1. Let P be a predicate

2. Let SEP be the property of being a set of things that satisfies P

3. Let SP be the property of satisfying P

4. Let NA be the property of not being in A

5. ∃A∃B(SEP(A)∧SP(B)∧NA(B))→∀A∃B(SEP(A)→SP(B)∧NA(B))

6. (5) is true since it is a validity in first order logic. Also, the antecedent is true. Thus, the consequent is true too.

7. ∀A∃B(SEP(A)→SP(B)∧NA(B))≡∀A(SEP(A)→∃SP(B)∧NA(B))

8. ∀A(SEP(A)→∃B(SP(B)∧NA(B)))≡∀A(¬∃B(SP(B)∧NA(B))→¬SEP(A))

9. ∀A(¬∃B(SP(B)∧NA(B))→¬SEP(A)) translates to: For all A, if there doesn’t exist B such that B has the property of satisfying P and B has the property of not being in A, then A doesn’t have the property of being a set of things that satisfy P.

10. (9) is relevant because of the following: Let P be the predicate not a member of itself. So, we have: For all A, if there doesn’t exist B such that it has the property of not being a member of itself and the property of not being in A, then A doesn’t have the property of being a set of things that aren’t members of themselves.

• would be cool for the slow of learning if you defined the difference Commented Jun 12 at 4:55
• How is SP(B) different from B? How is NA(B) different from ~B? Formula 5 is not valid in first order logic. SEP, SP, and NA are not parts of first order logic, so for the formula to be valid, it would have to be true for any predicates you substitute for those three names. Commented Jun 12 at 14:21
• I just realized that A and B are meant to represent predicates, so this isn't even first order logic. Commented Jun 12 at 14:27
• Also, you say that SEP, SP, and NA are "properties" which are predicates, but you use them syntactically as propositions. Commented Jun 12 at 14:38
• Voting to close until the question is clarified. Commented Jun 12 at 14:46

The original distiction between logical and epistemological contradictions was introduced by Ramsey (1926) (but it had already been hinted at by Peano in 1906):

"While logical contradictions involve mathematical or logical terms, like class, number, and hence show that our logic or mathematics is problematic, semantical contradictions involve, besides purely logical terms, notions like “thought”, “language”, “symbolism”, which, according to Ramsey, are empirical (not formal) terms."

According to this, Russell's Paradox (1901), as well as Burali-Forti's (1897), etc are "logical" ones while the Liar, Richard's paradox (1905), Berry paradox (1908), Grelling–Nelson paradox are "epistemological".

The paradoxes belonging to the second category, involving the notions of truth and self-reference, are also named "semantical antinomies", following Tarski's development of formal semantics during the 1930s.

See Alfred Tarski, The Semantic Conception of Truth and The Foundations of Semantics (1944): "For although the meaning of semantic concepts as they are used in everyday language seems to be rather clear and understandable, still all attempts to characterize this meaning in a general and exact way miscarried. And what is worse, various arguments in which these concepts were involved, and which seemed otherwise quite correct and based upon apparently obvious premises, led frequently to paradoxes and antinomies. It is sufficient to mention here the antinomy of the liar, Richard's antinomy of definability (by means of a finite number of words), and Grelling-Nelson's antinomy of heterological terms."

• I agree. I think Russell’s paradox is a semantic paradox because the notion of set while intuitive is ambiguous. Commented Jun 12 at 12:09
• Also, the notion of predicate, while intuitive is also ambiguous. Because it could either refer to a property or a relation which is not identical to a relational property. Commented Jun 12 at 12:11
• I also noticed that mathematicians don’t really distinguish between syntactic equivalence and semantic equivalence and that they are not identical. Commented Jun 12 at 12:14
• @AUTISTINC: The notion of set, being algebraically free, is not ambiguous. Indeed, type theories know that sets are types with equality. However, no first-order set theory can fully describe all sets (Gödel, Löwenheim-Skolem) and so the choice of axioms is relevant. Similarly, the notion of proposition ("predicate") is not ambiguous; propositions are merely truncated sets. Commented Jun 12 at 15:34
• No. I am using set and predicate in their natural language senses though. Commented Jun 12 at 16:14

It is a set-theoretic paradox. Syntactic paradox would imply it follows from certain syntactical rules. Assuming you treat the axioms of naïve set theory as syntactic, then Russell's paradox is a syntactic paradox.

I don't see the necessity of writing out a formal argument. This is a philosophy site and not a formal logic site. You might find Math.SE more jospitable to your style of thinking.

As for your headline question, it is both. We can write out Russell's paradox in natural language. This was done by Russell himself. This is the Barber Paradox and goes like this. A barber is the "one who shaves all those, and those only, who do not shave themselves". The question is, does the barber shave himself? If yes, then not; if not,then yes. So a paradox. And it is a semantic paradox because it relies on natural language.

On the other hand, any good book on logic/set theory will show how to formalise this paradox and how to get rid of it. This can be thought of as the syntactical version of the paradox.

• It relies on natural language, but it doesn't rely on the meaning of the words in the natural language. To show that the barber paradox is a paradox, we do not have to understand what the words "barber" or "shave" actually mean. Commented Jun 13 at 16:35
• @kaya3: No, but it makes it much easier to understand. The notion that formalism is the only true way to understand mathematics is dead in the water after Godel's theorem as pointed by Freeman Dyson. Only someone with a formalist fetish would think otherwise. Commented Jun 30 at 6:47