Using Boolean Algebraic manipulations, minimize the following functions to minimum number of literals in sum-of-products representation

Question: Using Boolean Algebraic manipulations, minimize the following functions to minimum number of literals in sum-of-products representation (Show your work clearly step by step indicating the used properties of Boolean Algebra):
a) A’C’+A’BC+B’C’
b) BC+AC’+AB+BCD
c) ABC+A’B’C+A’BC+ABC’+A’B’C’
d) ((CD)’+A)’+A+CD+AB
e) (A+C+D)(A+C+D’)(A+C’+D)(A+B’)
f) (Wx(Y’Z+YZ’)+W’x'(Y’+Z)(Y+Z’))’

ANswer:

Minimization of Boolean Functions:

a) A’C’+A’BC+B’C’

  1. Combine terms with shared literals: A’C’ + A’BC + B’C’ = A’C’ + BC’ + B’C’
  2. Apply the distributive property: (A’ + B)C’ + B’C’ = C'(A’ + B + B’) = C'(A’ + 1) = C’

Therefore, the minimized form is C’.

b) BC+AC’+AB+BCD

  1. Factor out common factors: BC + AC’ + AB + BCD = BC + AC’ + AB + B(CD)
  2. Identify redundant terms: BC + AC’ + AB + B(CD) is already covered by AB and BCD.

Therefore, the minimized form is AB + BCD.

c) ABC+A’B’C+A’BC+ABC’+A’B’C’

  1. Identify identical terms: ABC + A’B’C + A’BC + ABC’ + A’B’C’ contains two identical terms (ABC) and two others (A’B’C’).
  2. Remove redundant terms: ABC + A’B’C + A’BC + ABC’ + A’B’C’ = 2ABC + 2A’B’C
  3. Combine terms with shared literals: 2ABC + 2A’B’C = (A + A’)(BC + B’C)
  4. Apply the distributive property: (A + A’)(BC + B’C) = (A)(BC) + (A)(B’C) + (A’)(BC) + (A’)(B’C)
  5. Simplify: (A)(BC) + (A)(B’C) + (A’)(BC) + (A’)(B’C) = AC + AB’ + A’BC + A’B’C
  6. Identify redundant terms: AC is already covered by A’BC.

Therefore, the minimized form is AB’ + A’BC + A’B’C.

d) ((CD)’+A)’+A+CD+AB

  1. De Morgan’s theorem: ((CD)’+A)’+A+CD+AB = (C+D)(A’) + A + CD + AB
  2. Factor out common factors: (C+D)(A’) + A + CD + AB = (A’ + AB)(C+D) + CD
  3. Identify redundant terms: AB is already covered by A’B.

Therefore, the minimized form is A’D + CD.

e) (A+C+D)(A+C+D’)(A+C’+D)(A+B’)

  1. Simplify internal redundancies:
  • A+C+D = A+D (since C is included in D)
  • A+C+D’ = A+D’ (since C is included in D’)
  • A+C’+D = A+D (since C’ is included in D)
  1. Apply the distributive property: (A+D)(A+D’)(A+D)(A+B’) = (A^2 + AD)(A^2 + AD’)(A^2 + AD)(A^2 + AB’)
  2. Identify and eliminate redundancies:
  • A^2 = A
  • AD is included in AD’ and AB’

Therefore, the minimized form is AD’ + AB’.

f) (Wx(Y’Z+YZ’)+W’x'(Y’+Z)(Y+Z’))’

  1. De Morgan’s theorem: (Wx(Y’Z+YZ’)+W’x'(Y’+Z)(Y+Z’))’ = (W(Y’Z+YZ’))’ + (W’x'(Y’+Z)(Y+Z’))’
  2. Distributive property: [(W(Y’Z+YZ’))’ + (W’x'(Y’+Z)(Y+Z’))’] = W'(Y’Z+YZ’) + (W’x’)(Y’+Z)(Y+Z’)
  3. Simplify internal redundancies:
  • Y’Z + YZ’ = Y (Y’ and YZ’ are complementary)
  • W’x’ = W’ (since x is included in x’)
  • Y’ + Z = 1 (Y’ and Z are complementary)
  • Y + Z’ = 1 (Y and Z’ are complementary)

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