Equivalence of Definitions of Well-Founded Relation

Theorem

Let $\struct {S, \RR}$ be a Relational Structure.

The following definitions of the concept of Well-Founded Relation are equivalent:

Definition 1

$\RR$ is a well-founded relation on $S$ if and only if:

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

where $\O$ is the empty set.

Definition 2

$\RR$ is a well-founded relation on $S$ if and only if:

for every non-empty subset $T$ of $S$, $T$ has a minimal element.

Proof

By definition of minimal element:

$\forall y \in T: y \preceq x \implies x = y$

$(1)$ implies $(2)$

Let $\RR$ be a well-founded relation by definition $1$.

Then by definition:

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

So, let $T \subseteq S: T \ne \O$.

That is, $T$ is a non-empty subset of $S$.

It is asserted that there exists $z \in T \setminus \set z$ such that:

$\forall y \in T \setminus \set z: \tuple {y, z} \notin \RR$

This does not exclude the case that $\tuple {z, z} \in \RR$.

So if there exists $y \in T$ such that $\tuple {y, z} \in \RR$, then it must be the case that $y = z$

That is:

$\forall y \in T: \tuple {y, z} \in \RR \implies y = z$

That is, $z$ is a minimal element of $T$ under $\RR$.

So we have shown that:

for every non-empty subset $T$ of $S$, there exists an element $z$ in $T$ such that $z$ is a minimal element of $T$.

Thus $\RR$ is a well-founded relation by definition $2$.

$\Box$

$(2)$ implies $(1)$

Let $\RR$ be a well-founded relation by definition $2$.

Then by definition:

for every non-empty subset $T$ of $S$, $T$ has a minimal element.

That is, for every non-empty subset $T$ of $S$, there exists an element $z$ in $T$ such that:

$\forall y \in T: \tuple {y, z} \in \RR \implies y = z$

Hence:

$\forall y \in T \setminus \set z: \tuple {y, z} \in \RR \implies y = z$

But:

$\not \exists y \in T \setminus \set z: y = z$

and so it follows that:

$\forall y \in T \setminus \set z: \tuple {y, z} \notin \RR$

That is:

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

Thus $\RR$ is a well-founded relation by definition $1$.

$\blacksquare$