Simplifying a Sum of Products expression

Question

I'm having some trouble with reducing the Sum of Products expressions for some questions on an upcoming exam. Below is the table (which is correct) for the first part of the question, the second part is to simplify the SOP.

a b c y SOP 0 0 0 0 - 0 0 1 0 - 0 1 0 0 ¬A B ¬C 0 1 1 1 ¬A B C 1 0 0 0 - 1 0 1 1 A ¬B C 1 1 0 1 - 1 1 1 1 A B C

So this results in the SOP being: ¬AB¬C + ¬ABC + A¬BC + ABC

But I still need to simplify this into its most basic form.

Is there a method you can use indefinitely to reduce SOP expressions?

Answer 1

So we have, from the truth-table:

$$¬AB¬C + ¬ABC + A¬BC + ABC$$

By using the distributive law (DL), twice, we can further simplify the expression:

$$\begin{align}\color{blue}{¬AB}¬C + \color{blue}{¬AB}C + \color{red}{A}¬B\color{red}{C} + \color{red}{A}B\color{red}{C} & = \color{blue}{\lnot AB}(\lnot C + C) + \color{red}{AC}(\lnot B + B)\\ \\ &= \lnot AB(1) + AC(1)\\ \\ &= \lnot AB + AC\end{align}$$

Answer 2

As you have only three variables/inputs, maybe you can use Karnaugh Maps?

Answer 3

¬AB¬C + ¬ABC + A¬BC + ABC

¬AB(C + ¬C) + AC(¬B + B)

¬AB + AC

Similar SOP upto 3 variables can be solved using Karnaugh Maps and greater than 3 can be solved either by using the laws of Boolean algebra (mainly DeMorgans Law, Complimentary Law, Assosciative law etc) or a more complicated method called quine mccluskey method