First Principle of Transfinite Induction

Theorem
Let $A$ be a class.

Let $\preccurlyeq$ be a well-ordering on $A$.

Let $P$ be a property that satisfies the following condition:


 * For all $x \in A$, if $P$ holds for every $y \prec x$, then $P$ holds for $x$.

Then $P$ holds for all $x \in A$.

Proof
Let $s$ be the smallest element of $A$.

Then vacuously $P$ holds for every element $y \in A$ such that $y \prec s$.

Thus $P$ holds for $s$.

$P$ fails to hold for some $z \in A$.

The class of elements of $A$ for which $P$ does not hold is therefore non-empty.

We have that $A$ is a well-ordered class.

Hence there must be some smallest element $x \in A$ for which $P$ fails to hold.

This cannot be $s$ as we have seen above.

But for every $y \prec x$, we have that $P$ holds for $y$.

But then $P$ holds for $x$.

This contradicts our definition of $x$.

Hence there can be no such $z \in A$ for which $P$ fails to hold.