Simplify Boolean expression $A'B'C'D' + A'B'CD + A'B'CD' + AB'CD' + AB'C'D'$

Question

Using karnaugh map, I know that my final answer should be $B'D'+A'B'C$ But I cannot simplify this expression to that.

So far I got...

$A'B'C'D' + A'B'CD + A'B'CD' + AB'CD' + AB'C'D'$

$=A'B'C(D+D')+ A'B'C'D'+ AB'CD' + AB'C'D'$

$= A'B'C + B'C'D'(A'+A) + AB'CD'$

$= A'B'C + B'C'D' + AB'CD'$

$= A'B'C + B'D'(C'+AC)$

$= A'B'C + B'D'(A+C')(C+C')$

$= A'B'C + B'D'(A+C')$

What did I do wrong?

How can I get rid of that $A+C'$?

Answer 1

$A'B'C'D' + A'B'CD + A'B'CD' + AB'CD' + AB'C'D'$

$= A'B'CD + A'B'C'D' + A'B'CD' + AB'CD' + AB'C'D'$

$ = A'B'CD + A'B'D'(C + C') + AB'D' (C + C') = A'B'CD + A'B'D' + AB'D'$

$ = A'B'C + A'B'D' + AB'D' = A'B'C + B'D' \,$
(using CD + D' = C + D').