Proofs with Induction Imply Proofs Without Induction? Assume we can prove $\forall x P(x)$ in first order Peano Arithmetic (PA) using induction and modus ponens. Does this mean we can prove $\forall x P(x)$ from the other axioms of PA without using induction?
Given the induction axiom $(P(0) \land \forall x(P(x) \rightarrow P(Sx))) \rightarrow \forall x P(x)$ we must first prove $P(0) \land \forall x(P(x) \rightarrow P(Sx))$ using the other axioms of PA before we can deduce $\forall x P(x)$. This can be converted to $P(0) \land \forall x( \neg P(x) \lor P(Sx))$. We better not be able to prove $\exists x \neg P(x)$ from the other axioms so this reduces to $P(0) \land \forall x P(Sx)$. It seems reasonable if we can prove $P(0) \land \forall x P(Sx)$ from the other axioms without using induction we can prove $\forall x P(x)$ without induction.

The part: 
"This can be converted to $P(0) \land \forall x( \neg P(x) \lor P(Sx))$."
is correct.
The part: 
"We better not be able to prove $\exists x \neg P(x)$ from the other axioms so this reduces to $P(0) \land \forall x P(Sx)$."
is not. In general $\forall x(A(x)\lor B(x))$ is not equivalent to $\forall x A(x) \lor \forall x B(x)$.
This shows that your argument is not correct. One can also show that what it argues for is not true: It is easy to construct models of the rest of the axioms of first-order PA in which a sentence easily proved in first-order PA is false. 
The details depend on the exact formulation of first-order PA. For the version you linked to, note that the non-negative reals (or non-negative rationals, or non-negative numbers of the form $\frac{n}{2}$) with the usual interpretation of successor, addition, and multiplication form a model of the rest of the axioms if we throw the induction scheme away. The sentence $\exists x(x+x=S0)$ is true in all these models, but easily refutable in first-order PA.
