AQA 7517 · Paper 2 · 4.6.2

Logic Gates &
De Morgan's Laws

Boolean algebra · Truth tables · Circuit simplification · Section 4.6

WHAT YOU'LL LEARN

AND, OR, NOT, NAND, NOR, XOR · Truth tables · Boolean expressions · De Morgan's Laws

AQA SPEC LINK

4.6.2 — Logic gates, Boolean algebra, De Morgan's laws

Basic Gates

AND, OR, NOT Gates

AND Gate

A	B	A AND B
0	0	0
0	1	0
1	0	0
1	1	1

OR Gate

A	B	A OR B
0	0	0
0	1	1
1	0	1
1	1	1

NOT Gate

A	NOT A
0	1
1	0

More Gates

NAND, NOR, XOR Gates

NAND (NOT AND)

A	B	NAND
0	0	1
0	1	1
1	0	1
1	1	0

NOR (NOT OR)

A	B	NOR
0	0	1
0	1	0
1	0	0
1	1	0

XOR (Exclusive OR)

A	B	XOR
0	0	0
0	1	1
1	0	1
1	1	0

XOR = 1 when inputs are DIFFERENT. NAND and NOR are "universal gates" — can implement all other gates.

Boolean Expressions

Boolean Algebra Notation

AND: A · B or AB

OR: A + B

NOT: Ā (A-bar) or ¬A

XOR: A ⊕ B

Example expression: X = A · B + ¬C means X = (A AND B) OR (NOT C)

De Morgan's Laws

LAW 1 — NAND to NOR equivalent

¬(A · B) = ¬A + ¬B

NOT(A AND B) = NOT-A OR NOT-B

LAW 2 — NOR to NAND equivalent

¬(A + B) = ¬A · ¬B

NOT(A OR B) = NOT-A AND NOT-B

De Morgan's Example

Applying De Morgan's Laws

Simplify: ¬(A · ¬B)
Apply Law 1: = ¬A + ¬(¬B)
Double negation: ¬(¬B) = B
Result: ¬A + B

De Morgan's laws are used to simplify logic circuits, replace NAND/NOR with AND/OR/NOT, and simplify software conditionals. Essential for exam questions on Boolean simplification.

Boolean Laws

Key Boolean Algebra Laws

A · 0 = 0

AND with 0 always 0

A · 1 = A

AND with 1 is identity

A + 0 = A

OR with 0 is identity

A + 1 = 1

OR with 1 always 1

A · A = A

Idempotent law

¬(¬A) = A

Double negation

Circuit to Expression

Reading a Logic Circuit

Given inputs A, B, C:
Gate 1: AND → A · B
Gate 2: NOT → ¬C
Gate 3: OR takes Gate 1 and Gate 2 → (A · B) + ¬C

Build truth table: for each combination of A, B, C, evaluate the expression. 3 inputs = 2³ = 8 rows in truth table.

AQA Exam Style

Practice Question

AQA 7517 — Paper 2 Style

(a) Complete a truth table for the expression X = (A AND B) OR (NOT C) for all 8 combinations. [4]
(b) Using De Morgan's Law, simplify the expression: ¬(A + B). [1]
(c) Simplify ¬(¬A · B) using De Morgan's Law and Boolean laws. [2]

[7 marks]

4 marks

(a) 8 rows; A=0,B=0,C=0→1; A=0,B=0,C=1→0; A=0,B=1,C=0→1; A=0,B=1,C=1→0; A=1,B=0,C=0→1; A=1,B=0,C=1→0; A=1,B=1,C=0→1; A=1,B=1,C=1→1 [1 per correct half]

1 mark

(b) ¬A · ¬B (NOT-A AND NOT-B)

2 marks

(c) Apply Law 1: ¬(¬A) + ¬B [1] = A + ¬B [1] (double negation)

Summary

Key Points to Remember

AND: output 1 only if both inputs 1 · OR: output 1 if any input 1 · NOT: inverts

NAND = NOT-AND · NOR = NOT-OR · XOR: output 1 when inputs differ

De Morgan 1: ¬(A · B) = ¬A + ¬B

De Morgan 2: ¬(A + B) = ¬A · ¬B

NAND and NOR are universal gates — can build ANY logic function from just NAND or just NOR gates

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