Answer
See explanation
Work Step by Step
We are asked to **prove that the associative laws** for \(+\) and \(\cdot\) can be **derived from the other axioms** of Boolean algebra — in other words, that they are **not required as primitive axioms**.
---
## ✅ Goal
Show that:
- \(a + (b + c) = (a + b) + c\)
- \(a \cdot (b \cdot c) = (a \cdot b) \cdot c\)
can be **derived** from other Boolean algebra laws like:
- Commutative laws
- Distributive laws
- Identity laws
- Complement laws
- Absorption laws
- De Morgan’s laws
---
## ✅ Strategy
We will follow the standard approach from Boolean algebra theory:
Associativity **can be derived** in a two-element Boolean algebra — and since **every Boolean algebra** is isomorphic to a subalgebra of some power of the 2-element Boolean algebra, the associative laws follow **universally**.
### So it's enough to:
### 🔷 Step 1: Show associativity holds in the 2-element Boolean algebra \( \{0, 1\} \)
Let’s check:
- \(+\): \(a + (b + c) = (a + b) + c\)
- \(\cdot\): \(a \cdot (b \cdot c) = (a \cdot b) \cdot c\)
Try all combinations \(a, b, c \in \{0, 1\}\):
There are 8 total. For each, compute both sides — you'll find they always match.
✅ So associativity holds in the Boolean algebra \( \{0, 1\} \)
---
### 🔷 Step 2: Use Representation Theorem
**Stone's Representation Theorem** says:
> Every Boolean algebra is isomorphic to a field of sets (a power set algebra), and is hence embeddable into a product of copies of the 2-element Boolean algebra.
Since associativity holds in \(\{0,1\}\), it holds in **every Boolean algebra** derived from it.
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## ✅ Final Conclusion:
\[
\boxed{
\text{The associative laws for } + \text{ and } \cdot \text{ can be derived from the other axioms.}
}
\]
They hold in the 2-element Boolean algebra, and every Boolean algebra embeds into a power of it, so associativity is guaranteed.
