Simplify $xyz’w’+xy’zw+xyz’w+xy’z’w’+xy’z’w+zw’xy’$

Question

Simplify $xyz’w’+xy’zw+xyz’w+xy’z’w’+xy’z’w+zw’xy’$

( ` = not)

Using a karnaugh map I found that it's equal to $xy'+xz'$, but I can't find how to get there with simple algebraic simplifications. Help would be appreciated.

Answer 1

Let $$f(x,y,z,w) = xyz'w + xyz'w' + xy'zw + xy'zw' + xy'z'w + xy'z'w'.$$ Then,

\begin{align} f(x,y,z,w) &= x(yz'w + yz'w' + y'zw + y'zw' + y'z'w + y'z'w')\\ &= x( yz'(w + w') + y'z(w + w') + y'z'(w + w') )\\ &= x( yz' + y'z + y'z')\\ &= x( (y + y')z' + y'(z + z') )\\ &= x(z' + y'). \end{align} The steps are justified by distributivity (which apparently you are already used to) and the laws: $$a+a'=1\quad\text{and}\quad a1=a,$$ for each $a$, and the usual idempotence, commutativity and associativity.
I suppose you can fill in these justifications.