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ajprincess

  • 2 years ago

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

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  1. ajprincess
    • 2 years ago
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    |dw:1338094648240:dw| prove this using laws of algebra of propositions.

  2. ajprincess
    • 2 years ago
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    @amistre64, @jamesJ, @turingtest

  3. ajprincess
    • 2 years ago
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    plzzz help me.

  4. ajprincess
    • 2 years ago
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    |dw:1338095286775:dw| Except this other all are right.

  5. ajprincess
    • 2 years ago
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    ya sure. thanx. :) can u help me to prove it if u dnt mind.

  6. TuringTest
    • 2 years ago
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    not used to this kind of thing,; I'm not sure what the technically right approach would be start making assumptions I guess...

  7. ajprincess
    • 2 years ago
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    ha k. thanx. @jamesJ if u dnt can u plzzz help me with this question?

  8. ajprincess
    • 2 years ago
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    *mind

  9. JamesJ
    • 2 years ago
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    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

  10. ajprincess
    • 2 years ago
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    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.

  11. ajprincess
    • 2 years ago
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    @asnaseer can u plzz help me vth this if u dnt mind.

  12. ajprincess
    • 2 years ago
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    @JamesJ u too. can u help me plzzz

  13. JamesJ
    • 2 years ago
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    So you want to prove this now using propositions?

  14. ajprincess
    • 2 years ago
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    ya.

  15. ajprincess
    • 2 years ago
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    I tried bt was unable to get t.

  16. JamesJ
    • 2 years ago
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    I just constructed the truth table and the question as written here looks wrong. Could you please triple check it.

  17. ajprincess
    • 2 years ago
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    ha k.

  18. JamesJ
    • 2 years ago
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    I think it might be this, but you check it! \[ (\neg p\wedge(\neg q\wedge r))\vee(q\wedge r)\vee(p\wedge r)=r \]

  19. ajprincess
    • 2 years ago
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    ya that's it. sooo sorrry for the mistake.

  20. JamesJ
    • 2 years ago
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    ok, well you factor out ^ r from each of these terms and you have ( ~p^~q) v q v p ) ^ r

  21. JamesJ
    • 2 years ago
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    and ( ~p^~q) v q v p is always true. Hence the entire expression is equivalent to TRUE ^ r = r

  22. ajprincess
    • 2 years ago
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    bt hw do v knw that ( ~p^~q) v q v p is always true?

  23. JamesJ
    • 2 years ago
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    because ~p^~q = ~(p v q) and hence (~p^~q) v q v p = ~(p v q) v (p v q)

  24. JamesJ
    • 2 years ago
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    ...and X v ~X is always true.

  25. ajprincess
    • 2 years ago
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    I get t nw. thanxxxxxxxx a lot.

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