EDIT: Starting Over
I would like to see a formal proof in the language of ZFC of:
$\neg\exists S: \forall x:[x\in S \iff x\notin x]$
$\forall S: \exists x: x\notin S$ (added later)
Can someone list the formal proof here if it is not too long, or give me a reference, preferably online?
For 1, this is a derivation with Natural Deduction.
Consider a first-order language with a binary predicate $E(x,y)$ :
1) $(∃y)(∀x)(E(x,y) ↔ ¬E(x,x))$ --- assumed [b]
2) $(∀x)(E(x,s) ↔ ¬E(x,x))$ --- assumed [a] for $\exists$-elimination
3) $E(s,s) ↔ ¬E(s,s)$ --- from 2) by $\forall$-elimination
The formula in 3) is a contradiction, since $E(s,s) ↔ E(s,s)$. Thus, we have :
4) $\bot$ --- from 3)
5) $\bot$ --- from 4) by $\exists$-elimination, discharging [a] : $s$ does not occur in 5)
6) $(∃y)(∀x)(E(x,y) ↔ ¬E(x,x)) \to \bot$ --- from 1) and 5) by $\to$-introduction, discharging [a]
$\vdash \lnot (∃y)(∀x)(E(x,y) ↔ ¬E(x,x))$ --- from 6) by abbreviation : $\lnot p := p \to \bot$.
We have used no $\mathsf {ZFC}$ axioms.
Ref to :
Regarding 2, this formula is not valid.
Consider again the formula of first-order language with the binary predicate $E(x,y)$ :
$∀y∃x\lnot E(x,y)$.
If we interpret it into the domain $\mathbb N$ of natural numbers with $E$ interpreted as $\ge$, we have :
$∀y∃x(x < y)$
that is clearly not satisfied for $y=0$.
Thus, being not valid, it is not provable by rules of logic alone, i.e. :
$\nvdash ∀y∃x\lnot E(x,y)$.
In presence of the Axiom Schema of Separation, we can use the Russell's paradox argument to prove that the assumption :
$\exists y \forall x(x \in y)$
produces a contradiction. Simplifying a little :
1) $\exists y \forall x(x \in y)$ --- assumed [a]
2) $\forall x(x \in V)$ --- assumed [b] for $\exists$-elimination
3) $x \in V$ --- from 3) by $\forall$-elimination
4) $\mathsf {ZFC} \vdash \exists z \forall x (x \in z \leftrightarrow x \in V \land x \notin x)$ --- Axiom of Separation
5) $(x \in z \leftrightarrow x \in V \land x \notin x)$ --- from 4), assuming [c] : $\forall x (x \in z \leftrightarrow x \in V \land x \notin x)$ for $\exists$-introduction and by $\forall$-elimination
6) $(x \in z \leftrightarrow x \notin x)$ --- from 3) and 5) by tautological implication
7) $\exists z \forall x (x \in z \leftrightarrow x \notin x)$ --- from 6) by $\forall$-introduction : $x$ not free in assumptions [a], [b] or [c] and from 4) and 5) by $\exists$-elimination : $z$ not free in 7), discharging [c]
8) $\vdash \lnot \exists z \forall x (x \in z \leftrightarrow x \notin x)$ --- proved above
9) $\bot$ --- from 7) and 8)
10) $\bot$ --- from 1) and 2) by $\exists$-elimination : $V$ not free in 10), discharging [b]
11) $\exists y \forall x(x \in y) \to \bot$ --- from 1) and 10) by $\to$-introduction, discharging [a]
$\mathsf {ZFC} \vdash \lnot \exists y \forall x(x \in y)$ --- from 11) by abbreviation.
Thus, we have concluded with :
$\mathsf {ZFC} \vdash\forall y \exists x(x \notin y)$.
Two Proofs in ZFC that the Russell ``Set" is not a set.
(1) The axiom of regularity states: $\forall x ( x \neq \emptyset \rightarrow \exists y \in x\ ( x \cap y = \emptyset) )$. If we had a set $R = \{x: x \notin x\}$ then consider now $\{R\}$. We see that there must be an element of $\{R\}$ which is disjoint from $\{R\}$. Since the only element of $\{R\}$ is R, it must be that $R \cap \{R\} = \emptyset$. So, since $R \in \{R\}$, we cannot have $R \in R$ (by the definition of disjoint).
(2) If we had a set $R = \{x: x \notin x\}$ then $R \in R$ $\leftrightarrow$ $R \notin R$. By the bi-valence of 1st-order classical logic, we must have either $R \in R$ or $R \notin R$. Since in either we case we have contradictions, and contradictions are not coherent in 1st-order classical-logic, since they would entail the truth of every sentence in the language of ZFC, we reject the existence of $R$.
