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I am trying to simplify this equation: ( ' = NOT)

'AD+ACD+A'BC+ABCD

The result I get is when using demorgan's and other rules is:

DC+D'A+A'BC

However, I know that this answer is only partially complete since I know from using a karnaugh map that the correct answer is

D+A'BC.

I know that the law of common identities states that

A'B + B = A+B

and this makes sense to me, but I'm not sure how I can apply that law here. If common identities is not the way to go, could someone show me how to solve this problem?

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From the context, I suppose you're using $'$ as a prefix notation, as it is usual for $\neg$.
Note however that $'$ is normally used as a postfix notation, so as $A'$ with the same meaning as $\neg A$.

That said, you're claiming (with the usual notation) that using a Karnaugh map, you proved that $$A'D+ACD+AB'C+ABCD = D+AB'C.$$ But you must have got wrong somewhere, since applying with $A=D=1$ and $C=0$ (whatever $B$), this gives $0=1$.


By the way, you can get the other result using the identity that you mention like this: \begin{align} A'D+ACD+AB'C+ABCD &= A'D+ACD+AB'C\\ &= (A'+AC)D+AB'C\\ &= (A'+C)D +AB'C\\ &= A'D + CD + AB'C, \end{align} where the first equality follows from $ACD+ABCD=ACD$, the second and the fourth are distributivity and the third is the law you stated.

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