Solving this equation to be 3 literals

Question

I am trying to solve this to be 3 literals, but I keep getting errors:

Question:$(x'y' + z)' + z + xy + wz$

My answer:

$(x+y)z' + xy + z(1+w)\implies xz' + yz' + xy + z\implies zx + zz' + yz' + xy$

$\implies zx + yz' + xy \implies xyz + xzz' + xy \implies xyz + xy \implies yx (1 + z)\implies yx$

The right answer should be $x + y + z$

Answer 1

It is often quite useful to consider the negation of the given expression and then use

• $aa = a$
• $aa'=0$
• $a+ab = a(1+b) = a$

So, consider $E' = \left( (x'y' + z)' + z + xy + wz \right)'$: $$\begin{eqnarray*} E' & = & \left( (x'y' + z)' + z + xy + wz \right)' \\ & = & (x'y' + z)z'(xy)'(wz)' \\ & \stackrel{zz' = 0}{=}& x'y'z'(x'+y')(w'+z') \\ & \stackrel{x'x' = x' etc.}{=}& x'y'z' + x'y'z'w' \\ & = & x'y'z'(1+ w' ) \\ & = & x'y'z' \\ \end{eqnarray*} $$ $$\Rightarrow E = (x'y'z')' = x+y+z$$