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I am wondering if the following modifications of the Peano axioms result in a set of axioms equivalent to the Peano axioms, in the sense that any set of numbers satisfies these modified axioms if and only if they satisfy the Peano axioms:

  1. Delete the axiom of induction, namely "Let $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(S(n)$ is also true. Then $P(n)$ is true for every natural number $n$."

  2. Add the following axiom: "$n$ is a natural number only if it can be expressed as the sum of 2 natural numbers"

  3. (This may or may not be a modification) Define addition by defining $0 + 0 = 0$, and if $n+m = x$, then $S(n) + m = n+S(m) = S(x) $

where $S(n)$ is the successor of $n$.

I think this works because:

  1. We can guarantee that for any 2 numbers $n,m$ which are in a number system satisfying these axioms, that the addition of $n+m$ is defined- as because both $n$ and $m$ are, by the axiom in the second modification above, the result of addition, they are the result of repeated applications of the successor operation on $0$; therefore they are in the scope of the recursion contained in the addition definition.

  2. The principle of induction still holds: because every number in a system satisfying these axioms can be obtained by repeatedly applying the successor operations on $0$, due to how addition is defined for them as above, $P(0) $ and $P(n) \implies P(S(n))$ implies that for all $n, P(n)$.

Is this reasoning correct?

Princess Mia
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson May 24 '24 at 00:35

2 Answers2

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Consider the set of ordered pairs of natural numbers with a first coordinate no greater than the second coordinate, i.e.: $$\{(x,y)\mid (x,y) \in \mathbb{N}, x \leq y\}.$$ Let $(x,y) + (a,b) = (x+a,y+b)$ and $(x,y) \times (a,b) = (x\times a, y \times b)$. Let the successor function be defined by $S'((a,b)) = (a,S(b))$. Let $(x,y) < (a,b)$ be defined by $x < a \vee (x = a\wedge y < b)$. Then the set is a totally ordered set that satisfies your axioms but does not satisfy the induction axiom.

Edit: For a counterexample that satisfies the Robinson arithmetic, see this answer

Cameron Buie
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ioveri
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  • what is $S'$ here? – Princess Mia May 20 '24 at 01:15
  • @PrincessMia it is the successor function that I chose for the given set I defined, just to differentiate it from the successor function of natural numbers. – ioveri May 20 '24 at 01:35
  • Thanks a lot for this answer- I think the difference here with what I intended is that addition would be defined recursively like arturo magidin did in his response, i.e. "We define addition by saying what happens when we add 0 to 0 : you get 0 ; and if we know what happens to $n+m$, then define $n^{+}+m$ and $n+m^{+}$by letting them both be the successor of $n+m$". Here, the addition isn't defined recursively; but maybe it could be to achieve the same thing? So I am not sure whether this counterexample is indeed one of the type I wished for, although it could be – Princess Mia May 20 '24 at 03:08
  • @CameronBuie clearly $(x,y) + (0,0) = (x,y)$ and $(x,y) \times (1,1) = (x,y)$ with given definitions – ioveri May 20 '24 at 03:36
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    @CameronBuie oh that's my mistake, I meant to say $(x,y) + (a,b) = (x+a,y+b)$ and $(x,y) \times (a,b) = (x\times a, y \times b)$ – ioveri May 20 '24 at 04:02
  • I thought that might've been the idea! – Cameron Buie May 20 '24 at 04:08
  • (+1) This is a nice example. Out of curiosity, why the restriction that $x\leq y,$ rather than simply taking all $(x,y)\in\Bbb N^2$? It isn't obvious to me at the moment where that hypothesis would be needed. – Cameron Buie May 20 '24 at 18:53
  • @CameronBuie first order formulations of the Peano axioms .i.e. using first order induction scheme, sometimes require total ordering to be axiomatized because it can only be defined with the second order induction. So I gave additional restriction so that it would work with both versions. – ioveri May 21 '24 at 02:47
  • @PrincessMia "We define addition by saying what happens when we add 0 to 0 : you get 0 ; and if we know what happens to $n+m$, then define $n^++m$ and $n+m^+$ by letting them both be the successor of $n+m$". That is a recursive definition. I don't get the fuss with addition. The only difference is that with the second order induction, you can constructively create $+$ and prove that the domain of $+$ is the entire natural number set, whereas without it, you'll have to add $+$ as axioms. – ioveri May 21 '24 at 03:07
  • Interesting! And somewhat surprising, given that the corresponding strict order is of order type $\omega\cdot\omega$ in either case. – Cameron Buie May 21 '24 at 10:34
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    @PrincessMia The point of my addition is that is it is a solution to your axioms but isn't adequate to what you want. That's why I said it "satisfies your axioms but does not satisfy the induction axiom". Your "definition" simply isn't strong enough to rule out numbers that violate induction itself. Induction axiom is simply a lot stronger than you think. – ioveri May 21 '24 at 17:00
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    @PrincessMia here's a post that's related to what you might want. In short, induction is independent from other axioms of the Peano axioms, or ${PA}^-$. So it doesn't matter how you try to redefine addition or multiplication via axioms. https://math.stackexchange.com/questions/331733/is-induction-independent-of-the-other-axioms-of-pa – ioveri May 21 '24 at 17:17
  • It looks like, if we let $$\Bbb A:=\bigl{\langle x,y\rangle\in\Bbb N^2:x\leq y\bigr},$$ let $\oplus$ and $\odot$ be pointwise operations, let $\prec$ be the lexicographic order, let $S(x,y):=\langle x,y+1\rangle,$ and let $\mathbf{0}:=\langle 0,0\rangle$ and $\mathbf{1}:=S(\mathbf{0})=\langle 0,1\rangle,$ then the numbered axioms in my post below are satisfied except for 8, 14, and 20. If instead we take $\Bbb A:=\Bbb N^2$ but keep the rest the same, then 13 also fails to be satisfied (in addition to 8, 14, and 20), but the rest hold. Is that why you restricted it? Or for some other reason? – Cameron Buie Jul 02 '24 at 15:59
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Consider the set $\Bbb H:=\left\{\frac12 n\mid n\in \Bbb N\right\},$ where $\Bbb N$ is the usual set of natural numbers, so that $\Bbb H$ is a subset of the rational numbers that is closed under rational addition. Let $S:\Bbb H\to\Bbb H$ be given by $S(x):=x+1,$ where $+$ is the typical addition of rational numbers. This setup satisfies the following axioms:

  • $0\in\Bbb H$
  • $\forall x\in\Bbb H, S(x)\in\Bbb H$
  • $\forall x\in\Bbb H,S(x)\neq 0$
  • $\forall \langle x,y\rangle\in \Bbb H^2,S(x)=S(y)\implies x=y$

These are the Peano axioms except for the Induction Axiom. However, it does not satisfy the Induction axiom, as $\Bbb N$ is a proper subset which contains $0$ and is closed under the successor operation.

On the other hand, it satisfies the following axiom: $$\forall x\in\Bbb H,\exists y\in\Bbb H:\exists z\in\Bbb H:x=y+z.$$ This is true because $0+x=x$ for all $x\in\Bbb H,$ and (together with closure under addition) means that the axiom you're wanting to replace Induction with has been satisfied, so your replacement axiom won't do the job.

Added: In response to your edit, note that $\Bbb H$ also satisfies the following properties:

  • $0+0=0$
  • $\forall x\in\Bbb H,\forall y\in\Bbb H,S(x)+y=S(x+y)$
  • $\forall x\in\Bbb H,\forall y\in\Bbb H,S(x+y)=x+S(y)$

Added: Since you're interested in Peano Axioms without induction, consider the following list of axioms regarding a set $\Bbb A,$ a function $S:\Bbb A\to\Bbb A,$ elements $\mathbf{0}\in\Bbb A$ and $\mathbf{1}\in\Bbb A$ such that $\mathbf{1}=S(\mathbf{0}),$ two operations $\oplus$ and $\odot$ on $\Bbb A,$ and a binary relation $\prec$ and the corresponding reflexive relation $\preceq.$

  1. $\forall\langle x,y,z\rangle\in\Bbb A^3,(x\oplus y)\oplus z=x\oplus (y\oplus z)$
  2. $\forall\langle x,y\rangle\in\Bbb A^2,x\oplus y=y\oplus x$
  3. $\forall\langle x,y,z\rangle\in\Bbb A^3,(x\odot y)\odot z=x\odot (y\odot z)$
  4. $\forall\langle x,y\rangle\in\Bbb A^2,x\odot y=y\odot x$
  5. $\forall\langle x,y,z\rangle\in\Bbb A^3,x\odot (y\oplus z)=(x\odot y)\oplus (x\odot z)$
  6. $\forall x\in\Bbb A,x\oplus \mathbf{0}=x$
  7. $\forall x\in\Bbb A,x\odot \mathbf{0}=\mathbf{0}$
  8. $\forall x\in\Bbb A,x\odot \mathbf{1}=x$
  9. $\forall\langle x,y,z\rangle\in\Bbb A^3,x\prec y\wedge y\prec z\implies y\prec z$
  10. $\forall x\in\Bbb A,\neg[x\prec x]$
  11. $\forall\langle x,y\rangle\in\Bbb A^2,x\prec y\vee x=y\vee y\prec x$
  12. $\forall\langle x,y,z\rangle\in\Bbb A^3,x\prec y\implies x\oplus z\prec y\oplus z$
  13. $\forall\langle x,y,z\rangle\in\Bbb A^3,\mathbf{0}\prec z\wedge x\prec y\implies x\odot z\prec y\odot z$
  14. $\forall\langle x,y\rangle\in\Bbb A^2,x\prec y\implies\exists z\in\Bbb A:x\oplus z=y$
  15. $\mathbf{0}\prec\mathbf{1}\wedge\forall x\in\Bbb A,\mathbf{0}\prec x\implies \mathbf{1}\preceq x$
  16. $\forall x\in\Bbb A,\mathbf{0}\preceq x$
  17. $\forall x\in\Bbb A,\neg\bigl[S(x)=\mathbf{0}\bigr]$
  18. $\forall\langle x,y\rangle\in\Bbb A^2,S(x)=S(y)\implies x=y$
  19. $\forall\langle x,y\rangle\in\Bbb A^2,x\oplus S(y)=S(x\oplus y)$
  20. $\forall\langle x,y\rangle\in\Bbb A^2,x\odot S(y)=(x\odot y)\oplus x$

The assumptions that $\mathbf{0}\in\Bbb A$ and that $S:\Bbb A\to\Bbb A,$ along with the usual axioms of equality, the axiom of induction that you mentioned, and axioms 17 and 18 above, together comprise Peano's original (second-order) axiomatization of Peano Arithmetic, or $\mathsf{PA}.$ From those, the operation of $\oplus$ can be defined by 6 and 19 with induction (which allows $\odot$ to be defined by 7 and 20 with induction), and the relation $\prec$ (and then $\preceq$) can also be defined along the lines of 14.

On the other hand, if we add the single axiom $6$ to Peano's original axioms, then even without induction (which is needed in order to make $\oplus$ a total operation instead of a partial operation on $\Bbb A$), the statement $$\forall x,\Bigl(x\in\Bbb A\iff\exists\langle y,z\rangle\in\Bbb A^2:x=y\oplus z\Bigr)\tag{$\heartsuit$}$$ can be proved directly.

Now, consider the following:

Example:

  • Let $\Bbb A$ be the set whose elements are the polynomials in a single indeterminate (say $t$) with all coefficients in the integers, such that either every coefficient is $0$ or else the leading coefficient is positive. So, for example, $1t^5+-7$ is an element, but $-3t^3+10$ is not.
  • Let $\mathbf{0}$ be the polynomial with every coefficient equal to $0.$
  • Let $\mathbf{1}$ be the polynomial whose lowest degree coefficient is $1$ and the rest equal to $0.$
  • Let $\oplus $ be the usual polynomial addition operation.
  • Let the function $S$ be given by $S(x):=x\oplus \mathbf{1}.$
  • Let $\odot$ be the usual polynomial multiplication operation.
  • Let $\ominus$ be the usual polynomial subtraction operation.
  • Let $\preceq$ be defined by $x\preceq y$ if $y\ominus x\in\Bbb A.$
  • Let $\prec$ be defined by $x\prec y$ if $x\neq y$ and $x\preceq y.$

One can prove that the following are equivalent for any two polynomials $x,y\in\Bbb A:$

  • $x\prec y.$
  • $x\neq y,$ and on the highest degree term where the coefficients of $x$ and $y$ differ, the coefficient of $x$ is less than that of $y.$

From there, one can prove that $\Bbb A$ satisfies axioms $1$ through $20,$ as well as $(\heartsuit),$ and even the statement $$\forall x\in\Bbb A,x\neq \mathbf{0}\implies\exists\langle y,z\rangle\in\Bbb A^2:x=S(y\oplus z).$$ However, consider the set of polynomials whose coefficients are $0$ for all but (possibly) the lowest degree term, which is a proper submodel of $\Bbb A$ corresponding exactly to the usual natural numbers, and so induction fails.

Cameron Buie
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  • Thanks, is it not the case that because one 'modification' above was defining addition to result in only things which could be produced from repeated applications of the successor function to $0$, if we stipulate that every natural number must be the result of addition, it follows that the non-integer elements of $\mathbb{H}$ cannot actually be the result of addition, and fail the axiom with which I wanted to replace the axiom of induction? – Princess Mia May 20 '24 at 04:17
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    @PrincessMia: there is nothing in your axiom that limits the naturals to things which you can reach from $0$ by successor. In addition, your axiom 2 is second order logic, which PA avoids. The usual second order arithmetic induction is much simpler-it basically says all the naturals can be reached by successor from $0$. Second order logic does not have the nice proof properties that first order logic has, but this is a complicated subject. – Ross Millikan May 20 '24 at 04:52
  • @RossMillikan can you tell me where exactly the error is in the following reasoning: 1. every result of addition can be reached from $0$ by a successor- as we start off with $0+0 = 0$, and every other result of addition, $n+m$, is the successor of a 'previous' version of addition- it is either $S(S^{-1}(n) + m)$ or $S(n+ S^{-1}(m))$. So there is ultimately a chain of successors from $0$ to $n+m$. 2. Because every natural number is the result of addition by axiom 2, every natural number can be reached from $0$ by a successor. – Princess Mia May 20 '24 at 04:58
  • @PrincessMia: nothing you have said implies there are not naturals not reachable from $0$. The example with the half integers is a good one. They satisfy your axiom 2 using the usual addition function. You are hoping the only if will rule out numbers that are not the usual naturals, but we get to build a model with whatever we want as long as it satisfies the axioms. So $0+\frac 12=\frac 12$ and I have shown that $\frac 12$ is the sum of two naturals. – Ross Millikan May 20 '24 at 05:05
  • @RossMillikan isn't addition undefined for $1 \over 2$ though, according to the definition I have given above (using the successor function given in the post to which we are replying?) – Princess Mia May 20 '24 at 05:11
  • The usual definition of arithmetic has $0+x=x$ for all $x$, so $0+\frac 12=\frac 12$ is specified. It doesn't have an answer for $\frac 12+\frac 12$ but that means the model builder can do whatever is wanted and making it $1$ will make everything work. It is hard to make axioms exclude other elements, particularly in first order logic. This is a big topic in model theory, which I know little about. The usual PA axioms and yours both allow nonstandard naturals. You just have to make them work right and they can be there. – Ross Millikan May 20 '24 at 05:18
  • @PrincessMia, it seems like your idea may be salvaged in the following way. First of all, take $+$ to be a primitive notion, instead of trying to define it. Second give axioms about $+$ requiring the usual properties we want (definitely the closure and identity properties, and how it interacts with successors, but possibly other, too). Third, include the axiom that for any $x\neq 0,$ there exist $y,z$ such that $x=S(y+z).$ (cont'd) – Cameron Buie May 20 '24 at 19:01
  • @PrincessMia, I have finally gotten the chance to check whether the suggested fix will work. It turns out that it doesn't. See the edits to my answer for details, and let me know if you have any questions! – Cameron Buie Jun 25 '24 at 02:42