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Show that “\Gamma \models S \Rightarrow \Gamma \vdash S” entails “if \Gamma \nvdash P \And \sim P then \Gamma is satisfiable”
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So I think this is what we may want to do. Suppose for contradiction that \Gamma is not satisfiable. This means that \Gamma has no models. Now, fix some sentence P and let S \equiv P \wedge \neg P ...
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There are a few problems with what your definition of E_0. You haven't defined E_0 on the class of all tuples, just on the pairs (the 2-tuples). When should singletons be E_0-related? What ...
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Assuming that all the results you're invoking (for example that a maximal consistent set must contain either \phi or \neg\phi) are available, your argument looks correct but unnecessarily ...
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https://math.stackexchange.com/questions/661222/a-truth-definition-wrong-but-where
It seems that you are trying to inductively defined a class of the form S = \{(\phi, \vec{x}) : \phi \in \text{Form} \wedge \vec{x} \in V^{\omega} \wedge V \vDash \phi(\vec{x})\}. The trouble in ...
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https://math.stackexchange.com/questions/568236/comparing-models-through-partial-isomorphisms
As you have noticed, the condition implies that \mathfrak{A} and \mathfrak{B} satisfy the same existential sentences. It follows that they also satisfy the same universal sentences. And, of ...
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Certainly. Consider \cal L to be the language containing one binary relation symbol <. T is the theory stating that < is a linear order (irreflexive, transitive and total). \sigma is the ...
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