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liliy

discrete math:(attached)

  • one year ago
  • one year ago

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  1. liliy
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    is this right?

    • one year ago
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  2. josiahh
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    are you proving a tautology?

    • one year ago
  3. liliy
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    ya

    • one year ago
  4. Joseph91
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    p bar means negation p?

    • one year ago
  5. mathmate
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    It is not a tautology, use a->b <=> ~a or b and watch out for order of operations.

    • one year ago
  6. liliy
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    @Joseph91 , yes @mathmate , it is given that it is tautology, and you must prove it with equilence

    • one year ago
  7. liliy
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    see... you just need to PROVE it, .. but my teacher doesnt want the table..

    • one year ago
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  8. mathmate
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    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

    • one year ago
  9. mathmate
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    Do you want me to show all the work?

    • one year ago
  10. liliy
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    ya please

    • one year ago
  11. mathmate
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    ~p^(p or q) ->q ~p^p or ~p^q -> q

    • one year ago
  12. mathmate
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    ~p^(p or q) ->q ~p^p or ~p^q -> q (distributive property) ok so far?

    • one year ago
  13. liliy
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    yup

    • one year ago
  14. mathmate
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    ~p^p = F, and F or anything is anything, so ~p^q -> q ok so far @liliy ?

    • one year ago
  15. mathmate
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    @liliy still there?

    • one year ago
  16. liliy
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    ya

    • one year ago
  17. Joseph91
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    p implication q is equivalent to -pvq -[-p^(pvq)]vq -[-p^pv-p^q]vq -[Fv-p^q]vq -[-p^q]vq pvT=T

    • one year ago
  18. liliy
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    woah. wait.. let me digest that

    • one year ago
  19. mathmate
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    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.

    • one year ago
  20. liliy
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    got it! thanks

    • one year ago
  21. mathmate
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    You're welcome! :)

    • one year ago
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