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.
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.