# Equivalence of Definitions of Strictly Inductive Semigroup

## Theorem

The following definitions of the concept of **Strictly Inductive Semigroup** are equivalent:

### Definition 1

Let there exist $\beta \in S$ such that the only subset of $S$ containing both $\beta$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.

- $\exists \beta \in S: \forall A \subseteq S: \paren {\beta \in S \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$

Then $\struct {S, \circ}$ is a **strictly inductive semigroup**.

### Definition 2

Let there exist an epimorphism from $\struct {\N_{>0}, +}$ to $\struct {S, \circ}$.

Then $\struct {S, \circ}$ is a **strictly inductive semigroup**.

### Definition 3

Let $\struct {S, \circ}$ be such that either:

- $\struct {S, \circ}$ is isomorphic to $\struct {\N_{>0}, +}$

or:

- there exist $m, n \in \N_{>0}$ such that $\struct {S, \circ}$ is isomorphic to $\struct {\map {D^*} {m, n}, +^*_{m, n} }$

where $\struct {\map {D^*} {m, n}, +^*_{m, n} }$ is the restricted dipper semigroup on $\tuple {m, n}$.

Then $\struct {S, \circ}$ is a **strictly inductive semigroup**.

## Proof

### $(1)$ implies $(2)$

Let $\struct {S, \circ}$ be a strictly inductive semigroup by definition $1$.

Then by definition $1$:

Let there exist $\beta \in S$ such that the only subset of $S$ containing both $\beta$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.

- $\exists \beta \in S: \forall A \subseteq S: \paren {\beta \in S \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$

Then $\struct {S, \circ}$ is a **strictly inductive semigroup**.

This theorem requires a proof.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.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 `{{ProofWanted}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Thus $\struct {S, \circ}$ is a strictly inductive semigroup by definition $2$.

$\Box$

### $(2)$ implies $(1)$

Let $\struct {S, \circ}$ be a strictly inductive semigroup by definition $2$.

Then by definition $2$:

Let there exist an epimorphism from $\struct {\N_{>0}, +}$ to $\struct {S, \circ}$.

Then $\struct {S, \circ}$ is a **strictly inductive semigroup**.

This theorem requires a proof.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.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 `{{ProofWanted}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Thus $\struct {S, \circ}$ is a strictly inductive semigroup by definition $1$.

$\Box$

This needs considerable tedious hard slog to complete it.In particular: 2 iff 3To 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 `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.8$