**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**:

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## Minimization of Boolean Functions:

**a) A’C’+A’BC+B’C’**

**Combine terms with shared literals:**A’C’ + A’BC + B’C’ = A’C’ + BC’ + B’C’**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**

**Factor out common factors:**BC + AC’ + AB + BCD = BC + AC’ + AB + B(CD)**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’**

**Identify identical terms:**ABC + A’B’C + A’BC + ABC’ + A’B’C’ contains two identical terms (ABC) and two others (A’B’C’).**Remove redundant terms:**ABC + A’B’C + A’BC + ABC’ + A’B’C’ = 2*ABC + 2*A’B’C**Combine terms with shared literals:**2*ABC + 2*A’B’C = (A + A’)(BC + B’C)**Apply the distributive property:**(A + A’)(BC + B’C) = (A)(BC) + (A)(B’C) + (A’)(BC) + (A’)(B’C)**Simplify:**(A)(BC) + (A)(B’C) + (A’)(BC) + (A’)(B’C) = AC + AB’ + A’BC + A’B’C**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**

**De Morgan’s theorem:**((CD)’+A)’+A+CD+AB = (C+D)(A’) + A + CD + AB**Factor out common factors:**(C+D)(A’) + A + CD + AB = (A’ + AB)(C+D) + CD**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’)**

**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)

**Apply the distributive property:**(A+D)(A+D’)(A+D)(A+B’) = (A^2 + AD)(A^2 + AD’)(A^2 + AD)(A^2 + AB’)**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’))’**

**De Morgan’s theorem:**(Wx(Y’Z+YZ’)+W’x'(Y’+Z)(Y+Z’))’ = (W(Y’Z+YZ’))’ + (W’x'(Y’+Z)(Y+Z’))’**Distributive property:**[(W(Y’Z+YZ’))’ + (W’x'(Y’+Z)(Y+Z’))’] = W'(Y’Z+YZ’) + (W’x’)(Y’+Z)(Y+Z’)**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)