This is in response to Adrewj Andrej Bauer's comment on Ricky Derner post, as the answer does not seem to fit in the comment section.
Suppose we take out the Induction Schema and replace it with: $$(SI)\quad \forall k\Bigl(\bigl(\forall n(n\lt k\rightarrow \phi(n))\bigr)\longrightarrow \phi(k)\Bigr)\Rightarrow \forall m(\phi(m)).$$
Now, consider the theory that has the first four Peano Axioms and (SI), instead of the usual Induction schema. The statement "Every natural number is either $0$ or a successor" is a theorem under the usual Peano Axioms, but is not a theorem under this modified axiomatic system. To see this, take $\omega+\omega$ as a model. It satisfies the first four axioms using the usual ordinal successor function, and it satisfies the "strong induction" schema (SI) as well. However, $\omega$ itself, as an element of $\omega+\omega$, is neither a successor nor $0$. (Think of having two copies of the natural numbers, a "red" copy going first and a "blue" copy going second; then the "blue $0$" is neither $0$ nor a successor; you can apply regular induction to the proposition "$n$ is red", but the set you obtain is not all of your set of numbers).
So the two theories are not equivalent. (SI) is a theorem in Peano Arithmetic, but regular induction is not a theorem in the theory that has the first four Peano Axioms and (SI).
However, if you take the first four Peano Axioms, you add "Every number is either $0$ or a successor" as a "fourth-and-a-half" axiom, and then you take (SI) instead of the usual Induction Schema, then you can prove the usual Induction Schema as a theorem in this system; so the two systems (usual Peano Axioms, and first four Axioms plus the "fourth-and-a-half" axiom plus (SI)) are equivalent. If you want (SI) to be equivalent to regular induction, you need a bit more than just the first four Peano Axioms.
