Reformulation of Russell's Paradox? Imagine the set D of all logical propositions that are true if and only if they are false.
Is it proven that this is the same thing as the set of all sets that are not members of themselves?
Could there possibly be a logical proposition which is true iff it's false, even if we don't allow sets to be self-referential?
Edit:
From a naive view, it looks like the claim is that it should be impossible to construct a statement that negates itself.
So for example, the set D1 of all sets that are not members of themselves apart from D1, but D1 is not a member of itself, is a perfectly acceptable set.
And the set D2 of all sets that are not members of themselves apart from D2, and also contains D2 as a member, is another perfectly acceptable set. We can do whatever logic we like on both of those sets.
The problem only comes if we are allowed to state "the set of all sets that are not members of themselves" and use that without qualification. The problem is that we can say that, and then when we say it we are troubled.
I can define a square circle as a circle that has no corners and also has four corners. This is a paradox because if it has no corners then it can't have four corners, and vice versa. But there is no real problem here because there are no square circles to trouble me. I can make up all the bad definitions I want, and nobody will care.
But they do care about the set that's a member of itself iff it is not a member of itself.
Why is THIS paradoxical entity important when none of the others are important?
Why is the solution to the problem not merely to declare that statements which are true iff they are false are badly-formed statements that we will not perform logic on?
 A: Russell’s Paradox tells us that unrestricted comprehension is not a principle that is suitable for set theory. Unrestricted comprehension was an axiom of Frege’s set theory that allows one to assert the existence of a set containing all and only the objects that have some well-defined property $\varphi$. Russell realized that this principle is inconsistent with first-order reasoning. That is, let $\varphi(x)=x \not \in x$. Then, Frege’s unrestricted comprehension gives us a set $A$ that must contain itself, but cannot. If $A$ doesn’t contain itself, then it satisfies $\varphi$ and must be a member of itself. If $A$ contains itself, then it must not contain itself, since members of $A$ don’t contain themselves.
More generally, Russell’s Paradox exemplifies the first-order theorem $\forall x \exists y (Rxy \iff Ryy)$, for an arbitrary two-place predicate “$R$”. Frege’s original axiom yields a formula logically equivalent to the the negation of this theorem where “$R$” is defined as “$\in$”, namely $\exists x \forall y (Rxy \iff \neg Ryy)$.
