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tech / sci.math / How Russell's Paradox is Avoided by Using the Subset Axiom in DC Proof

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* How Russell's Paradox is Avoided by Using the Subset Axiom in DC ProofDan Christensen
`- Re: How Russell's Paradox is Avoided by Using the Subset Axiom in DC ProofY Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y E S

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How Russell's Paradox is Avoided by Using the Subset Axiom in DC Proof

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Subject: How Russell's Paradox is Avoided by Using the Subset Axiom in DC Proof
From: Dan_Chri...@sympatico.ca (Dan Christensen)
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 by: Dan Christensen - Sat, 24 Dec 2022 00:11 UTC

Statement of Russell's Paradox: Contrary what was once thought, there does NOT exist a set y such that ALL(a):[a in y <=> ~a in a]. Assuming it does exist, leads a contradiction.

PROOF

Suppose to the contrary....

1. EXIST(y):[Set(y) & ALL(a):[a in y <=> ~a in a]]
Premise

Applying Existential Specification...

2. Set(y) & ALL(a):[a in y <=> ~a in a]
E Spec, 1

3. Set(y)
Split, 2

4. ALL(a):[a in y <=> ~a in a]
Split, 2

Apply this definition to y itself, we obtain the contradiction...

5. y in y <=> ~y in y
U Spec, 4

As required:

6. ~EXIST(y):[Set(y) & ALL(a):[a in y <=> ~a in a]]
Conclusion, 1

Now, if y was a SUBSET of another set x, then the problem disappears!

PROOF

Let x be any set.

1. Set(x)
Premise

Applying the Subset Axiom...

2. EXIST(sub):[Set(sub) & ALL(a):[a in sub <=> a in x & ~a in a]]
Subset, 1

3. Set(y) & ALL(a):[a in y <=> a in x & ~a in a]
E Spec, 2

4. Set(y)
Split, 3

5. ALL(a):[a in y <=> a in x & ~a in a]
Split, 3

Applying this definition to y itself...

6. y in y <=> y in x & ~y in y
U Spec, 5

7. [y in y => y in x & ~y in y] & [y in x & ~y in y => y in y]
Iff-And, 6

8. y in y => y in x & ~y in y
Split, 7

9. y in x & ~y in y => y in y
Split, 7

Suppose to the contrary that...

10. y in y
Premise

11. y in x & ~y in y
Detach, 8, 10

12. ~y in y
Split, 11

13. y in y & ~y in y
Join, 10, 12

By contradiction..

14. ~y in y
Conclusion, 10

Now, suppose to the contrary...

15. y in x
Premise

16. y in x & ~y in y
Join, 15, 14

17. y in y
Detach, 9, 16

18. y in y & ~y in y
Join, 17, 14

By contradiction...

19. ~y in x
Conclusion, 15

Joining results...

20. ~y in x & ~y in y
Join, 19, 14

21. Set(y) & ALL(a):[a in y <=> a in x & ~a in a]
& [~y in x & ~y in y]
Join, 3, 20

As required:

22. ALL(x):[Set(x)
=> EXIST(y):[Set(y) & ALL(a):[a in y <=> a in x & ~a in a]
& [~y in x & ~y in y]]]
Conclusion, 1

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Re: How Russell's Paradox is Avoided by Using the Subset Axiom in DC Proof

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Subject: Re: How Russell's Paradox is Avoided by Using the Subset Axiom in DC Proof
From: yyyyyyyy...@gmail.com (Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y E S)
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 by: Y Y Y Y Y Y Y Y Y Y - Sat, 7 Jan 2023 09:35 UTC

G00d mornin9 ................................................................

Are You goin9 to do something 9reat t0day ?

Wishing You a great day........

A NN GGGGGG EEEEEEEEEEEEEEEEEE LLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

On Saturday, December 24, 2022 at 2:11:43 AM UTC+2, Dan Christensen wrote:
> Statement of Russell's Paradox: Contrary what was once thought, there does NOT exist a set y such that ALL(a):[a in y <=> ~a in a]. Assuming it does exist, leads a contradiction.
>
> PROOF
>
> Suppose to the contrary....
>
> 1. EXIST(y):[Set(y) & ALL(a):[a in y <=> ~a in a]]
> Premise
>
> Applying Existential Specification...
>
> 2. Set(y) & ALL(a):[a in y <=> ~a in a]
> E Spec, 1
>
> 3. Set(y)
> Split, 2
>
> 4. ALL(a):[a in y <=> ~a in a]
> Split, 2
>
> Apply this definition to y itself, we obtain the contradiction...
>
> 5. y in y <=> ~y in y
> U Spec, 4
>
> As required:
>
> 6. ~EXIST(y):[Set(y) & ALL(a):[a in y <=> ~a in a]]
> Conclusion, 1
>
> Now, if y was a SUBSET of another set x, then the problem disappears!
>
> PROOF
>
> Let x be any set.
>
> 1. Set(x)
> Premise
>
> Applying the Subset Axiom...
>
> 2. EXIST(sub):[Set(sub) & ALL(a):[a in sub <=> a in x & ~a in a]]
> Subset, 1
>
> 3. Set(y) & ALL(a):[a in y <=> a in x & ~a in a]
> E Spec, 2
>
> 4. Set(y)
> Split, 3
>
> 5. ALL(a):[a in y <=> a in x & ~a in a]
> Split, 3
>
> Applying this definition to y itself...
>
> 6. y in y <=> y in x & ~y in y
> U Spec, 5
>
> 7. [y in y => y in x & ~y in y] & [y in x & ~y in y => y in y]
> Iff-And, 6
>
> 8. y in y => y in x & ~y in y
> Split, 7
>
> 9. y in x & ~y in y => y in y
> Split, 7
>
> Suppose to the contrary that...
>
> 10. y in y
> Premise
>
> 11. y in x & ~y in y
> Detach, 8, 10
>
> 12. ~y in y
> Split, 11
>
> 13. y in y & ~y in y
> Join, 10, 12
>
> By contradiction..
>
> 14. ~y in y
> Conclusion, 10
>
> Now, suppose to the contrary...
>
> 15. y in x
> Premise
>
> 16. y in x & ~y in y
> Join, 15, 14
>
> 17. y in y
> Detach, 9, 16
>
> 18. y in y & ~y in y
> Join, 17, 14
>
> By contradiction...
>
> 19. ~y in x
> Conclusion, 15
>
> Joining results...
>
> 20. ~y in x & ~y in y
> Join, 19, 14
>
> 21. Set(y) & ALL(a):[a in y <=> a in x & ~a in a]
> & [~y in x & ~y in y]
> Join, 3, 20
>
> As required:
>
> 22. ALL(x):[Set(x)
> => EXIST(y):[Set(y) & ALL(a):[a in y <=> a in x & ~a in a]
> & [~y in x & ~y in y]]]
> Conclusion, 1
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

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