Answer
See explanation
Work Step by Step
We are asked to prove:
\[
\textbf{6a. } \overline{0} = 1 \quad \text{ and } \quad \textbf{6b. } \overline{1} = 0
\]
We’ll prove both using the **definition of complement** in a Boolean algebra:
For any element \( a \in B \), its complement \( \overline{a} \) satisfies:
\[
a + \overline{a} = 1 \quad \text{and} \quad a \cdot \overline{a} = 0
\]
---
## ✅ Part 6a: Prove \( \overline{0} = 1 \)
We must verify that \( 1 \) satisfies the complement laws for \( 0 \):
1. \( 0 + 1 = 1 \) ✅
2. \( 0 \cdot 1 = 0 \) ✅
Since both conditions are met, \( 1 \) is the complement of \( 0 \), and complements are **unique**, so:
\[
\boxed{\overline{0} = 1}
\]
---
## ✅ Part 6b: Prove \( \overline{1} = 0 \)
Check if \( 0 \) satisfies the complement laws for \( 1 \):
1. \( 1 + 0 = 1 \) ✅
2. \( 1 \cdot 0 = 0 \) ✅
Again, both conditions are met, and complements are unique:
\[
\boxed{\overline{1} = 0}
\]
---
### ✅ Final Answer:
\[
\boxed{
\begin{aligned}
\textbf{(a)} &\ \overline{0} = 1 \\
\textbf{(b)} &\ \overline{1} = 0
\end{aligned}
}
\]
