Boolean algebra lets you rewrite a logic expression into a simpler one that gives exactly the same truth table, so the circuit needs fewer gates. This lesson sets out the laws, De Morgan's laws in particular, and works through simplifications step by step.
In Boolean algebra A.B means A AND B, A + B means A OR B and a bar over a term means NOT, so A is NOT A. The laws of Boolean algebra (identity, null, idempotent, inverse, commutative, associative, distributive, absorption) let an expression be rewritten without changing its truth table. De Morgan's laws say A.B = A + B and A + B = A.B: break the bar and change the operator. Simplifying an expression means fewer gates, so a circuit that is cheaper, smaller, faster and uses less power. A simplification can always be checked by comparing the truth tables of the original and the result.
Fig. 1: a NAND gate does the same job as an OR gate with both inputs inverted, and a NOR gate the same as an AND gate with both inputs inverted
Section 01
The laws of Boolean algebra.
Each law below is true for every combination of inputs, which you can check with a truth table. Most look like ordinary algebra; the ones marked absorption and redundancy do not, and they do most of the work in exam simplifications.
Law
Rule
Identity
A.1 = A
A + 0 = A
Null (annulment)
A.0 = 0
A + 1 = 1
Idempotent
A.A = A
A + A = A
Inverse (complement)
A.A = 0
A + A = 1
Double complement
A = A
Commutative
A.B = B.A
A + B = B + A
Associative
(A.B).C = A.(B.C)
(A + B) + C = A + (B + C)
Distributive
A.(B + C) = A.B + A.C
A + B.C = (A + B).(A + C)
Absorption
A + A.B = A
A.(A + B) = A
Redundancy
A + A.B = A + B
A.(A + B) = A.B
De Morgan
A.B = A + B
A + B = A.B
Your turn
Use a truth table to show that A + A.B = A.
A = 0, B = 0: 0 + 0 = 0. A = 0, B = 1: 0 + 0 = 0. A = 1, B = 0: 1 + 0 = 1. A = 1, B = 1: 1 + 1 = 1. The output column 0, 0, 1, 1 is the same as the A column, so A + A.B = A.
Examiner trap
In Boolean algebra A + A = A, not 2A, and A.A = A, not A squared. There are only two values, 0 and 1.
Section 02
De Morgan's laws.
De Morgan's laws turn a NAND into an OR of inverted inputs and a NOR into an AND of inverted inputs. They work for any number of inputs, which matters because circuits may include gates with three or more inputs.
Law 1:A.B = A + B. The bar over an AND breaks into separate bars, and the AND becomes OR.
Law 2:A + B = A.B. The bar over an OR breaks into separate bars, and the OR becomes AND.
Three inputs:A.B.C = A + B + C and A + B + C = A.B.C.
Going the other way: any OR can be written with an AND and bars, A + B = A.B, which is how a circuit can be built from NAND gates only.
Worked example De Morgan, then absorb
Simplify X = A.B + A + C.
X = A.B + A + C
= (A + B) + A.CDe Morgan on each bracket
= A + B + A.Cdouble complement: the double bar on A cancels
= A + A.C + Bcommutative: reorder
= A + C + Bredundancy: A + NOT A.Y = A + Y
= A + B + Creorder the answer
Answer: X = A + B + C
Your turn
Use De Morgan's laws to show that A + B = A.B.
A + B = A.B (De Morgan, law 2) = A.B (double complement).
Examiner trap
The commonest error is to break the bar but forget to change the operator. A + B is notA + B: try A = 1, B = 0, which gives 0 on the left and 1 on the right.
Stage 1 · Learn
Check what the sim just showed you
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which expression is equivalent to A.B?
Quick check+10 XP
What does A + A.B simplify to?
Quick check+10 XP
What is A.A?
Quick check+10 XP
Why is it worth simplifying a logic circuit?
Section 03
Simplifying step by step.
Expand brackets, collect terms, then look for the patterns X + X = X, X + X = 1, X + X.Y = X and X + X.Y = X + Y. Write one law per line so each step earns credit. Check your answer in the lab: build the original and the simplified expression and compare their truth tables.
X = B.(A + A) + A.B = B + A.B (inverse, identity) = A + B (redundancy: B + NOT B.A = B + A). The circuit needs one OR gate instead of three AND gates, two NOT gates and an OR gate.
Examiner trap
An answer with no working rarely earns full marks. Show each law you use, and do not combine several laws in one line.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 3 and Paper 4 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Which is the simplest form of A.B + A.B + A.B?
Exam style+20 XP
Which expression is equivalent to A + B?
Exam style+20 XP
Which law is used in the step A.(B + C) = A.B + A.C?
Exam style+20 XP
What is the simplest form of (A + B).(A + B)?
Skill unlocked
Boolean algebra, mastered.
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