Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find a Boolean product of the Boolean variables x, y,and z, or their complements, that has the value 1 if and only if

a)x=y=0, z=1

b)x=0, y=1, z=0

c)x=0, y=z=1

d)x=y=z=0

Short Answer

Expert verified

The Boolean products are

  1. The product is\(\overline x \cdot \overline y \cdot z\).
  2. The product is\(\overline x \cdot y \cdot \overline z \).
  3. The product is\(\overline x yz\).
  4. The product is \(\overline x \overline y \overline z \).

Step by step solution

01

Definition

The complements of an elements \(\overline 0 = 1\) and \(\overline 1 = 0\).

The Boolean sum + or \(OR\) is 1 if either term is 1.

The Boolean product (.) or \(AND\) is 1 if both terms are 1.

02

(a) Find the result.

Here \(x = 0,{\rm{ }}y = 0,{\rm{ }}z = 1\).

If a Boolean variable is 0, the complement of the Boolean variable is 1.

\(\begin{array}{l}\overline x = 1\\\overline y = 0\\z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot \overline y \cdot z = 1\).

Thus, the Boolean product is \(\overline x \cdot \overline y \cdot z\).

03

(b) Determine the result. 

Here \(x = 0,{\rm{ }}y = 1,{\rm{ }}z = 0\).

If a Boolean variable is 0, the complement of the Boolean variable is 1.

\(\begin{array}{l}\overline x = 1\\y = 1\\\overline z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot y \cdot \overline z = 1\).

Thus, the Boolean product is \(\overline x y\overline z \).

04

(c) Evaluate the result.

Here \(x = 0,{\rm{ }}y = 1,{\rm{ }}z = 1\).

If a Boolean variable is 0,the complement of the Boolean variable is 1.

\(\begin{array}{c}\overline x = 1\\y = 1\\z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot y \cdot z = 1\).

Thus, the Boolean product is \(\overline x yz\).

05

(d) Find the result. 

Here \(x = 0,{\rm{ }}y = 0,{\rm{ }}z = 0\).

If a Boolean variable is 0,the complement of the Boolean variable is 1.

\(\begin{array}{c}\overline x = 1\\\overline y = 1\\\overline z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot \overline y \cdot \overline z = 1\).

Therefore, the Boolean product is \(\overline x \overline y \overline z \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(a)\) What does it mean for a set of operators to be functionally complete\(?\)

\(b)\)Is the set \(\{ {\bf{ + }}, \cdot \} \) functionally complete\(?\)

\(c)\)Are there sets of a single operator that are functionally complete\(?\)

Use NAND gates to construct circuits with these outputs.

\(\begin{array}{l}{\bf{a)}}\overline {\bf{x}} \\{\bf{b)x + y}}\\{\bf{c)xy}}\\{\bf{d)x}} \oplus {\bf{y}}\end{array}\)

Let \({\bf{x}}\) and \({\bf{y}}\) belong to \(\left\{ {{\bf{0,1}}} \right\}\). Does it necessarily follow that \({\bf{x = y}}\) if there exists a value \({\bf{z}}\) in \(\left\{ {{\bf{0,1}}} \right\}\) such that,

\(\begin{array}{l}{\bf{a) xz = yz?}}\\{\bf{b) x + z = y + z?}}\\{\bf{c) x}} \oplus {\bf{z = y}} \oplus {\bf{z?}}\\{\bf{d) x}} \downarrow {\bf{z = y}} \downarrow {\bf{z?}}\\{\bf{e) x}}|{\bf{z = y}}|z{\bf{?}}\end{array}\)

A Boolean function \({\bf{F}}\) is called self-dual if and only if \({\bf{F}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}} \right){\bf{ = }}\overline {{\bf{F}}\left( {{{{\bf{\bar x}}}_{\bf{1}}}{\bf{, \ldots ,}}{{{\bf{\bar x}}}_{\bf{n}}}} \right)} \).

Show that cells in a \({\bf{K}}\)-map for Boolean functions in five variables represent minterms that differ in exactly one literal if and only if they are adjacent or are in cells that become adjacent when the top and bottom rows and cells in the first and eighth columns, the first and fourth columns, the second and seventh columns, the third and sixth columns, and the fifth and eighth columns are considered adjacent.

Show that the relation \( \le \) is a partial ordering on the set of Boolean functions of degree \(n\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free