Integer induction without infinity In ZFC minus infinity (let us call this T), one can still define ordinals, and
then define integers as ordinals all of whose members are zero or successor
ordinals. 
I am looking for a formula $\psi$ such that


*

*T proves $\psi(0)$

*T proves $\psi(n)\Rightarrow \psi(n+1)$ for every "integer" (in the above sense ) $n$,

*T does not prove the (obviously true) $\forall \ \text{integer}\ n, \psi(n)$.
This would be a typical situation of $\omega$-incompleteness. This is probably 
well-known, but I cannot remember where it is explained.
 A: Suppose that $\psi$ is such formula, and let $(M,E)$ be a model of $T$ in which $\forall n(n\text{ is an integer}\rightarrow\psi(n))$ is false.
Consider $A=\{k\mid M\models\lnot\psi(k)\land k\text{ is an integer}\}$, then this class cannot have an $E$-minimal element. If $M\models k=\min A$, then of course $k\neq 0$, since $T\vdash\varphi(0)$, so $k=n+1$, but $M\models\psi(n)\land(\psi(n)\rightarrow\psi(n+1))$, so $M\models\psi(k)$.
Therefore $A$ is a class without a least element. This is a contradiction, since if $M\models k\in A$, then $M\models k\cap A\text{ is a set linearly ordered by }E\text{ and without a least element}$, which is a contradiction to the fact that $M$ satisfies the axiom of foundation.

Note that the assumption that $T\vdash\psi(0)\land\forall k(k\text{ is an integer}\land\psi(k)\rightarrow\psi(k+1))$ was necessary here.
On the other hand, if you only require that $T$ proves this for meta-integers, then something like $n$ encodes a proof for contradiction from $T$ is an example for a statement like that.
