# DeMorgan's Law with Boolean Algebra

So I'm studying for an Assembly Language final tomorrow and I'm trying to simplify the following expression using Boolean Algebra. Here are the steps I've written so far, am I safe in assuming that this is a proper process and answer? The answer my teacher wrote seems to get to the same result, but his application of DeMorgan's Law seems incorrect.

\begin{align}(x'+y)'(x+y)' & = (x''y')(x'y') \\ & = (xy')(x'y') \\ & = xy'x'y' \\ & = y'(xx') \\ & = y' \end{align}

My teacher's process is this:

$$(x'+y)'(x+y)'$$ $$= (x'' + y')(x' + y')$$ $$= (x + y')(x' + y')$$ $$= xx' + xy' + x'y' + y'y'$$ $$= 0 + xy' + x'y' + y'y'$$ $$= xy' + x'y' + y'y'$$ $$= xy' + x'y' + y'$$ $$= y'$$

He claims "$(x'' + y')(x' + y')$" is done using DeMorgan's Law, but is that correct? Wouldn't it actually the first step in my process?

• Shouldn't your process end $y'(xx') = y' 0 = 0$? And you should probably say if $+$ means 'or' or 'xor'. – Nick Matteo Dec 10 '14 at 4:41

You are right that your teacher has used De Morgan's law incorrectly and so his answer is also wrong. Your solution in the last step is wrong. You must know that $x\bar x=0$ and then $0\bar y=0$.