I often read that arithmetic in first order logic has problems and you really want to do it in second order logic.
However, aren't the Zermelo–Fraenkel axioms written down in the language of first order logic?
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2$\begingroup$ Yes, but most mathematicians don't check to see if what they're doing can be formalized in ZF. $\endgroup$– Qiaochu YuanCommented Jun 28, 2012 at 21:04
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$\begingroup$ @NickKidman: What some might call problems of first-order logic, I prefer to call features, indeed useful features. $\endgroup$– André NicolasCommented Jun 28, 2012 at 21:11
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$\begingroup$ @Andre: Did you pioneer the known hi-tech slogan "it's not a bug, it's a feature!"? :-) $\endgroup$– Asaf Karagila ♦Commented Jun 28, 2012 at 21:13
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2$\begingroup$ @AsafKaragila: Why pioneer when you can steal? $\endgroup$– André NicolasCommented Jun 28, 2012 at 21:14
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1 Answer
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Note that ZFC is a theory strong enough to prove second-order arithmetics. So if you agree to take ZFC as your foundational point, taking second-order PA for arithmetic should not pose any problems.
This is one of the reasons set theory is a good foundational basis for mathematics, since it allows second-order (and higher) to work via first-order formulas in the universe of set theory.
Further reading:
answered Jun 28, 2012 at 21:04
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$\begingroup$ Thanks, interesting reads. I find the "second order theory worked out in first order theory" still to be mysterious however. Is seems that the only thing one really loses by considering the first order alternative is determination of interpretation or something like that. Then I find it hard to see why one would be able to say "we know the statement in this theory - so we know it in the other": Identifying two statements in two different theories (which differ in a way) with each other sounds fishy to me. Might be a philosophical thing though. $\endgroup$ Commented Jun 28, 2012 at 21:55