Test: Minimization of Boolean Functions - 2 - Computer Science Engineering (CSE) MCQ

Given: X Z’ + (Y + Y’Z) + XY
= XZ’ + (Y + Z) + XY
= XZ’ + Y + Z + XY
= (XZ’ + Z) + (Y + XY)
= (X + Z) + Y (1 + X)
= X + Y + Z.

We have, x+x’y’z+yz
= x+y’z+yz [since, x+x’y’z=x+y’z]
= x+z(y’+y)
= x + z.

Y = A’B’C’ + A’BC’ + A’BC + ABC’

After simplification we get

Y = A'C' (B' + B) + A’BC + ABC’

Y = A'C' + A’BC + ABC’

Y = A' (C' + BC) + ABC'

Y = A' (C' + B) (C' + C) + ABC'

Y = A'C' + A'B + ABC'

Y= A'C' + B (A' + AC')

Y = A'C' + B (A' + A) (A' + C')

Y = A'C' + A'B + BC'

(X + Z)(X + XZ’) + XY + Y [Original Expression]
= (x + z)X(1 + Z’) + XY + Y [Distributive]
= (X + Z)X + XY + Y [Complement, Identity]
= (X+Z)X + Y(X+1) [ Distributive]
= (X+Z)X + Y [Idempotent]
= XX + XZ + Y [Distributive]
= X + XZ + Y [Identity]
= X(1+Z) + Y
= X + Y [Idempotent].

Given: MN(M + N’) + M(N + N’)
= MN(M+N’) + M.1
= MNM + MNN’ + M
= MN + 0 +M
= M(N + 1)
= M.

Given: Y’ (X’ + Y’) (X + X’Y)
= Y’(X’ + Y’)(X + Y)
= (X’Y’ + Y’)(X + Y)
= (XX’Y’ + X’Y’Y + XY’ + YY’)
= XY’.

E(E + F) + DE + D(E + F)

=E.E + E.F + D.E + D.E + D.F

=E + E.F + D.E + D.F

=E (1+F) + D.E + D.F         As per Annulment Law : 1 + A = 1

=E + D.E + D.F                   As per Identity Law : A.1 =  A

=E(1+D) + D.F

=E + D.F  ( Answer )

Given XY’+X’+Y’X’ = Y’(X+X’) + X’ = Y’.1 + X’ = X’ + Y’ = (XY)’ [De Morgan’s law].

Given, F(A,B,C) = (A+BC’)(AB’+C)
= (AAB’ + BC’AB’ + AC + BC’C)
= (AB’ + 0 + AC + 0)
= A(B’ + C).