Answer
See explanation
Work Step by Step
We are asked to justify the steps in the Boolean algebra proof of:
\[
\textbf{2. For all } a \in B,\quad a + 1 = 1
\]
---
### ✅ Proof Breakdown:
\[
\begin{align*}
a + 1 &= a + (a + \overline{a}) \quad \text{(a)} \\
&= (a + a) + \overline{a} \quad \text{(b)} \\
&= a + \overline{a} \quad \text{(c)} \\
&= 1 \quad \text{(d)}
\end{align*}
\]
---
### ✅ Fill in reasons:
- **(a)**: Use **Identity for 1**: \(1 = a + \overline{a}\) from **Example 6.4.2**
- **(b)**: Apply **Associative Law** of + (from Boolean algebra definition)
- **(c)**: Use **Idempotent Law** (Theorem 6.4.1 #4a): \(a + a = a\)
- **(d)**: Use **Uniqueness of 1** (Theorem 6.4.1 #2): \(a + \overline{a} = 1\)
---
### ✅ Final Boxed Justification:
\[
\boxed{
\begin{aligned}
\text{(a)} &\ \text{Identity: } 1 = a + \overline{a} \\
\text{(b)} &\ \text{Associative Law} \\
\text{(c)} &\ \text{Idempotent Law: } a + a = a \\
\text{(d)} &\ \text{Uniqueness of 1: } a + \overline{a} = 1
\end{aligned}
}
\]
