Principle of Finite Induction/Proof 2

Theorem

Let $n_0 \in \Z$ be given.

Let $\Z_{\ge n_0}$ denote the set:

$\Z_{\ge n_0} = \set {n \in \Z: n \ge n_0}$

Let $S \subseteq \Z_{\ge n_0}$ be a subset of $\Z_{\ge n_0}$.

Suppose that:

$(1): \quad n_0 \in S$

$(2): \quad \forall n \ge n_0: n \in S \implies n + 1 \in S$

Then:

$\forall n \ge n_0: n \in S$

That is:

$S = \Z_{\ge n_0}$

Proof

The validity of the material on this page is questionable. In particular: This only takes on board a subset of $\N$, where we need a subset of $\Z$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Consider $\N$ defined as a naturally ordered semigroup.

The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result.

$\blacksquare$