That is not the case!y point is that ZF exclude an axiom of the empty set because the empty set concept is a theorem within the framework of ZF, not an axiom.
The theorem is that if there exists a set, there exists an empty set. IOW:
(Ex) => (EyAz:!(z is in y))
The catch is that the axiom of the empty set is the only ZFC axiom capable of proving Ex. (Note: you can't even state the axiom of infinity unless the empty set is known to exist)
Let me restate this to emphasize what this means:
Without the axiom of the empty set, ZFC cannot prove the existance of even a single set.
I am asserting:Could you specify what you mean? And where are the contradictions with my definitions?
Az: [ (z = 0) <=> (Ay: ~(y in z)) <=> (AT: z U T = T) <=> (AT: z x T = z) ]
(where 0 is as I defined it in my previous post)
I'm not sure why you ask about constradictions with your definitions; I think your definition (if I interpret you correctly) is equivalent to the usual definition, but is redundant, complicated, and confusing.