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:
 * $$\forall s \in S: \left({\forall t \in S: t \preceq s \implies t \in T}\right) \implies s \in T$$

Then $$T = S$$.

Proof
Suppose $$T \ne S$$.

Then $$S \setminus T \ne \varnothing$$, where $$S \setminus T$$ denotes set difference.

Let $$s$$ be the minimal element of the non-empty set $$S \setminus T$$.

Such an element will always exist because $$S \setminus T \subseteq S$$ from Set Difference Subset and the definition of well-ordered set.

Let $$m$$ be the minimal element of $$S$$.

By definition, $$m \preceq s$$ and so by the properties of $$T$$ it follows that $$m \in T$$.

Now $$s \ne m$$ as we have defined $$s$$ such that $$s \notin T$$.

But we have chosen $$s$$ so that $$t \preceq s \implies t \in T$$.

So by hypothesis, $$s \in T$$ and so $$s \notin S \setminus T$$.

So from this contradiction we see that $$S \setminus T = \varnothing$$ and so $$S = T$$.