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I have written a formal proof of the theorem:

$\forall U \exists r(\forall a(a\in r \leftrightarrow (a\in U \wedge a\notin a)) \wedge r\notin U \wedge r\notin r)$

See: http://www.dcproof.com/SeparationAxiom.htm

(Somewhat non-standard notation: & = $\wedge$, | = $\vee$)

It seems you can select a subset much like the so-called Russell Set from any set $U$ without obtaining a contradiction. How can I be certain?

Edit: Apparently you can't be certain. See my own answer below.

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    Let me just finish this discussion with the remarked that if I could write a proof for the inconsistence of ZF, I would probably submit it as a dissertation and skip all those years of extensive study instead. Good for me, I cannot write such proof so I can spend a few more years as a student meddling with ZF.2012-06-26

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I'll rewrite my answer.

First, the question is vague. You are saying:

I have proven a theorem $T$ in some axiom system. How do I show that $T$ does not lead to a contradiction, given that another statement, $T'$, that looks superficially similar, does lead to a contradiction.

If you want us to check your proof, which was perhaps your goal, given that you link to your proof, we'd have to know what axioms you are using. That said, the theorem you say you have proven follows pretty much directly from most axioms for set theory. Given that you haven't asked us explicitly to verify your proof, and that is a lot of work, I will not be attempting that.

Given that you have proven the theorem in some axiom system, it technically "leads to a contradiction" if and only if the original axiom system leads to a contradiction. Given your comments, you are not asking about the consistency of the original axioms, so I will skip this.

Perhaps you mean, "How do I prove this does not directly lead to a contradiction?" The only way to do that is to define "directly," which is actually fairly hard, and is impossible without some explicit axioms.

Finally, perhaps you mean "Why does $T'$ lead to a contradiction, but $T$ doesn't?" This is not a formal question, but it could be a request for clarification of the differences between the two statements that causes on to yield a contradiction but not the other.

So let's write $T'$:

$\exists R: \forall a: a\in R \iff a\notin a$

Why does this yield a contradiction? Because if such an $R$ exists, we can ask "Is $R\in R$?" and we get the usual contradiction: $R\in R \iff R\notin R$.

However, your statement, $T$, does not have such a paradox, because it can be rephrased as:

$\exists R:R\notin R \land R\notin U \land (\forall a: a\in R \iff (a\in U \land a\notin a))$

Where $U$ is some set. But the reason there is no contraction is the simple addition of that condition on $R$: $a\in U$. There is no contradiction because the requirement "$a\in U$" gives you an "out." That "out" is why the $T$ does not lead to a contradiction, but $T'$ does. It is the difference between $a\notin a$ and $a\notin a \land a\in U$. Once you have that condition, $a\in U$, you lose the contradiction.

Note that $R$ depends on $U$. An $R$ for one $U$ is not equal to the $R$ for another $U$. There is not one universal set $R$ which applies to every $U$ - the conditions, in particular, make it clear that $R\subset U$. (This was why I thought you might be confusing the difference between $\forall\exists$ nd $\exists\forall$ in my early answer - a common mistake.)

As I mentioned in the comments, you re-gain the contradiction if you assert the existence of a universal set: $\exists V:\forall a: a\in V$. But that just means that, in this axiom system, that assertion is invalid - you have essentially proven that the universal set does not exist in your axiom system (or your axiom system is inconsistent.)

But again, it is not clear what you are asking. You haven't asked a well-formed question, and most of the answers and comments posted here have been based on efforts to deduce your meaning based on common types of confusion that plague newcomers to set theory and logic. Are you looking for some formal argument? Are you looking just for clarification/elucidation? Are you looking for a review of your proof? Or are you reaching for something else that none of us have quite grasped yet?

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    I think you might want to start considering the possibility that the word you are writing don't actually mean what you think they mean. I'm not trying to be snide, but the initial question is vague (how do I show a theorem I've proven doesn't lead to a contradiction the way an unrelated statement leads to a contradiction?) Several people have tried to guide you into clarifying your question, using common understanding of what you might be trying to ask, but you seem to be missing the plot. Were you just asking us to check your proof?2012-06-25
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As far as I can tell, your question is whether the theory (in the language whose only nonlogical symbol is the binary relation-symbol $\in$) consisting solely of the axiom schema of separation is consistent. This is obvious: it has the empty structure as a model. Another simple model (if your logic doesn't allow empty structures, say) is the one-element structure $\{\varnothing\}$ (with $\varnothing \in \varnothing$ defined to be false).

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    Chris, see my first comment to you above.2012-06-25
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If it's not bad form to answer your own question...

Strictly speaking, if, after postulating the existence of $U$, I did get a contradiction from my construction of $r$, then, even if the axioms of this system were consistent, no sets would exist! (This system doesn't actually assume the existence of any sets, not even the empty set.)

Bottom line, we have no proof that such a contradiction is impossible. Such a contradiction would be problematic not only for the system I use here, but for the standard ZFC set theory as well. With over a century of extensive research behind the ZFC theory, however, such a contradiction seems highly unlikely.

Thanks all for your, ummm.... patience.