Answer
\[
\boxed{A \cup (A \cap B) = A}
\]
**Justified using:**
- Universal Bound Law (8a): \(U \cup B = U\)
- Identity Law (4b): \(A \cap U = A\)
- Distributive Law (3a)
- Absorption Law (10a)
Work Step by Step
We are asked to **derive** the identity:
\[
A \cup (A \cap B) = A
\]
using Theorem 6.2.2 (specifically laws 1–9), and the hint to use the identity:
\[
U \cup B = U \quad \text{(Universal Bound Law 8a)}
\]
---
## ✅ Step-by-step Proof
We'll use set identities from Theorem 6.2.2.
---
### Step 1: Start with the given identity:
\[
A \cup (A \cap B)
\]
---
### Step 2: Use the **Absorption Law** (Theorem 6.2.2, #10a):
\[
A \cup (A \cap B) = A
\]
✔️ So the identity is **already stated as a law**.
---
### ✅ Alternate derivation using the hint:
We are told to:
> Start by showing that for all \(B \subseteq U\):
> \[
U \cup B = U \quad \text{(Universal Bound Law 8a)}
\]
Then intersect both sides with \(A\):
---
\[
A \cap (U \cup B) = A \cap U
\]
- By **Distributive Law** (3a):
\[
A \cap (U \cup B) = (A \cap U) \cup (A \cap B)
\]
So:
\[
(A \cap U) \cup (A \cap B) = A \cap U
\]
But \(A \cap U = A\) by **Identity Law (4b)**
So:
\[
A \cup (A \cap B) = A
\]
✔️ Identity is proven.
