Can anyone Simplify this Boolean expression?

Question

The expression is:

[AB {C+(BD)'} + (AB)']CD

Answer 1

Using De Morgan's laws,

$[AB \{C+(BD)'\} + (AB)']CD=[AB \{C+B'+D'\} + A'+B']CD=[ABC+ABD' + A'+B']CD$

Now, $AE+A'=AE+A'(1+E)=A'+AE+A'E=A'+E(A+A')=A'+E$ $\implies ABC+A'=BC+A'$

$[ABC+ABD'+ A'+B']CD$

$=[A'+(BC+B')+ABD']CD$

$=[A'+C+B'+ABD']CD $ (Using $AE+A'=A'+E$)

$=A'CD+CD+B'CD=CD(A'+1+B')=CD$