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Apr 13, 2017 at 12:19 history edited CommunityBot
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S Aug 23, 2015 at 5:20 history bounty ended drhab
S Aug 23, 2015 at 5:20 history notice removed drhab
Aug 23, 2015 at 5:20 vote accept drhab
Aug 20, 2015 at 16:06 answer added Emil Jeřábek timeline score: 6
Aug 20, 2015 at 0:22 answer added Giraffro timeline score: 7
Aug 19, 2015 at 14:36 comment added drhab @JonMarkPerry I am not asking for that. My question: can this be proved without making use of axiom REG?. Btw, in a collection of axioms redundancy is allowed to appear. No need to take away the label "axiom". (I have to go now, but will be back later).
Aug 19, 2015 at 14:28 comment added JMP if an axiom can be proved using a collection of axioms, then it is not an axiom.
Aug 19, 2015 at 14:25 comment added drhab @JonMarkPerry What do you mean with AOR?
Aug 19, 2015 at 14:14 comment added JMP @drhab; if AOR is an axiom, then what you are asking is impossible.
Aug 19, 2015 at 7:55 comment added drhab @JonMarkPerry 'My' definition of the axiom of regularity: $\forall a\left[\exists x\; x\in a\Rightarrow\exists b\in a\forall x\in a\; x\notin b\right]$.
Aug 19, 2015 at 6:03 comment added JMP how have you defined 'axiom'
Aug 17, 2015 at 10:34 comment added bof This must be a dumb question, but can you prove (without regularity) that a proper class must have an infinite subset? If $C$ were a proper class with no infinite subsets, then the class of all (finite) subsets of $C$ would be well-founded under reverse inclusion, but would have no "minimal" element.
Aug 17, 2015 at 6:54 history edited drhab CC BY-SA 3.0
edited body
S Aug 16, 2015 at 13:38 history bounty started drhab
S Aug 16, 2015 at 13:38 history notice added drhab Draw attention
Aug 16, 2015 at 13:37 history edited drhab CC BY-SA 3.0
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Aug 14, 2015 at 12:22 review First posts
Aug 14, 2015 at 12:47
Aug 14, 2015 at 12:19 history asked drhab CC BY-SA 3.0