Suppose you have $A$ iff $B$ iff $C$.

If you assume $A$ to be true to prove $B$, $B$ to be true to prove $C$, and $C$ to be true to prove $A$, then doesn't that imply you've assumed $A$ to be true to prove $A$?

I ask because of the method of proof in https://proofwiki.org/wiki/Equivalence_of_Well-Ordering_Principle_and_Induction.

  • 13
    $\begingroup$ You are not proving any of them. You are proving that they are equivalent. $\endgroup$ Commented Feb 17, 2019 at 20:10
  • 3
    $\begingroup$ The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you? $\endgroup$
    – Asaf Karagila
    Commented Feb 18, 2019 at 8:25
  • $\begingroup$ @Karagila because of the given context. $\endgroup$ Commented Feb 19, 2019 at 17:39

2 Answers 2


Yes, you are absolutely right the proof $A \implies B \implies C \implies A$ only shows that $A,B,C$ are equivalent. So if one of them is true, all the others are true and if one is false all the others are false. However, the proof you linked doesn't try to show that the principle of mathematical/complete induction is true or the principle of well ordering is true, it only shows that they are equivalent. In fact, in the usual framework of mathematics these are taken as axioms. The proof shows that you only need to assume one of them as an axiom and you get the other for free.

  • $\begingroup$ Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately. $\endgroup$
    – Remellion
    Commented Feb 18, 2019 at 2:49
  • $\begingroup$ @Jannik Pitt Suppose we have A iff B and A iff C. Does this mean that B iff C? $\endgroup$
    – ado sar
    Commented Jan 12, 2020 at 21:58
  • $\begingroup$ @user599310 Yes. $\endgroup$ Commented Jan 13, 2020 at 8:15

Yes. But, this helps math, it gives a bunch of statements, that we either can all dismiss at once with a disproof of one, or accept if we accept or prove the others. Goldbach's conjecture, is sufficient to proves Bertrand's postulate. But, Bertrand's postulate is not sufficient to prove Goldbach's conjecture. In this case, we can't use the proof of Bertrand's postulate to prove Goldbach's conjecture. But we could use ANY proof of Goldbach's conjecture, to prove Bertrand's postulate again possibly in a new way. IFF means we could prove any statement in the chain and the other's follow.


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