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Math Help - Easy doubt on second-order logic

  1. #1
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    Easy doubt on second-order logic

    Why does the negation of the operator "For all" results in the operator "There exists", and vice-versa?
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  2. #2
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    Re: Easy doubt on second-order logic

    This happens in first-order logic as well. Basically, it's common sense. You can prove that $M\models\neg\forall x\,P(x)$ iff $M\models\exists x\,\neg P(x)$, but this still relies on the feature of informal, meta-level quantifiers that if it is not the case that a property holds for all elements, then it does not hold for some element, i.e., its negation holds for that element.
    Last edited by emakarov; June 24th 2014 at 06:09 PM.
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