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Completeness Theorem says: $\Gamma \models \phi \longrightarrow \Gamma \vdash \phi$

And from definition of satisfaction: $\neg(\Gamma \models \phi) \longleftrightarrow \Gamma \models \neg\phi$

Now see the following and please tell where I am wrong in it:

$(\Gamma \models \phi \longrightarrow \Gamma \vdash \phi)$ implies $\Bigg(\neg(\Gamma \vdash \phi) \longrightarrow \neg(\Gamma \models \phi)\Bigg)$

Now $\neg(\Gamma \models \phi) \longleftrightarrow \Gamma \models \neg\phi$

Hence $\neg(\Gamma \vdash \phi) \longrightarrow \Gamma \models \neg\phi$

And then $\Gamma \models \neg\phi \longrightarrow \Gamma \vdash \neg\phi$

Hence we get $\neg(\Gamma \vdash \phi) \longrightarrow \Gamma \vdash \neg\phi$

Really this can't be true. Because it says all theories are complete becasue completeness theorem holds for each theory where I am exactly wrong in the arguement, I am not able to figure out.

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  • $\begingroup$ The completeness in the "completeness theorem" refers to the logic, rather than the theory. $\endgroup$ – Asaf Karagila Apr 10 '16 at 4:59
  • $\begingroup$ okay I got your point. I'll edit it now @AsafKaragila $\endgroup$ – Sushil Apr 10 '16 at 5:11
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For a structure $\mathfrak M$ and a sentence $\phi$ we do have $$ \neg(\mathfrak M\vDash \phi) \iff \mathfrak M\vDash \neg \phi $$

But this does not imply that $$ \neg(\Gamma\vDash \phi) \iff \Gamma\vDash \neg \phi $$ for a theory $\Gamma$, because the notation $\Gamma\vDash$ hides an implicit quantification over all structures that satisfy $\Gamma$, and this quantification does not commute with the negation.

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