Well-Founded Induction

Theorem
Let $\left({A,\prec}\right)$ be a foundational relation.

Let $\prec^{-1} \left({ x }\right)$ denote the preimage of $x$ for each $x \in A$.

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

Suppose $\forall x \in A: \left({ \prec^{-1} \left({ x }\right) \subseteq B \implies x \in B }\right)$. (1)

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
Assume, to the contrary, that $A \not \subseteq B$.

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

By Well-Founded Relation Determines Minimal Elements, $A \setminus B$ must have some $\prec$-minimal element.


 * $\displaystyle \exists x \in \left({ A \setminus B }\right): \left({ A \setminus B }\right) \cap \prec^{-1} \left({ x }\right) = \varnothing$, so $A \cap \prec^{-1} \left({ x }\right) \subseteq B$.

Since $\prec^{-1} \left({ x }\right) \subseteq A$,
 * $\prec^{-1} \left({ x }\right) \subseteq B$.

Thus, by hypothesis (1), $x \in B$.

But this contradicts the fact that $x \in \left({A \setminus B}\right)$.

Therefore, we are forced to conclude that $\left({A \setminus B}\right) = \varnothing$ and $A \subseteq B$.

Therefore, $A = B$.

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