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Thread: entailment - interpretation

  1. #1
    Dec 2010

    entailment - interpretation

    Hello All

    To prove the following, I have tried resolution with success, but I need to do it using semantics of interpretation; that every interpretation that satisfies the first sentence satisifies the second. I am not sure on how to go about this ....... Your help is extrememly appreciated.

    F(x) iff P(x) or (G(x) and ~N(x))

    G(x) and ~N(x) implies F(x)

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  2. #2
    Senior Member
    Dec 2010
    Consider any model
    $\displaystyle \mathcal{M} \models \forall x (F(x) \leftrightarrow (P(x) \vee (G(x) \wedge \lnot N(x))))$

    Using semantics of satisfaction this means, for any valuation $\displaystyle \mu (x) = v $
    $\displaystyle \mathcal{M}[\mu (x) = v]} \models F(x) \leftrightarrow (P(x) \vee (G(x) \wedge \lnot N(x)))$

    Again by semantics of satisfaction, this means that $\displaystyle \mathcal{M}[\mu (x) = v]$ models a truth table for the logical expression shown (of course, rigorously you have to decompose the semantics of all the logical connectives one by one). If the model interpretation is $\displaystyle I$, then:
    $\displaystyle v\in I(F) $ iff ($\displaystyle v\in I(P) $ or ($\displaystyle v\in I(G) $ and not $\displaystyle v\in I(N) $))

    When the truth table of $\displaystyle v\in I(...)$ for the above expression evaluates to true, you can show that
    $\displaystyle v\in I(G) $ and (not $\displaystyle v\in I(N) $) implies $\displaystyle v\in I(F) $ also evaluates to true.

    This means
    $\displaystyle \mathcal{M}[\mu (x) = v]} \models (G(x) \wedge \lnot N(x)) \rightarrow F(x)$

    since the valuation for x was arbitrary
    $\displaystyle \mathcal{M} \models \forall x ((G(x) \wedge \lnot N(x)) \rightarrow F(x))$
    Last edited by snowtea; Dec 18th 2010 at 08:38 PM.
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