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If we take a first-order theory (like $\mathsf{ZFC}$, or $\mathsf{ZFC}$ plus some additional axioms) as the foundation of mathematics, does that imply that mathematical reasoning (theorems, proofs, algorithms, etc.) must be expressible in first-order logic (and associated rules of inference) in order to be "legitimate"? (not that you necessarily have to actually write it all out in FOL syntax, but that you have confidence that you could do so). Is there any sort of consensus around whether such "first-orderizability" is deemed a requirement of "legitimate" mathematical reasoning?

Some comments/observations related to the question:

In Mathematical Logic by Ebbinghaus et. al. (2nd ed.), section VII.2 states that, given a foundation of set theory with axioms $\boldsymbol{\Phi_0}$ (and the first-order language of set theory denoted by $L^S$), then:

Mathematically provable propositions have formalizations which are derivable from $\boldsymbol{\Phi_0}$. Thus it is in principle possible to imitate all mathematical reasoning in $L^S$ using the rules of the sequent calculus. In this sense, first-order logic is sufficient for mathematics.

But then later (section IX.2) they say:

The class of torsion groups cannot be characterized in first-order logic. But we can axiomatize this class if we add to the group axioms the "formula" $\forall x (x \equiv e \lor x \circ x \equiv e \lor x \circ x \circ x \equiv e \lor \ldots)$. Thus we gain expressive power when allowing infinite disjunctions and conjunctions. Such formations are characteristic of the so-called infinitary languages...This leads to the system $\mathcal{L}_{\omega_{1},\omega}$

This seems like a direct contradiction of the first quote (from VII.2). If a notion like "let $\mathfrak{G}$ be a torsion group" can't be expressed in (finitary) first-order logic, then it's hard to see how one can argue that this language is "sufficient for all mathematics".

More generally, the following considerations seem to weigh against restricting mathematical reasoning to arguments expressible in standard (finitary) first-order logic ("FOL"):

  1. Since an infinitary language like $L_{\omega_{1},\omega}$ is an extension of FOL, it seems like one could take $L_{\omega_{1},\omega}$ (instead of FOL) to be the language of a set theory like $\mathsf{ZFC}$ (the axioms remain valid statements of the language). In that case it would seem at least plausible that with a more expressive language (i.e. $L_{\omega_{1},\omega}$) having become available, one might be able to use it to formally prove statements which are true in every model of $\mathsf{ZFC}$ yet are not provable in FOL (maybe not even expressible in FOL).
  2. This post re "non-first-orderizability" on Terence Tao's blog (see especially last paragraph staring with "It seems to me that first order logic is limited") appears to take the position that mathematical reasoning need not be "first-orderizable", or even formalizable at all in any known formal language (see Tao's comments regarding "mathematical English").

Update/Clarification:

I realize that different axiom systems differ as to which first-order statements they can prove. So potentially statement $S$ is not formally derivable as a consequence of first-order axioms $\boldsymbol{\Phi}$, but $S$ is a formal consequence of some stronger first-order axiom system $\boldsymbol{\Psi}$.

But what I have in mind is a different issue, namely the notion of a statement in “mathematical English” (to borrow Tao’s phrase) which, due to its logical “structure”, cannot in any way even be expressed as a formula in first-order logic (regardless of what axioms or ontological assumptions one stipulates).

  • Such “inherently non-first-orderizable” statements are what Tao seems to be describing in the last paragraph of the blog post, starting with “first order logic is limited by the linear…nature of its sentences” and following up with “this does not fully capture all of the dependency tree of variables” and “subtleties may appear when one deals with large categories”.
  • A concrete example might be something equivalent to a formula involving infinite disjunction in $L_{\omega_{1}, \omega}$: given a well-defined infinite collection of (finite) sub-formulas $\Phi_{n}$, imagine the (infinite) formula $\Phi_{0}(x) \land (\Phi_{1}(x) \lor \Phi_{2}(x) \lor \Phi_{3}(x) \lor \ldots)$ Surely such a statement is not, in general, expressible in (standard, finitary) first-order logic (unless somehow it can always be made first-order-expressible by “upgrading” to a stronger theory? like upgrading from $\mathsf{ZFC}$ to Kelley-Morse set theory, or something like that?)
NikS
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  • How do we decide what qualifies as a “mathematical theorem”? – Porky Apr 01 '24 at 07:44
  • I presume it to mean: a statement which is shown to be true in every model of one’s chosen foundational axiom system (such as $\mathsf{ZFC}$). The wrinkle there regards what it means for a statement to be “shown” to be true…which is kind of the crux of my question. – NikS Apr 01 '24 at 07:57
  • No. However, second-order descriptions are often seen as presuming an underlying theory of sets/relations/properties, and many set theorists want to avoid that presumption. – Corbin Apr 01 '24 at 17:56

2 Answers2

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Yes, when someone says that ZFC is a suitable foundation for mathematics, that means that all standard mathematical notions can be expressed in the language of ZFC using first-order logic and all standard mathematical arguments can be written as derivations from ZFC using first-order logic.

For your particular example of torsion groups, there is a first-order formula $\psi(x)$ with one parameter in the language of set theory such that, for any $x,$ $\psi(x)$ holds iff $x$ is a torsion group.

The statement that torsion groups can't be characterized in first-order logic means something different, namely:
There is no first-order sentence $\varphi$ in the language of group theory such that, for every group $G,$ $G$ is a torsion group iff $G$ satisfies the sentence $\varphi.$
Note that $\varphi$ here is required to be in the language of group theory; it can only quantify over elements of the group (not subsets of the group, or natural numbers, or anything else), and it can't refer to any functions or relations other than the multiplication operation of the group.

So how would you write the original formula $\psi$ in the language of set theory? The formula $\psi(x)$ would say something like the following (you'd have to expand all the circumlocutions fully, of course, but, although that's a bit unwieldy, it's routine to do):

  1. $x$ is an ordered pair $\langle G, \circ\rangle.$

  2. $\circ$ is a function mapping $G\times G$ to $G$ satisfying the axioms for a group.

  3. For every $a\in G,$ there exists a natural number $n$ and a function $f: n+1\to G$ such that:

(i) $f(0) = a;$

(ii) $(\forall k\lt n)(f(k)\circ a = f(k+1));$

(iii) $f(n)$ is the identity element of the group $\langle G, \circ\rangle.$

This is a first-order formula in the language of set theory, but you can see that it is not a formula in the language of group theory; it talks about objects other than elements of $G,$ and it involves quantifying over natural numbers (and also functions, or finite tuples from $G).$

(For anybody who hasn't seen the usual set-theoretic convention regarding natural numbers, I should mention that $\psi$ as written above uses the formulation that for any natural number $m,$ we have $m = \{k \mid k \lt m\}.)$

  • I see, that makes sense. But does it mean that one should read Terence Tao’s aforementioned blog post as advocating a philosophy (somewhat) contrary to the one you describe? He seems to be endorsing the “legitimacy” of non-first-orderizable reasoning (or maybe I’m just misunderstanding him?) – NikS Apr 02 '24 at 01:29
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    @NikS It looks to me like Tao, in the blog post, is pointing out that arguments involving proper classes can involve subtle points that are easily missed -- which I agree with. But I'm not sure how he would propose to address this by extending first-order logic (rather than simply changing the axiomatization being used within first-order logic), or how such an extension could be put on a solid foundational footing. I agree that it sounds like he's thinking of something specific that he has chosen not to elaborate on. Maybe he talks about this elsewhere? – Mitchell Spector Apr 02 '24 at 01:57
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    FYI -- you can nest (i-iii) under list item #3 by adding three spaces at the beginning of each of those lines. (I tried to suggest an edit that does this, but StackExchange doesn't allow proposed edits that modify only whitespace.) – ruakh Apr 02 '24 at 17:06
  • @ruakh Thank you; I appreciate the comment. I remembered there was a way to do it but didn’t want to bother looking it up since it looked OK and it was just a one-off use. – Mitchell Spector Apr 02 '24 at 17:51
  • Torsion groups aside, 'm unaware of any first-order theory of the real numbers that doesn't first postulate the natural numbers. And any model of such a theory (with countably many symbols) will have a countable model, certainly not what one would think of as the real numbers. Also, lets not forget that Goedel has shown that 2nd order logic "does not admit an effective, sound, and complete proof calculus" (see https://en.wikipedia.org/wiki/Higher-order_logic). – Tony Dean Apr 06 '24 at 02:38
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    This answer addresses the question as I described it, so I’ll mark it as “accepted” (and post follow-ups as a separate question) – NikS Apr 14 '24 at 00:21
  • Thanks. By the way, I’m not saying that all legitimate mathematics has to be first-order. But it seems to me that more complicated logics generally can’t be foundational, since they depend on set theory for their semantics. – Mitchell Spector Apr 14 '24 at 04:17
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The question of whether first-order logic is sufficient for all of mathematics is different from the question of whether first-orderizability is a requirement for legitimate mathematical reasoning.

Possibly first-order logic (in the sense of first-order set theory, as explained by Mitchell Spector) is sufficient for all of mathematics. This does not at all imply that whether some is a piece of legitimate mathematical reasoning depends on its first-orderizability.

For example, people were already proving theorems of the form

For each $x$ in [some domain] there are only finitely many $y$'s in [some domain] such that [some property holds for $x$ and $y$].

a long time before anyone invented ZFC or even first-order logic. So, if you want to claim that first-orderizability is necessary for legitimate mathematics, you and thereby claiming that for several centuries it was unknown or indeterminate whether such theorems are legitimate pieces of mathematics. This, to me, is non-sense. People didn't have to wait until someone invented first-order set theory to see that

Every polynomial which is not identically zero has finitely many roots.

is a legitimate piece of mathematics. (Yes, this theorem can be expressed as a first-order claim in the language of set theory, but the point is that its status as a legitimate piece of mathematics does not depend on this.)

Pilcrow
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    I think OP's question was how to reconcile the idea that first-order ZFC could be a suitable foundation for mathematics with the fact that torsion groups can't be characterized in a first-order way. The key point is that the languages involved are different in those two statments (as in my answer). – Mitchell Spector Apr 01 '24 at 19:28
  • I also don't think anyone has ever claimed that first-order logic was a prerequisite to doing mathematics; as you pointed out, people were doing mathematics long before first-order logic was invented. The relevant claim is that first-order ZFC can be a suitable foundation for all of mathematics. (continued...) – Mitchell Spector Apr 01 '24 at 19:35
  • ... It's hard to see how second-order logic (or some infinitary logic) could be a suitable foundation for mathematics, because those logics have to rest on set theory. But it's also true that even first-order logic depends on Peano arithmetic or some other foundation for discrete mathematics. So there are a number of issues here. However, OP's question had to do with the more straightforward issue of the language being used in the two claims (first-order ZFC being a foundation for all math vs. torsion groups not being characterizable in a first-order way) which look contradictory at first. – Mitchell Spector Apr 01 '24 at 19:57
  • @MitchellSpector Part of the OP's question is indeed based on this essentially linguistic confusion, and you do a good job addressing this. But they also explicitly ask (in the title and twice in the body of the question) whether all mathematical reasoning must be first-orderizable in order to count as "legitimate mathematical reasoning". – Pilcrow Apr 02 '24 at 01:41
  • @MitchellSpector "I also don't think anyone has ever claimed that first-order logic was a prerequisite to doing mathematics" – well, you and I might know this, but it seems like the OP doesn't, hence my answer. I added my answer because while your answer clarifies the terminological issue, it doesn't dispell the idea that mathematics must be first-orderizable to count as genuine mathematics. My point is not that there is non-first-orderizable mathematics, it's simply that being translatable into FOL is not a criterion for "genuine mathematics". – Pilcrow Apr 02 '24 at 01:43
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    I certainly agree that there's a deeper question there -- I just think that OP was actually stuck on that other issue. (For what it's worth, I don't think ZF exhausts our mathematical intuition, but trying to go beyond first-order logic for a foundation seems problematic to me.) (By the way, Pilcrow, I can't seem to tag you -- is your name being misinterpreted because it's the same as the name of a character?) – Mitchell Spector Apr 02 '24 at 01:48
  • (To Pilcrow) It looked to me like OP was questioning whether first-order ZFC was actually sufficient for "normal" mathematics (like torsion groups), and there's no doubt that it is. By "normal" math here, I mean math as it is done today by a typical mathematician doing research outside mathematical logic. Even when people stretch things via, say, Grothendieck universes, they're just adding a large cardinal axiom to ZFC; they're not extending first-order logic. (Higher-order logics are quite interesting; I just don't see how they could be foundational due to their dependence on set theory.) – Mitchell Spector Apr 02 '24 at 02:12
  • Regarding the last sentence of your answer: If one seriously commits to ZFC (or ZFC plus some additional first-order axioms) as the foundational theory of mathematics, then per the answer by @MitchellSpector, if some statement about polynomials was found to not be formally inferable from the axioms in FOL, isn’t it then by definition not part of “genuine mathematics”? Unless and until you change your foundational theory, things which your foundations can’t accommodate have to be “out of bounds”, right? Otherwise your alleged foundations aren’t really “foundational”. – NikS Apr 02 '24 at 03:36
  • That's what the additional axioms are for (or one of the things that they're for). If, for example, you take ZFC as your foundational theory, you won't be able to prove the consistency of ZFC within that theory (assuming it really is consistent -- but if it weren't consistent, it wouldn't be a suitable foundational theory anyway). So you can add Con(ZFC) to your axiomatization -- or you can add as an axiom that there's a strongly inaccessible cardinal. ZFC as a foundational theory is merely intended to formalize math as it's currently carried out by normal research mathematicians. (1/2) – Mitchell Spector Apr 02 '24 at 07:20
  • From a Platonist perspective, as additional facts about the mathematical universe become accepted, the theory ought to be expanded or modified to include those. By Gödel, mathematics will never be completed; it's a growing process, a never-ending program. But the problem isn't with first-order logic; the limitations are with whatever axioms you currently realize are true. And there's no way around this. (2/2) – Mitchell Spector Apr 02 '24 at 07:21
  • @NikS While the belief that "all interesting mathematics can be done within ZFC" is common, in my experience there are very few people who are ZFC-fanatics to the extent that they believe that "if you are not working in ZFC, then you are not doing mathematics". It's plain as day that people who are working on, for example, the axiom of determinacy are doing mathematics, even though this axiom is incompatible with ZFC. – Pilcrow Apr 02 '24 at 20:16
  • @NikS How would you feel about it if some constructivist mathematician told you that you are not doing genuine mathematics because you are not working in their favorite system of Heyting arithmetic minus $\Omega_{4}$-induction plus Glazunov's axiom of constructive anti-determinacy? Presumably this would feel silly to you. The claim that you are not doing mathematics unless you are working in ZFC is equally silly. It would take a genuine ideological fanatic to think that you're not doing mathematics unless you accept the Axiom of Foundation and reject the existence of measurable cardinals. – Pilcrow Apr 02 '24 at 20:16
  • @MitchellSpector : I realize that adding new axioms can expand the set of formally provable first-order statements. But what I have in mind is something different, namely “inherently non-first-orderizable” statements, i.e. statements which are not even *expressible* as a formula in first-order logic (see Update/Clarification that I added to the question). Can such an “inherently non-first-orderizable” statement be considered “legitimate mathematics” despite its non-first-orderizability? (your earlier comments kind of imply an answer of “no”, but I don’t want to assume too much…) – NikS Apr 04 '24 at 06:12
  • To be a bit more specific: Obviously for a logician specializing in infinitary languages, non-first-orderizable statements are the norm in their work (thus “legitimate”). But if a proof in, say, number theory involved some statement(s) which were “inherently non-first-orderizable” (in the sense described in “Update/Clarification” above), might the proof be regarded as not quite legit due to its non-first-orderizable nature? – NikS Apr 04 '24 at 08:11
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    In addition to the reason @MitchellSpector gave for not using richer logics than first-order as a foundation for mathematics, there's also the fact that most such logics don't have complete deductive systems; there's no way to tell whether an alleged deduction is correct. This applies in particular to systems like second-order logic that include a notion of "genuine" set and to systems that that include a notion of "genuine" natural number. (There are exceptions, like adding the quantifier "for uncountably many x", but that doesn't look useful for a foundation.) – Andreas Blass Jan 09 '25 at 18:25