The induction schema of Peano Arithmetic is standardly given as the universal closure of $\phi(0)\land \forall x (\phi(x)\rightarrow \phi(x+1)) \rightarrow \forall x\phi(x)$. However, since the language of arithmetic has a name for every standard number, it is not obvious (to a beginner like me) why parameters are necessary in the induction schema; why not restrict to the case where $x$ is the only free variable in $\phi$? 


* Does having parameters in the induction schema really make the system stronger, and, if so, how is that proven?
* Are there natural theorems that can only or most easily be proven using the stronger system?
* Is the weaker system of any interest?



Parameters in arithmetic induction axiom schemas
The two theories are equivalent. To see this, let's assume that we have the parameter-free induction, and suppose that $\phi(x,y)$ is a formula with two free variables. Suppose we have a model $M$, satisfying the parameter-free induction, and there is a $b\in M$ such that $\phi(x,b)$ obeys the hypothesis of the induction scheme with parameter $b$, but not the conclusion. I claim that there is a least such $b$ in $M$. The reason is that the collection of such $b$ violating the induction scheme for $\phi(x,b)$ with parameter $b$ is a parameter-free definable subset of $M$, since this property is expressible, but the parameter-free induction scheme proves that every nonempty definable set $B$ has a least member, because otherwise the assertion $\psi(x)$ expressing that $x$ is below all members of $B$ would be inductive and hence universal, contrary to $B$ being nonempty. So there is a least such $b$.


In particular, the least such $b$ is actually definable, and so we do not actually need it as a parameter after all, and so the induction scheme for $\phi(x,b)$ follows by the parameter-free induction scheme. So there can be no such $b$ for which the parameter-induction scheme fails.


Thus, the theory of parameter-free induction implies the full theory you mention. 


