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If we consider the sentences in ZFC:

EM that express that exist a model of ZFC.

SM that express that exist a standard transitive model of ZFC.

OM that express that exist a $\omega$-model of ZFC.

By Godel second incompleteness theorem we know that $\mathsf{ZFC}\nvdash \mathsf{EM}$

What is known about:

$\mathsf{ZFC+EM}\nvdash \mathsf{OM}$ ?

or

$\mathsf{ZFC+OM}\nvdash \mathsf{SM}$ ?

or

$\mathsf{ZFC+EM}\nvdash \mathsf{SM}$ ?

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  • $\begingroup$ For the second one: Caicedo's comment here shows that $\textsf{ZFC}+\textsf{SM}\vdash\text{Con}(\textsf{OM})$, so $\textsf{SM}$ is consistency-wise strictly stronger than $\textsf{OM}$. In particular, $\textsf{ZFC}+\textsf{OM}\not\vdash\textsf{SM}$. math.stackexchange.com/questions/33688/… $\endgroup$ Feb 2 '17 at 22:34
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All three statements hold, as mentioned in a couple of other answers on Math.SE. The second and third one follows from Caicedo's answer here, in which he shows that $\operatorname{ZFC}+\operatorname{SM}\vdash\text{Con}(\operatorname{OM})$. For the first one, as $\text{Con}(\operatorname{ZFC})$ is absolute between $\omega$-models as mentioned in Trevor's answer here, we get that $\operatorname{OM}$ is consistency-wise strictly stronger than $\operatorname{EM}$, yielding that $\operatorname{ZFC}+\operatorname{EM}\not\vdash\operatorname{OM}$.

We thus have a strict consistency-wise ordering $\operatorname{EM}<\operatorname{OM}<\operatorname{SM}$.

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  • $\begingroup$ thank you! ,now I understand $\endgroup$
    – luis
    Feb 2 '17 at 23:27

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