Boolean Algebra - reducing a function

Question

Let $$f(w,x,y,z) = w'x'y'z' + w'x'yz' + wx'yz'$$

How can you reduce it to: $$x'z'(w' +y)$$

Thanks!

Answer 1

By the distributive property ("in reverse") applied multiple times, we have

$$\begin{align} f(w,x,y,z) &= w'x'y'z' + w'x'yz' + wx'yz' \\\\ &= x'z'(w'y' + w'y + wy) \\\\ & = x'z'(w'(y'+ y) + wy) \\\\ & = x'z'(w' + wy)\\\\ & = x'z'(w' +y)\end{align}$$

Answer 2

$$\begin{align}w'x'y'z' + \color{blue}{w'x'yz'} + wx'yz'&=w'x'y'z' + \color{blue}{w'x'yz'} +\color{blue}{w'x'yz'} + wx'yz' \\~\\&=(w'x'y'z' + \color{blue}{w'x'yz'}) +(\color{blue}{w'x'yz'} + wx'yz')\\~\\&= w'x'z' + x'yz'\\~\\&=x'z'(w'+y)\end{align}$$