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liliy

  • 2 years ago

discrete math:(attached)

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  1. liliy
    • 2 years ago
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    is this right?

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  2. josiahh
    • 2 years ago
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    are you proving a tautology?

  3. liliy
    • 2 years ago
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    ya

  4. Joseph91
    • 2 years ago
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    p bar means negation p?

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

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

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

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  8. mathmate
    • 2 years ago
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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

  9. mathmate
    • 2 years ago
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    Do you want me to show all the work?

  10. liliy
    • 2 years ago
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    ya please

  11. mathmate
    • 2 years ago
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    ~p^(p or q) ->q ~p^p or ~p^q -> q

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

  13. liliy
    • 2 years ago
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    yup

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

  15. mathmate
    • 2 years ago
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    @liliy still there?

  16. liliy
    • 2 years ago
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    ya

  17. Joseph91
    • 2 years ago
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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

  18. liliy
    • 2 years ago
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    woah. wait.. let me digest that

  19. mathmate
    • 2 years ago
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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.

  20. liliy
    • 2 years ago
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    got it! thanks

  21. mathmate
    • 2 years ago
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    You're welcome! :)

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