[FOM] consistency and completeness in natural language

[Torkel Franzen]

Neil says:

>	Assume that the formal system S is consistent. Then the
>	G"odel-sentence (x)A(x) for S is not provable S. 
>
>	[The proof of this unprovability result makes no mention at all of
>	truth. It is a purely syntactic result.]
>
>        Each instance A(_n) is of the form not-B(_n), where B(_n)
>	represents "n is a proof, in S, of (x)A(x)". If, for any n, B(_n) 
>	were a theorem of S, then there would be a proof, in S, of (x)A(x). 
>	But there is no proof, in S, of (x)A(x). Hence B(_n) is not a
>	theorem	of S (for any n). But every p.r. statement is provable or 
>	refutable in S. And for each n, B(_n) is p.r. Hence, for every n,
>	B(_n) is refutable in S; whence, A(_n) [i.e., not-B(_n)] is
>	provable in S.
>
>	[Note that we have thus far confined ourselves to syntax, talking 
>	only about provability in S. We have not yet made any use of
>	the notion of truth (in the intended model). We are arguing in the
>	metalanguage with regard to the formal system S. Now, for the
>	first time, we bring in the notion of truth:]
>
>	Now assume that every p.r. sentence provable in S is true in N
>	(i.e., the intended model of the natural numbers). 
>	
>	So for every n, A(_n) is true in N. 
>
>	Note further that our metalinguistic quantifications "for every n"
>	are intended to range over the natural numbers,	which form the
>	domain of N. 
>
>	So now, by the semantical rule for the universal quantifier, it
>	follows (in the metalanguage) that the universal sentence (x)A(x)
>	is true in N. That is, the independent G"odel-sentence for S is
>	true in N. 

>Unlike Torkel, I see no vitiating or "baffling" circularity in this piece
>of metalinguistic reasoning.

  I don't see any circularity either, but how about sheer baffling
pointlessness? By standard properties of "true", (x)A(x) is true if
and only if every A(_n) is true, A(_n) is true if and only if B(_n) is
not true, and B(_n) is true if and only if n is a proof in S of
(x)A(x). Since (x)A(x) is not provable in S given that S is
consistent, it follows that (x)A(x) is true if S is consistent. We
can't get more semantic than this.  Why the peculiar added bells and
whistles?

---
Torkel Franzen