How do I finish the last few steps of this simplification of a boolean expression?

Question

Say I want to expand the following

$$\overline{(x' + y)(x + y')},$$

The correct answer is given through for example the following procedure

$$ \overline{(x' + y)(x + y')} = \overline{x'+y} + \overline{x + y'} = x''y'+ x'y'' = xy' + x'y.$$

I should too be able to solve it in this way, no? I get stuck on the last step though.

$$\overline{(x' + y)(x + y')} = \overline{x'x + x'y' + xy + yy'} = \overline{x'y' + xy} =\ ? $$

EDIT: I have mixed notation even though I shouldn't have. $x'$ means the same thing as $\bar{x}$ in this context, negation.

Answer 1

$$\cdots=(x'y')'(xy)'=(x+y)(x'+y')=xx'+xy'+x'y+yy'=xy'+x'y$$

Answer 2

According to Demorgan's Theorem, for boolean simplification,
Law 1: A.B = $\bar{A}$+$\bar{B}$
Law 2: A+B = $\bar{A}$.$\bar{B}$.

So,

$\overline{x′y′+xy}$ = ($\overline{x'y'}$).($\overline{xy}$) = (x"+y")(x'+y') = x"x' + x"y' + y"x' + y"y' = xy' + x'y