Question

Using Boolean algebra, simplify the following Boolean functions.

a. ABC + A’BC + (AB’C)’

b. AB’ + A’B + AB + A’B’

c. A’(A+B) + (B+A)(A+B’)

Create a truth table, Karnaugh map and show the simplified the expression.

a. A’B + B

Answer 1

a. ABC + A’BC + (AB’C)’
  = (A+A')BC + (AB’C)’
  = (1)BC + (AB’C)’
  = BC + (AB’C)’
  = BC + (A'+B+C')
  = B(C+1) + A'+C'
  = B(1) + A'+C'
  = B + A' + C'

b. AB’ + A’B + AB + A’B’
  = A(B’+B) + A’B + A’B’
  = A(1) + A’B + A’B’
  = A + A’B + A’B’
  = A + A’(B + B’)
  = A + A’(1)
  = A + A’
  = 1

c. A’(A+B) + (B+A)(A+B’)
  = A’A+A'B + (B+A)(A+B’)
  = 0+A'B + (B+A)(A+B’)
  = A'B + (B+A)(A+B’)
  = A'B + (BA+BB'+AA+AB’)
  = A'B + (BA+0+AA+AB’)
  = A'B + BA+AA+AB’
  = A'B + BA+A+AB’
  = A'B + BA + A(1+B’)
  = A'B + BA + A(1)
  = A'B + BA + A
  = A'B + (B+1)A
  = A'B + (1)A
  = A'B + A

a. A’B + B

A	B	A’B + B
F	F	F
F	T	T
T	F	F
T	T	T

So, From the truth table we can say that A’B + B = B