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I am trying to come up with a model of first order Peano Arithmetic (PA) where induction fails. Let $PA^{-IND}$ have the same axioms as PA except the first order induction axiom schema is replaced with its negation. I need to show there exists a predicate, $P$, that makes first order induction false. $P$ must satisfy $P(0) \land \forall x(P(x) \rightarrow P(Sx))$, yet $\forall x(P(x))$ is false. We can prove multiplication is commutative using double induction on $P(x,y)= (xy=yx)$. Why Does Induction Prove Multiplication is Commutative?

Consider the 2x2 matrices $M_2(N)$ with the standard definitions for matrix addition and multiplication. Let the zero matrix be $0$ and the identity matrix be $S0$. $\forall x(Sx=x+S0)$ is a theorem of $PA^{-IND}$. Matrix multiplication is not commutative, yet we can prove multiplication is commutative for all the successors of $0$. Would $M_2(N)$ be a model of $PA^{-IND}$?

The negation of first order induction says there exists a predicate: $P(0) \land \forall x(P(x) \rightarrow P(Sx)) \land Ex( Not(Px))$

This looks like quantification over predicates but it isn't. The induction schema requires us to add an infinite number of axioms to the language. The negation of induction only requires the addition of a single axiom. Unlike PA, $PA^{-IND}$ has a finite number of axioms. I have simply added a new predicate to the language.

I am using the axioms of PA given by Wikipedia for First Order Arithmetic. These axioms use induction to prove commutativity. Without induction, these axioms are very weak. They don't even require addition to be commutative. I would be interested in any model of $PA^{-IND}$.

Community
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Russell Easterly
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    What is the negation of a schema? – Asaf Karagila Mar 16 '13 at 02:06
  • You are quantifying over predicates. That is not a first order quantification anymore. – Asaf Karagila Mar 16 '13 at 02:38
  • I agree. We do assume the induction schema is true for all predicates. This is a quantification over predicates. – Russell Easterly Mar 16 '13 at 02:47
  • @Russell Easterly: in the Wikipedia article, $\text{PA}^-$ is the list of 15 axioms listed in http://en.wikipedia.org/wiki/Peano_axioms#Equivalent_axiomatizations – Carl Mummert Mar 16 '13 at 03:01
  • Thanks. I need to learn more about semi-rings. Finding a model of semi-rings where induction fails looks much harder. – Russell Easterly Mar 16 '13 at 03:07
  • I gave an example in my answer – Carl Mummert Mar 16 '13 at 03:08
  • By the way, in first order theory, you can't say $\exists P$. – Thomas Andrews Mar 16 '13 at 03:19
  • Also, are you using the definition of $PA^-$ on Wikipedia, or what you defined above: Peano minus induction. Because $PA^-$ on Wikipedia takes away induction but adds in a lot of axioms that would otherwise be proved as theorems if we had induction available... – Thomas Andrews Mar 16 '13 at 03:29
  • I changed the question to avoid confusion. I am using the axioms above the axioms for $PA^-$. $PA^{-IND}$ has the negation of first order induction as an axiom. The negation of induction is not first order as several people have pointed out. – Russell Easterly Mar 16 '13 at 04:01
  • @Russell Easterly: the new question is more confusing to me. You write, "a predicate, $P$, that satisfies first order induction yet ∀x(Px) is false." What does it mean for a predicate to satisfy first-order induction? It seems to me that we could take ''any'' nonstandard model and interpret the new symbol $P$ to only hold of the standard numbers in that model - that must not be what you have in mind. – Carl Mummert Mar 16 '13 at 12:31
  • @Carl Mummert: I mean a predicate that satisfies $P(0) \land \forall x(P(x) \rightarrow P(Sx))$ and also satisfies $\exists x(Not(Px))$. – Russell Easterly Mar 16 '13 at 19:04

6 Answers6

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A structure that satisfies the axioms of Peano arithmetic except for induction is $\mathbb N[x]$, the set of all polynomials with natural number coefficients, setting the successor $S(p) = p + 1x^0$ and zero being $0x^0$.

It has a zero, successor is closed under the set, it is injective. If you think of the Peano axioms as constraints on a directed graph, this is the natural structure to arrive at:

  • $0 \in \mathbb N \land \lnot \exists x \in \mathbb N~:~ S(x) = 0$ - there is a source
  • $x \in \mathbb N \implies S(x) \in \mathbb N$ - there are no sinks
  • $S$ is an injective function - there are no forks in the graph

All that remains is to establish that the graph is connected, which induction does.

So you can have multiple zeros, like $\mathbb N[x]$.

Or you can have a disconnected graph, such as using $\mathbb N \cup \mathbb Z \cup \mathbb Z' \cup \mathbb Z'' \dots$ (treating each pair of the previous sets as pairwise disjoint, but otherwise maintaining their usual properties).

Or some combination of both of the above.

DanielV
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You have to be careful about what you mean by $\text{PA}^-$. The convention I expect is that $\text{PA}^-$ is the finite set of axioms for a discretely ordered semiring - which includes the axiom that multiplication is commutative. This set of axioms is typical when we want to look at systems of arithmetic with limited induction. For example, it is the set of axioms in Kaye's book The Structure of Models of Peano Arithmetic.

The original five axioms proposed by Peano, which only mention the successor operation, but not addition, multiplication, or order, are not sufficient for first-order Peano Arithmetic, because they are not sufficient to define addition or multiplication in first-order semantics.

On the other hand, some authors use an abbreviated set of axioms for Peano arithmetic. For example, Mendelson's logic text uses a smaller set of axioms that do not mention the order relation. These work because he only considers them in conjunction with the induction scheme. It seems to me that $M_2(\mathbb{N})$ is a model of this smaller set of axioms, but I would not call them $\text{PA}^-$.

Even when we look at the full set of axioms for $\text{PA}^-$, we can give an example of a model that does not satisfy induction. Tennenbaum's theorem shows that no nonstandard model of Peano arithmetic is computable. But the set of polynomials over $\mathbb{Z}$ with positive leading coefficient (together with the $0$ polynomial, of course) are a computable nonstandard model of $\text{PA}^-$, so induction must fail in that model, although I am not sure which instance of the schema fails. I believe the proof only relies on a finite number of instances of induction, so that would give a short list of candidates.

Cameron Buie
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Carl Mummert
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  • What about defining the exponential function? Might that require induction and not be possible in the structure that is the set of polynomials over $\mathbb{Z}$ with positive leading coefficient? – Quinn Culver Mar 16 '13 at 02:51
  • Good point - if $\alpha$ is nonstandard in that model then ${\alpha^k : k \in \omega}$ is cofinal, so $\alpha^\alpha$ cannot exist in the model. – Carl Mummert Mar 16 '13 at 02:59
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    A simple example to show induction fails in the ring $\mathbb{Z}[x]$. Let the predicate $P(n)$ be $\exists_k : (n = 2k) \vee (n=2k+1)$. Then $P(1)$ is true, and $P(n) \implies P(n+1)$, but $P(x)$ is false. – David E Speyer Mar 16 '13 at 23:36
  • @DavidSpeyer Why is $P(x)$ false? – YoTengoUnLCD Jan 30 '16 at 06:24
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    @YoTengoUnLCD: because $x$ is neither odd nor even in that ring. – Carl Mummert Jan 30 '16 at 11:15
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If you drop the induction principle it will become easy to find a model so that the other axioms are fulfilled. Consider the set $\mathbb N\cup \{\infty\}$ where $\sigma(\infty) = \infty$ and $\sigma$ is defined as normal for $\mathbb N$. Then the other axioms are fulfilled, but the induction principle doesn't work. Consider for example the statement $\phi(x): x\in\mathbb N$. If $\phi(n)$ we clearly have $\phi(\sigma(n))$ and we have $\phi(0)$, but still $\phi(k)$ is not valid over the entire set since $\neg\phi(\infty)$ (that is the induction principle doesn't work for this set).

skyking
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  • Isn't one of the axioms that no number is its own successor? – No Name Jul 05 '22 at 05:56
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    @NoName No, because it can be proven using the other axioms: First we have that $\sigma(0)\ne0$ by axiom. And then if $\sigma(n)\ne n$ it follows that $\sigma(\sigma(n))\ne\sigma(n)$ from injectivity of $\sigma$ and by induction we conclude that $\sigma(x)\ne x$ for all $x\in\mathbb N$ – skyking Jul 05 '22 at 13:45
  • Ah, OK thanks. I had mistaken injectivity for its implication – No Name Jul 05 '22 at 17:14
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This is not a first-order approach, but you may be able to adapt it to PA.

If we grant the existence of at least one Dedeknd-infinite set $X$, the principle of mathematical induction can be shown to be independent of the other Peano Axioms (the set-theoretic version).

Since $X$ is Dedekind-infinite, there exists a function $f: X\to X$ such that $f$ is injective and not surjective.

Since $f$ is not surjective, we have $x_0 \in X$ such that $\forall x\in X:f(x)\neq x_0$

We can construct subset $n\subset X$ such that $n=\{x_0, f(x_0), f(f(x_0)), \cdots \}$ and $(n,f,x_0)$ satisfies all the Peano axioms (PA1-5):

We can also construct a successor function $g:n\to n$ such that $(n,g,f(x_0))$ satisfies axioms PA1-4, but not induction.

$$g(x) = \begin{cases} x_0, & \text{if $x=x_0$} \\ f(x), & \text{otherwise} \end{cases}$$

See detailed formal proof.

Dan Christensen
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  • This is a valid answer, not sure why someone downvoted it. – DanielV Jan 20 '16 at 22:18
  • I have a pretty good idea. – Dan Christensen Jan 20 '16 at 22:47
  • Do you have a summary of the axiom/rules-inferences/unquanitified constants for DC-proof? I'd be interested in reading about the logic and design choices you made in creating your software. – DanielV Jan 20 '16 at 22:59
  • @DanielV I have tried without success to separately post the user manual online. You can download my software along with the manual at no cost at my website http://www.dcproof.com – Dan Christensen Jan 20 '16 at 23:04
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    I didn't downvote but probably the reason is that you didn't explain any bit of how you do the construction, and answers should be as self-contained as possible according to the rules and goals of this site. In particular, in this case everyone talking about independence knows everything you said in your post so it does not even add anything. – user21820 Jan 30 '16 at 03:47
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[Note, I'm using your definition, Peano minus induction. Carl pointed me to the wikipedia page, where $PA^-$ is defined as something quite different. There, it is Peano minus induction plus a lot of the basic theorems we normally prove by induction.]

With induction, we can prove that every natural number is either zero, or the successor of a natural number. Indeed, this is the easiest usage of induction possible. (It's my favorite induction proof, because $P(0)$ is clearly true, and $P(Sn)$ is true, whether $P(n)$ is true or not.)

There are some simple examples where the other axioms hold but this most basic induction theorem is not true.

Thomas Andrews
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  • One such example is the closure of $\omega + \omega $ under ordinal addition and ordinal multiplication. – Carl Mummert Mar 16 '13 at 02:48
  • I was think non-negative rationals, but that works too, @CarlMummert. – Thomas Andrews Mar 16 '13 at 02:52
  • But I was just looking for the proof that induction was independent, not looking for specifically non-commutativity of multiplication. – Thomas Andrews Mar 16 '13 at 02:53
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    $\text{PA}^-$ has the property that every model has an initial segment isomorphic to $\omega$, so $\mathbb{Q}^+$ can't be a model of that theory. It will be a model of some fragments of PA even smaller than $\text{PA}^-$, of course. – Carl Mummert Mar 16 '13 at 03:12
  • @CarlMummert which axiom is not satisfied by the non-negative rationals? – Thomas Andrews Mar 16 '13 at 03:17
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    The axiom that there is no number between 0 and 1. $\text{PA}^-$ is the set of axioms for discretely ordered semirings - see my answer. – Carl Mummert Mar 16 '13 at 03:18
  • Ah, the question defines the axioms very differently as Peano without the induction hypothesis. This is something else entirely. It's really "Peano, minus induction, plus a lot of the most basic theorems we normally prove by induction." @CarlMummert – Thomas Andrews Mar 16 '13 at 03:23
  • Right - Peano arithmetic includes much more than just induction and the 5 Peano axioms for successor. Even with the full induction scheme in the language with only successor, it is not possible to define addition and multiplication in first-order semantics (e.g. it is not possible to define multiplication in Presberger arithmetic). – Carl Mummert Mar 16 '13 at 03:26
  • Yes, but you can prove addition is commutative, associative, etc. So yeah, I was seeing it as "minimal Peano for addition and multiplication." Aside: Even though I've read the proof, I still don't "get" why exponentiation can be defined with a Diophantine equation. That seems mystifying to me. @CarlMummert – Thomas Andrews Mar 16 '13 at 03:33
  • @ThomasAndrews So the closure of $w+w$ under ordinal multiplication and addition satisfies the axioms of PA- (as given in the link of Carl)? – Vivaan Daga Oct 19 '21 at 17:35
  • Answer will be much appreciated! – Vivaan Daga Oct 20 '21 at 06:35
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I think a simple example of PA negating induction would be the set $\mathbb N \cup C$ where $C$ is some (any) cycle in the sense that some symbol $c \in C$, and for every $x \in C$, $Sx \in C$, and $SSSSSSSx = x$.

  • A cycle either wouldn't have a zero or it would have a fork, defying the axiom that successor is an injective function. Induction actually does apply to integers $\pmod p$. – DanielV Jan 20 '16 at 21:59
  • @DanielV Where does this answer mention that some element of the cycle is reachable from the naturals? – No Name Jul 05 '22 at 05:59
  • @NoName You are right I must have misread it six years ago. Maybe I took "any" too literally. – DanielV Jul 05 '22 at 06:11