Is the Russell's Paradox Resolved in Predicate Calculus?

  • Context:
  • Thread starter Thread starter solakis1
  • Start date Start date
  • Tags Tags
    Calculus Proof
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 1K views
solakis1
Messages
407
Reaction score
0
Prove (formall
y) in predicate calculus :

$\neg\exists y\,\forall x\,(x\in y\leftrightarrow \neg x\in x)$.
 
Physics news on Phys.org
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
 
Last edited by a moderator: