Simplifying for boolean algebra

Question

I have this boolean expression.

$$(abcd') + (abc'd) + (ab'cd) + (ab'c'd') + (a'bcd) + (a'bc'd') + (a'b'c'd) + (a'b'cd')$$

However, I am required to simplify it as much as possible and I am not allowed to do it with karnaugh maps. I have to use basic boolean rules as well as distributive, commutative and associative laws. I am allowed to use the xor expression. For now, based on a truth table, I know that there is a xor somewhere within the simplified expression.

($+$ for or, $xy$ is for x and y, $'$ for complement)

Answer 1

HINT: With an XOR ($\oplus$), you can do:

$abcd'+abc'd=ab(cd'+c'd)=ab(c \oplus d)$

Also, something like $cd + c'd'$ is equivalent to $c \oplus d'$

So, play around with that ...