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liliy Group TitleBest ResponseYou've already chosen the best response.0
is this right?
 2 years ago

josiahh Group TitleBest ResponseYou've already chosen the best response.0
are you proving a tautology?
 2 years ago

Joseph91 Group TitleBest ResponseYou've already chosen the best response.0
p bar means negation p?
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
It is not a tautology, use a>b <=> ~a or b and watch out for order of operations.
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
@Joseph91 , yes @mathmate , it is given that it is tautology, and you must prove it with equilence
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
see... you just need to PROVE it, .. but my teacher doesnt want the table..
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.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
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Do you want me to show all the work?
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
~p^(p or q) >q ~p^p or ~p^q > q
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
~p^(p or q) >q ~p^p or ~p^q > q (distributive property) ok so far?
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
~p^p = F, and F or anything is anything, so ~p^q > q ok so far @liliy ?
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
@liliy still there?
 2 years ago

Joseph91 Group TitleBest ResponseYou've already chosen the best response.0
p implication q is equivalent to pvq [p^(pvq)]vq [p^pvp^q]vq [Fvp^q]vq [p^q]vq pvT=T
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
woah. wait.. let me digest that
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.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.
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
got it! thanks
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
You're welcome! :)
 2 years ago
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