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I mean here a logic in the sense of a language and semantics. By strongly axiomatizable I mean strongly sound and strongly complete. So I'm basically asking if there is a particular deductive system for which strong completeness and soundness hold, but compactness fails.

Obviously the deductive system cannot be finitary, for if it were then compactness would follow from completeness. But does it follow immediately that a deductive system containing an infinitary rule prevents compactness? I can see that it must but don't see how to 'prove' it, without using induction up to at least the ordinal associated with the infinitary rule.

Hence, a sub-question: Is there a standard way to 'prove' such a result?

For motivation, I know for example that infinitary logics are weakly complete and not compact.

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    Oh I see. I thought you were referring to infinitary languages. Then the general answer to your question is: If there are theorems whose every proof has infinite size (and only then) compactness fails while soundeness and completeness hold.2011-05-26

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First, I think we can always define a semantics for any deduction system such that the strong completeness and soundness will hold by taking the Lindenbaum algebra (models are made of constant and semantics is defined over the corresponding algebra of provably equivalent classes of formulas). And there are deduction systems which are not compact so that would answer the question.

An interesting case would be the $\omega$-logic with the infinitary $\omega$ rule:

$\{\varphi(n)\}_{n\in\omega} \vdash \forall x \ \varphi(x) $