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I know the first-order version of ZFC, but not second-order ZFC. Can anyone explain how axioms (and other things) of second-order ZFC differ from the first-order version?

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Second-order logic allows us to quantify over subcollections of the universe. This allows us to write things like "For every class which contains all the ordinals ..." within the theory, whereas in first-order logic these sort of sentences are actually schema of sentences (where class becomes a definable class).

It is important to point out that there are two [common] ways to interpret second-order logic, the first is with Henkin semantics where the classes are allowed to be definable classes only; this results in something resembling first-order logic. The second way is to full semantics which allows the classes to be any collection.

In second-order ZFC instead of the replacement schema we have one axiom which says that every function whose domain is a set then its range is a set.

For example we don't have countable models of second-order ZFC when interpreted in full-semantics, because those "hide" a very [first-order] undefinable bijection between $\omega$ and their universe. Second-order replacement can see this bijection, so the model cannot satisfy replacement (The range of a function whose domain is a set is everything).

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    $\begingroup$ This is not a very thorough explanation... There is no mention of the difference in the language, and in the second paragraph full second-order semantics are being assumed, while second-order ZFC could also be studied in Henkin semantics. $\endgroup$ Feb 22, 2013 at 11:42
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    $\begingroup$ Isn't second-order ZFC in Henkin semantics logically equivalent to first-order ZFC? $\endgroup$
    – Zhen Lin
    Feb 22, 2013 at 11:44
  • $\begingroup$ @Carl, yes you are correct. I didn't write a long answer for two reasons: it seems the OP didn't try to search on their own first, and more importantly I am writing from my iPhone. :-) $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 11:44
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    $\begingroup$ @Carl: I should also add that this is not the first time you are trying to prod me into improving my answer instead of posting your own, and I am certain that many of these times you could have given a much better answer. $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 11:51
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    $\begingroup$ @Carl: Now by a computer I added a bit (and correct an iPhone generated typo, and made room for man-made typos instead). I still think that you should write something yourself as well. $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 12:37

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