Equivalence of Definitions of Strictly Well-Founded Relation

$(1)$ $(2)$
By definition, $y \in T$ is a strictly minimal element (under $\RR$) of $T$ :


 * $\forall z \in T: z \not \mathrel \RR y$

Thus it is seen that definition $1$ means exactly the same as definition $2$.

$(1)$ iff $(3)$
Let $\RR$ be a Strictly Well-Founded Relation by definition $1$.

From Strictly Well-Founded Relation is Well-Founded, $\RR$ is a well-founded relation on $S$.

From Strictly Well-Founded Relation is Antireflexive, $\RR$ is antireflexive.

Thus $\RR$ is a Strictly Well-Founded Relation by definition $3$.

$(3)$ implies $(2)$
Let $\RR$ be a Strictly Well-Founded Relation by definition $3$.

Then by definition:
 * $\RR$ is a well-founded relation on $S$


 * $\RR$ is antireflexive.

Thus we have by definition of well-founded relation on $S$:
 * $\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in T \setminus \set z: \tuple {y, z} \notin \RR$

But because $\RR$ is antireflexive:
 * $\tuple {z, z} \notin \RR$

Hence it follows that:


 * $\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in \paren {T \setminus \set z} \cup \set z: \tuple {y, z} \notin \RR$

That is, from Set Difference Union Second Set is Union:


 * $\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in \paren {T \cup \set z}: \tuple {y, z} \notin \RR$

But as $\set z \subseteq T$, we have from Union with Superset is Superset that $T \cup \set z = T$.

Hence:


 * $\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in T: \tuple {y, z} \notin \RR$

Thus $\RR$ is a Strictly Well-Founded Relation by definition $2$.