How would you simplify the following boolean expression $(!A B)+(B !C)+(BC)+(A !B !C)$? How would you simplify the following boolean expression $(!A B)+(B !C)+(BC)+(A !B !C)$?
I factorised B and managed to get $B(!A+!C+C)+(A !B !C) = B+(A !B !C)$, but I do not know how to continue.
Using a K-map, I managed to get the result of $B+A!C$ and I am trying to achieve the same result using regular identities and laws of boolean algebra. By the way, sorry for poor formatting, but I do not know how I could paste an expression from word to make it look better and easier to read.

Actually, there is a non-intuitive (doesn't hold in ordinary algebra) Boolean algebra law that can be applied here: distributivity of disjunction over conjunction expressed as follows
$$x+yz=(x+y)(x+z).$$
Using your notation $x=B$, $y=!B$, $z=A!C$. Hence we have
$$B+!BA!C=(B+!B)(B+A!C)=1\cdot(B+A!C)=B+A!C.$$
