Boolean expressions simplification I have found these boolean expressions based on a truth table:

*

*$(abc'd') + (ab'c'd) + (a'bcd') + (a'b'cd)$
(EDITED for 2))


*$(abcd') + (abc'd) + (ab'cd) + (ab'c'd') + (a'bcd) + (a'bc'd') + (a'b'c'd) + (a'b'cd')$
However, I need to simplify them. (' for complement, + for or, xy is for x and y)
 A: $1. ~~(abc'd')+(ab'c'd) \iff  ac'(bd'+b'd),$ and $(a'bcd')+ (a'b'cd) \iff a'c(bd'+b'd)$, so this yields $(ac'+a'c)(bd'+b'd) \iff (a \oplus c)(b \oplus d)$.
$2.$  Factor as in problem $1$ using $ab'$ and $a'b$, and then $ab$ and $a'b'$, and you'll end up with $(a \oplus b) \oplus (c \oplus d)$.
A: Using $\oplus$ to denote the XOR operation.
For the first one:
\begin{align}
&(ab{c'}d') + ({a}b'{c'}d) + ({a'}b{c}d') + ({a'}b'{c}d)&\\\
&{ac'}{(bd'+b'd)}+{a'c}{(bd'+b'd)}&\text{Factor}\\
&ac'{(b\oplus d)}+a'c{(b\oplus d)}&\text{Use }\oplus\\
& {(b\oplus d)}(ac'+a'c)&\text{Factor}\\
&(b\oplus d)(a\oplus c)&\text{Use }\oplus
\end{align}
For the second one:
\begin{align}&{
(abcd') + (abc'd) + (ab'cd) + (ab'c'd') + (a'bcd) + (a'bc'd') + (a'b'c'd) + (a'b'cd')}&\\
&{ab(cd'+c'd)+cd(ab'+a'b)+a'b'(cd'+c'd)+c'd'(ab'+a'b)}&\text{Factor}\\
&{ab(c\oplus d)+cd(a\oplus b)+a'b'(c\oplus d)+c'd'(a\oplus b)}&\text{Use }\oplus\\
&{(c\oplus d)(ab+a'b')+(a\oplus b)(cd+c'd')}&\text{Factor}\\
&{(c\oplus d)(a\oplus b)'+(a\oplus b)(c\oplus d)'}&\text{Use }\oplus\\
&{(c\oplus d)\oplus(a\oplus b)}&\text{Use }\oplus
\end{align}
