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Simplifying Boolean Expressions

Knowing the Boolean laws is useful, but learning to work a problem is a different skill. On this page we will repeatedly take an expression apart, identify a useful law, make one legal change, and continue until nothing simpler remains.

Our preferred NOT notation: ~A means NOT A. You may also see ¬A or an overbar such as A. They mean the same thing.

A short working toolbox

Do one recognizable thing at a time. When learning, do not compress three or four algebraic moves into one line. Writing the intermediate steps makes mistakes much easier to find.

Worked example 1 — identity

A · 1 + 0
Step 1: A · 1 = A.
A + 0
Step 2: A + 0 = A.
A

Worked example 2 — absorption

A + A·B
Step 1: Recognize the absorption form A + A·B.
A

There is no need to know whether B is 0 or 1. If A is 1, the whole OR expression is already 1. If A is 0, the product A·B is also 0.

Worked example 3 — factor, complement, identity

A·B + A·~B
Step 1: Factor out A.
A·(B + ~B)
Step 2: A variable OR its complement is 1.
A·1
Step 3: AND with 1 leaves A unchanged.
A

Worked example 4 — distribute, then eliminate a complement

(A+B)(A+~B)
Step 1: Use the Boolean distributive identity (A+B)(A+C) = A + B·C.

Here C is ~B, so:

A + B·~B
Step 2: B·~B = 0.
A + 0
Step 3: A+0=A.
A

Worked example 5 — a useful absorption variation

A + ~A·B
Step 1: Use X + YZ = (X+Y)(X+Z).
(A+~A)(A+B)
Step 2: A+~A=1.
1·(A+B)
Step 3: AND with 1 changes nothing.
A+B

Worked example 6 — De Morgan, then absorption

~(A+B) + A
Step 1: Apply De Morgan.
~A·~B + A
Step 2: Reorder the OR terms.
A + ~A·~B
Step 3: Use the same form as the previous example: A + ~A·X = A+X.
A + ~B

Worked example 7 — two parenthesized terms

(A+B)(~A+B)
Step 1: Reorder each sum so the common term B is easy to see.
(B+A)(B+~A)
Step 2: Use (X+Y)(X+Z)=X+YZ.
B + A·~A
Step 3: A·~A=0.
B + 0
Step 4: B+0=B.
B

Worked example 8 — three variables

A·B + ~A·B
Step 1: Factor out B.
B(A+~A)
Step 2: A+~A=1.
B·1
Step 3: B·1=B.
B

Worked example 9 — a longer expression

A·B + A·~B + ~A·B
Step 1: Combine the first two terms by factoring A.
A(B+~B) + ~A·B
Step 2: B+~B=1.
A + ~A·B
Step 3: From Worked Example 5, A + ~A·B = A+B.
A+B

Worked example 10 — verify the result with a truth table

Algebra gives us a shorter expression, but a truth table can independently check that we did not change the function.

Check:

A·B + A·~B = A
AB~BA·BA·~BOriginalA
0010000
0100000
1010111
1101011

The last two columns match in every row, so the simplification is confirmed.

Two useful ways to prove equivalence:
1. Simplify one expression into the other by Boolean algebra.
2. Build a truth table and show that the final output columns match on every row.

Now try some yourself

Work each problem on paper before opening its solution. Try to write the law you used beside each step.

Problem 1

A · 1
Show solution
A · 1 = A

Identity law.

Problem 2

A + 0
Show solution
A + 0 = A

Identity law.

Problem 3

A + A
Show solution
A + A = A

Idempotent law.

Problem 4

A · ~A
Show solution
A · ~A = 0

Complement law.

Problem 5

A + A·B
Show solution
A

Absorption: A + A·B = A.

Problem 6

A·B + A·~B
Show solution
A(B+~B) = A·1 = A

Factor A, use complement, then identity.

Problem 7

(A+B)(A+~B)
Show solution
A + B·~B = A+0 = A

Problem 8

A + ~A·B
Show solution
(A+~A)(A+B) = 1(A+B) = A+B

Problem 9

~(A+B)
Show solution
~A·~B

De Morgan's law.

Problem 10

~(A·B)
Show solution
~A + ~B

De Morgan's law.

Problem 11

(A+B)(~A+B)
Show solution
B + A·~A = B+0 = B

Problem 12

A·B + ~A·B
Show solution
B(A+~A) = B·1 = B

Problem 13

A·B + A·~B + ~A·B
Show solution
A(B+~B)+~A·B
A+~A·B
A+B

Problem 14

A(A+B+C)
Show solution
A

This is an extended absorption form: if A is 0 the product is 0; if A is 1, the parenthesis already contains A and is therefore 1.

Problem 15

A + ~A·B + ~A·~B
Show solution
A + ~A(B+~B)
A + ~A
1

Problem 16 — prove by truth table

Show that these two expressions are equivalent:

A + ~A·B
A+B
Show solution

Evaluate both expressions for 00, 01, 10, and 11. Both produce the pattern 0, 1, 1, 1, so they are equivalent.

What matters most

The goal is not merely to memorize a list of laws. It is to learn to recognize forms: a common factor, a variable beside its complement, an absorption pattern, or a NOT over a parenthesized expression. With practice, long Boolean expressions begin to look like a series of small familiar moves.

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