# Boolean simplification

So I am giving this expression D +B’C’ + CD’ +A B’C and I ask to simplify it

When working through it I get

D+B'C'+CD'+AB'C D'(A'B'+CD'+AB) D'(A'B'+A(B'+B)) D'(A'B'+AC') D'(B'+A)

Am I on the right track or am I completely missing the point?

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You have made a mistake. If you set $(A,B,C,D) = (F,T,F,T)$ then the first formula will evaluate to $T$, the second to $F$. – copper.hat Oct 10 '12 at 18:27
You have changed the formula. However you are missing the point, the second expression if far more complicated, not simpler. – copper.hat Oct 10 '12 at 18:34

Try using the following rules: $A + \overline{A }B = A + B$, and $A + A B = A$ (and convince your self that they are true).

Then you could do \begin{eqnarray} D + \overline{B} \overline{C} + C\overline{D} +A \overline{B} C & = & D + C + \overline{B} \overline{C}+A \overline{B} C \\ & = & D + C + \overline{B} + A \overline{B} C \\ & = & D + C + \overline{B} \end{eqnarray}

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Alright excuse my ignorance but what rule is that? Also after each step a variable was dropped. Why is that? i.e. D',C',A. Where do they go? Are they get canceled out? – Leo Oct 10 '12 at 18:49
Thank you by the way. – Leo Oct 10 '12 at 18:54
There are 3 reductions above: #1 $D + C\overline{D} = D + C$, #2 $C + \overline{B}\overline{C} = C + \overline{B}$, and #3 $C+A \overline{B} C = C$. Only the variable $A$ was dropped. The notion of 'simplified' is not necessarily obvious (here it is fairly clear). When implementing digital circuits, simpler may mean reduced number of gates (less important nowadays) or reduced logic depth (important for speed). – copper.hat Oct 10 '12 at 19:11
Okay, I get it! Thanks copper.hat. – Leo Oct 10 '12 at 20:15