B^c) - (B \cap C^c)
= (A \cap B^c) \cap (B \cap C^c)^c
\quad \text{(by 12)}
\]
---
### Step 2: Apply **De Morgan’s Law (9a)**
\[
(B \cap C^c)^c = B^c \cup C
\]
So:
\[
(A \cap B^c) \cap (B^c \cup C)
\]
---
### Step 3: Distribute using **Distributive Law (3b)**
\[
= [(A \cap B^c) \cap B^c] \cup [(A \cap B^c) \cap C]
\]
Now simplify each part:
- \((A \cap B^c) \cap B^c = A \cap B^c\) (idempotent/intersection is associative)
- \((A \cap B^c) \cap C\) is a subset of \(A \cap B^c\)
So the **union**:
\[
A \cap B^c \cup \text{(subset of } A \cap B^c) = A \cap B^c
\]
Thus:
\[
(A - B) - (B - C) = A \cap B^c = A - B
\]
✅ Proven.
