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

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|dw:1338094648240:dw| prove this using laws of algebra of propositions.
plzzz help me.

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|dw:1338095286775:dw| Except this other all are right.
ya sure. thanx. :) can u help me to prove it if u dnt mind.
not used to this kind of thing,; I'm not sure what the technically right approach would be start making assumptions I guess...
ha k. thanx. @jamesJ if u dnt can u plzzz help me with this question?
How have you proved these sorts of things in the past? The most straight-forward 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
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.
@asnaseer can u plzz help me vth this if u dnt mind.
@JamesJ u too. can u help me plzzz
So you want to prove this now using propositions?
I tried bt was unable to get t.
I just constructed the truth table and the question as written here looks wrong. Could you please triple check it.
ha k.
I think it might be this, but you check it! \[ (\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge r)=r \]
ya that's it. sooo sorrry for the mistake.
ok, well you factor out ^ r from each of these terms and you have ( ~p^~q) v q v p ) ^ r
and ( ~p^~q) v q v p is always true. Hence the entire expression is equivalent to TRUE ^ r = r
bt hw do v knw that ( ~p^~q) v q v p is always true?
because ~p^~q = ~(p v q) and hence (~p^~q) v q v p = ~(p v q) v (p v q)
...and X v ~X is always true.
I get t nw. thanxxxxxxxx a lot.

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