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Math Help - Reversing quantifiers

  1. #1
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    Reversing quantifiers

    Any hints on the strategy to derive ∃y∀x(Ax & By) from ∀x∃y(Ax & By) in fitch-format? I've been stuck on this one for weeks.
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  2. #2
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    Re: Reversing quantifiers

    "There exists x (property of x)" can be changed to "not For all x, not (property of x)." Same thing in converse. Is that enough of a hint?
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  3. #3
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    Re: Reversing quantifiers

    Quote Originally Posted by Annatala View Post
    "There exists x (property of x)" can be changed to "not For all x, not (property of x)." Same thing in converse.
    Yes, but I am not sure how this helps here.

    Starting from \forall x\exists y\,(Ax\land By), we can instantiate x with an arbitrary term t (e.g., a fresh variable) and get a y such that At\land By, in particular, By. So, we are left to show \forall x\,(Ax\land By). Fix an arbitrary x, instantiate the universal variable in \forall x\exists y\,(Ax\land By) with x and thus get Ax.
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  4. #4
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    Re: Reversing quantifiers

    Emakarov, I'm not sure it's that easy to do in a fitch-format derivation. A violation of rules involving the eigen-variable occur. So, for example, I can't do this:
    ∀x∃y(Ax & By)
    ∃y(Az & By)
    -Az&Bw
    -Az
    -∀xAx

    ('-' indicates a sub-derivation). The last line in the sub-derivation is a violation. I can't universally quantify anything that's free in an assumption.
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  5. #5
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    Re: Reversing quantifiers

    Here is a derivation in tree format.

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  6. #6
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    Re: Reversing quantifiers

    Quote Originally Posted by emakarov View Post
    Yes, but I am not sure how this helps here.
    My bad. I don't understand this question at all, although I thought I did.
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