Simplify boolean expression, so close but yet so far

Question

$$F(w,x,y,z) = w'y' + w'z + x' + yz + y'z'$$

The simplest form is apparently $w'y' + x' + yz + y'z'$, but for the life of me, I cannot figure out how, no matter what trick I use.

Answer 1

Given $$F(w,x,y,z) = w'y' + \color{red}{w'z} + x' + yz + y'z',\quad\quad (*)$$ we want to show that $F(w,x,y,z)$ reduces to $\;w'y' + x' + yz + y'z'.$

For this, it suffices to prove that $\color{red}{w'z}$ is covered by other terms in $(*).$
Rewrite $$\color{red}{w'z}=w'zxy+w'zx'y+w'zxy'+w'zx'y'.$$ Note that $w'zxy$ is covered by $yz,$ and similarly find in $(*)$ a covering term for each of $w'zx'y,\; w'zxy',\; w'zx'y'.$

Answer 2

The $yz+y'z'$ terms take care of all cases where $y$ and $z$ have the same truth value, so the other terms only matter if $y$ and $z$ have different truth values. In that case we have $z=y'$, so $w'y'=w'z$ and you only need one of those equivalent (in context) terms.