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Simplifying Boolean Expressions

Use the laws step by step, naming each one, then prove the result with a truth table.

A Method That Works

1 Look for a common factor you can take out (distributive law). 2 Look for pairs like B + B (= 1) or B \cdot B (= 0) — complement law. 3 Remove anything multiplied by 1 or added to 0 (identity law), and use the null law (A + 1 = 1). 4 Watch for absorption : A + A \cdot B = A . 5 Write the law you used on every line. 6 Check your final answer with a truth table.

Worked example

Simplify X = A·B + A·B̄

Step Expression Law used 1 A \cdot B + A \cdot B Start 2 A \cdot (B + B) Distributive (take out A) 3 A \cdot 1 Complement: B + B = 1 4 A Identity: A \cdot 1 = A Answer: X = A — two AND gates, a NOT and an OR are replaced by a single wire .

Worked example

Simplify X = A̅·B + A·B + A·B̄

Step Expression Law used 1 A \cdot B + A \cdot B + A \cdot B Start 2 A \cdot B + A \cdot B + A \cdot B + A \cdot B Idempotent: A \cdot B = A \cdot B + A \cdot B 3 B \cdot (A + A) + A \cdot (B + B) Distributive (twice) 4 B \cdot 1 + A \cdot 1 Complement (twice) 5 A + B Identity, then commutative Answer: X = A + B . The trick in step 2 — writing a term twice — is allowed because P = P + P .

Always prove your simplification. Build a truth table with a column for the original expression and a column for your simplified one. They must match in every row.

X = 0

Try it:

Six gates — yet the output column is exactly A OR B. One OR gate would do the same job.

The unsimplified circuit for X = A̅\cdotB + A\cdotB + A\cdotB̄.

Practice

Simplify X = A + A\cdotB\cdotC.

A Method That Works

1 Look for a common factor you can take out (distributive law). 2 Look for pairs like B + B (= 1) or B \cdot B (= 0) — complement law. 3 Remove anything multiplied by 1 or added to 0 (identity law), and use the null law (A + 1 = 1). 4 Watch for absorption : A + A \cdot B = A . 5 Write the law you used on every line. 6 Check your final answer with a truth table.

Worked example

Simplify X = A·B + A·B̄

Step Expression Law used 1 A \cdot B + A \cdot B Start 2 A \cdot (B + B) Distributive (take out A) 3 A \cdot 1 Complement: B + B = 1 4 A Identity: A \cdot 1 = A Answer: X = A — two AND gates, a NOT and an OR are replaced by a single wire .

Worked example

Simplify X = A̅·B + A·B + A·B̄

Step Expression Law used 1 A \cdot B + A \cdot B + A \cdot B Start 2 A \cdot B + A \cdot B + A \cdot B + A \cdot B Idempotent: A \cdot B = A \cdot B + A \cdot B 3 B \cdot (A + A) + A \cdot (B + B) Distributive (twice) 4 B \cdot 1 + A \cdot 1 Complement (twice) 5 A + B Identity, then commutative Answer: X = A + B . The trick in step 2 — writing a term twice — is allowed because P = P + P .

Checking with a Truth Table

Always prove your simplification. Build a truth table with a column for the original expression and a column for your simplified one. They must match in every row.

X = 0

Try it:

Six gates — yet the output column is exactly A OR B. One OR gate would do the same job.

The unsimplified circuit for X = A̅\cdotB + A\cdotB + A\cdotB̄.

Practice

Simplify X = A + A\cdotB\cdotC.

Step	Expression	Law used
1	$(A + B) \cdot (A + B)$	Start
2	$A + B \cdot B$	Distributive (OR form): $(A + B) (A + C) = A + B \cdot C$
3	$A + 0$	Complement: $B \cdot B = 0$
4	$A$	Identity

A	B	$A \cdot B$	$A \cdot B$	$A \cdot B$	Original	$A + B$
0	0	0	0	0	0	0
0	1	1	0	0	1	1
1	0	0	0	1	1	1
1	1	0	1	0	1	1

Step	Expression	Law used
1	$(A + B) \cdot (A + B)$	Start
2	$A + B \cdot B$	Distributive (OR form): $(A + B) (A + C) = A + B \cdot C$
3	$A + 0$	Complement: $B \cdot B = 0$
4	$A$	Identity

Step	Expression	Law used
1	$A \cdot B + A \cdot B + A \cdot B$	Start
2	$A \cdot B + A \cdot B + A \cdot B + A \cdot B$	Idempotent: $A \cdot B = A \cdot B + A \cdot B$
3	$B \cdot (A + A) + A \cdot (B + B)$	Distributive (twice)
4	$B \cdot 1 + A \cdot 1$	Complement (twice)
5	$A + B$	Identity, then commutative