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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.