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Date: Fri, 28 Jan 2022 09:02:09 -0800 (PST)
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Subject: Re: Resolving Burse's Paradox
From: Mostowski Collapse <bursejan@gmail.com>
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Now use your function axiom, and Russells constructions
of ~(x e N), to prove that there is a function that satisfies
your axiom, but disagrees with your conclusion.

Hence your system gets inconsistent.

Dan Christensen schrieb am Freitag, 28. Januar 2022 um 16:58:06 UTC+1:
> On Friday, January 28, 2022 at 10:36:13 AM UTC-5, Dan Christensen wrote: 
> > It was a simple mistake of a faulty definition of a function. Take, for example, an identity function f on N. It could properly be defined as follows: 
> > 
> > ALL(a):ALL(b):[a in N & b in N => [f(a)=b <=> a=b]] 
> > 
> > (Yes, it can be defined more succinctly, but this will be best demonstrate the error in Burse's Paradox.) 
> > 
> > What is f(x) here if x is NOT an element of N? It is not defined here. Simple as that. Jan Burse on the other hand, might have defined it as follows: 
> > 
> > ALL(a):ALL(b):[f(a)=b <=> a in N & b in N & a=b] 
> > 
> > Looks reasonable to the untrained eye, but if x is not element of N, we must conclude that f(x)=/=x. (Proof left as an exercise to the reader.)
> Oh, alright! Here it is: 
> 
> 1. ALL(a):ALL(b):[f(a)=b <=> a in n & b in n & a=b] 
> Axiom 
> 
> 2. ~t in n 
> Premise 
> 
> 3. ALL(b):[f(t)=b <=> t in n & b in n & t=b] 
> U Spec, 1 
> 
> 4. f(t)=t <=> t in n & t in n & t=t 
> U Spec, 3 
> 
> 5. [f(t)=t => t in n & t in n & t=t] 
> & [t in n & t in n & t=t => f(t)=t] 
> Iff-And, 4 
> 
> 6. f(t)=t => t in n & t in n & t=t 
> Split, 5 
> 
> 7. t in n & t in n & t=t => f(t)=t 
> Split, 5 
> 
> 8. ~[t in n & t in n & t=t] => ~f(t)=t 
> Contra, 6 
> 
> 9. ~[~[~t in n | ~t in n] & t=t] => ~f(t)=t 
> DeMorgan, 8 
> 
> 10. ~~[~~[~t in n | ~t in n] | ~t=t] => ~f(t)=t 
> DeMorgan, 9 
> 
> 11. ~~[~t in n | ~t in n] | ~t=t => ~f(t)=t 
> Rem DNeg, 10 
> 
> 12. ~t in n | ~t in n | ~t=t => ~f(t)=t 
> Rem DNeg, 11 
> 
> 13. ~t in n | ~t in n 
> Arb Or, 2 
> 
> 14. ~t in n | ~t in n | ~t=t 
> Arb Or, 13 
> 
> 15. ~f(t)=t 
> Detach, 12, 14 
> 
> 16. ALL(a):[~a in n => ~f(a)=a] 
> Conclusion, 2 
> 
> Dan