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ajprincess
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prove\[(\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge q)=r\]
 2 years ago
 2 years ago
ajprincess Group Title
prove\[(\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge q)=r\]
 2 years ago
 2 years ago

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ajprincess Group TitleBest ResponseYou've already chosen the best response.0
dw:1338094648240:dw prove this using laws of algebra of propositions.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
@amistre64, @jamesJ, @turingtest
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
plzzz help me.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
dw:1338095286775:dw Except this other all are right.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
ya sure. thanx. :) can u help me to prove it if u dnt mind.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
not used to this kind of thing,; I'm not sure what the technically right approach would be start making assumptions I guess...
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
ha k. thanx. @jamesJ if u dnt can u plzzz help me with this question?
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
How have you proved these sorts of things in the past? The most straightforward way to show this would be write out a truth table with eight rows, one row for each of the truth value combinations of p, q and r
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
i am ok with proving it using truth table. this is one of my past paper question. they hav askd to prove t using laws of algebra of propositions. bt I am nt ued to it.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
@asnaseer can u plzz help me vth this if u dnt mind.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
@JamesJ u too. can u help me plzzz
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
So you want to prove this now using propositions?
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
I tried bt was unable to get t.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
I just constructed the truth table and the question as written here looks wrong. Could you please triple check it.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
I think it might be this, but you check it! \[ (\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge r)=r \]
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
ya that's it. sooo sorrry for the mistake.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
ok, well you factor out ^ r from each of these terms and you have ( ~p^~q) v q v p ) ^ r
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
and ( ~p^~q) v q v p is always true. Hence the entire expression is equivalent to TRUE ^ r = r
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
bt hw do v knw that ( ~p^~q) v q v p is always true?
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
because ~p^~q = ~(p v q) and hence (~p^~q) v q v p = ~(p v q) v (p v q)
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.3
...and X v ~X is always true.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
I get t nw. thanxxxxxxxx a lot.
 2 years ago
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