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If we have two axioms $A$ and $B$, what exactly is meant by axiom $A$ being weaker than axiom $B$? This question is a follow-up to A weaker Axiom of Infinity?

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We are working within a certain theory $T$ and we are contemplating adding Axiom $A$ or Axiom $B$. Then (in the context of $T$) Axiom $A$ is (strictly) weaker than $B$ if from $T$ together with $B$ we can derive $A$, but from $T$ together with $A$ we cannot derive $B$. Often there is a tacit assumption that $T$ is consistent.

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  • $\begingroup$ Could you have a look at math.stackexchange.com/questions/472045/… ? $\endgroup$ – Dan Christensen Aug 20 '13 at 21:32
  • $\begingroup$ I have not written out details, which is the only way one can be sure, but if one can define the naturals, one can prove the traditional Axiom of Infinity. $\endgroup$ – André Nicolas Aug 20 '13 at 21:56
  • $\begingroup$ In your opinion, is the proposed axiom at the above link weaker than than the usual ZFC Axiom of Infinity? It seems to me that every structure satisfying the ZFC axiom will also satisfy the requirements for the proposed axiom. But not every structure satisfying the proposed axiom will satisfy the requirements for the ZFC axiom (since $x\cup\{x\}\ne x$). $\endgroup$ – Dan Christensen Aug 21 '13 at 3:53
  • $\begingroup$ The issue though is whether given the existence of your $X$, we can prove there is a set $Y$ (certainly not necessarily equal to $X$) satisfying the ordinary Axiom of Infinity. $\endgroup$ – André Nicolas Aug 21 '13 at 4:12
  • $\begingroup$ Isn't the issue really that not every set that satisfies the proposed axiom will satisfy the requirements of the ordinary AOI? For example, some set with $S(a)=a$ for some $a\in X$ will satisfy the requirements of the proposed axiom, but such a set can never satisfy the requirements of the ordinary AOI since $ a\cup \{a\}\ne a$ for all $a$. $\endgroup$ – Dan Christensen Aug 21 '13 at 13:54
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Axiom A is said to be weaker than axiom B if axiom B implies axiom A.

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    $\begingroup$ So being weaker is usually thought in non-strict way, so $A$ is weaker than $A$. $\endgroup$ – user87690 Aug 20 '13 at 18:58

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