Answer
See explanation
Work Step by Step
We are asked to prove:
\[
\textbf{4. For all } a \in B,\quad a \cdot 0 = 0
\]
without using Theorem 6.4.1 directly (unless previously proved), but using Boolean algebra axioms and earlier exercises.
---
### ✅ Proof:
Let \(a \in B\). We want to show:
\[
a \cdot 0 = 0
\]
---
### **Step 1**: Use the identity:
\[
0 + 0 = 0 \quad \text{(from previous exercise or idempotent law)}
\]
Now consider:
\[
a \cdot 0 = a \cdot (0 + 0)
\]
---
### **Step 2**: Apply the **Distributive Law**:
\[
a \cdot (0 + 0) = a \cdot 0 + a \cdot 0
\]
---
### **Step 3**: Simplify:
\[
a \cdot 0 = a \cdot 0 + a \cdot 0
\]
So:
\[
x = x + x \quad \text{where } x = a \cdot 0
\]
---
### **Step 4**: Apply the **Uniqueness of 0** (Exercise 2 from earlier):
If \(x = x + x\), then \(x = 0\)
This implies:
\[
a \cdot 0 = 0
\]
---
### ✅ Final Answer:
\[
\boxed{a \cdot 0 = 0}
\quad \text{(proved using distributive law and uniqueness of 0)}
\]
