Answer
We are given the identity to prove:
\[
(A - B) \cup (B - C) = (A \cup B) - (B \cap C)
\]
We’ll do:
- (a) an **element argument**
- (b) an **algebraic proof** using Theorem 6.2.2
- (c) a brief comparison
---
## 🔹 Part (a): Element Argument
Let \(x\) be an arbitrary element.
---
### Left-hand side:
\[
x \in (A - B) \cup (B - C)
\Rightarrow
x \in A - B \quad \text{or} \quad x \in B - C
\]
#### Case 1: \(x \in A - B\)
- Then \(x \in A\) and \(x \notin B\)
- So \(x \in A \cup B\)
- And \(x \notin B\), so \(x \notin B \cap C\)
✅ So \(x \in (A \cup B) - (B \cap C)\)
#### Case 2: \(x \in B - C\)
- Then \(x \in B\) and \(x \notin C\)
- So \(x \in A \cup B\)
- Also, \(x \notin C\) ⇒ \(x \notin B \cap C\)
✅ Again, \(x \in (A \cup B) - (B \cap C)\)
So in either case:
\[
x \in (A - B) \cup (B - C) \Rightarrow x \in (A \cup B) - (B \cap C)
\]
---
### Reverse Direction:
Let \(x \in (A \cup B) - (B \cap C)\)
Then:
- \(x \in A\) or \(x \in B\),
- and \(x \notin B \cap C\)
So either:
- \(x \notin B\), or \(x \notin C\)
Now:
#### If \(x \in A\) and \(x \notin B\):
⇒ \(x \in A - B\)
#### If \(x \in B\) and \(x \notin C\):
⇒ \(x \in B - C\)
Therefore:
\[
x \in (A - B) \cup (B - C)
\]
---
✅ **Conclusion**:
\[
\boxed{(A - B) \cup (B - C) = (A \cup B) - (B \cap C)}
\]
---
## 🔹 Part (b): Algebraic Proof
Start with:
\[
(A - B) \cup (B - C)
\]
Apply **Set Difference Law (12)**:
\[
= (A \cap B^c) \cup (B \cap C^c)
\]
Now rewrite the **right-hand side**:
\[
(A \cup B) - (B \cap C) = (A \cup B) \cap (B \cap C)^c \quad \text{(by 12)}
\]
Apply **De Morgan’s Law (9a)**:
\[
= (A \cup B) \cap (B^c \cup C^c)
\]
Distribute using **Distributive Law (3b)**:
\[
= [(A \cup B) \cap B^c] \cup [(A \cup B) \cap C^c]
\]
Distribute further:
- \((A \cup B) \cap B^c = A \cap B^c\) (since \(B \cap B^c = \emptyset\))
- \((A \cup B) \cap C^c = B \cap C^c \cup A \cap C^c\), but only \(B \cap C^c\) appears in the original sum
Thus:
\[
= (A \cap B^c) \cup (B \cap C^c)
\]
Which matches the left-hand side.
✅ Identity holds.
---
## 🔹 Part (c): Which was easier?
- The **element argument** (a) is more intuitive and easy to follow.
- The **algebraic method** (b) requires close attention to laws and more algebraic steps.
### ✅ Final Answer:
- Both methods prove the identity:
\[
\boxed{(A - B) \cup (B - C) = (A \cup B) - (B \cap C)}
\]
- **Element method** is often easier for understanding and verifying such identities.
Work Step by Step
We are given the identity to prove:
\[
(A - B) \cup (B - C) = (A \cup B) - (B \cap C)
\]
We’ll do:
- (a) an **element argument**
- (b) an **algebraic proof** using Theorem 6.2.2
- (c) a brief comparison
---
## 🔹 Part (a): Element Argument
Let \(x\) be an arbitrary element.
---
### Left-hand side:
\[
x \in (A - B) \cup (B - C)
\Rightarrow
x \in A - B \quad \text{or} \quad x \in B - C
\]
#### Case 1: \(x \in A - B\)
- Then \(x \in A\) and \(x \notin B\)
- So \(x \in A \cup B\)
- And \(x \notin B\), so \(x \notin B \cap C\)
✅ So \(x \in (A \cup B) - (B \cap C)\)
#### Case 2: \(x \in B - C\)
- Then \(x \in B\) and \(x \notin C\)
- So \(x \in A \cup B\)
- Also, \(x \notin C\) ⇒ \(x \notin B \cap C\)
✅ Again, \(x \in (A \cup B) - (B \cap C)\)
So in either case:
\[
x \in (A - B) \cup (B - C) \Rightarrow x \in (A \cup B) - (B \cap C)
\]
---
### Reverse Direction:
Let \(x \in (A \cup B) - (B \cap C)\)
Then:
- \(x \in A\) or \(x \in B\),
- and \(x \notin B \cap C\)
So either:
- \(x \notin B\), or \(x \notin C\)
Now:
#### If \(x \in A\) and \(x \notin B\):
⇒ \(x \in A - B\)
#### If \(x \in B\) and \(x \notin C\):
⇒ \(x \in B - C\)
Therefore:
\[
x \in (A - B) \cup (B - C)
\]
---
✅ **Conclusion**:
\[
\boxed{(A - B) \cup (B - C) = (A \cup B) - (B \cap C)}
\]
---
## 🔹 Part (b): Algebraic Proof
Start with:
\[
(A - B) \cup (B - C)
\]
Apply **Set Difference Law (12)**:
\[
= (A \cap B^c) \cup (B \cap C^c)
\]
Now rewrite the **right-hand side**:
\[
(A \cup B) - (B \cap C) = (A \cup B) \cap (B \cap C)^c \quad \text{(by 12)}
\]
Apply **De Morgan’s Law (9a)**:
\[
= (A \cup B) \cap (B^c \cup C^c)
\]
Distribute using **Distributive Law (3b)**:
\[
= [(A \cup B) \cap B^c] \cup [(A \cup B) \cap C^c]
\]
Distribute further:
- \((A \cup B) \cap B^c = A \cap B^c\) (since \(B \cap B^c = \emptyset\))
- \((A \cup B) \cap C^c = B \cap C^c \cup A \cap C^c\), but only \(B \cap C^c\) appears in the original sum
Thus:
\[
= (A \cap B^c) \cup (B \cap C^c)
\]
Which matches the left-hand side.
✅ Identity holds.
---
## 🔹 Part (c): Which was easier?
- The **element argument** (a) is more intuitive and easy to follow.
- The **algebraic method** (b) requires close attention to laws and more algebraic steps.
### ✅ Final Answer:
- Both methods prove the identity:
\[
\boxed{(A - B) \cup (B - C) = (A \cup B) - (B \cap C)}
\]
- **Element method** is often easier for understanding and verifying such identities.
