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prove\[(\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge q)=r\]
 one year ago
 one year ago
prove\[(\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge q)=r\]
 one year ago
 one year ago

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ajprincessBest ResponseYou've already chosen the best response.0
dw:1338094648240:dw prove this using laws of algebra of propositions.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
@amistre64, @jamesJ, @turingtest
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
dw:1338095286775:dw Except this other all are right.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ya sure. thanx. :) can u help me to prove it if u dnt mind.
 one year ago

TuringTestBest 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...
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ha k. thanx. @jamesJ if u dnt can u plzzz help me with this question?
 one year ago

JamesJBest 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
 one year ago

ajprincessBest 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.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
@asnaseer can u plzz help me vth this if u dnt mind.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
@JamesJ u too. can u help me plzzz
 one year ago

JamesJBest ResponseYou've already chosen the best response.3
So you want to prove this now using propositions?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
I tried bt was unable to get t.
 one year ago

JamesJBest 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.
 one year ago

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

ajprincessBest ResponseYou've already chosen the best response.0
ya that's it. sooo sorrry for the mistake.
 one year ago

JamesJBest 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
 one year ago

JamesJBest 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
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
bt hw do v knw that ( ~p^~q) v q v p is always true?
 one year ago

JamesJBest 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)
 one year ago

JamesJBest ResponseYou've already chosen the best response.3
...and X v ~X is always true.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
I get t nw. thanxxxxxxxx a lot.
 one year ago
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