Why is a Skolem theory model complete?
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Let $Delta$ be a Skolem theory. Let $M$ be a model of $Delta$ and $N$ a substructure of $M$. Then we need to proof that for every $L_N$-sentence $phi$ we have the equivalence $N models phi iff M models phi$. By the Tarski-Vaught test it suffices to find an $a in N$ for all $L_N$-sentences of the form $exists x phi(x)$ such that $M models phi(a)$. I'm stuck at finding this $a$. Because $Delta models exists x phi(x) tophi(c)$, it would be sufficient to proof that $N models Delta$, but I'm not sure whether that is true or how to otherwise approach the problem.
logic model-theory
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
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add a comment |
$begingroup$
Let $Delta$ be a Skolem theory. Let $M$ be a model of $Delta$ and $N$ a substructure of $M$. Then we need to proof that for every $L_N$-sentence $phi$ we have the equivalence $N models phi iff M models phi$. By the Tarski-Vaught test it suffices to find an $a in N$ for all $L_N$-sentences of the form $exists x phi(x)$ such that $M models phi(a)$. I'm stuck at finding this $a$. Because $Delta models exists x phi(x) tophi(c)$, it would be sufficient to proof that $N models Delta$, but I'm not sure whether that is true or how to otherwise approach the problem.
logic model-theory
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
$endgroup$
1
$begingroup$
$NmodelsDelta$ is part of the definition of $Delta$ being model complete. (In other words you don’t need all substructures of $M$ to be elementary, obly those which are also models of $Delta.$)
$endgroup$
– spaceisdarkgreen
Jan 16 at 17:46
add a comment |
$begingroup$
Let $Delta$ be a Skolem theory. Let $M$ be a model of $Delta$ and $N$ a substructure of $M$. Then we need to proof that for every $L_N$-sentence $phi$ we have the equivalence $N models phi iff M models phi$. By the Tarski-Vaught test it suffices to find an $a in N$ for all $L_N$-sentences of the form $exists x phi(x)$ such that $M models phi(a)$. I'm stuck at finding this $a$. Because $Delta models exists x phi(x) tophi(c)$, it would be sufficient to proof that $N models Delta$, but I'm not sure whether that is true or how to otherwise approach the problem.
logic model-theory
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
$endgroup$
Let $Delta$ be a Skolem theory. Let $M$ be a model of $Delta$ and $N$ a substructure of $M$. Then we need to proof that for every $L_N$-sentence $phi$ we have the equivalence $N models phi iff M models phi$. By the Tarski-Vaught test it suffices to find an $a in N$ for all $L_N$-sentences of the form $exists x phi(x)$ such that $M models phi(a)$. I'm stuck at finding this $a$. Because $Delta models exists x phi(x) tophi(c)$, it would be sufficient to proof that $N models Delta$, but I'm not sure whether that is true or how to otherwise approach the problem.
logic model-theory
logic model-theory
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
asked Jan 16 at 16:57
Pel de PindaPel de Pinda
882416
882416
1
$begingroup$
$NmodelsDelta$ is part of the definition of $Delta$ being model complete. (In other words you don’t need all substructures of $M$ to be elementary, obly those which are also models of $Delta.$)
$endgroup$
– spaceisdarkgreen
Jan 16 at 17:46
add a comment |
1
$begingroup$
$NmodelsDelta$ is part of the definition of $Delta$ being model complete. (In other words you don’t need all substructures of $M$ to be elementary, obly those which are also models of $Delta.$)
$endgroup$
– spaceisdarkgreen
Jan 16 at 17:46
1
1
$begingroup$
$NmodelsDelta$ is part of the definition of $Delta$ being model complete. (In other words you don’t need all substructures of $M$ to be elementary, obly those which are also models of $Delta.$)
$endgroup$
– spaceisdarkgreen
Jan 16 at 17:46
$begingroup$
$NmodelsDelta$ is part of the definition of $Delta$ being model complete. (In other words you don’t need all substructures of $M$ to be elementary, obly those which are also models of $Delta.$)
$endgroup$
– spaceisdarkgreen
Jan 16 at 17:46
add a comment |
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$begingroup$
$NmodelsDelta$ is part of the definition of $Delta$ being model complete. (In other words you don’t need all substructures of $M$ to be elementary, obly those which are also models of $Delta.$)
$endgroup$
– spaceisdarkgreen
Jan 16 at 17:46