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Math Help - predicate forms implications

  1. #1
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    predicate forms implications



    Attempt:

    (a) Let I be an interpetation, and let v be an I-assignment.
    Suppose that I \models_v A. We want I \models_v \exists x A. And we know that I \models_v \exists x A iff there is some d \in D with I \models_{v \frac{d}{x}}A. But how do we show that we have such a d?


    (b) Suppose I \models_v \exists x Ax, I have to show that I \not{\models_v} Ax. But I'm confused because assuming I \models_v \exists x Ax means there is some d in the domain with I \models_{v \frac{d}{x}} Ax. What could I do? Are there better ways to show that the implication doesn't hold?
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  2. #2
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    Re: predicate forms implications

    Again, one should use the property that if x is not free in B, then I\models_{v\frac{d}{x}}B iff I\models_v B.

    For (a), suppose that v maps x to d, i.e., v=v'\frac{d}{x} for some smaller v'. Then I\models_v A means I\models_{v'\frac{d}{x}}A, so I\models_{v'}\exists x\,A and by the property above, I\models_v\exists x\,A.

    For (b), suppose that I\models_v\exists x\,Ax, which means that I\models_{v\frac{d}{y}}Ay for some d. However, v does not have to map x to this particular d, so there is no reason for I\models_v Ax. You should construct a counterexample.
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  3. #3
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    Re: predicate forms implications

    Quote Originally Posted by emakarov View Post
    For (b), suppose that I\models_v\exists x\,Ax, which means that I\models_{v\frac{d}{y}}Ay for some d. However, v does not have to map x to this particular d, so there is no reason for I\models_v Ax. You should construct a counterexample.
    To give a counter example, can I be an interpetation holding iff the variable x is even? Then I \models_v \exists x A means that I \models_{v \frac{d}{x}}Ax for some d which is even. On the right hand side of the implication we have Ax without any quantifiers before it. So this x doesn't have to be d, it could be an odd number e. Is this correct?


    P.S. Normally, when we give counter-examples for implications involving 2 variables we use an interpetation such that A^I(m,n) holding iff m<n, then it's easy to use natural numbers as the domain to get a counter example. But here since here we have a single variable I'm not sure what kind of counter-example to use...
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  4. #4
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    Re: predicate forms implications

    Let the carrier of I be natural numbers and A^I(m) hold iff m is even. Let also v map a single variable x to 3. Then I\models_v\exists x\,Ax, but I\not\models_v Ax.
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