Well-Founded Induction

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Theorem

Let $\struct {A, \RR}$ be a strictly well-founded relation.

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

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

Suppose that:

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


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 $A \nsubseteq B$.

Then $A \setminus B \ne 0$.

By Strictly Well-Founded Relation determines Strictly Minimal Elements, $A \setminus B$ must have some strictly minimal element under $\RR$.

Then:

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

so:

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

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

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

Thus, by hypothesis $(1)$:

$x \in B$

But this contradicts the fact that:

$x \in A \setminus B$

By Proof by Contradiction it follows that:

$A \setminus B = \O$

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.


Sources