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:
 * 1) The minimal element of $$S$$ is an element of $$T$$;
 * 2) For any $$s \in S$$, if $$\forall t \in S: t \preceq s \implies t \in T$$, then $$s \in T$$.

Then $$T = S$$.

Proof
Suppose $$T \ne S$$.

Then $$S - T \ne \varnothing$$.

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

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

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

By condition $$1$$ above, $$s \ne m$$ as $$m \in T$$.

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

So by $$2$$ above, $$s \in T$$ and so $$s \notin S - T$$

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