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Cambridge IGCSE 0478 · Topic 10 · 10.2

Logic Circuits &
Simplification

Circuit Diagrams · Writing Expressions from Circuits · Simplification Rules

CSZoneCambridge IGCSE Computer Science 0478
Logic Gate Symbols

Drawing Logic Circuits

AND gate: D-shape; flat back on input side. Output is 1 only when A=1 AND B=1.
OR gate: curved shield shape. Output is 1 when A=1 OR B=1.
NOT gate: triangle with a circle (bubble). Single input. Inverts the signal.
NAND: AND gate + bubble at output. NOR: OR gate + bubble at output.
XOR: curved OR gate with extra curved line. Output 1 only when inputs DIFFER.
Cambridge 0478 uses standard IEC/IEEE gate symbols — you must recognise and draw them correctly in the exam
Reading a Logic Circuit

Expression from Circuit → Truth Table

// Example circuit: A, B are inputs
// NOT A feeds into an AND gate with B
// AND output feeds into OR gate with C

// Step 1: write intermediate expressions
Wire 1 = NOT A
Wire 2 = (NOT A) AND B

// Step 2: write final expression
X = ((NOT A) AND B) OR C

// Step 3: complete truth table
// (list all combos of A, B, C = 8 rows)
// work left to right through the expression
Work through the circuit left to right — label each wire with its expression, then combine to get the output
Boolean Simplification Rules

De Morgan's Laws & Key Identities

De Morgan's Theorem:
NOT(A AND B) = NOT A OR NOT B
NOT(A OR B) = NOT A AND NOT B

Identity laws: A AND 1 = A  ·  A OR 0 = A
Annihilation: A AND 0 = 0  ·  A OR 1 = 1
Idempotent: A AND A = A  ·  A OR A = A
Complement: A AND NOT A = 0  ·  A OR NOT A = 1
De Morgan's laws are the most commonly tested — know both forms and how to apply them to simplify expressions
Exam Practice

Have a go at this question

Cambridge IGCSE 0478 style
A logic circuit has inputs A and B. The output X is: X = NOT(A OR B). (a) Complete a truth table for this circuit. (b) Using De Morgan's Law, write an equivalent expression for X without using NOT directly on the bracket.
5 marks
(a) Truth table: A=0,B=0→X=1; A=0,B=1→X=0; A=1,B=0→X=0; A=1,B=1→X=0 [2]. (b) By De Morgan's Law: NOT(A OR B) = NOT A AND NOT B [1], so X = NOT A AND NOT B [1]. This is equivalent to a NOR gate [1].
Key Takeaways

What to Remember

AND = D-shape; OR = curved shield; NOT = triangle + bubble; NAND/NOR = AND/OR + output bubble
Read circuit left to right; label each wire; combine to form final Boolean expression
De Morgan's: NOT(A AND B) = NOT A OR NOT B; NOT(A OR B) = NOT A AND NOT B
Simplification reduces circuit complexity — fewer gates needed saves cost and power