Given the Boolean function: F(x, y,z)=x' y + xyz', derive an algebraic expression for the complement of F. Express in sum-of-products form.

a. (x'y + xyz')' = xy' + xz + x'y' + y' + y'z'
b. (x'y + xyz')' = x'y' + xz + x'y' + y' + y'z
c. (x'y + xyz')' = xy' + xz + x'y' + yy' + y'z
d. (x'y + xyz')' = xy' + xz + x'y' + y' + y'z

d. (x'y + xyz')' = xy' + xz + x'y' + y' + y'z

Explanation:

F = x'y + xyz'

F' = (x'y + xyz')'     ,     DeMorgan's

= (x'y)'(xyz')'

= (x+y')(x'+y'+z)     ,     Distributive Property

= xx' + xy' + xz + y'x' + y'y' + y'z     ,     Redundancy Law: AA' = 0

= 0 + xy' + xz + y'x' + y' + y'z          ,     Redundancy Law: AA = A

= xy' + xz + y'x' + y' + y'z  

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