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I am interested in mathematical systems where first order induction (IND) fails. One example is ring theory + $\forall x(Sx=x+1)$. Non-commutative rings are models of this theory. Induction proves $\forall x\forall y(xy=yx)$ which is false in these models.

Consider the theory $ZF + IND + 0=\{\} + \forall x(S(x)=x \bigcup \{x\})$.

Let $P(x) = Not(0 \in x) \lor \exists y(y \in x \land Not(S(y) \in x)$.

$P(0)$ is true because $0$ is not an element of $0$. $\forall x(P(x) \rightarrow P(S(x)))$ is also true. Notice $P(\omega ) \rightarrow P(S(\omega ))$ is true because $P(\omega )$ is false.

$\forall x(Not(0 \in x) \lor \exists y(y \in x \land Not(S(y) \in x)$ is the negation of the Axiom of Infinity as given by Wikipedia. Does this prove ZF+IND is inconsistent? If so, is this well known?

MyUserIsThis
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Russell Easterly
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    First-order induction is a schema, and it requires some sort of a successor operation which invariably means that something is well-ordered. Secondly, if you assume that there are no inductive sets then obviously assuming the infinity axiom would be inconsistent. You can't eat the cake and have it too. – Asaf Karagila Mar 13 '13 at 01:56
  • I use a standard definition of successor. Where do I assume there are no inductive sets? I just add the first order induction schema to the axioms of ZF. – Russell Easterly Mar 13 '13 at 02:02
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    What does it mean that induction proves xy=yx for rings?? – zyx Mar 13 '13 at 02:04
  • But why? ZF proves second-order Peano is consistent. It proves that every inductive set either has a limit point or is $\omega$. – Asaf Karagila Mar 13 '13 at 02:05
  • Induction proves multiplication is commutative. This means first order induction is independent of the axioms of ring theory because there are commutative rings and non-commutative rings. – Russell Easterly Mar 13 '13 at 02:13
  • I am giving an example of a system where induction proves something that is not true. see http://math.stackexchange.com/questions/318441/can-an-axiom-schema-be-independent – Russell Easterly Mar 13 '13 at 02:31
  • Induction requires definitions for the constant $0$ and the successor function. For ring theory I use $\forall x(Sx=x+1)$ as the definition of successor. There are lots of rings where induction fails. – Russell Easterly Mar 13 '13 at 03:28
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    It would have been nice to display the proof on any of the occasions when it was requested, but apparently what was meant is that xy=yx is a property that the theory of rings can prove is true for x=y=0 and closed under successor in x or y, so that if induction is present, the property holds for all x,y. – zyx Mar 13 '13 at 08:00
  • Only one question, is $x\epsilon y$ some standard language in set theory instead of $x\in y$?, because it took some time to really understand that $P(x)$. – MyUserIsThis Mar 15 '13 at 19:57
  • @MyUserIsThis: No. Just my bad Latex skills. – Russell Easterly Mar 16 '13 at 19:21
  • @RussellEasterly Then I've proposed an edition, the $\LaTeX$ for $\in$ is \in – MyUserIsThis Mar 16 '13 at 21:10

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To answer your question, yes, this does prove that "$ZF$ plus induction" is inconsistent, although I find the phrasing somewhat odd.

The induction "axiom" essentially states that the universe in question is an inductive set. As you have stated it here, $A \vDash INF$ it can be reduced to the statement $A \subseteq \omega$. Clearly this is not true given the axiom of infinity, but it also fails in $ZF$ minus the axiom of infinity (Let $P(x)$ be the statement $x \ne \{ \{ \varnothing \} \}$).

Proof that $A \vDash IND \iff A \subseteq \omega$, given that $A$ is nonempty and transitive:

Let $X$ be the set of all elements in $A$ that are not successors of some other element. If for some nonempty $x$, $x \in X$, then let $P(a)$ be the statement $a \ne x$.

Clearly $P(\varnothing)$, and for any $a$, $P(S(a))$, since $x$ is not a successor. So by induction $A \vDash \forall a: a \ne x$ (since the successor function is absolute with respect to $A$) and $A \vDash x \notin V$, so $x \notin A$. This is a contradiction, so $X \subseteq \{ \varnothing \} \subseteq \omega$. Since $X$ generates $A$ under the successor operation, and $\omega$ is closed under the successor operation, $A \subseteq \omega$.

Conversely, suppose $A \subseteq \omega$, with $A$ nonempty and transitive. Then $A$ is an ordinal (under von Neumann's definition), so this reduces to a statement of induction on the natural numbers.

A.S
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  • Thanks. ZF+IND would be inconsistent even if we restrict first order induction to von Neumann's ordinals. We would still be able to prove $\forall x(x \neq \omega )$. – Russell Easterly Mar 14 '13 at 01:30
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    Yes. It might be worthy of note that a corresponding transfinite induction is actually a provable theorem in ZFC. See $\epsilon$-induction which follows from the Axiom of Regularity. – A.S Mar 14 '13 at 04:08
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Why first-order induction? Using ZF-like axioms, I can prove that, given an injective function $f: s\rightarrow s$ and a $1\in s$ such that $\forall a\in s (f(a)\neq 1)$, there exists a unique subset of $s$ on which all the Peano Axioms hold, including second order induction with $f$ as the successor function, and $1$ as the "first element." In this way, the principle of mathematical induction can actually be derived. Therefore, I would think that if ZF is consistent, then second order induction would also be consistent.

See my formal proof (247 lines) at http://www.dcproof.com/ProofByInduction.html

Dan Christensen
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