I am confused by this sentence in the Wikipedia article for "Well-ordering theorem":
…the well-ordering theorem is equivalent to the axiom of choice, in the sense that either one together with the Zermelo–Fraenkel axioms is sufficient to prove the other, in first order logic (The same applies to Zorn's Lemma.) . In second order logic, however, the well-ordering theorem is strictly stronger than the axiom of choice: from the well-ordering theorem one may deduce the axiom of choice, but from the axiom of choice one cannot deduce the well-ordering theorem
Isn't second-order logic stronger than first-order logic? How is it that the equivalence proof in first-order logic is no longer valid in second-order logic?
Best Answer
If you follow the references given in the Wikipedia article, you will find out that the context of this theorem is very different.
While a lot of mathematics is done inside models of $\sf ZFC$ with first-order logic (and so we can make statements about high order logic inside the model). However one can use second-order logic (or rather some systems of second-order logic) as a foundation for mathematics. That is, we no longer work in $\sf ZFC$, we work in a context of second-order logic.
In certain systems which include the axiom of choice, the well-ordering principle is not provable. However without using the axiom of choice it is not hard to show that the well-ordering principle still implies the axiom of choice.
This is essentially theorem 5.4 which you can find on page 107 in the book by Shapiro.