Principle of Mathematical Induction/Well-Ordered Set

Theorem
Let $\left({S, \preceq}\right)$ be a well-ordered set.

Let $T \subseteq S$ be a subset of $S$ such that:
 * $\forall s \in S: \left({\forall t \in S: t \prec s \implies t \in T}\right) \implies s \in T$

Then $T = S$.

Proof
With a view to obtaining a contradiction, suppose that $T \ne S$.

From Set Difference is Subset, $S \setminus T \subset S$.

From Set Difference with Proper Subset, $S \setminus T \ne \varnothing$.

By the definition of a well-ordered set, there exists a smallest element $s$ of $S \setminus T$.

As $s \in S$, it follows from the definition of $T$ that:
 * $\forall t \in S: t \prec s \implies t \in T$

But then $s \in T$ by hypothesis, contradicting the definition of $s$.

Hence the result, by Proof by Contradiction.