How can I simplify this boolean expression? 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?
 A: $D=A'BCD+AB'C'D+AB'CD'+AB'CD+ABC'D'+ABC'D+ABCD'+ABCD$
$= A(B'C'D + B'CD' + B'CD + BC'D' + BC'D + BCD' + BCD) + A'BCD$
$= A(B(C'D'+C'D+CD'+CD) + B'(C'D + CD' + CD))+A'BCD$
$= A(B(C(D'+D)+C'(D+D')) + B'(C(D'+D)+C'D))+A'BCD$
$=A(B(C*1+C'*1)+B'(C*1+C'D))+A'BCD$
$=A(B + B'(C+C'D))+A'BCD$
$= AB + AC + AC'D +A'BCD$
$=AB + AC + AD + A'BCD$
$= AB + AC + AD + BCD$
The trick is in the last 3 lines to think about what is covered in the previous cases. For example, if you have $AB + AB'C$, then if $A$ and $B$ are correct, you have 1 anyway. Or in other words you could write this as $A(BC + BC' + B'C)$. I hope my point gets clear here, when in doubt, a table will always help.
A: $$A’BCD+AB’C’D+AB’CD’+\color{red}{AB’CD}+ABC’D’+\color{green}{ABC’D}+\color{blue}{ABCD’}+\color{orange}{ABCD} = \text{(Idempotence x 6)}$$
$$A’BCD+AB’C’D+AB’CD’+\color{red}{AB’CD+AB’CD}+ABC’D’+\color{green}{ABC’D+ABC’D}+\color{blue}{ABCD’+ABCD’}+\color{orange}{ABCD +ABCD+ABCD+ABCD}= \text{(Commutation)}$$
$$\color{red}{ABC’D’+ABC’D+ABCD’+ABCD}+\color{green}{AB’CD’+AB’CD+ABCD’+ABCD} +\color{blue}{AB’C’D +ABC’D+AB’CD+ABCD}+\color{orange}{A’BCD+ABCD}= \text{(Adjacency)}$$
$$\color{red}{ABC’+ABC}+\color{green}{AB’C+ABC} +\color{blue}{AB’D+ABD}+\color{orange}{BCD}= \text{(Adjacency)}$$
$$\color{red}{AB}+\color{green}{AC} +\color{blue}{AD}+\color{orange}{BCD}$$
(note: the colors help connect lines 1 and 2, and also lines 3,4,5, but the colors do not reveal how line 2 goes to line 3: that step is really just a matter of reordering all the terms by Commutation)
