Let $M$ be a model over the $\Sigma$ signature. Let $\Delta$ be a theory of model $M$. Let $\phi$ be a sentence over $\Sigma$ signature.
$M \models \phi$ Does it implies $\Delta \models \phi$.
Why?
Models and their theories.
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logic
first-order-logic
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1Recall that the *theory* of a model $M$ is the set of all sentences true in $M$. – 2017-01-27
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0Recall that for a set $\Delta$ of sentences, $\Delta \vDash \phi$ iff every model of $\Delta$ is also a model of $\phi$. – 2017-01-27
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0Yes, it is obvious. Just $\phi \in \Delta$ so, $\Delta \models \phi$, right? – 2017-01-27
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0***The*** theory of a model $M$ is the set of all sentences true in $M.$ Not sure if that's what you mean by "***a*** theory of model $M.$ – 2017-01-27
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0ok. So, if $\Delta$ has only one model $M$ then if $M \models \phi$ then $\Delta \models \phi$. But, if $\Delta$ has more model than 1 then it isn't true. – 2017-01-27
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0If $M$ is infinite, then "$\Delta$ has only one model $M$" is not possible. – 2017-01-27
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0What do you mean? Do you mean that: If $\Delta$ has a model that is infinite it means that there is ifinite models? Why? – 2017-01-27