Simplify this boolean algebra?

Question

$$ \begin{align} &\lnot x_1(x_2\land\lnot x_3\lor x_3)\lor x_1(\lnot x_2\land\lnot x_3\lor x_2\land x_3)\\ &=\lnot x_1\land x_2\land\lnot x_3\lor\lnot x_1\land x_3\lor x_1\land\lnot x_2\land\lnot x_3\lor x_1\land x_2\land x_3 \end{align} $$

Is it possible to simplify the above?

@Kaster: I don't see anywhere x1x2, but why you ask about the meaning? Is the way I write those wrong? What's the correct way? — Luka, May 08 '14 at 08:16
@Luka you have something like $x_1 (...)$, so I wonder what's the difference between $x_1(...)$ and $x_1 * (...)$ — Kaster, May 08 '14 at 08:38

score 1 · Accepted Answer · answered May 11 '14 at 12:55

You can further simplify by using Karnaugh map method. Before that, you must transform your expression into its CDNF, so you get: $$\neg x_1 x_2 \neg x_3 \vee \neg x_1 x_2 x_3 \vee \neg x_1 \neg x_2 x_3 \vee x_1 \neg x_2 \neg x_3 \vee x_1 x_2 x_3.$$

Finally, the simplified expression looks like this: $$x_1 \neg x_2 \neg x_3 \vee \neg x_1 x_2 \vee \neg x_1 x_3 \vee x_2 x_3.$$

Simplify this boolean algebra?

1 Answers1