# Simplifying expressions containing boolean logic

I have an expression below:

$$A'B'C'D' + A'B'C'D + A'BC'D' + A'BC'D + ABCD$$

I want to simplify this. I've been using a program which gives the following simplified form:

$$ABCD + A'C'$$

I'm not entirely sure how it came to this. Here's how I've been trying it:

\begin{align}A'B'C'D' + A'B'C'D + A'BC'D' + A'BC'D + ABCD && \mathrm{Initial} \\ A'B'C'D' + ABCD + A'B'C'D + A'BC'D' + A'BC'D && \mathrm{Rearrange} \\ 0 + A'B'C'D + A'BC'D + A'BC'D && \mathrm{Cancel\ Opposite\ Terms} \\ 0 + A'C'D + A'BC'D && \mathrm{B\ and\ B'\ Cancel} \\ A'C'D + A'BC'D && \mathrm{End\ Result} \\ A'C'D(1 + B) && \mathrm{Potentially\ going\ further?}\end{align}

Have I made a mistake in here? Can the final step be done, and if so, what does it signify for the purpose of circuitry?

Yes, you cannot "Cancel Opposite Terms" the way that you've shown, consider studying DeMorgan's Law to see why. Instead, think about using the law of Boolean Distribution:

$$A^\prime B^\prime C^\prime D^\prime + A^\prime B^\prime C^\prime D + A^\prime B C^\prime D^\prime + A^\prime B C^\prime D + ABCD$$

$$= (A^\prime B^\prime C^\prime)( D^\prime + D) + (A^\prime B C^\prime)( D^\prime + D) + ABCD$$

$$= A^\prime B^\prime C^\prime + A^\prime B C^\prime + ABCD$$ $$= A^\prime C^\prime (B^\prime + B) + ABCD$$ $$= A^\prime C^\prime + ABCD$$

As required. This uses distribution, definition of complementation, and properties of the boolean product identity.

• Just so I don't make another question, is this correct? $A'B'C'D' + A'B'CD + ABC'D' + ABCD = (A'B')(CD+C'D') + (AB)(CD + C'D') = (A'B' + AB) = 1$? – gator Oct 23 '16 at 23:09
• No, that is not correct. Please look up DeMorgan's Law, and you'll see that $(CD)^\prime = C^\prime + D^\prime \neq C^\prime D^\prime$. – Larry B. Oct 24 '16 at 1:17

Yes, you made a mistake. The 'Cancel Opposite Terms' cancels $A'B'C'D'$ and $ABCD$, but these are not opposite .. the opposite of $ABCD$ would be $A'+B'+C'+D'$.

You also need to be careful with 'Cancelling':, e.g. if you had $A + A' + B$ you don't get $0 + B$ and thus $B$, but rather $1 + B$, and thus $1$

Also, the 4th term from line 2 has a $D'$, but on line 3 this has become $D$