Re: incompleteness of first-order logic

Jan Burse wrote: 
M is a model. Let Th M = {p : M |= p}.
Show that Th M is a complete theory, that is to say, Th M |/= p => Th M
|= not p.

Sorry, do you mean by p an arbitrary sentence.
Then it is true in the propositional case.
But false in the predicate logic case.

The reason is that in your univers U of M
you might have elements which might not
have terms in your language.

Thus for example it could be that Th M |/=
forall y R(c,y), and Th M |/= exists y
not R(c,y).

Must think about such an M, let me see in a
next E-mail.
.

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