The simplified form of the Boolean expression (AB′+AB′C+AC)(A′C′+B′) is:

Concept

Boolean algebra uses absorption X+XY=X, complement XX′=0, idempotence XX=X, and the identity X+1=1.

To simplify a product of sums, first absorb redundant terms, then distribute, reduce every product, and independently check equivalence.

Application

1. Let F=(AB′+AB′C+AC)(A′C′+B′).

2. Absorb AB′C into AB′: AB′+AB′C=AB′(1+C)=AB′. Therefore F=(AB′+AC)(A′C′+B′).

3. Distribute the two factors: F=AB′A′C′+AB′B′+ACA′C′+ACB′.

4. Reduce each product: AB′A′C′=0 because AA′=0; AB′B′=AB′ because B′B′=B′; ACA′C′=0 because AA′=0; and ACB′=AB′C.

5. Thus F=AB′+AB′C=AB′(1+C)=AB′ by absorption.

Cross-check

• If A=0, the first factor is 0, so F=0; AB′ is also 0.

• If A=1, the factors reduce to (B′+C) and B′, so F=(B′+C)B′=B′+CB′=B′; AB′ also reduces to B′.

Result

The simplified expression is AB′.