User:Dfeuer/Definition:Foundational Relation

Definition
Let $\prec$ be a relation on a class $A$.

Classically, the intuitionistic definition is equivalent to the Smullyan and Fitting one. Intuitionistically, they appear to be incomparable, but the intuitionistic definition (explicitly) allows well-founded induction to work, which is the main reason to define a foundational relation.

Intuitionistic Definition
Suppose that for every subclass $B \subseteq A$:


 * $\biggl({ \forall x \in A:

\Bigl({ \bigl({ \forall y \in A: (y \prec x \implies y \in B) }\bigr) \implies x \in B }\Bigr) }\biggr) \implies B = A$

Then $\prec$ is foundational.

Smullyan &amp; Fitting's Definition
$\prec$ is well-founded iff every non-empty subclass $B$ of $A$ has a minimal element.

Intuitionistic (classically) implies SF
Let $B$ be a non-empty subclass of $A$.

Then $A \setminus B \ne A$.

Thus $\exists x \in A: \Bigl({ \bigl({ \forall y \in A: (y \prec x \implies y \in A \setminus B) }\bigr) \land x \notin A \setminus B }\Bigr)$.

But $x \notin A \setminus B$ implies that $x \in B$, And $y \prec x \implies y \in A \setminus B$ is the same (for elements of $A$) as $y \in B \implies y \nprec x$, so:

$\exists x \in A: \Bigl({ \bigl({ \forall y \in A: (y \in B \implies y \nprec x) }\bigr) \land x \in B }\Bigr)$

That is, $x$ is a minimal element of $B$.