Equivalence of weak induction

Let $P(n)$ be an open statement about $n\in \mathbb{N}$. In the weak induction, beside the base case, it is stated:

$\forall k\in \mathbb{N}(P(k)\implies P(k+1))$$\qquad (i)$

But, in this page right in the beginning, it states differently, according to my understanding (with $a=1$)

$((\exists k\in \mathbb{N}) P(k))\implies P(k+1)$$\qquad (ii)$

Are the statements (i) and (ii) equivalent? If yes, why?

Solution 1:

Consider the following sentence: "If something is a cube, then it is large"

At first sight, you might be inclined to translate this to:

$\exists x \ Cube(x) \to Large(x)$

but this has a problem: the $x$ in $Large(x)$ is a free variable, and as such it has lost its connection to the $x$ in $(Cube(x)$

Adding it to the scope of the existential quantifier:

$\exists x \ (Cube(x) \to Large(x))$

doesn't help, because that sentence would be vacuously true as soon as there is something that is not a cube ... which is clearly not what we want either.

Instead, you should realize that by 'some', we really just mean 'any one', i.e. you need to use a universal. That is, what the sentence means is: "Anything that is a cube is large" ... and now you see that the correct formalization is:

$\forall x \ (Cube(x) \to Large(x))$

The same is going on in your case.

On the linked page, it states:

if P(k) is true for some integer k≥a, then P(k+1) is also true.

Ignoring the $a$, it looks as if this translates to (as you suggest):

$((\exists k\in \mathbb{N}) P(k))\implies P(k+1)$$\qquad (ii)$

but as with the earlier example, the problem is that in this formula, the $k$ in $P(k+1)$ ends up being free, whereas it should clearly refer to the very same $k$ used in $P(k)$.

What the authors try to say is that "if some number $k$ has property $P$, then its successor (i.e. $k+1$) has property $P$ as well", and as with the earlier example, we now realize that this translates to your i):

$\forall k\in \mathbb{N}(P(k)\implies P(k+1))$$\qquad (i)$

So no, your given statements i) and ii) are not equivalent, but the authors didn't mean to express ii) in the first place: what they say also translates to i)