Simplify $(\overline A+\overline{AB})\oplus B$.
Using Demorgan's Law that $A\oplus B = \overline AB+A\overline B$, I got an answer of $AB+\overline B$.
This looked as if it was in simplest terms, but it actually wasn't.
If $f(A,B)=AB+\overline B$, then $f(0,0)=1$, $f(1, 0)=1$, $f(0,1)=0$, and $f(1,1)=1$. All $4$ possible cases for $A$ and $B$ are handled.
But, the same results are acquired if $f(A,B)=A+\overline B$. And I would consider $A+\overline B$ to be more simpler than $AB+\overline B$.
How would you get from $AB+\overline B$ to $A+\overline B$?
This is actually a very common occurrence when doing Boolean Algebra, so there is a known equivalence for that:
Reduction
$P + P'Q= P + Q$
Not all texts have reduction though (surprisingly, because again, this pattern often occurs), so you can derive it from more basic equivalences as follows:
$$P + P' Q \overset{Distribution}{=}$$
$$(P + P') (P + Q) \overset{Complement}{=}$$
$$1 (P + Q) \overset{Identity}{=}$$
$$P + Q$$