Equality of two circular sets

Given the definition of subsets and equality of sets:

• A $\subset$ B, if x $\epsilon$ A $\rightarrow$ x $\epsilon$ B for every set x.
• A = B, if A $\subset$ B and B $\subset$ A

Why is it impossible to decide whether two circular sets I = {I} and J = {J} are equal.

I mean, the way is see it is that I is not an element of J, since only J is an element of J, so the two circular sets are not equal.

What's wrong in my reasoning?

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You're assuming I≠J to get that I is not an element of J. – Ricky Demer Sep 8 '10 at 11:07
Right, and to verify this I have to perform the same operation again and I will end up with the same problem. That makes sense. – berater Sep 8 '10 at 11:24
Note that x={x} is explicitly forbidden by the ZF axiom of regularity (en.wikipedia.org/wiki/Axiom_of_regularity). But if decide that you are talking of another binary relation, that you still denote $\in$, but it is not the usual "is an element of" of ZF set theory, and X={X} may happen, then you also can have many such "self-singletons", why not. In particular, deciding if 2 sets are equal can't be decided iterating the procedure of checking if their elements are equal, like in your example. – Pietro Majer Sep 8 '10 at 11:30
@berater: you could as well check that $I=J$. It would reduce to checking $J=I$ again, which makes as much sense than the converse. – Benoît Kloeckner Sep 8 '10 at 11:45
There is some relevant discussion of non-well-founded sets and the relevant axioms of foundation, anti-foundation etc at another MO question: mathoverflow.net/questions/33282/can-we-have-aa – Peter LeFanu Lumsdaine Sep 8 '10 at 13:55