Determining if two statements are equivalent, logical sense.

Question

I am confused, I am working with proofs and I have the following statement to work with

$\forall n\in\mathbb{N},P(n) \implies P(n+1)$

1. I have a second statement $\forall n\in\mathbb{N}, P(n)\Rightarrow \forall n\in\mathbb{N},P(n+1)$

I thought they were equivalent, but now I am confused because I am asked to use parenthesis to express the second statement in a much 'closer' way to the first statement... which imo means that I am looking at two logically distinct statements, but why?

And also, 2. is there any way I can rewrite $\forall n\in\mathbb{N},P(n+1)$ without addition?

I have 6 proofs to work with that relay on this main question I am asking; therefore, I need your help trying to understand this...

Can someone place some explanation in natural language of why they aren't equivalent, I was reading them like this:

If theres an $n, P(n)$, and the second one, if theres an $n$ then $P(n)$ then there is an $n (p + 1)$ but sounds too awkward and I don't get what I'm saying.

Answer 1

I think that the correct reading of :

$\forall n \in \mathbb N,P(n) \rightarrow P(n+1)$

must be :

$\forall n (P(n) \rightarrow P(n+1))$ --- (*)

omitting the specification "$\in \mathbb N$", for simplicity.

As said above, it is the standard expression of the inductive step of an inductive proof (where $P(0)$ is the basis step).

The formula :

$\forall n P(n) \rightarrow \forall n P(n+1)$

is equivalent to :

$\forall n P(n) \rightarrow \forall n (n > 0 \rightarrow P(n))$ ---(§).

The two formulae are not equivalent.

Consider as $P(x)$ the predicate : $(x=0)$.

With this interpretation, from (*) we have :

$\forall n(n=0 \rightarrow n+1=0)$

which is false, because for $n=0$ we have that $(n=0 \rightarrow n+1=0)$ is false [$0=0$ is true, while $0+1 = 0$ is false; then apply truth-table for $\rightarrow$]. Thus, it is not true that $(n=0 \rightarrow n+1=0)$ is true for all $n$.

On the other hand, from (§) we have :

$\forall n(n=0) \rightarrow \forall n (n > 0 \rightarrow n=0)$

which is trivially true, because $\forall n(n=0)$ is false.