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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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  1. ajprincess Group Title
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    |dw:1338094648240:dw| prove this using laws of algebra of propositions.

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

    • 2 years ago
  3. ajprincess Group Title
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    plzzz help me.

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

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

    • 2 years ago
  6. TuringTest Group Title
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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...

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

    • 2 years ago
  8. ajprincess Group Title
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    *mind

    • 2 years ago
  9. JamesJ Group Title
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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

    • 2 years ago
  10. ajprincess Group Title
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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.

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

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

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

    • 2 years ago
  14. ajprincess Group Title
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    ya.

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

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

    • 2 years ago
  17. ajprincess Group Title
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    ha k.

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

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

    • 2 years ago
  20. JamesJ Group Title
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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

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

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

    • 2 years ago
  23. JamesJ Group Title
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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)

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

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

    • 2 years ago
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