# Well-Founded Induction

## Theorem

Let $\struct {A, \prec}$ be a foundational relation.

Let $\prec^{-1} \sqbrk x$ denote the preimage of $x$ for each $x \in A$.

Let $B$ be a class such that $B \subseteq A$.

Suppose that:

$(1): \quad \forall x \in A: \paren {\prec^{-1} \sqbrk x \subseteq B \implies x \in B}$

Then:

$A = B$

That is, if a property passes from the preimage of $x$ to $x$, then this property is true for all $x \in A$.

## Proof

Aiming for a contradiction, suppose $A \not \subseteq B$.

Then $A \setminus B \not = 0$.

By Well-Founded Relation is Strongly Well-Founded, $A \setminus B$ must have some $\prec$-minimal element.

Then:

$\exists x \in \paren {A \setminus B}: \paren {A \setminus B} \cap \prec^{-1} \sqbrk x = \O$

so:

$A \cap \prec^{-1} \sqbrk x \subseteq B$

Since $\prec^{-1} \sqbrk x \subseteq A$:

$\prec^{-1} \sqbrk x \subseteq B$

Thus, by hypothesis $(1)$:

$x \in B$

But this contradicts the fact that:

$x \in \paren {A \setminus B}$

By Proof by Contradiction it follows that:

$\paren {A \setminus B} = \O$

and so:

$A \subseteq B$

Therefore:

$A = B$

$\blacksquare$

## Also see

Well-Ordered Induction, a weaker theorem that does not require the Axiom of Foundation.