Principle of Mathematical Induction/Naturally Ordered Semigroup

Theorem
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $T \subseteq S$ such that $0 \in T$ and $n \in T \implies n \circ 1 \in T$.

Then $T = S$.

Proof
Aiming for a contradiction, suppose that $T \subset S$.

That is, $T$ is a proper subset of $S$:
 * $T \ne S$

Let $T' = S \setminus T$.

Then by Set Difference with Proper Subset:
 * $T' \ne \varnothing$

By axiom $NO1$, $S$ is well-ordered.

By definition of well-ordered set, it follows that $T'$ has a smallest element $x$.

By definition of $T$:
 * $0 \in T$

and so by definition of $T'$:
 * $0 \notin T'$

so:
 * $0 \prec x$

By Sum with One is Immediate Successor in Naturally Ordered Semigroup:
 * $1 \preceq x$

By the definition of a naturally ordered semigroup:
 * $\exists y \in S: y \circ 1 = x$

Again by Sum with One is Immediate Successor in Naturally Ordered Semigroup:
 * $y \prec x$

We have that $x$ is the smallest element of $T'$ and $y \prec x$.

Therefore:
 * $y \notin T'$

and so
 * $y \in T$

But from the definition of $T$:
 * $y \in T \implies y \circ 1 = x \in T$

But then by the definition of $T'$:
 * $x \in T' \implies x \notin T$

From this contradiction, it follows that:
 * $T = S$

Also see

 * Principle of Mathematical Induction


 * Second Principle of Mathematical Induction for Naturally Ordered Semigroup