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In Peano arithmetic (first order), we first define natural numbers using a successor function and Peano axioms, then we define addition (and multiplication), and then, we define inequality, as:

$a\leq b\leftrightarrow\exists c\left(a+c=b\right)$

Is there any way to define inequality first, directly from the successor function and Peano axioms? (I mean, if I don't need addition for my purpose, why define it?).

Bram28
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Aris Makrides
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4 Answers4

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In a precise sense, the answer is no. Namely, let $PA_{succ}$ be the set of PA-theorems in the language containing only the symbol for the successor function; then we can show:

There are models of $PA_{succ}$ with no definable linear ordering.

In particular, this means that there is no first-order formula using only successor which PA proves defines a linear ordering.

Specifically, consider the structure (in the language of successor only) $\mathbb{N}+\mathbb{Z}+\mathbb{Z}$. This is a model of $PA_{succ}$ (this takes a bit of work, but isn't hard), but has no definable linear ordering: consider any automorphism swapping the two $\mathbb{Z}$-parts.

(A bit more thought also shows that there is no formula in the language of successor alone which defines a linear ordering in the standard model $\mathbb{N}$; the key ingredient is the proof that $PA_{succ}$ is complete. And in fact thinking along these lines ultimately shows that no model of $PA_{succ}$ has a definable linear ordering.)

Noah Schweber
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  • For the asker's benefit, I'll note that this answer addresses the technical notion of first-order definability; "$\le$" is not first-order definable over PA[succ], just as neither are $+$ or $·$. However, if the asker wants a 'recursive definition' of "$\le$" in a way that allows one to recover the usual properties of the ordering based on the other axioms of PA[succ], then my answer or Bram28's answer provide possible ways. – user21820 Nov 16 '19 at 19:06
  • @user21820 Note that I explicitly mentioned first-order logic in my answer. – Noah Schweber Nov 16 '19 at 19:06
  • Yup; it's just to eliminate the usual confusion that a lot of people have due to having been told that one can recursively define $+$ and $·$ over PA[succ]. It could be due to a misinterpretation of the phrase "recursive definition". – user21820 Nov 16 '19 at 19:08
  • @user21820 That's a good point. – Noah Schweber Nov 16 '19 at 19:08
  • I upvoted anyway, and I realize that the asker may have been confused and thought the definition of "$\le$" over the usual PA is like the 'definition' of $+,·$ over PA[succ], which it certainly is not. – user21820 Nov 16 '19 at 19:17
  • @user21820 They are careful to specify that they mean first-order PA - I think they're just being a bit cavalier about its presentation. – Noah Schweber Nov 16 '19 at 19:18
  • Well Aris said "then we define addition (and multiplication), and then, we define inequality", but the first "define" is different from the second in precisely the senses we are talking about. So I think it's good for Aris to see both our answers, and our comments. =) – user21820 Nov 16 '19 at 19:21
  • @user21820 Sure - that's what I meant by "cavalier." :) – Noah Schweber Nov 16 '19 at 19:22
  • Thank you all for your answers. I understand that addition is not definable in a language with signature $\left\langle 0,S\right\rangle$ , where $S$ is the successor function. With the recursive definition, we add to the language a non logical symbol, “$+$”, and then we get a signature $\left\langle 0,S,+\right\rangle$ . That is because, there isn't any expression containing only logical symbols, which is equivalent with an expression of the form “$x+y=z$”. From Noah Schweber's answer, I understand that “$\leq$” is not definable either. – Aris Makrides Nov 17 '19 at 20:48
  • I don't fully understand the example, though. @NoahSchweber, I don't know a lot about model theory yet, so I would appreciate if you add an explanation of what you mean by the structure $\mathbb{N}+\mathbb{Z}+\mathbb{Z}$. Also, what if $\forall x\left(x\in\mathbb{N}\right)$? Is there again a model with no definable linear ordering? Or, if “$x\leq y$” holds only for $x,y\in\mathbb{N}$, is “$\leq$” definable then? – Aris Makrides Nov 17 '19 at 20:49
  • @user21820, do you mean I should use the term “definition”, only when a new symbol is introduced merely as an abbreviation of an existing expression? – Aris Makrides Nov 17 '19 at 20:51
  • @ArisMakrides "what if $\forall x(x\in\mathbb{N})$? Is there again a model with no definable linear ordering?" The linear order on $\mathbb{N}$ isn't definable in $(\mathbb{N}; S)$ (note that we don't need to include $0$ since it's definable from $S$ alone). In fact it's even worse: no model of $PA_{succ}$ has a definable linear ordering. – Noah Schweber Nov 17 '19 at 20:56
  • @NoahSchweber: So, the set $\mathbb{N}$ is linearly ordered in one sense, but we can not define an ordering relation? I mean, that is what Peano axioms do, right? They define a series $a,b,c,...$ of objects, in witch one comes after the other. Can you explain it a bit more? – Aris Makrides Nov 17 '19 at 21:11
  • @ArisMakrides The usual ordering $\le$ is not first-order definable in the structure $(\mathbb{N};S)$. It is definable in the structure $(\mathbb{N}; S,+)$, but that's a different structure; similarly it is definable in $(\mathbb{N}; S)$ by a second-order formula, but that's a different logic. There are different kinds of definitions out there - that's sort of the point of logic - and in the context of first-order logic, the ordering on the naturals is not definable from the successor operation alone. – Noah Schweber Nov 17 '19 at 21:59
  • @ArisMakrides: Indeed in logic it is conventional to use "define" only for definitorial expansion where you add a symbol to abbreviate an expression over the original system. Outside logic, you may commonly see people use the phrase "recursive definition", because they are working in a foundational system that supports recursion, where you can define a sequence by specifying a recursive relation between each term and the preceding terms. – user21820 Nov 18 '19 at 08:55
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One way to axiomatize "$\le$" (better not to call it "defining $\le$") that plays very nicely with induction is:

$a≤b ⇔ a=0 ∨ ∃c,d ( c≤d ∧ S(c) = a ∧ S(d) = b )$.

Do not forget that, whether you have addition or not, you must be able to prove that $≤$ is a strict total order. That is what truly matters.

user21820
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  • @NoahSchweber: This is obviously not a definition in the usual sense, and you know that I know it. I took it as granted that the asker meant it in the sense of axiomatizing "$\le$". – user21820 Nov 16 '19 at 19:00
  • Fair, I suppose I was being uncharitable. – Noah Schweber Nov 16 '19 at 19:00
  • @NoahSchweber: I'll edit anyway to clarify the terminology just in case someone else gets confused (I admit your concern has some validity). – user21820 Nov 16 '19 at 19:01
  • I've added an answer explaining my concern, and how that can lead to a negative answer to the question. – Noah Schweber Nov 16 '19 at 19:03
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Let's assume you have the typical axioms for Successor:

$$\forall x \ s(x) \neq 0$$

$$\forall x \forall y \ (s(x) = s(y) \to x = y)$$

and the Axiom Scheme of Induction, which states that for any formula $\varphi(x)$, we have:

$$(\varphi(0) \land \forall x (\varphi(x) \to \varphi(s(x))) \to \forall x \ \varphi(x)$$

Then, if you add:

$$\forall x \forall y (x < y \leftrightarrow (s(x) = y \lor \exists z (y = s(z) \land x < z))) \tag{*}$$

you can prove all of the following:

$$\forall x \ x < s(x)$$

$$\forall x \ \neg x < 0$$

$$\forall x \neg \exists y (x < y \land y < s(x))$$

$$\forall x \ \neg x < x \text{ (irreflexivity)}$$

$$\forall x \forall y (s(x) < s(y) \to x < y)$$

$$\forall x \forall y \forall z ((x < y \land y < z) \to x < z) \text{ (transitivity)}$$

$$\forall x \forall y (x < y \to \neg y < x) \text{ (asymmetry)}$$

$$\forall x \forall y (x = y \lor x < y \lor y < x) \text{ (trichotomy)}$$

So, you can prove all kinds of important and elementary facts about $<$ by adding that one statement to the basic axioms about $s$.

Of course, you would define $x \leq y$ simply as $x < y \lor x = y$ to get results regarding $\leq$, including:

$$\forall x \ x \leq x \text{ (reflexivity)}$$

$$\forall x \forall y \forall z ((x \leq y \land y \leq z) \to x \leq z) \text{ (transitivity)}$$

$$\forall x \forall y (x \leq y \lor y \leq x) \text{ (totality)}$$

Also, once you do add the typical axioms for addition:

$$\forall x \ x+0=x$$

$$\forall x \forall y \ x + s(y) = s(x+y)$$

then you can derive the 'standard' definition of inequality in terms of addition:

$$\forall x \forall y (x \leq y \leftrightarrow \exists z \ x + z = y)$$

In sum: Yes, we can have an alternative definition of smaller than or inequality that allow you to prove important things about it without using the axioms of addition.

Bram28
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  • Nice and detailed. I didn't check, but I trust you did so I upvoted. =) – user21820 Nov 16 '19 at 19:06
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    @user21820 Thanks, and yes, I have the fully checked proofs :) – Bram28 Nov 16 '19 at 19:13
  • Wouldn't ∀x∀y(x<y↔s(x)=y∨s(x)<y) work as well? If not, why? – ByteEater Aug 12 '22 at 11:46
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    @ByteEater That wouldn’t work. Note that your statement would be true if we interpret $<$ as a total relationship, i.e. if $x<y$ for any $x$ and $y$ … meaning that it doesn’t capture much of interest at all … certainly not many of the things we want to be true of $<$ – Bram28 Aug 15 '22 at 00:42
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You could try something like$$a\le b\iff Sb\not\leq a\land Sa=b\lor Sa\le b.$$

J.G.
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