Computer Science
Reduce the following Boolean expression into their simplest forms:
- {(CD)' + A} + A + C.D + A.B
- A.{B + C (A.B + A.C)'}
Answer
{(CD)' + A} + A + C.D + A.B
{(CD)' + A} + A + C.D + A.B
= C' + D' + A + A + C.D + A.B [De-Morgan's Law]
= C' + D' + A + C.D + A.B [∵ A+A=A]
= A(1 + B) + C' + D' + C.D
= A + C' + D' + C.D [∵ 1+B=1]
= (A + C' + D' + C).(A + C' + D' + D) [Distributive Law]
= (A + D' + 1).(A + C' + 1) [∵ C'+C=1, D'+D=1]
= 1.1 [∵ A+D'+1=1, A+C'+1=1]
= 1
A.{B + C (A.B + A.C)'}
A.{B + C(A.B + A.C)'}
A.{B + C[(A.B)'.(A.C)']} [De-Morgan's Law]
A.{B + C[(A' + B').(A' + C')]} [De-Morgan's Law]
A.{B + C[A'A' + A'C' + A'B' + B'C']} [Distributive Law]
A.{B + C[A' + A'C' + A'B' + B'C']} [∵ A'A'=A']
A.{B + C[A'(1 + C' + B') + B'C']}
A.{B + C[A'.1 + B'C']} [∵ 1 + C' + B' = 1]
A.{B + A'C + B'CC'} [Distributive Law]
A.{B + A'C + 0} [∵ CC'=0]
AB + AA'C [Distributive Law]
AB [∵ AA'C=0]
Related Questions
Give the dual of the following:
(A' B) + (C' 1) = (A' + C) (B + C)Convert the following cardinal expression into its canonical form and reduce it using Boolean laws:
F(L, M, O, P) = π(0, 2, 8, 10)
If A = 1, B = 0, C = 1 and D = 1, find its
- maxterm
- minterm
Prove the following Demorgan's laws using laws of boolean algebra:
(a) (A + B)' = A'.B'
(b) (A.B)' = A' + B'