Simplify the boolean expression $(xy')\cdot(x'+ y')'$.
Here is my work:
$(xy')\cdot(x'+ y')' = [(x' + y'') · (x' + y')]'$ de Morgans rule (b)
$= [(x' + y) · (x' + y')]'$ involution
$= (x' + y)' + (x' + y')'$ de Morgans rule (b)
$= (x''·y') + (x''·y'')$ de Morgans rule (a)
$= (x·y') + (x·y)$ involution
$= x(y' + y)$ distributive rule (b)
$= x(1)$ complement rule (a)
$= x$ identity rule (b)
A bit more briefly:
$$[ (xy')' * (x'+y') ]' \equiv (xy') + (x'+y')'\tag{DeMorgan's}$$
$$\equiv (xy') + (xy)\tag{DeMorgan's}$$
$$\equiv x(y' + y)\tag{Distributive property}$$
$$\equiv x*1 \tag{complement rule}$$
$$\equiv x \tag {identity rule}$$