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I was told in a previous answer (but dont remember by whom but he then didnt answer back), that the hyperarithmetical hierarchy is second order arithmetic but not second order logic. Is this so? what precludes to interpret those formulas with full semantics? do the sets defined by a given statement differ with the two different semantics (full vs Henkin) ) or they are the same? The motivation for this question has to do with some more complicated question that I plan to make in the short future

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Do you mean the analytical hierarchy (e.g $\Sigma^1_4$, $\Pi^1_8$)? The hyperarithmetical hierarchy is about transfinitely iterated Turing jumps, and as such it only classifies sets that are already $\Delta^1_1$. – Carl Mummert Feb 27 '13 at 12:37
My concern is related to the analytical hierarchy, but the specific response was on the hyperaritmethical ones, that is why I limited the question to it. – Wolphram jonny Feb 27 '13 at 13:25
up vote 4 down vote accepted

All of the classical hierarchies of sets and formulas are defined in a way that does not depend on the semantics being used. They are either defined syntactically, or in terms of a single fixed model, while the only role of the semantics is to limit which models can be used.

Moreover, "second-order logic" includes both Henkin and full semantics. Unlike first-order logic where there is a single standard semantics, second-order logic has two different semantics (plus a few inessential variations of these two) and neither is assumed just by the term "second-order logic".

I found the previous question and answer. The point being made there is that the analytical hierarchy is not defined for arbitrary languages - it is specific to the language of second-order arithmetic. If you had a different language, say the second-order language of graphs, you could define a hierarchy of formulas, but it would not be the "analytical hierarchy".

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Thanks, now I got it! – Wolphram jonny Feb 27 '13 at 13:28

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