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 parameter $y$. 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. First, I claim that there is a least such $b$ in $M$. This The reason is proved by 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, since because otherwise the formula assertion $\psi(x)$, asserting \psi(x)$expressing that$x$is below every such$y$, that is, below any$y$for which$\phi(x,y)$satisfies the hypothesis all members of the induction scheme but not the conclusion,$B$would itself be inductive and hence universal; but we assumed it was not, contrary to$B$being nonempty. ThusSo there is a least such$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$for which the parameter-induction scheme fails. Thus, the theory of parameter-free induction implies the full theory you mention. 1 The two theories are equivalent. To see this, let's assume that we have the parameter-free induction, and suppose that$\phi(x,y)$is a formula with parameter$y$. Suppose we have a model$M$, satisfying the parameter-free induction, and there is a$b\in M$such that$\phi(x,b)$obeys the hypothesis of the induction scheme, but not the conclusion. First, I claim that there is a least such$b$in$M$. This is proved by the parameter-free induction, since otherwise the formula$\psi(x)$, asserting that$x$is below every such$y$, that is, below any$y$for which$\phi(x,y)$satisfies the hypothesis of the induction scheme but not the conclusion, would itself be inductive and hence universal; but we assumed it was not. Thus, the least such$b$is actually definable, and so we do not need it as a parameter after all, and so the induction scheme for$\phi(x,b)\$ follows by the parameter-free induction scheme.