# entailment - interpretation

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• Dec 18th 2010, 05:09 PM
domad
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)

Thanks
• Dec 18th 2010, 08:22 PM
snowtea
Consider any model
$\mathcal{M} \models \forall x (F(x) \leftrightarrow (P(x) \vee (G(x) \wedge \lnot N(x))))$

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

Again by semantics of satisfaction, this means that $\mathcal{M}[\mu (x) = v]$ 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 $I$, then:
$v\in I(F)$ iff ( $v\in I(P)$ or ( $v\in I(G)$ and not $v\in I(N)$))

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

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

since the valuation for x was arbitrary
$\mathcal{M} \models \forall x ((G(x) \wedge \lnot N(x)) \rightarrow F(x))$