Original expression: $$D=A’BCD+AB’C’D+AB’CD’+AB’CD+ABC’D’+ABC’D+ABCD’+ABCD$$
This is the simplest form I could find: $$B\cdot C\cdot D+A\cdot(B'\cdot D+B \cdot C'+CD')$$ It is equivalent to the solution. This is the solution I got from an online calculator: $$A \cdot B + A \cdot C + A \cdot D + B \cdot C \cdot D$$
How can I do this? Do I need to take a step back and then simplify more?