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Math Help - Predicate Forms

  1. #1
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    Predicate Forms



    Here's my attempt so far in proving the equivalence:

    The question says "x does not occur freely in A" (which means it is either the variable immediately following a quantifier or it is in the scope of some quantifier involving x). Let I be an interpetation with domain on D, and v be an I-assignment. Suppose I \models_v \exists x (A \vee B). We wish to show that I \models_v (A \vee \exists x B). I don't know if this is correct but I'm considering two cases depending on wether A^I(d) holds for all d \in D:

    Case 1: A^I(d) holds for all d \in D. So I \models_{v \frac{d}{x}} A so I \models_{v \frac{d}{x}} (A \vee \exists x B). Is this correct?

    Case 2: There is some d \in D such that A^I(d) does not hold... And we assumed that there is a d such that I \models_{v\frac{d}{x}} (A \vee B). But how can I complete this part?

    I greatly appreciate any help on how to prove this question...
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  2. #2
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    Re: Predicate Forms

    The exact details depend on the definition of I\models A, but here are some remarks.

    What is A^l(d)? The formula A does not depend on x, so there is no sense in providing d to it.

    The proof of equivalence should not require considering whether A^I(d) holds for all d. One should use the fact that if x is not free in A, then I\models_{v\frac{d}{x}}A iff I\models_{v}A. So, if I \models_v \exists x (A \vee B), then there is a d\in D such that I\models_{v\frac{d}{x}}A\lor B. If I\models_{v\frac{d}{x}}A, then I\models_{v}A and so I \models_v A\lor\exists x\, B. If I\models_{v\frac{d}{x}}B, then I\models_{v}\exists x\,B and so I \models_v A\lor\exists x\, B.
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