Principle of Mathematical Induction/Naturally Ordered Semigroup/General Result

Theorem

Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.

Let $p \in S$.

Let $T \subseteq S$ such that:

$x \in T \implies p \preceq x \land \paren {x \in T \implies x \circ 1 \in T}$

Then:

$S \setminus S_p \subseteq T$

where:

$\setminus$ denotes set difference

$S_p$ denotes the set of all elements of $S$ preceding $p$.

Proof

Let $S_p$ be the set of all elements of $S$ preceding $p$:

$S_p = \set {x \in S: x \prec p}$

Let $T' = T \cup S_p$.

Then the set $T'$ satisfies the conditions of the Principle of Mathematical Induction for a Naturally Ordered Semigroup.

From that result:

$T' = S$

By Set Difference with Union is Set Difference:

$S \setminus S_p = T \setminus S_p$

By Set Difference is Subset:

$T \setminus S_p \subseteq T$

completing the proof.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Corollary $16.5.1$