[Math] Simplify $A’B’C’D’ + A’B’CD’ + A’BCD’ + ABCD’ + AB’CD’$

boolean-algebra

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?

Best Answer

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.

Related Question