Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ketz

  • 2 years ago

f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent

  • This Question is Closed
  1. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you use the identity, A+A' = 1 so, f=x'y'z+(x'y'z)' = 1 (where x'y'z can be considered as A)

  2. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but if i proceed by expanding the bracket! I'm not getting the answer!!!!

  3. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    expanding the bracket ? like x'y'z +xyz' ? no need for that, you can use A=x'y'z and that A+A'=1 you can also verify it using truth table.

  4. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeh i know but i want to proceed by expanding the bracket using de morgan's laws. and am not getting the answer. can anyone try it by this method!!

  5. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let me try on, left of + x'y' = (x+y)' ---> x'y'z =(x+y)'z on right of + (x'y'z)' = (x+y+z') hmm

  6. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    f=x'y'z+(x'y'z)' x'y'z + x + y + z' y'z + x + y + z' x + y'z + y + z' x + y + z + z' x + y + 1 1

  7. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    in 2ns line use this rule : x+x'y = x+y

  8. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can you prove the law x+x'y=x+y???

  9. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    By Boolean Algebra!

  10. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    did u try to prove already... it wont be difficult i think

  11. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No

  12. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How can you prove or simplify it by boolean algebra??

  13. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    do u knw distributive property for AND logic ?

  14. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    x + (y.z) = (x+y).(x+z)

  15. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily

  16. ketz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok got it!

  17. ganeshie8
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    great :)

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.