Simplifying the boolean expression $AB+BC'D'+AC+AD$

Question

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.

Answer 1

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}

Answer 2

• $AB=ABCD+ABCD'+ABC'D+ABC'D'$
• $BC'D'=ABC'D'+A'BC'D'$
• $AC=ABCD+ABCD'+AB'CD+AB'CD'$
• $AD=ABCD+ABC'D+AB'CD+AB'C'D$

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.