Statement equivalent to Induction Principle How to prove that the following statement is equivalent to Induction principle?(I'v alredy done that IP $\Rightarrow$ Statement)
For every non empty $A\subset \mathbb{N}$ we have $A-s(A) \neq \emptyset $ where $s$ is the injective function in Peano Axioms($s(n)=n+1$)
 A: Assume the statement, and suppose that $1\in A$, $n\in A\to s(n)\in A$, and $A\ne\mathbb{N}$. Let $B=\mathbb{N}\setminus A$; clearly $B\ne\varnothing$, so $A\setminus s[A]\ne\varnothing$, so $B\setminus s[B]\ne\varnothing$. Fix $b\in B\setminus s[B]$. 
You have as an axiom that $\mathbb{N}\setminus s[\mathbb{N}=\{1\}$, and $b\ne 1$ (since $1\in A=\mathbb{N}\setminus B$), so $b\in s[\mathbb{N}]$, i.e., $b=s(n)$ for some $n\in\mathbb{N}$. Now $b\notin s[B]$, so $n\notin B$, and therefore $n\in A$. But then the hypothesis on $A$ ensures that $s(n)\in A$, i.e., that $b=s(n)\in A\cap B=\varnothing$, which is absurd. This contradiction shows that in fact $A$ must be all of $\mathbb{N}$.
(Your book introduces the natural numbers in a somewhat unusual way, which accounts for the confusion in the answers and comments..)
A: It's not equivalent to the induction principle.
The induction principle essentially says that applying the successor function to the smallest natural number eventually produces every natural number; i.e., it says that $\mathbb{N}=\{0, s(0), s^2(0), s^3(0), \dots\}$.  The statement you give, on the other hand, says that every nonempty subset $A \subset \mathbb{N}$ has an element that is not the successor of any element of $A$:
$$A-s(A)\neq\emptyset \equiv \exists_{n\in A}[n \notin s(A)]\equiv\exists_{n\in A}\forall_{m\in A}[n\neq s(m)].$$
But this is strictly weaker than the induction principle.  For instance, it still holds if you take $s(n)=n+2$, or indeed any increasing injection on the natural numbers.  There will still be a smallest element (in the ordinary sense) in any nonempty $A\subset\mathbb{N}$, whose predecessor (in the sense of $s^{-1}$) will either not exist or not be in $A$.  But if $s(n)=n+2$, the induction principle no longer holds.
