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