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

Question

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

I'm not quite sure how to interpret this (how to parenthesize this).

On one hand it could be $\forall n \in \mathbb{N}, (P(n) \implies \forall n \in \mathbb{N}, P(n+1))$, following operator precedence. If you were to negate this it becomes kind of weird though...

But it could also be $(\forall n \in \mathbb{N}, P(n)) \implies (\forall n \in \mathbb{N}, P(n+1))$.

Answer 1

The usual convention is that [see Enderton, page 78] :

$¬, ∀, ∃$ apply to as little as possible. For example, [...]

$∀xα→β$ is $(∀xα→β)$, and not $∀x(α→β)$.

Thus, the correct reading of the formula without parentheses is :

$\forall nP(n) \rightarrow ∀nP(n+1)$ --- (A).

The "alternative" reading as :

$\forall n(P(n) \rightarrow ∀nP(n+1))$ --- (B)

is not equivalent to the former.

We have that [see Enderton, page 122] :

$(∃xβ→α)↔∀x(β→α)$, if $x$ does not occur free in $\alpha$.

Thus, (B) means :

$\exists nP(n) \rightarrow ∀nP(n+1)$.

Answer 2

The second one makes sense : if you know that for all n something is true, then you know that for all "n+1" it's also true. The first one would mean that if something is true for a given n, then it's true for any n...Weird.

Answer 3

The statement is a tautology. Certainly if $P(n)$ is true for all natural numbers, $P(n+1)$ is true as well.

The intended meaning that I gather is wanting to show an inductive proof, that is if $P(n)$ is true, then $P(n+1)$ is also true.