Is the Russell's Paradox Resolved in Predicate Calculus?

[solakis1]

Prove (formall
y) in predicate calculus :

$\neg\exists y\,\forall x\,(x\in y\leftrightarrow \neg x\in x)$.

 

[MarkFL]

Prove (formall
y) in predicate calculus :

$\neg\exists y\,\forall x\,(x\in y\leftrightarrow \neg x\in x)$.

Please post the solution you have ready.

 

[solakis1]

Please post the solution you have ready.

Spoiler

\(\displaystyle \neg\exists y\forall x(x\in y\Longleftrightarrow\neg(x\in x))\)

Proof:

1) \(\displaystyle \exists y\forall x(x\in y\Longleftrightarrow\neg(x\in x))\)........Hypothesis for RAA

2)\(\displaystyle \forall x(x\in y\Longleftrightarrow\neg(x\in x))\).......Hypothesis for existential elimination (EE)

3)\(\displaystyle y\in y\Longleftrightarrow\neg(y\in y)\).........2,Universal elimination (UE)

4) \(\displaystyle y\in y\)..............hypothesis for RAA

5) \(\displaystyle y\in y\Longrightarrow\neg(y\in y)\)......... 3,elimination of double implication (<=>E)

6) \(\displaystyle \neg(y\in y)\).................4,5 M.Ponens

7) \(\displaystyle y\in y\wedge \neg(y\in y)\)............4,6 addition introduction (& I)

8) \(\displaystyle \neg(y\in y)\)...............4 to 7 RAA

9) \(\displaystyle \neg( y\in y)\Longrightarrow y\in y\)..........3,<=> E

10) \(\displaystyle y\in y\)................ 8,9 M.Ponens

11) \(\displaystyle A\wedge \neg A\)...............8,10 CONTRAD

12) \(\displaystyle A\wedge \neg A\)......1,2 to 11 EE

13) \(\displaystyle \neg\exists y\forall x(x\in y\Longleftrightarrow\neg(x\in x))\)..........1 to 12 RAA