OCR H446 · A Level Computer Science · ~15 min read
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Boolean Algebra Laws
Boolean algebra provides rules for manipulating and simplifying logical expressions. These laws allow you to reduce complex circuits to simpler equivalents with fewer gates — saving cost, power, and space.
Identity Laws
A AND 1 = AA · 1 = A
A AND 0 = 0A · 0 = 0
A OR 0 = AA + 0 = A
A OR 1 = 1A + 1 = 1
Idempotent Laws
A AND A = AA · A = A
A OR A = AA + A = A
Complement Laws
A AND NOT A = 0A · ¬A = 0
A OR NOT A = 1A + ¬A = 1
NOT (NOT A) = A¬(¬A) = A (Double Negation)
Commutative Laws
A AND B = B AND AA · B = B · A
A OR B = B OR AA + B = B + A
Associative Laws
(A AND B) AND C = A AND (B AND C)(A · B) · C = A · (B · C)
(A OR B) OR C = A OR (B OR C)(A + B) + C = A + (B + C)
Distributive Laws
A AND (B OR C) = (A AND B) OR (A AND C)A · (B + C) = A·B + A·C
A OR (B AND C) = (A OR B) AND (A OR C)A + (B · C) = (A+B) · (A+C)
Absorption Laws
A AND (A OR B) = AA · (A + B) = A
A OR (A AND B) = AA + (A · B) = A
De Morgan's Laws
De Morgan's Laws are critical for simplification and for converting between different gate types. They come in two forms:
De Morgan's First Law
NOT (A AND B) = NOT A OR NOT B¬(A · B) = ¬A + ¬B
In words: "NOT of (A AND B)" = "(NOT A) OR (NOT B)". Break the bracket, flip AND to OR.
De Morgan's Second Law
NOT (A OR B) = NOT A AND NOT B¬(A + B) = ¬A · ¬B
In words: "NOT of (A OR B)" = "(NOT A) AND (NOT B)". Break the bracket, flip OR to AND.
Memory aid: "Break the line, change the sign" — when you move the NOT inside the brackets, AND becomes OR and OR becomes AND.
Verification by Truth Table
Verify De Morgan's First Law: ¬(A · B) = ¬A + ¬B
A
B
A·B
¬(A·B)
¬A
¬B
¬A + ¬B
Equal?
0
0
0
1
1
1
1
✓
0
1
0
1
1
0
1
✓
1
0
0
1
0
1
1
✓
1
1
1
0
0
0
0
✓
Both columns match in every row — the law is verified.
Simplification Example
Simplify: X = A · B + A · ¬B + ¬A · B
Step 1: Factor using distributive law: A · B + A · ¬B = A · (B + ¬B) = A · 1 = A (complement law)
Step 2: Remaining: A + ¬A · B
Step 3: Distributive (OR): A + ¬A · B = (A + ¬A) · (A + B) = 1 · (A + B) = A + B
So X = A · B + A · ¬B + ¬A · B simplifies to X = A + B
Exam tip: When asked to simplify a Boolean expression, work step by step, citing the law used at each step. Common exam sequence: spot the distributive law to factor, then apply complement/identity laws to reduce. Always verify by checking a row of the truth table after simplification.
Exam tip: De Morgan's laws are essential for converting NAND/NOR to equivalent AND/OR/NOT circuits. ¬(A·B) is a NAND gate — De Morgan says this equals ¬A + ¬B. Examiners often ask you to apply De Morgan's law once or twice in a simplification chain.
⚠ Common Mistakes
Applying De Morgan's incorrectly: ¬(A + B) ≠ ¬A + ¬B. The sign MUST flip: ¬(A + B) = ¬A · ¬B (AND, not OR).
Forgetting to distribute the NOT to BOTH variables: ¬(A · B) ≠ ¬A · B — the NOT must apply to both A and B.
Double negation forgotten: ¬(¬A) = A. When NOT is applied twice, it cancels out.
Using ordinary algebra distributive law incorrectly: A + B·C ≠ (A+B)·(A+C) in normal algebra, but it IS valid in Boolean algebra (this trips students up).
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Worksheet — 1.4.3b Boolean Algebra and De Morgan's Laws
8 questions · 22 marks · instantly marked
Q1Simplify the following using Boolean laws, stating the law used at each step: X = A · A + A · 0[3 marks]
✓ Mark scheme
A · A = A (Idempotent law) [1]. A · 0 = 0 (Identity law) [1]. X = A + 0 = A (Identity law: A + 0 = A) [1]. Final simplified expression: X = A.
Q2Apply De Morgan's First Law to simplify: X = ¬(A · B). Write the equivalent expression.[2 marks]
✓ Mark scheme
De Morgan's First Law: ¬(A · B) = ¬A + ¬B [1]. In words: NOT of (A AND B) equals (NOT A) OR (NOT B). The AND becomes OR, and NOT distributes to each individual variable [1].
Q3Apply De Morgan's Second Law to: X = ¬(A + B + C). Write the simplified expression.[2 marks]
✓ Mark scheme
De Morgan's Second Law (extended): ¬(A + B + C) = ¬A · ¬B · ¬C [1]. NOT distributes inside the brackets, and each OR becomes AND [1]. De Morgan applies to any number of variables: NOT of an OR becomes AND of NOTs.
Q4Simplify: X = A · (A + B). State which Boolean law applies.[2 marks]
✓ Mark scheme
X = A · (A + B) = A. Absorption law (or show working: distribute → A·A + A·B = A + A·B → A(1+B) = A·1 = A) [1]. The absorption law states A · (A + B) = A — the A "absorbs" the B [1].
Q5Verify De Morgan's Second Law ¬(A + B) = ¬A · ¬B by completing a truth table. Show all intermediate columns.[4 marks]
✓ Mark scheme
Truth table with columns: A, B, A+B, ¬(A+B), ¬A, ¬B, ¬A·¬B [1 for correct columns]. Rows: (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 ✓ [2 for all 4 rows correct]. Conclusion: ¬(A+B) and ¬A·¬B columns are identical for all inputs → law verified [1].
Q6Simplify step by step: X = ¬(¬A · ¬B). Show all steps with laws cited.[4 marks]
✓ Mark scheme
Step 1: Apply De Morgan's First Law: ¬(¬A · ¬B) = ¬(¬A) + ¬(¬B) [1]. Step 2: Apply Double Negation (¬¬A = A): = A + B [1]. So X = A + B — a simple OR gate [1]. Verification: this is the NOR-of-NOTs, which equals OR. Connecting with universal gate theory: ¬(¬A · ¬B) is how you build an OR gate from NAND gates (De Morgan's reveals the equivalence) [1].
Q7Simplify: X = A · B + ¬A · B. Show your working and state the law used.[3 marks]
✓ Mark scheme
Step 1: Factor out B using distributive law: A·B + ¬A·B = B·(A + ¬A) [1]. Step 2: A + ¬A = 1 (Complement law) [1]. Step 3: B · 1 = B (Identity law) [1]. Final answer: X = B.
Q8Express ¬(A + B) in an alternative form using De Morgan's Law, then state which type of gate this expression represents.[3 marks]
✓ Mark scheme
De Morgan's Second Law: ¬(A + B) = ¬A · ¬B [1]. The original expression ¬(A + B) — NOT of an OR — represents a NOR gate [1]. The equivalent expression ¬A · ¬B — AND of two NOTs — shows that a NOR gate is equivalent to inverting both inputs then AND-ing them, or alternatively: (NOT A) AND (NOT B) [1].
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Mini Test — 1.4.3b Boolean Algebra
10 questions · 10 marks · 10 minutes
5 MCQ + 5 short answer
⏱10:00
10 marks
Section A — Multiple Choice
Q1What does De Morgan's First Law state?
Q2Simplify A · 0.
Q3Simplify A + NOT A.
Q4Which law states: A · (A + B) = A?
Q5Simplify NOT(NOT A).
Section B — Short Answer
Q6Apply De Morgan's Second Law to: ¬(P + Q). Write the equivalent expression.
Mark scheme¬(P + Q) = ¬P · ¬Q. De Morgan's Second Law: NOT of an OR becomes AND of NOTs. Break the bracket, change OR to AND, distribute NOT to each variable. [1 mark]
Q7Simplify: A · B + A · B (using an idempotent law).
Mark schemeA·B + A·B = A·B (Idempotent law: X + X = X, where X = A·B). Anything OR'd with itself equals itself. Result: A · B. [1 mark]
Q8State the simplified form of: A · NOT A.
Mark schemeA · NOT A = 0. Complement law: A AND NOT A is always 0 (contradiction — a variable cannot be both true and false simultaneously). This is used to eliminate contradictory terms in simplification. [1 mark]
Q9What is the memory aid for De Morgan's Laws?
Mark scheme"Break the line, change the sign" — when you move NOT inside the brackets (breaking the overline in bar notation), AND becomes OR and OR becomes AND. All variables also become individually negated. [1 mark]
Q10Simplify: X = A · 1 + B · 0.
Mark schemeA·1 = A (identity law). B·0 = 0 (identity law). X = A + 0 = A (identity: A+0=A). Final answer: X = A. [1 mark]