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Topic: A finite set of all naturals
Replies: 32   Last Post: Aug 15, 2013 4:07 PM

 Messages: [ Previous | Next ]
 Peter Percival Posts: 2,623 Registered: 10/25/10
Re: A finite set of all naturals
Posted: Aug 12, 2013 7:40 AM

Ben Bacarisse wrote:
> fom <fomJUNK@nyms.net> writes:
> <snip>

>> Here is the proof he provides when pressed to produce:
>>

>
> I could not resist taking just a little peek... That post claims to be
> "written in the language of arithmetic L(PA)" but it contains this
> definition early on:
>
> Def-03a: even1(x) <-> Ey[x=y+y]
> Def-03b: even2(x) <-> Ey[x=2*y]
> Def-03c: even(x) <-> (even1(x) \/ even2(x))
>
> Does this mean that Ax[even1(x)=even2(x)] is not a theorem of the system
> being used, or does it mean that the author is inclined to use overly
> complex definitions? Either way, I didn't find it an encouraging start.
>
> Maybe I don't understand the notation, because just a few lines further
> on I see:
>
> Def-05c: aGC(x) <->
> (even(x) /\ (SS0<x)) -> Ap1p2[prime(p1) /\ prime(p2) /\
> (p1+p2<x \/ x<p1+p2)]
>
> but isn't Ap1p2[prime(p1) /\ prime(p2) /\ (p1+p2<x \/ x<p1+p2)] simply
> false?
>
> <snip>

Yes, since not every natural number is a prime, and a conjunct claims
otherwise, the whole is false. Perhaps

Ap1p2[[prime(p1) /\ prime(p2)] -> (p1+p2<x \/ x<p1+p2)]

was meant.

--
Nam Nguyen in sci.logic in the thread 'Q on incompleteness proof'
on 16/07/2013 at 02:16: "there can be such a group where informally
it's impossible to know the truth value of the abelian expression
Axy[x + y = y + x]".

Date Subject Author
8/12/13 Ben Bacarisse
8/12/13 Peter Percival
8/12/13 fom
8/12/13 Ben Bacarisse
8/12/13 namducnguyen
8/12/13 namducnguyen
8/12/13 antani
8/12/13 namducnguyen
8/13/13 Marshall
8/13/13 quasi
8/13/13 namducnguyen
8/13/13 quasi
8/13/13 namducnguyen
8/13/13 namducnguyen
8/13/13 quasi
8/13/13 namducnguyen
8/14/13 quasi
8/14/13 namducnguyen
8/14/13 quasi
8/14/13 namducnguyen
8/15/13 namducnguyen
8/15/13 Virgil
8/15/13 namducnguyen
8/15/13 antani
8/13/13 antani
8/13/13 Peter Percival
8/13/13 Ben Bacarisse
8/13/13 namducnguyen
8/14/13 Peter Percival
8/14/13 Shmuel (Seymour J.) Metz