Boolean Algebra Simplification

Question

I have to simplify

A'BC' + A'B'C + A'BC + ABC

My result was

A'BC

Is this correct?

Answer 1

Edited: $$ A'B C' + A'B'C + A'BC + ABC \\ =A'(B C' + B'C) + (A' + A)BC\\= A'(B C' + B'C) + BC$$ OK I simplified this. It is still not what you have. You can see that your answer is wrong by plugging in $\{A,B,C\} = \{1,1,1\}$. Your final expression evaluates to false (i.e. $0$), but the original one evaluates to $(0+0+0+1) = 1 \Rightarrow$ true.

Answer 2

We will use the Karnaugh map technique: We make a chart and put $\times$es into the four slots that correspond to the four terms in your original CNF formula:

$$\begin{array}{c|cccc} & AB & AB' & A'B' & A'B \\ \hline C & \times & & \times & \times \\ C' & & & & \times \end{array} $$

In the Karnaugh map technique, we look for adjacent pairs of $\times$, lines of four, or blocks of four, and we imagine that the left and right ends of the chart are connected. There are no lines or blocks of four, but we can cover the four $\times$ with the three pairs $A'B$ (the vertical pair on the right), $A'C$ (the horizontal pair un the upper-right corner), and $BC$ (the horizontal pair wrapping from the upper right to the upper left corner). So the original formula is equivalent to $$A'B + A'C + BC.$$

If desired, we can factor out the $A'$ from the first two terms to obtain $A'(B+C) + BC$, or the $C$ from the second two terms to obtain $A'B + (A' + B)C$.

If your answer $A'BC$, were correct, only one box would be filled, so your answer is not correct.