anonymous
  • anonymous
discrete math:(attached)
Mathematics
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anonymous
  • anonymous
discrete math:(attached)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
is this right?
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anonymous
  • anonymous
are you proving a tautology?
anonymous
  • anonymous
ya

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anonymous
  • anonymous
p bar means negation p?
mathmate
  • mathmate
It is not a tautology, use a->b <=> ~a or b and watch out for order of operations.
anonymous
  • anonymous
@Joseph91 , yes @mathmate , it is given that it is tautology, and you must prove it with equilence
anonymous
  • anonymous
see... you just need to PROVE it, .. but my teacher doesnt want the table..
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mathmate
  • mathmate
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
mathmate
  • mathmate
Do you want me to show all the work?
anonymous
  • anonymous
ya please
mathmate
  • mathmate
~p^(p or q) ->q ~p^p or ~p^q -> q
mathmate
  • mathmate
~p^(p or q) ->q ~p^p or ~p^q -> q (distributive property) ok so far?
anonymous
  • anonymous
yup
mathmate
  • mathmate
~p^p = F, and F or anything is anything, so ~p^q -> q ok so far @liliy ?
mathmate
  • mathmate
@liliy still there?
anonymous
  • anonymous
ya
anonymous
  • anonymous
p implication q is equivalent to -pvq -[-p^(pvq)]vq -[-p^pv-p^q]vq -[Fv-p^q]vq -[-p^q]vq pvT=T
anonymous
  • anonymous
woah. wait.. let me digest that
mathmate
  • mathmate
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.
anonymous
  • anonymous
got it! thanks
mathmate
  • mathmate
You're welcome! :)

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