Boolean Algebra simplifcation

Question

I have an expression I need to simplify for a class assignment, yet I simply can not figure out how to apply the rules in this case. Can someone put me on the right direction?

w’x’y’z + w’xy’z + wxy’z + wx’y’z + w’x’yz

Answer 1

Final steps hidden in spoiler quotes. Disclaimer: there may be a faster way to see this, but these are the steps that jumped out at me to follow in the order I noticed them

$w'x'y'z+w'xy'z+wxy'z+wx'y'z+w'x'yz$

$=w'y'z(x'+x) + wy'z(x+x')+w'x'yz~~~~$ by applying rules of distributivity

$=w'y'z+wy'z+w'x'yz~~~~$ by the fact that $x'+x=1$

$=(w'+w)y'z+w'x'yz~~~~$ by distributivity

$=y'z+w'x'yz~~~~$ by $w'+w=1$

.

$=z(y'+wx'y)~~~~$ by distributivity

.

$=z(y'+wx')~~~~$ by $a+a'b=a+b$