# 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:

$T' = S$
$S \setminus S_p = T \setminus S_p$
$T \setminus S_p \subseteq T$

completing the proof.

$\blacksquare$