[FOM] A proof that cannot be formalized in ZFC

laureano luna laureanoluna at yahoo.es
Wed Oct 10 14:42:07 EDT 2007

Alex Blum wrote:

>Why does it follow that if there is no model of ZFC
>in which con(ZFC)
> is 
>true that the negation of con(ZFC) has a proof in
>Doesn't this assume that ZFC is  complete?

No, it doesn't.

For any first order theory T, T has a model iff T is
consistent (here I had previously mentioned Henkin's
1949 result but Martin Davis has reminded me of the
fact that this was already in Gödel's 1930
completeness theorem); for any theory T and
any sentence S, T+S is inconsistent iff the negation
of S is a logical consequence of T.

Now, since first order logic is complete, ZFC proves
whatever is a logical consequence of its axioms. So,
if ZFC+Con(ZFC) has no model, then ZFC proves the
negation of Con(ZFC).
Laureano Luna

Sé un Mejor Amante del Cine                         
¿Quieres saber cómo? ¡Deja que otras personas te ayuden!

More information about the FOM mailing list