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How do you simply the following equation? $X = A'B'C'D' + A'B'CD' + A'BCD' + ABCD' + AB'CD'$

Here is what I did:

$\begin{eqnarray} X & = & A'B'C'D'+A'CD'(B'+B) + ACD'(B+B') \\ & =& A'B'C'D'+CD'(A'+A) \\ & = & D'(A'B'C'+C) \end{eqnarray}$

Is this correct?

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    @clueless you need to know that $C + C' Y = C + Y$ for any $Y$.2012-08-23

3 Answers 3

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It looks great. The one improvement that could be made is that the $C'$ is redundant, owing to an identity:

$ZY'+Y=Z+Y$

You can deduce this using the absorbtion law $ZY+Y=Y$, and the complementary law $Y+Y'=1$.

Intuitively, when adding part of $Z$ outside of $C$ to $C$, you may as well add all of $Z$ to $C$, because the part already inside $C$ will be abosorbed anyway.

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    @clueless Taking $Y=C$ and $Z=A'B'$, yes.2012-08-23
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we can simplify it much $(A'B'C'+C)=(C+C')*(C+A'B')$ we can proof it if we open brackets,we get
$(C+C')*(C+A'B')=C+A'B'C+A'B'C'=C+A'B'C'$

so finally we get $D'*(C+A'B')=C*D'+A'B'D'$ because $C+C'=1$

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    good lucks friend,you are welcome2012-08-23
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As all above answer pointed towards Absorption Law which is

ZY′+Y=Z+Y 

So here you can apply this law -

 = > D′(A′B′C′+C)  = > D′(A′B′+C)  = > D′A′B′+D′C 

which is your answer

Also you can verify this by K-Map that, this is the simplest form possible of this Boolean expression.

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