Rules of Boolean Algebra

AND Operations (·)

0·0 = 0 A·0 = 0

1·0 = 0 A·1 = A

0·1 = 0 A·A = A

1·1 = 1 A·A' = 0

OR Operations (+)

0+0 = 0 A+0 = A

1+0 = 1 A+1 = 1

0+1 = 1 A+A = A

1+1 = 1 A+A' = 1

NOT Operations (')

0' = 1 A'' = A

1' = 0

•Associative Law

–(A·B)·C = A·(B·C) = A·B·C

–(A+B)+C = A+(B+C) = A+B+C

•Distributive Law

–A·(B+C) = (A·B) + (A·C)

–A+(B·C) = (A+B) · (A+C)

•Commutative Law

–A·B = B·A

–A+B = B+A

•Precedence

–AB = A·B

–A·B+C = (A·B) + C

– A+B·C = A + (B·C)

•De Morgan's Theorem

–(A·B)' = A' + B' (NAND)

–(A+B)' = A' · B' (NOR)

•Redundance Law

–A + AB = A

–A (A + B) = A

–0 + A = A
–0 A = 0

–1 + A = 1
–1 A = A

A' A=0–A+A=1
–

A+A'B=A+B–

A(A'+B)=AB–

• Identity Law

–A + A = A
–A A = A

– AB+AB'=A

–(A+B)(A+B')=A

How to use Boolean algebra to simplify a sum of products expression

1. Try to find variables that can be factored out in such a way as to leave

terms that will simplify to zero or one. For example AB’ + AB =

A(B’+B) = A(1)=A.

2. If it helps you recognize patterns, rearrange the order of the terms in

the expression to group terms that have common variables together.

3. Keep the second distributive law in mind. It is used a great deal and

situations where it would help are often missed by people new to

Boolean expression simplification.

                                    A+BC = (A+B)(A+C)

The derivation of A+AB

A+AB     = A(1+B)

                = A(1)

                = A

The derivation of A+A’B

A+A’B    = (A+AB)+A’B

                = (AA+AB)+A’B

                = AA+AB+AA’+A’B

                = AA+AB+A’B

                = (A+A’)(A+B)

    = 1× (A+B)

                = (A+B)

The derivation of (A+B)(A+C)

(A+B)(A+C) = A+AC+AB+BC

                      = A(1+C)+AB+BC

                      = A×1+AB+BC

                      = A×1+ BC

                      = A+BC