Boolean Expression Simplification

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SUMMARY

This discussion focuses on the simplification of Boolean expressions using laws of Boolean algebra. The first expression, (x'y' + xy' + x'y), simplifies to y' using the consensus theorem and absorption laws. The second expression, (p + q'p)(p + qr), ultimately simplifies to p, demonstrating the application of the idempotent law and the principle of cancellation in Boolean algebra. Participants emphasize the importance of recognizing cancellation opportunities and applying relevant laws without a strict order.

PREREQUISITES
  • Understanding of Boolean algebra laws, including consensus theorem and absorption laws.
  • Familiarity with simplification techniques for Boolean expressions.
  • Knowledge of idempotent law and cancellation principles in Boolean algebra.
  • Ability to manipulate and factor Boolean expressions effectively.
NEXT STEPS
  • Study the consensus theorem in Boolean algebra.
  • Learn about the absorption laws and their applications in simplification.
  • Explore the idempotent law and its implications for Boolean expressions.
  • Practice simplifying complex Boolean expressions using various techniques.
USEFUL FOR

Students and professionals in computer science, electrical engineering, and anyone involved in digital logic design or optimization of Boolean functions.

mad_monkey_j
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Homework Statement


Simplify the following expressions
(x'y'+xy'+x'y)

(p+q'p)(p+qr)


Homework Equations


Laws of boolean algebra

The Attempt at a Solution


For the first one I've found many ways to solve it... something isn't right here
(x'y'+xy'+x'y)
=x'(y'+y)+xy'
=x'+xy'
=x'+x+y'
=y'

OR

=y'(x'+x)+x'y
=y'+x'y
=y'+x'+y
=x'
I'm definantly doing something wrong here... probably making up laws :biggrin:

For the second I'm not sure if the last line is even possible
(p+q'p)(p+qr)
=pp+pqr+ppq'+q'qr
=p+pqr+pq'+0
=p+pqr
=p? Can I do that with pqr having 3 variables?
 
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mad_monkey_j said:
=x'+xy'
=x'+x+y'
=y'
That is wrong. Apply:

a+a'b=a+b

For the second question, replace the fist factor by p+q'p=p. You made a mistake in the derivation, but the result is p.

ehild
 
Ahh so the first is
=x'+y'

and with the second if I change it to:
p(p+qr)
won't I end up with
=p+pqr in the end anyway?

Also how do you know what rule you're supose to use? Is there a specific order or is it just whatever you notice first?
 
mad_monkey_j said:
and with the second if I change it to:
p(p+qr)
won't I end up with
=p+pqr in the end anyway?

I said the result was correct. You made a mistake as the last term was q'qpr, but it canceled because of q'q.

mad_monkey_j said:
Also how do you know what rule you're suppose to use? Is there a specific order or is it just whatever you notice first?

There are no general recipes.
Use what you notice first, but it helps if you notice when something cancels, for example E+a=E, a+a'=E, aa'=0. Factor out what you can, so a+ab=a(E+b)=a. But it is not a rule; sometimes you need to apply other tricks.

ehild
 
ehild said:
I said the result was correct. You made a mistake as the last term was q'qpr, but it canceled because of q'q.



There are no general recipes.
Use what you notice first, but it helps if you notice when something cancels, for example E+a=E, a+a'=E, aa'=0. Factor out what you can, so a+ab=a(E+b)=a. But it is not a rule; sometimes you need to apply other tricks.

ehild

Ahh fair enough, thanks for the help.
 

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