Answer
\[
\boxed{
(A - B) \cup (B - A) = (A \cup B) - (A \cap B)
}
\]
**Justified using:**
- Set Difference Law (12)
- De Morgan’s Law (9)
- Distributive Law (3)
- Complement Law (5)
Work Step by Step
We are asked to prove the set identity:
\[
(A - B) \cup (B - A) = (A \cup B) - (A \cap B)
\]
using algebraic laws from **Theorem 6.2.2** (Set Identities).
---
## ✅ Step-by-step Algebraic Proof:
### Left-hand side:
\[
(A - B) \cup (B - A)
\]
**Use the Set Difference Law (12):**
\[
= (A \cap B^c) \cup (B \cap A^c)
\quad \text{(by 12)}
\]
---
### Now apply **Distributive Law (3b)** in reverse:
Let’s try to write this as a single difference.
But instead, go to the **right-hand side** and show it simplifies to the same thing.
---
## ▶️ Right-hand side:
\[
(A \cup B) - (A \cap B)
\]
**Apply Set Difference Law (12):**
\[
= (A \cup B) \cap (A \cap B)^c
\quad \text{(by 12)}
\]
**Apply De Morgan’s Law (9b):**
\[
= (A \cup B) \cap (A^c \cup B^c)
\quad \text{(by 9b)}
\]
**Apply Distributive Law (3b):**
\[
= [(A \cup B) \cap A^c] \cup [(A \cup B) \cap B^c]
\quad \text{(by 3b)}
\]
Now simplify each term:
---
### Term 1: \((A \cup B) \cap A^c\)
**Use Distributive Law (3b):**
\[
= (A \cap A^c) \cup (B \cap A^c)
\quad \text{(by 3b again)}
\]
- \(A \cap A^c = \emptyset\) (by 5b)
- So:
\[
= \emptyset \cup (B \cap A^c) = B \cap A^c
\]
---
### Term 2: \((A \cup B) \cap B^c\)
Similarly:
\[
= (A \cap B^c) \cup (B \cap B^c)
= (A \cap B^c) \cup \emptyset
= A \cap B^c
\]
---
### Putting it together:
\[
(A \cup B) - (A \cap B)
= B \cap A^c \cup A \cap B^c
= (A \cap B^c) \cup (B \cap A^c)
\]
Now recall:
\[
A - B = A \cap B^c \quad \text{and} \quad B - A = B \cap A^c
\]
So:
\[
(A - B) \cup (B - A) = (A \cup B) - (A \cap B)
\quad \text{✅ Proven}
\]
