Find complement of given expression using Boolean algebra

Question

Find the complement of $xy(y+y'z) + x'z$

I solve this problem but get 3 different solution by using 3 different methods (rules of boolean algebra)

Solution that I get each time

1. $\quad x'z'+xy'+y'z'$
2. $\quad xy+xyz'$
3. $\quad y'z'$

Answer 1

Let us give the above expression as $f(x,y,z)$, and exploit the laws of Boolean Algebra. First, we will simplify the above expression. $$ \begin{split} f(x,y,z) & = xy(y+y'z) + x'z & \space \space (\text{Distributivity})\\ & = xyy +xyy'z + x'z & \space \space (\text{Idempotence, Complement}) \\ & = xy +x'z & \end{split} $$

Its complement is given taking the complement of the expression above. $$ \begin{split} f'(x,y,z) & = (xy +x'z)' & \space \space (\text{Distributivity})\\ & = (xy)'(x'z)' & \space \space (\text{De Morgan}) \\ & = (x' + y')(x + z') & \space \space (\text{De Morgan, Double Negation}) \\ & = xx' + x'z' + xy' + y'z' & \space \space (\text{Distributivity}) \\ & = x'z' + xy' + y'z' & \space \space (\text{Complement}) \\ & = (x+z)' + xy' + (y+z)' & \space \space (\text{De Morgan}) \\ \end{split} $$

Assuming you are required to leave the answer in the canonical conjunctive normal form, point 1 which you have written is the correct answer. The trick is to simplify the expression, then take the compliment of the expression.