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Math Help - Prove/disprove using logical using logical arguments

  1. #1
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    Prove/disprove using logical using logical arguments

    Prove or disprove the following using a logical argument.
    The universe of discourse for x,y,z is integers.



    \forall [p(x) \lor q(x)]   \Leftarrow \forall p(x) \lor \forall q(x)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aaronrj View Post
    Prove or disprove the following using a logical argument.
    The universe of discourse for x,y,z is integers.



    \forall [p(x) \lor q(x)]   \Leftarrow \forall p(x) \lor \forall q(x)
    I don't think what you wrote makes sense. plus writing the arrow backwards looks odd.

    do you mean (\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)] ?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    I don't think what you wrote makes sense. plus writing the arrow backwards looks odd.

    do you mean (\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)] ?
    I looked at all his posts. I don't think he wants help. He seems to be making fun of people by posting rubbish on purpose.
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    Quote Originally Posted by Jhevon View Post
    do you mean (\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)] ?
    Yes Jhevon, that is right, our substitute instructor corrected that homework problem today. I am in the same class as the original poster, so I assure you, he is not just posting rubbish!
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    Quote Originally Posted by inferno View Post
    Yes Jhevon, that is right, our substitute instructor corrected that homework problem today. I am in the same class as the original poster, so I assure you, he is not just posting rubbish!
    Your friend is lacking the courtesy in acknowledging Jhevon.
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  6. #6
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    Quote Originally Posted by inferno View Post
    Yes Jhevon, that is right, our substitute instructor corrected that homework problem today. I am in the same class as the original poster, so I assure you, he is not just posting rubbish!

    It's a theorem of first-order predicate logic.
    Proof?
    First thing to notice is that the antecedent is a disjunction.
    Informally, think proof by cases.
    Or formally, say in an NDS, think VE.
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