Boolean algebra simplification

Question

I have the equation (CD)+(BC'D'+A'B'C'D)+(C'D') and I can't seem to simplify it enough for my homework. I've tried multiple ways of simplifying the equation and I keep ending with an extra variable D.

Answer 1

First of all, note that $B+1=1,$ so that $$BC'D'+C'D'=(B+1)C'D'=C'D'.$$ Likewise, $1+A'B'=1,$ so since $C+C'=1,$ then $$\begin{align}CD+A'B'C'D &= (1+A'B')CD+A'B'C'D\\ &= CD+A'B'CD+A'B'C'D\\ &= CD+A'B'(C+C')D\\ &= CD+A'B'D.\end{align}$$ That, together with commutativity lets you simplify it to $$CD+A'B'D+C'D'.$$ Is that the answer you were looking for?

Answer 2

There are only four variables here, so let's make a Karnaugh map.

$$ \begin{array}{r|cccc|} & AB & AB' & A'B' & A'B \\ \hline CD & X & X & X & X\\ CD' & \\ C'D' & X & X & X & X\\ C'D & & & X & \\ \hline \end{array} $$

While you're making the map, you'll notice that the $BC'D'$ boxes are already filled in, because the $C'D'$ boxes are all filled by the $C'D'$ term. So the $BC'D'$ term was entirely redundant, and can be dropped. Even if you don't eventually show the Karnaugh map to your teacher, that's a useful thing to notice; it's the first thing Cameron Buie mentioned in his answer.

Now when you look at the map the two horizontal lines jump out at you. Unfortunately they aren't adjacent, or they could be turned into a single very simple term. (For example, the middle two rows are simply $D'$.) As it is, the two horizontal rows are $CD$ and $C'D'$.

This leaves only one box left over, at $A'B'C'D$. We could write it that way, but we get a simpler expression by amalgamating it with an adjacent box. If we use the one above, we get $A'B'C'$, and our final answer is then $$CD + C'D' + A'B'C'.$$

If we chose to amalgamate $A'B'C'D$ with the box below, we would get the equally simple equivalent solution $$CD + C'D' + A'B'D$$ which Cameron Buie found.

The form of the Karnaugh map makes clear that there's nothing simpler to find.