Why induction is defined as an implication instead of an "if and only if" statement? The formal definition of induction, taken from wikipedia, is written as
$$\forall P.\,[[P(0)\land \forall (k\in \mathbb {N} ).\,[P(k)\implies P(k+1)]]\implies \forall (n\in \mathbb {N} ).\,P(n)]$$
But then, it is this version true?
$$\forall P.\,[[P(0)\land \forall (k\in \mathbb {N} ).\,[P(k)\implies P(k+1)]]\iff \forall (n\in \mathbb {N} ).\,P(n)]\tag{1}$$
If it is not true can you show me why? (I cant found a counterexample.) Thank you.

The background of this question is that if $(1)$ is true then the proof for the equivalence of weak induction ($W$) and strong induction ($S$) is really trivial.
Let the shortened theorem of strong induction to be $$S\implies \forall (n\in \mathbb {N} ).\,P(n)$$
and by $(1)$ the definition of weak induction $$W\iff \forall (n\in \mathbb {N} ).\,P(n)$$
The case $S\implies W$ is trivial but now the case $W\implies S$ is trivial too because we can verify that 
$$\forall (n\in \mathbb {N} ).\,P(n)\implies S$$
 A: Of course it is true, but the $\Longleftarrow$ direction is trivially true as a matter of logic, so writing $\Longrightarrow$ directs the reader's focus to the interesting direction.
If $\forall n\, P(n)$ then clearly $P(0)$ is true, and clearly also any formula of the form $\cdots\Rightarrow P(k+1)$ will be true.
A: The implication from left to right is the key property that the ordering of the natural numbers satisfies.  (In fact, a second-order version of induction uniquely characterizes the set of natural numbers.)
But the implication from right to left holds trivially for any structure at all: If $A$ is any set, $a_0$ is any member of $A,$ and $f\colon A\to A$ is any function, then we have
$$(\forall P) \Big(\big((\forall x\in A)\,P(x)\big) \implies \big(P(a_0)\land (\forall x\in A)\,\big(P(x)\implies P(f(x))\big)\Big).$$
So this reversed implication tells us nothing interesting about $\mathbb{N}.$
A: Induction is 1/2 of the definition of $\forall$ for natural numbers.  It is the rule that let's you introduce $\forall$.  It is:
$$\frac{P(0),~ (P(n) \vdash P(n + 1))}{\forall m~P(m)}$$
Note that it is not $\forall n P(n) \implies P(n+1)$.  In induction, $n$ is treated as a free variable.  How exactly would you prove $\forall n P(n) \implies P(n+1)$ if you don't already have a rule for introducing $\forall$?  Induction is the rule for introducing $\forall$ so it can't depend on it already being introduced.
The rule for eliminating $\forall$ is (roughly) :
$$\frac{\forall m~P(m)}{P(n)}$$
So there are 2 reasons it is not an equivalence:  first, induction is (usually) an inference, not an axiom, and inferences are 1 directional.  Second, the reverse direction isn't useful, the elimination rule is what you want for the reverse direction.
