Answer
See explanation
Work Step by Step
We are asked to prove the following set identity using an **element argument**:
\[
(A - B) \cup (C - B) = (A \cup C) - B
\]
Assume all sets are subsets of a universal set \(U\).
---
### ✅ Goal
Show that both sides of the equation contain **exactly the same elements** by showing:
1. \((A - B) \cup (C - B) \subseteq (A \cup C) - B\)
2. \((A \cup C) - B \subseteq (A - B) \cup (C - B)\)
---
### 🔹 Part 1: Left ⊆ Right
Let \(x \in (A - B) \cup (C - B)\).
Then either:
- \(x \in A\) and \(x \notin B\),
or
- \(x \in C\) and \(x \notin B\).
In either case:
- \(x \in A \cup C\) (since \(x\in A\) or \(x\in C\)),
- \(x \notin B\)
Thus:
\[
x \in (A \cup C) - B
\]
✅ So:
\[
(A - B) \cup (C - B) \subseteq (A \cup C) - B
\]
---
### 🔹 Part 2: Right ⊆ Left
Let \(x \in (A \cup C) - B\).
Then:
- \(x \in A \cup C\)
- \(x \notin B\)
Since \(x \in A \cup C\), it means:
- \(x \in A\), or
- \(x \in C\)
Now combine with \(x \notin B\):
- If \(x \in A\) and \(x \notin B\), then \(x \in A - B\)
- If \(x \in C\) and \(x \notin B\), then \(x \in C - B\)
Therefore,
\[
x \in (A - B) \cup (C - B)
\]
✅ So:
\[
(A \cup C) - B \subseteq (A - B) \cup (C - B)
\]
---
### ✅ Final Conclusion:
Since both inclusions hold:
\[
\boxed{(A - B) \cup (C - B) = (A \cup C) - B}
\]
This completes the proof.
