Editing Mathematical induction (section)

====Equivalence with ordinary induction====
Complete induction is equivalent to ordinary mathematical induction as described above, in the sense that a proof by one method can be transformed into a proof by the other. Suppose there is a proof of <math>P(n)</math> by complete induction. Then, this proof can be transformed into an ordinary induction proof by assuming a stronger inductive hypothesis. Let <math>Q(n)</math> be the statement "<math>P(m)</math> holds for all <math>m</math> such that <math>0\leq m \leq n</math>"—this becomes the inductive hypothesis for ordinary induction. We can then show <math>Q(0)</math> and <math>Q(n + 1)</math> for <math>n \in \mathbb N</math> assuming only <math>Q(n)</math> and show that <math>Q(n)</math> implies <math>P(n)</math>.<ref>{{cite web |title=Proof:Strong induction is equivalent to weak induction |url=https://courses.cs.cornell.edu/cs2800/wiki/index.php/Proof:Strong_induction_is_equivalent_to_weak_induction |website=[[Cornell University]] |access-date=4 May 2023}}</ref>

If, on the other hand, <math>P(n)</math> had been proven by ordinary induction, the proof would already effectively be one by complete induction: <math>P(0)</math> is proved in the base case, using no assumptions, and <math>P(n+1)</math> is proved in the induction step, in which one may assume all earlier cases but need only use the case <math>P(n)</math>.