# Why is better to work with first-order logic than with second-order logic? [duplicate]

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Why is better to work with first-order logic than with second-order logic?

In the case of Peano's axioms the second-order version is categorical, but the first-order is not. Besides, with the second-order version the operations: addition, multiplication and exponentiation can be defined, but with the first-order version the cannot. However, in books they prefer and promote the use the first-order version. Why is the first-order version of Peano's axioms better to describe the true nature of natural numbers?

## marked as duplicate by Asaf Karagila♦, Claude Leibovici, mau, user127.0.0.1, Davide GiraudoFeb 20 '14 at 9:19

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• The advantage of first-order is that there is a useful notion of formal proof. The "true nature" is certainly much better captured by second-order. – André Nicolas Feb 20 '14 at 4:32

## 1 Answer

Start with the second question:

Why is the first-order version of Peano's axioms better to describe the true nature of natural numbers?

Careful! First order PA [PA1] is, as said, not categorical, and has non-standard models (models which are not isomorphic to the 'true' natural numbers, because -- as well as having a 'zero' and its 'successors' -- they have additional, non-standard, elements that come 'after' the zero and its successors). Since first-order PA can't distinguish its minimal model which does just have a zero and its successors from the models with unintended surplus content, it doesn't pin down "the true nature of natural numbers", and can't be sold as doing so.

So, it might naturally be asked, why do we at least start doing formal arithmetic by investigating PA1 rather than looking at some arithmetic PA2 with a second-order logic? More generally,

Why is it better to work with first-order logic than with second-order logic?

But remember: second-order semantic consequence [with "full semantics"] can't be recursively axiomatized -- i.e. there is no nice formal deductive system where we can effectively decide what is an axiom, and what is a well-constructed proof etc., such that a conclusion can be formally deduced from some axioms in the proof system just if the conclusion semantically follows from the axioms. Which means that any effectively axiomatised formal deductive system of second-order arithmetic capture all the semantic consequences of the second-order Peano axioms. It will in fact be negation incomplete. So in one good sense it won't be a complete theory of the nature of the numbers either.

True, an axiomatized second order arithmetic will be stronger than PA1, will prove more, so why not go straight to discuss such a stronger theory (it will tell us more about the numbers even if it doesn't fully pin down their "true nature")? Well, for natural pedagogic reasons, we'll want to start with the simpler theory! So it is entirely natural to start our investigations of arithmetic by (1) investigating PA1 and (2) showing by contrast that second-order arithmetic with full semantics is not axiomatizable [two standard ingredients of a first course] before (3) looking at axiomatizable versions of PA2 of various strengths which extend PA1.

• Let me add an additional foundational motivation for studying first-order arithmetic. For statements that only quantify over numbers (e.g. Fermat's last theorem, the twin primes conjecture), it is natural to ask whether they can be proved using only the resources of first-order arithmetic (or even some weak set of axioms for first-order arithmetic). This idea - of trying to find ontologically minimal theories to prove mathematical results - has a long history in foundations of mathematics. – Carl Mummert Feb 20 '14 at 12:12
• I have trouble with the line -any effectively axiomatized version of second-order arithmetic is in the same boat as PA1- because as far I know, PA2 is categorical. Is there any contradiction here? – Chilote Feb 20 '14 at 17:59
• @Chilote Yes that WAS stupid. I've revised. – Peter Smith Feb 20 '14 at 18:37
• @Chilote: PA2 is categorical in full second-order semantics, but the incompleteness theorem still applies to any effective axiom system for PA2. The incompleteness theorem applies to an extremely broad family of effective formal systems, not just to ones that might be considered first-order logic. Separately, any axiom system for PA2 (even ineffective ones, like the set of all true sentences of second-order arithmetic) have nonstandard models. The whole point of "full second order semantics" is that we just ignore those models, which is why PA2 becomes categorical in those semantics. – Carl Mummert Feb 21 '14 at 4:02
• Can you explain what do you mean by "second-order semantic consequence [with "full semantics"] can't be recursively axiomatized"? I understand that recursively axiomatizing a theory means extracting recursive set of sentences that forms that theory under $\vdash$closure. I do not understand, though, what it means to recursively axiomatize second-order semantic consequence. – Trismegistos Oct 1 '14 at 19:01