Re: incompleteness of first-order logic

Thank you.
Anyway, I wish to see a counterexample with
T |/- p true
but
T |- not p false

Jan Burse wrote:
> Li Yi wrote:
> > 1.
> > Give a concrete counterexample to:
> > For any theory T and sentence p, T |/- p => T |- not p.
> Assume completeness, i.e. T |= A iff
> T |- A. Then take for example the
> void Theory T. It is neither T |- p
> nor T |- ~p.
>
> Because you can find M1={p} with
>
> M1 |= T and not M1 |= ~p
>
> Hence it is not T |- p.
>
> And M2={~p} with
>
> M2 |= T and not M2 |= p
>
> Hence it is not T |- ~p.
>
> > 2.
> > 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.
> This is not true. Th M = {p : M |= p} is
> not a complete theory.
>
> Take for example M2 from above, then Th M2
> is the void theory, and as was shown above,
> the void theory is not complete.
>
> Bye

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