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On the wikipedia page for ZFC: link

It says that ZFC is formally a theory in logic. And I guess all of mathematics relies on logic, but we use logic informally when creating mathematical arguments.(This seems to be done in all textbooks.)

Does this mean that mathematics can live or exist without formal logic? As have been said on this site mathematics existed before formal logic. So is it correct to say that we can build up analysis, algebra etc., with ZFC and logical arguments, but not using formal first order logic?

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  • $\begingroup$ Mathematics did live and exist without formal logic for thousands of years. However, in modern times, mathematicians became aware that perhaps we ought to study the fundamentals in order to be certain that mathematics won't collapse under our feet one day. Thus formal logic was born in an attempt to ensure that all our previous results (and future) are indeed sound. $\endgroup$
    – Arthur
    Commented Aug 29, 2016 at 14:18

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Norman Megill addresses this here: http://us.metamath.org/mpeuni/mmset.html#axioms

Note. Books sometimes make statements like "(essentially) all of mathematics can be derived from the ZFC axioms." This should not be taken literally—there's not much you can do with those 7 axioms by themselves! The authors are assuming that you will combine the ZFC axioms with logic (that is, the axioms and rules of propositional and predicate calculus). Between ZFC axioms and logic there is a total of 20 axioms and 2 rules in our system.

If you start working with these it quickly becomes clear that you need the logic axioms to do much of anything.

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  • $\begingroup$ Thank you very much. Does this mean that if we just argue by using the rules of the axioms given, we are good? I have seen first order logic very complicated with having to define an alphabet etc., but in this article it means that all we have to do is follow those axioms when reasoning? Also there is one thing about the predicate axioms I do not quite understand, they talk about beeing a "member of", $\in$, in axiom 13 and 14, but when working with predicate calculus we do not know if we have sets yet, so how does that make sense? $\endgroup$
    – user119615
    Commented Aug 29, 2016 at 16:44
  • $\begingroup$ @user119615 In propositional calculus you don't have sets. In predicate calculus you do, but you know very little about it -- all you know is what you get in axioms 13 and 14, that equality means that things contain and are contained by the same things. $\endgroup$
    – Charles
    Commented Aug 29, 2016 at 16:56
  • $\begingroup$ @user119615 Yes, all you have to do is follow those axioms (or ones substantially similar). But that doesn't obviate the need for an alphabet -- even for propositional calculus you need to know which symbols you can use and which ways you can string them together to make wffs. $\endgroup$
    – Charles
    Commented Aug 29, 2016 at 16:58
  • $\begingroup$ Can you please say what would be considered the alphabet in the case on the site? $\endgroup$
    – user119615
    Commented Sep 2, 2016 at 6:43
  • $\begingroup$ @user119615 I don't know what "the case on the site" is, but usually the alphabet would consist of variables (usually x, x', x'', etc.), logical symbols like $\forall$ and $\vee$, set symbols like $\in$ and $=$, and so on. $\endgroup$
    – Charles
    Commented Sep 3, 2016 at 0:39

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