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3$\begingroup$ "C" already denotes the axiom of choice, so it's not the best abbreviation for another axiom. $\endgroup$– John BentinCommented Jul 22, 2012 at 18:58
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$\begingroup$ Fair enough. I'll change it to "Ca". $\endgroup$– Mike BattagliaCommented Jul 23, 2012 at 6:54
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$\begingroup$ This is not an answer, but a repetition of my linked questions already stated in <mathoverflow.net/questions/28389>. 1/ Is it possible to dispense with axiom 4 (the axiom of pairing from ZF(C)) to develop the Tarski-Grothendieck (TG) set theory? 2/ Does anybody know how to prove the 16 equivalences between conditions of axioms A and A' of Tarski ? Gérard Lang $\endgroup$– Gérard LangCommented Aug 8, 2012 at 21:36
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