Simplifying the boolean expression $AB+BC'D'+AC+AD$ I'd like to simplify the expression
$$AB+BC'D'+AC+AD$$
Logically, I understand why the AB term isn't needed, if both A and B are true, then at least one of the other terms will always be true, making the AB term redundant. However, I cannot for the life of me apply the laws of boolean algebra to actually simplify it.
 A: Hint:
The trick is notice that $A$ and $B$ implies $C$ and $C'$ and $D$ and $D'$ which is impossible, so we will use $C$ or $C'$ and $D$ or $D'$ to cancel that term, and the basic idea is use Identity law and Negation law: $$P=P(1)=P(Q+Q')$$
Answer:
\begin{align}
&AB+BC'D'+AC+AD\\
=&AB(1)+BC'D'+AC+AD\tag*{Identity law}\\
=&AB(C+C')+BC'D'+AC+AD\tag*{Negation law}\\
=&ABC+ABC'+BC'D'+AC+AD\tag*{Distributive law}\\
=&BAC+AC+ABC'+BC'D'+AD\tag*{Reordering}\\
=&AC+ABC'+BC'D'+AD\tag*{Absorption law}\\
=&AC+ABC'(1)+BC'D'+AD\tag*{Identity law}\\
=&AC+ABC'(D+D')+BC'D'+AD\tag*{Negation law}\\
=&AC+ABC'D+ABC'D'+BC'D'+AD\tag*{Distributive law}\\
=&AC+BC'AD+AD+ABC'D'+BC'D'\tag*{Reordering}\\
=&AC+AD+BC'D'\tag*{Absorption law}\\
\end{align}
If you have Consensus law:
\begin{align}
&AB+BC'D'+AC+AD\\
=&AB+BC'D'+A(C+D)\tag*{Distributive law}\\
=&AB+BC'D'+A(C'D')'\tag*{De Morgan's law}\\
=&BC'D'+A(C'D')'\tag*{Consensus law}\\
=&BC'D'+A(C+D)\tag*{De Morgan's law}\\
=&BC'D'+AC+AD\tag*{Distributive law}\\
\end{align}
A: *

*$AB=ABCD+ABCD'+ABC'D+ABC'D'$

*$BC'D'=ABC'D'+A'BC'D'$

*$AC=ABCD+ABCD'+AB'CD+AB'CD'$

*$AD=ABCD+ABC'D+AB'CD+AB'C'D$
Now observe that every term on RHS of the first equality is present already as a term on RHS of at least one of the other equalities.
So idempotency tells us that by a summation of the terms on RHS we can leave them out without affecting the end result.
Then of course by a summation of the terms on LHS we can leave $AB$ out without affecting the end result.
