My question is about exercise 1.6 in "Propositional and Predicate Calculus" by D. Goldrei: "Explain how the above variant of the method of proof by mathematical induction follows from the principle of mathematical induction. [Hint: You might wish to exploit the well-order property of $\Bbb N$.]" The "above variant" refers to strong induction, the "principle of mathematical induction" to weak induction.
Assuming my proof is correct, it only needs the well-order property to show the validity of strong induction:
Let $P$ be a subset of $\Bbb N$ with $0 \in P$ and the property that for all $n \in \Bbb N$, if $k \in P$ for all $k \leq n$, then $k \in P$ for all $k \leq n+1$. Assume there exists a non-empty set $S$ containing the natural numbers not in $P$. By the well-order property, $S$ has a least natural number $s_0$. Since $0 \in P$, $s_0 \neq 0$. Thus, there is a $t \in \Bbb N$ such that $t+1=s_0$. Notice $t \notin S$ because $t<s_0$, meaning $t \in P$. More specifically, $k \in P$ for all $k \leq t$. According to the assumption about $P$, this implies $t+1=s_0 \in P$ and, thus, $S=\emptyset$ and $P=\Bbb N$.
How would I incorporate both the weak induction and the well-order property, as hinted in the exercise, when proving strong induction?