Chapter 6 - Set Theory - Exercise Set 6.4 - Page 381: 3

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We are asked to justify the steps in the Boolean algebra proof of: \[ \textbf{3. For all } a, b \in B,\quad (a + b) \cdot a = a \] --- ### ✅ Proof Breakdown: \[ \begin{align*} (a + b) \cdot a &= a \cdot (a + b) \quad \text{(a)} \\ &= a \cdot a + a \cdot b \quad \text{(b)} \\ &= a + a \cdot b \quad \text{(c)} \\ &= a \cdot 1 + a \cdot b \quad \text{(d)} \\ &= a \cdot (1 + b) \quad \text{(e)} \\ &= a \cdot 1 \quad \text{(f)} \\ &= a \quad \text{(g)} \end{align*} \] --- ### ✅ Fill in Reasons: - **(a)**: **Commutative Law** of multiplication (\(x \cdot y = y \cdot x\)) — Theorem 6.2.2 (1b) - **(b)**: **Distributive Law** — Theorem 6.2.2 (3a) - **(c)**: **Idempotent Law** — Theorem 6.4.1 (4a): \(a \cdot a = a\) - **(d)**: **Identity Law** for multiplication — Theorem 6.4.1 (5a): \(a = a \cdot 1\) - **(e)**: **Distributive Law** (in reverse): factoring \(a\) — Theorem 6.2.2 (3a) - **(f)**: **Universal Bound Law** — Theorem 6.4.1 (5a): \(1 + b = 1\) - **(g)**: **Identity Law** — Theorem 6.4.1 (5a): \(a \cdot 1 = a\) --- ### ✅ Final Boxed Justification: \[ \boxed{ \begin{aligned} \text{(a)} &\ \text{Commutative Law of } \cdot \\ \text{(b)} &\ \text{Distributive Law} \\ \text{(c)} &\ \text{Idempotent Law: } a \cdot a = a \\ \text{(d)} &\ \text{Identity Law: } a = a \cdot 1 \\ \text{(e)} &\ \text{Distributive Law (factoring)} \\ \text{(f)} &\ \text{Universal Bound Law: } 1 + b = 1 \\ \text{(g)} &\ \text{Identity Law: } a \cdot 1 = a \end{aligned} } \]