1
$\begingroup$

Simplify the boolean expression, into the “Canonical Conjunctive Normal Form":

$$x_0\overline{x_0} + x_1\overline{x_1} + \overline{x_2}+x_3$$


Here's my attempt:

$$\begin{align} x_0\overline{x_0} + x_1\overline{x_1} + \overline{x_2}+x_3 &\equiv (x_0\overline{x_0} + x_1) \cdot (x_0\overline{x_0} + \overline{x_1}) + \overline{x_2} + x_3\\ &\equiv (x_0 + x_1) \cdot (\overline{x_0} + x_1) \cdot (x_0 + \overline{x_1}) \cdot (\overline{x_0}+ \overline{x_1})+\overline{x_2}+x_3 \\ &\equiv (\overline{x_2} + x_3+(x_0 +x_1) \cdot (\overline{x_0} + x_1)) \cdot (\overline{x_2} + x_3+(x_0 +x_1) \cdot (\overline{x_0} + \overline{x_1})) \\ &\equiv (\overline{x_2}+x_3+x_0+x_1) \cdot (\overline{x_2}+x_3+\overline{x_0}+x_1)\cdot (\overline{x_2}+x_3+x_0+x_1) \cdot (\overline{x_2}+x_3+ \overline{x_0} + \overline{x_1}) \end{align}$$

I know I can simplify the first and third terms, but let's assume that this is my final answer. Is my working out correct? I'm a little unsure because I've never really worked with terms with multiple conjugations and disjunctions at one time.

$\endgroup$
  • $\begingroup$ $x_0\overline{x_0} = 0$. Ditto for the second product in the sum. That leaves you with $\overline{x_2}+ x_3$. $\endgroup$ – amWhy Jun 8 at 12:12
1
$\begingroup$

Note that $x_0\overline{x_0}$ is equal to either $1\cdot 0$ or $0\cdot 1$, both of which equal $0$. Similarly, $x_1\overline{x_1}=0$.

So what remains is $\overline{x_2} + x_3$, which is indeed in conjunctive normal form (aka, a product of sums). Here, we have just one factor in that product.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.