Induction on finite subset of natural numbers Can we use induction to prove that a statement $P(n)$ is true for all $n \in \mathbb{N} $ such that $n \leq s$, where $s \in \mathbb{N}$? Specifically, in the second induction step, is it enough to show that
$$P(n) \implies P(n+1),$$
by assuming $n<s$ and $n+1 \leq s$?
 A: Yes, you can. It is equivalent to using regular induction (the version that proves that something is true for all natural numbers) to prove that the statement
$$Q(n) = ((n\leq s)\implies P(n))$$
is true for all natural numbers.
A: Yes, in the sense that it will prove the claim. Though it is important to understand that such a scenario is not a proof by induction, since it does not require to invoke the principal of induction in order to prove such a claim. The principal of induction takes the claims $P(1)$ and $P(n)\implies P(n+1)$ for all $n\ge 1$, which is a recipe for infinitely many proofs, one for each $n$, and accepts that that recipe for infinitely many proofs actually constitutes a single finite proof. The case that you are considering, while technically similar, is different. Here you establish that $P(1)$ and that $P(n)\implies P(n+1)$ for all $1\le n< s$. This is a recipe for creating a single proof, that of $P(n)$ for all $1\le n\le s$. For the given $s$ is given, we know just how to construct a proof of $P(s)$, i.e., $P(1)$ is true. Since $P(1)\implies P(2)$, it follows that $P(2)$ is true, and so on. The "and so on" here only hides a finite portion of the proof, so it is short-hand. In actual proof by induction such a "and so on" hides a portion of proof of variable length, depending on the $n$ being considered. The latter is a strictly stronger principal than the former. 
