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

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

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

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

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

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

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

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

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