CE2704 · Digital Logic Design
Theme 2 · Boolean algebra & logic gates
Boolean theorems & simplification
The algebra rules that let you shrink an expression — fewer gates, less power, faster circuits. And the intuition to trust what a synthesis tool produces.
Built from first principles.
Before you start
What you need first
- Boolean basics & gates — AND/OR/NOT and their notation.
- Truth tables — to check any identity in a minute.
What you'll be able to do
- Apply the postulates (identity, null, idempotent, complement).
- Use the theorems (commutative, associative, distributive, absorption).
- Simplify an expression step by step to fewer gates.
Postulates — the starting facts
The rules everything builds on. The first three behave much like ordinary arithmetic:
| Rule | AND form | OR form |
|---|---|---|
| Identity | A·1 = A | A+0 = A |
| Null | A·0 = 0 | A+1 = 1 |
| Idempotent | A·A = A | A+A = A |
Three more involve NOT:
| Rule | Form | Why |
|---|---|---|
| Complement (AND) | A·A′ = 0 | a signal can't be 1 and 0 at once |
| Complement (OR) | A+A′ = 1 | a signal is either 1 or 0 — no third option |
| Double negation | (A′)′ = A | inverting twice returns the original |
Duality: every AND rule has a matching OR rule — swap
AND↔OR and 0↔1. That halves what you memorise. And any identity can be checked with a small
truth table.
Theorems — the rules you'll lean on
Two are exactly like ordinary algebra:
| Theorem | AND form | OR form |
|---|---|---|
| Commutative | A·B = B·A | A+B = B+A |
| Associative | (A·B)·C = A·(B·C) | (A+B)+C = A+(B+C) |
Two are more powerful than they look:
| Theorem | Form 1 | Form 2 |
|---|---|---|
| Distributive | A·(B+C) = A·B + A·C | A + B·C = (A+B)·(A+C) |
| Absorption | A·(A+B) = A | A + A·B = A |
The shocker: the OR-form distributive
A + B·C = (A+B)·(A+C) is false in ordinary math
(\(3 + 2\cdot4 \neq (3+2)(3+4)\)) but true in Boolean algebra. Absorption
says a term swallowed by a broader one disappears: whenever A is 1, A + A·B is
1 regardless of B.
📐 Worked example
Simplify
Simplify F = A·B + A·B′ + A′·B
That's three AND terms and two ORs — about seven gate inputs. Apply one rule per step:
1Factor A from the first two terms (distributive):
$$ F = A\cdot(B + B') + A'\cdot B $$
2\(B+B' = 1\) (complement), and \(A\cdot 1 = A\) (identity):
$$ F = A + A'\cdot B $$
3Use \(A + A'\cdot B = A + B\) (distributive OR-form, then complement + identity):
$$ F = A + B $$
Payoff: seven gate inputs collapse to a single 2-input OR
— one gate instead of five. Smaller, cooler, faster.
🔭 Why bother in 2026, when a synthesis tool does this automatically? Because
you need this intuition to debug, to verify, and to understand what the
tool produced.
✏️ Try it yourself
Simplify each to its shortest form: (a) A·1 + 0 (b) A + A·B
(c) (A+B)·A (d) (A+1)·(B+0)
(a) A·1 = A, A+0 = A → A.
(b) absorption → A.
(c) absorption variant → A.
(d) A+1 = 1 (null), B+0 = B, 1·B → B.
Recap — the whole topic on one screen
| Group | Rules you own now |
|---|---|
| Postulates | identity · null · idempotent · complement · double-negation |
| Theorems | commutative · associative · distributive · absorption |
| Most useful for shrinking | identity, complement, absorption |
| Duality | swap AND↔OR and 0↔1 to get the partner rule |