Axiom of Separation & Russell's Paradox I've looked through the discussions of Russell's paradox and the Axiom of Separation to aid in my question but I'm still stuck. Here's the problem:
The Axiom of Separation, I'm told, allows us to define, for every set $A$, the set $\{x|x \in A \land x \notin x\}.$ Call this $S$.
I need to show that from the assumption that either $S \in S \lor S \notin S$ what follows is that $S \notin A.$
But I'm having trouble not getting the contradiction.
Here's what I've done. Assuming that $S \in S$, that means that, given the definition of $S$, that $S\in A \land S\notin S.$ So it's true that $S\in A$ and that $S \notin S$. But if the latter is true, then either $S\notin A \lor S\in S$ is true. But this contradicts what my original assumption.
Does anyone have tips about how to go about the proof? I'm missing something, I just don't know what I'm doing wrong. So, I cannot seem to infer that no contradiction but only that $S \notin A$ follows from my assumption.
 A: I assume the setup of your question is intended to be

Let $A$ be a set. Define $S = \{ x \mid x \in A \wedge x \notin x \}$....

In the argument that follows, you want the contradiction you arrived at! The argument you give is predicated on the assumption that $S \in S$. By derivating a contradiction, you conclude the assumption is false; that is, you conclude $S \notin S$.
No further contradiction follows; when expanding out the meaning of $S \notin S$ and simplifying, you ultimately arrive at the conclusion $S \notin A$.
A: Hint: You have set A and subset S of A such that $\forall x:[x\in S \iff x\in A \land x\notin x]$. First, prove by contradiction that $A\notin S$. Then prove by contradiction that $S\notin S$. Finally, prove that $S\notin A$.
A: The issue is to show that the Axiom of Separation avoid the reproduction of the Russell's Paradox.
The "Russell's argument" applied with Separation amounts to concluding with the innocuous $S \notin A$.
By Separation, we have that, for any set $A$, the set:

$S = \{ x \mid x \in A \text { and } x \notin x \}$

exists.
Assume that $S \in A$, and reason by cases.
Firts possibility (i): $S \in S$. 
By definition of $S$, we have $S \in \{ x \mid x \in A \text { and } x \notin x \}$. The syntax of $\{ x \mid \varphi(x) \}$ is so that $a \in \{ x \mid \varphi(x) \}$ iff $\varphi(a)$.
Thus, we have that $S \in A$ and $S \notin S$, and thus we have:

if $S \in S$, then $S \notin S$.

Consider nowe (ii): if $S \notin S$, we have that $S \in A $ and $S \notin S$.
So $S$ satisfies the condition of the above definition of $S$, i.e. (again from $\varphi(a)$ to $a \in \{ x \mid \varphi(x) \}$) we have $S \in \{ x \mid x \in A \text { and } x \notin x \}$, that means: $S \in S$.
Nowe we have proved:

if $S \notin S$, then $S \in S$.

From the two results above, we conclude with:

$S \in S$ iff $S \notin S$,

and this is a contradicition.
But this is the general schema of a proof by contradiction: having assumed $S \in A$ and having derived a contradiction, we conclude that our original assumption is untenable, and we have proved the denial of our assumption:


$S \notin A$.


