Strictly Increasing Sequence of Natural Numbers

Theorem
Let $$\mathbb{N}^*$$ be the set of natural numbers without zero: $$\mathbb{N}^* = \left\{{1, 2, 3, \ldots}\right\}$$.

Let $$\left \langle {n_r} \right \rangle$$ be a sequence in $\mathbb{N}^*$.

Let $$\left \langle {n_r} \right \rangle$$ be strictly increasing.

Then $$\forall r \in \mathbb{N}^*: n_r \ge r$$.

Proof
This is to be proved by induction on $$r$$.

For all $$r \in \mathbb{N}^*$$, let $$P \left({r}\right)$$ be the proposition $$n_r \ge r$$.

Basis for the Induction

 * When $$r = 1$$, it follows that $$n_1 \ge 1$$ as $$\mathbb{N}^*$$ is bounded below by $$1$$.

Thus $$P(1)$$ is true.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So our induction hypothesis is that $$n_k \ge k$$.

Then we need to show that $$n_{k+1} \ge k+1$$.

Induction Step
This is our induction step:

Suppose that $$n_k \ge k$$.

Then as $$\left \langle {n_r} \right \rangle$$ is strictly increasing, $$n_{k+1} > n_k \ge k$$

From Precedes Next (which applies because the Natural Numbers are a Naturally Ordered Semigroup), we have $$n_{k+1} > k \Longrightarrow n_{k+1} \ge k+1$$.

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall r \in \mathbb{N}^*: n_r \ge r$$.