Editing Axiom of regularity (section)

==The axiom of dependent choice and no infinite descending sequence of sets implies regularity==
Let the non-empty set ''S'' be a counter-example to the axiom of regularity; that is, every element of ''S'' has a non-empty intersection with ''S''. We define a binary relation ''R'' on ''S'' by <math display="inline">aRb :\Leftrightarrow b \in S \cap a</math>, which is entire by assumption. Thus, by the axiom of dependent choice, there is some sequence (''a<sub>n</sub>'') in ''S'' satisfying ''a<sub>n</sub>Ra<sub>n+1</sub>'' for all ''n'' in '''N'''. As this is an infinite descending chain, we arrive at a contradiction and so, no such ''S'' exists.