# 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).
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