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Nov 6, 2021 at 22:29 comment added shane.orourke Also if $d$ satisfies S1, so does $-d$.
Sep 29, 2021 at 8:28 comment added user3840170 S2 is not a good definition. It leaves gcd(0, 0) undefined where it should be 0.
Sep 29, 2021 at 6:04 comment added mrtaurho @LSpice Ah, I see. Then I most likely misremembered something here. Thanks for clearing that up!
Sep 28, 2021 at 23:14 comment added LSpice @mrtaurho, re, certainly I agree that induction for $\omega + \omega$ is stronger than well ordering for $\omega$ (though not than well ordering for $\omega + \omega$, I think?). But I think injectivity of the successor function (subsingleton preimages) is exactly about unique predecessors. Unique successors is just the statement that the successor function is a function.
Sep 28, 2021 at 21:45 comment added mrtaurho @LSpice This version only guarentees the existence of a unique successor; not predecessor (or I'm confusing things right now). The example $\omega+\omega$ stands regardless, which satisfies well-ordering but not induction.
Sep 28, 2021 at 21:15 history edited LSpice CC BY-SA 4.0
Mild proofreading and TeX
Sep 28, 2021 at 21:14 comment added LSpice @mrtaurho, re, I'm not sure what are the usual Peano axioms, but, in the formulation with which I'm familiar (and that Wikipedia uses), injectivity of the successor map is taken as one of the axioms.
Sep 28, 2021 at 21:00 comment added mrtaurho As far as I remember, well-ordering is not able to prove the full strength of induction as witnessed by $\omega+\omega$. The crucial problem is that "every natural number has a unique predecessor" requires more than the usual Peano Axioms with Induction replaces by Well-Ordering.
S Sep 28, 2021 at 16:53 history answered Aeryk CC BY-SA 4.0
S Sep 28, 2021 at 16:53 history made wiki Post Made Community Wiki by Aeryk