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

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

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

liliy
 2 years ago
Best 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

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

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

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

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

mathmate
 2 years ago
Best 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 ?

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

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

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1So 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.
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