- #76

Hurkyl

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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.