MHB Is the Russell's Paradox Resolved in Predicate Calculus?

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The discussion centers on proving the statement $\neg\exists y\,\forall x\,(x\in y\leftrightarrow \neg x\in x)$ in predicate calculus, which relates to Russell's Paradox. Participants are encouraged to provide formal proofs demonstrating that no such set can exist where an element is a member of itself if and only if it is not a member of itself. The focus is on the implications of this statement for set theory and the resolution of the paradox. The thread emphasizes the need for rigorous formalization in predicate calculus to address the paradox effectively. Engaging with this proof is crucial for understanding foundational issues in mathematics.
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Prove (formall
y) in predicate calculus :

$\neg\exists y\,\forall x\,(x\in y\leftrightarrow \neg x\in x)$.
 
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solakis said:
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.
 
MarkFL said:
Please post the solution you have ready.
$$\neg\exists y\forall x(x\in y\Longleftrightarrow\neg(x\in x))$$

Proof:

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

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

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

4) $$ y\in y$$..............hypothesis for RAA

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

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

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

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

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

10) $$ y\in y$$................ 8,9 M.Ponens

11) $$ A\wedge \neg A$$...............8,10 CONTRAD

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

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