Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

liliy

  • 3 years ago

discrete math:(attached)

  • This Question is Closed
  1. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is this right?

    1 Attachment
  2. josiahh
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you proving a tautology?

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

    ya

  4. Joseph91
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    p bar means negation p?

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

    It is not a tautology, use a->b <=> ~a or b and watch out for order of operations.

  6. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Joseph91 , yes @mathmate , it is given that it is tautology, and you must prove it with equilence

  7. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    see... you just need to PROVE it, .. but my teacher doesnt want the table..

    1 Attachment
  8. mathmate
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You're right, it is a tautology, I made a mistake in one of the terms. Still, start with the distributive property: ~a^(a or b) <=> ~a^a or ~a^b then use ~a and ~b <=> F and then ~a or b <=> b That leaves you with ~p^q -> q and you'll get it after that using a->b <=> ~a or b

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

    Do you want me to show all the work?

  10. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ya please

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

    ~p^(p or q) ->q ~p^p or ~p^q -> q

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

    ~p^(p or q) ->q ~p^p or ~p^q -> q (distributive property) ok so far?

  13. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup

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

    ~p^p = F, and F or anything is anything, so ~p^q -> q ok so far @liliy ?

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

    @liliy still there?

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

    ya

  17. Joseph91
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    p implication q is equivalent to -pvq -[-p^(pvq)]vq -[-p^pv-p^q]vq -[Fv-p^q]vq -[-p^q]vq pvT=T

  18. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    woah. wait.. let me digest that

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

    So Joseph91 has completed the proof. The last step T is from ~q or q = T after substituting ~(~p^q) by p or ~q due to de morgan's theorem.

  20. liliy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    got it! thanks

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

    You're welcome! :)

  22. 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