# Simplify $A'B'C'D' + A'B'CD' + A'BCD' + ABCD' + AB'CD'$

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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Relevant: Karnaugh maps. –  user2468 Aug 23 '12 at 14:12
you have done everything fine,but you could simplify it more,please see my answer ,you need to know that $c'*c=0$ and $c'+c=1$ –  dato datuashvili Aug 23 '12 at 14:22
@clueless you need to know that $C + C' Y = C + Y$ for any $Y$. –  user2468 Aug 23 '12 at 14:27

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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I'm sorry. I did an edit. Could you please not use $X$ since it might confuse the OP (s/he already has an $X$ in the question). –  user2468 Aug 23 '12 at 14:34
@JenniferDylan Oh thanks! Yeah I'll switch that. –  rschwieb Aug 23 '12 at 14:35
does that means that (A′B′C′+C) = (A'B'+C) because of the absorbtion and complementary law? –  clueless Aug 23 '12 at 14:45
@clueless Taking $Y=C$ and $Z=A'B'$, yes. –  rschwieb Aug 23 '12 at 15:42

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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hmm i do not under this c+a′b′c+a′b′c′=c+a′b′c′ can u explain? –  clueless Aug 23 '12 at 14:26
sure ,first what does equal to $c+a'b'c$?take c out of brackets,you get $c(1+a'b')$ right? but $1+k$ in boolean algebra is equal to $1$ regardles any value of $k$ –  dato datuashvili Aug 23 '12 at 14:30
oh I see it now that you write it that way thanks alot –  clueless Aug 23 '12 at 14:34
good lucks friend,you are welcome –  dato datuashvili Aug 23 '12 at 14:35

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


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

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