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The set $\Phi$ of S-formula is infinite, how to define $\mathcal{I} \vDash \Phi$? And how to define $\Phi \vDash \varphi$?

If $\Phi\vdash\varphi$, we only need a subset of $\Phi$ to derive $\varphi$. But in model-theroy, do we need every element of $\Phi$ to have $\Phi \vDash \varphi$? If this is true why it is not symmetric between the model-theroy and the sequent calculus where the completeness and soundness lemma $\Phi \vdash \varphi \Leftrightarrow\Phi \vDash \varphi$ holds?

Arctic Char
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Nebula Libro
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    What's wrong with 'every element of $\Phi$ is satisfied in $\mathcal I$'? – Berci Nov 05 '21 at 18:07
  • If $\Phi \vdash \varphi$, we only need a subset of $\Phi$ to derive $\varphi$. But in model-theroy, do we need every element of $\Phi$ to have $\Phi \vDash \varphi$? If this is true why it is not symmetric between the model-theroy and the sequent calculus where the completeness and soundness lemma $\Phi \vdash \varphi \Leftrightarrow \Phi \vDash \varphi$ holds. Thanks for your answering. – Nebula Libro Nov 05 '21 at 18:22
  • @NebulaLibro The compactness theorem says that if $\Phi\models \varphi$, then there is a finite subset $\Delta\subseteq \Phi$ such that $\Delta\models \varphi$. One what to prove this is as a consequence of the completeness theorem, as you suggest. – Alex Kruckman Nov 05 '21 at 18:53
  • @AlexKruckman I get it. Thank you very much. I have not learnt compactness theorem yet, it seems that a long way in logic waits me to go through. – Nebula Libro Nov 05 '21 at 19:00

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This is addressing the issue as clarified in your comments above rather than the body of the question itself:

The definition of $\models$ is very unlike that of $\vdash$ ... at least, at first glance. In particular, while "$\Phi\vdash\varphi$" is trivially equivalent to "For some finite subset $\Phi_0\subseteq\Phi$, we have $\Phi_0\vdash\varphi$," this is not the case for $\models$. Or rather, this same fact is true but it is not at all trivial: it is the compactness theorem.

Going beyond that, the complete coincision of $\vdash$ and $\models$ - that is, the soundness and completeness theorems - should in my opinion at least be extremely surprising; specifically, we should expect $\models$ to be vastly more complicated than it actually winds up being.

Note that more obvious symmetry between $\vdash$ and $\models$ would make this fact less surprising - so in a sense, the answer to your question "why it is not symmetric between the model-theroy and the sequent calculus" is "because it's more exciting that way!" (This is somewhat tongue-in-cheek, admittedly.)

Noah Schweber
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