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

Question

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?

Answer 1

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.

Answer 2

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 $C+C'=1$

Answer 3

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.