# Boolean simplification of y(x' + (x+y)')

I've been trying to perform Boolean simplification on the following expression: $y(x' + (x+y)')$

So far, my steps have been:

$y(x' + (x+y)')$

$y(x' + x'y')$

$y(x' + x')(x' + y')$

$(yx')(x' + y')$

I have no idea where to go from here. I feel as if I've complicated this. I know the end result is supposed to be $x'y$.

• I believe your third step there is where you are going off target. – IntegrateThis Oct 8 '17 at 1:40
• @IntegrateThis The third step is valid (it's an instance of Distibution) ... but indeed not helpful – Bram28 Oct 8 '17 at 2:51

$x' + x'y'$ simplifies to $x'$.