3.1.2 Boolean algebra: laws, duality, De Morgan and algebraic simplification

0 · Recall Warm-up

Answer from memory first. Answers are in the Answer Vault at the bottom.

Which gate is 1 only when all inputs are 0? from Lesson 3.1.1
Truth-table rows for n variables? from Lesson 3.1.1
2's-complement negation rule? from Lesson 1.1.1

1 · The Hook

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A circuit for AB + AB' needs two AND gates, an inverter and an OR gate. But AB + AB' is just A, which needs no gates at all. Boolean algebra is the set of rules that spots such savings on paper. GATE's "minimisation by algebraic technique" is exactly this, and the laws are also behind K-maps and NAND/NOR conversions in later lessons.

2 · Predict First

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Is A + AB equal to A? Try all four values of A and B.

Make a guess in your head. The answer opens Section 3.

3 · Core Concept

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Predict First — the answer

Yes. If A = 1 both sides are 1; if A = 0 both sides are 0. B never matters. (This is the absorption law.)

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In plain words

Boolean algebra is ordinary algebra with only two numbers, 0 and 1, where + means OR and · means AND. Most rules look familiar (A + 0 = A, A · 1 = A), a few are new (A + A = A, A + 1 = 1, A + BC = (A + B)(A + C)). Any law can be checked with a truth table, so you never have to trust it blindly.

1. The laws

Law	OR form	AND form
Identity	A + 0 = A	A · 1 = A
Null (dominance)	A + 1 = 1	A · 0 = 0
Idempotent	A + A = A	A · A = A
Complement	A + A' = 1	A · A' = 0
Involution	(A')' = A
Commutative	A + B = B + A	AB = BA
Associative	(A + B) + C = A + (B + C)	(AB)C = A(BC)
Distributive	A + BC = (A + B)(A + C)	A(B + C) = AB + AC
Absorption	A + AB = A	A(A + B) = A
Redundant literal	A + A'B = A + B	A(A' + B) = AB
De Morgan	(A + B)' = A'B'	(AB)' = A' + B'
Consensus	AB + A'C + BC = AB + A'C	(A + B)(A' + C)(B + C) = (A + B)(A' + C)

The first distributive law, A + BC = (A + B)(A + C), has no counterpart in ordinary arithmetic. Check it: if A = 1 both sides are 1; if A = 0 both sides are BC.

2. Duality

The dual of an expression swaps + ↔ · and 0 ↔ 1, leaving every variable (and every ' ) as it is. Every law's dual is also a law: that is why the table has two columns. Dual of A + 0 = A is A · 1 = A.

3. Complementing a function (generalised De Morgan)

To get F', swap + ↔ · and complement every literal (keep brackets so the grouping is preserved). F = AB + C' → F' = (A' + B')C.

4. Simplifying with the laws (algebraic minimisation)

AB + AB' = A(B + B') = A · 1 = A.
AB + AB' + A'B = A + A'B = A + B (redundant literal).
(A + B)(A + C) = A + BC (distributive).
A'B'C + A'BC + AB'C + ABC = A'C(B' + B) + AC(B' + B) = A'C + AC = C.
AB + A'C + BC = AB + A'C (consensus: BC is covered by the other two).

5. XOR identities

A ⊕ B = A'B + AB'; A ⊕ 0 = A; A ⊕ 1 = A'; A ⊕ A = 0; A ⊕ A' = 1; XOR is commutative and associative. (A ⊕ B)' = A ⊕ B' = A' ⊕ B = A'B' + AB.

4 · Visual

A B	A + B	(A + B)'	A'	B'	A'B'
0 0	0	1	1	1	1
0 1	1	0	1	0	0
1 0	1	0	0	1	0
1 1	1	0	0	0	0

Proof of De Morgan's law by truth table: the two bold columns agree in every row.

5 · Worked Examples

Example 1 Easy

Simplify AB + AB'.

Solution

A(B + B') = A.

Example 2 Medium

Simplify (A + B)(A + C).

Solution

Distributive law: A + BC.

Example 3 GATE

Find the complement of F = AB + C'.

Solution

Swap the operations and complement the literals: F' = (A' + B')C.

6 · GATE Traps

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The dual does not complement variables; the complement does.
A + BC = (A + B)(A + C) is valid in Boolean algebra (not in arithmetic).
A + A'B simplifies to A + B, not to B.
When complementing, keep the original grouping with brackets: AB + C' → (A' + B')·C.

7 · Memory Hook

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"Absorb, distribute, De Morgan, consensus. Dual swaps the operators; complement also flips the letters."

8 · Connect the Dots

Chapter 3.1 is complete. Algebra works, but it needs insight to know which law to apply. Chapter 3.2 writes every function in a standard form (a list of the rows where it is 1), the starting point for the systematic methods of Phase 4.

9 · Self-Test

GATE format. Try every question before opening the vault.

Q1MCQ1 markAbsorption

A + AB simplifies to:

(A) AB
(B) A + B
(C) B
(D) A

Q2MCQ1 markRedundant literal

AB + AB' + A'B simplifies to:

(A) AB
(B) A + B
(C) 1
(D) A

Q3MCQ2 marksComplement

The complement of F = A(B + C') is:

(A) A'(B'C)
(B) A + B'C
(C) A' + B'C
(D) A' + B' + C

Q4MCQ1 markDuality

The dual of A + A'B = A + B is:

(A) A(A' + B) = AB
(B) A'(A + B') = A'B'
(C) A(A' + B) = A + B
(D) A' + AB' = A' + B'

Q5NAT2 marksSimplification

A'B'C + A'BC + AB'C + ABC simplifies to a single literal. How many of the 8 rows of (A, B, C) make it 1?

Numerical answer: ________

10 · Rule Card

Identity, null, idempotent, complement, involution, commutative, associative.
Distributive both ways: A(B + C) = AB + AC and A + BC = (A + B)(A + C).
Absorption A + AB = A; redundant literal A + A'B = A + B; consensus AB + A'C + BC = AB + A'C.
De Morgan (A + B)' = A'B', (AB)' = A' + B'. Complement: swap operators and complement literals.
Dual: swap + ↔ · and 0 ↔ 1 only. XOR: A ⊕ B = A'B + AB'.

11 · Concept Ledger

Unlocked in this lesson. From here on, these terms may be used.

Boolean algebra lawsidentity and null lawsidempotent lawcomplement lawdistributive lawsabsorptionredundant literal ruleDe Morgan's lawsconsensus theoremdualitycomplement of a functionalgebraic simplificationXOR identities

12 · Answer Vault

Answer Vault

Try the questions first. Open when you are ready to check.

Recall Warm-up

NOR.
2ⁿ.
Invert all bits and add 1.