Answer
See explanation
Work Step by Step
We are given the set \( S = \{0, 1\} \) with operations \(+\) and \( \cdot \) defined by the tables, and asked to verify that this structure forms a **Boolean algebra**.
---
## ✅ Part (a): Show properties hold
Let’s check each listed property using the tables:
---
### (i) **Commutative Law for \(+\)**
Check that:
\[
a + b = b + a \quad \text{for all } a, b \in S
\]
From the table:
- \(0 + 0 = 0\), \(0 + 1 = 1\), \(1 + 0 = 1\), \(1 + 1 = 1\)
Clearly, \(a + b = b + a\) — ✅
---
### (ii) **Commutative Law for \( \cdot \)**
Check that:
\[
a \cdot b = b \cdot a
\]
From the table:
- \(0 \cdot 0 = 0\), \(0 \cdot 1 = 0\), \(1 \cdot 0 = 0\), \(1 \cdot 1 = 1\)
✅ Commutative — values match both directions.
---
### (iii) **Associative Law for \(+\)**
We must check:
\[
a + (b + c) = (a + b) + c
\quad \text{for all } a, b, c \in S
\]
There are only 8 combinations of \((a, b, c)\), and checking each using the table confirms they are equal. ✅
---
### (iv) **Associative Law for \( \cdot \)**
Same check as above for multiplication:
\[
a \cdot (b \cdot c) = (a \cdot b) \cdot c
\]
Try all 8 combinations — all hold. ✅
---
### (v) **Distributive Law of \( \cdot \) over \(+ \)**
Check:
\[
a \cdot (b + c) = (a \cdot b) + (a \cdot c)
\]
Example: \(1 \cdot (0 + 1) = 1 \cdot 1 = 1\), and \((1 \cdot 0) + (1 \cdot 1) = 0 + 1 = 1\)
Check all combinations — all valid. ✅
---
### (vi) **Distributive Law of \(+\) over \( \cdot \)**
Check:
\[
a + (b \cdot c) = (a + b) \cdot (a + c)
\]
Again try values like:
- \(1 + (0 \cdot 1) = 1 + 0 = 1\)
- \((1 + 0) \cdot (1 + 1) = 1 \cdot 1 = 1\)
Try all combinations — ✅ they match.
---
## ✅ Part (b): Identity Elements
- **Identity for \(+\)**: Is there \(0\) such that \(a + 0 = a\)? ✅ yes.
- **Identity for \( \cdot \)**: Is there \(1\) such that \(a \cdot 1 = a\)? ✅ yes.
So:
\[
\boxed{\text{0 is identity for } +,\quad 1 \text{ is identity for } \cdot}
\]
---
## ✅ Part (c): Complements
Define:
- \( \overline{0} = 1 \)
- \( \overline{1} = 0 \)
We want to show:
- \(a + \overline{a} = 1\)
- \(a \cdot \overline{a} = 0\)
Try:
- \(0 + \overline{0} = 0 + 1 = 1\)
- \(1 + \overline{1} = 1 + 0 = 1\)
- \(0 \cdot \overline{0} = 0 \cdot 1 = 0\)
- \(1 \cdot \overline{1} = 1 \cdot 0 = 0\)
✅ So complement laws hold.
---
## ✅ Final Conclusion
All axioms are satisfied:
\[
\boxed{
(S = \{0, 1\}, +, \cdot) \text{ is a Boolean algebra}
}
\]
