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

  • Thread starter Thread starter solakis1
  • Start date Start date
  • Tags Tags
    Calculus Proof
AI Thread Summary
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.
solakis1
Messages
407
Reaction score
0
Prove (formall
y) in predicate calculus :

$\neg\exists y\,\forall x\,(x\in y\leftrightarrow \neg x\in x)$.
 
Mathematics 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:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
6
Views
2K
Replies
7
Views
3K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
7
Views
2K
Replies
6
Views
1K
Replies
1
Views
2K
Back
Top