I am reading a book about logic and in the proof of the completeness theorem there is a point I would like to clarify.
The part that is troubling me is the "easy" part, so basicly that $T \vdash F$ implies that $T \vDash F$.
In the book the author uses the natural deduction system. To prove this he takes $M$ to be a model of $T$ and he supposes that we have a proof of $F$ using the axioms of $T$ and proceeds by induction on the size of the proof.
In the induction he proves things for the rules associated to the symbols $\rightarrow, \land, \forall, \exists$ and says that we can use similar arguments for $\lor$, so far so good until here everything seems legit. But after this he concludes the proof, and my problem is that he did not do any details for the symbol $\neg$ which bothers me. After this the author says that we obtain the following corollary: If $T$ has a model then $T \nvdash \bot$. This bothers me because when I try to prove the $\neg$ part in the completeness theorem I end up proving the corollary first. I will explain my idea bellow.
The rules I am using for $\neg$ are the following:
Rule 1: From $\Gamma, A \vdash \bot$ we obtain $\Gamma \vdash \neg A$
Rule 2: From $\Gamma \vdash \neg A$ and $\Gamma \vdash A$ we obtain $\Gamma \vdash \bot$
Rule 3: From $\Gamma, \neg A \vdash \bot$ we obtain $\Gamma \vdash A$
And basicly I want to consider the last rule used in the proof is one of the three rules and show that the induction works. This is how I am thinking about doing it:
If the last rule is 2:
We may use our induction to prove that we must have that $M \vDash \neg A$ and $M \vDash A$ which is impossible by definition of $\vDash$. Thus this proves that if $T$ has a model then $T \nvdash \bot$. (this is the mentioned corollary above) Is this reasoning correct?
If the last rule is 1:
If we suppose that $M \vDash A$ then we would have $M \vDash T \cup \lbrace A \rbrace$, which means that $M$ is also a model for $T \cup \lbrace A \rbrace$. But this is not possible because we have $T \cup \lbrace A \rbrace \vdash \bot$ which means that $T \cup \lbrace A \rbrace$ does not have any model by the above discussion. Thus we cannot have $M \vDash A$, in other words we have $M \vDash \neg A$.
If the last rule used is 3 it is almost the same as for 1.
Could someone tell me if this reasoning is correct? Maybe I am over complicating simple things.
I think I finally understood the solution. So first of all as pointed out in the comments what I am trying to understand is the soundness theorem and not the completeness theorem.
The details I wrote for rule 2 is correct however I thought this also proves that if $T$ has a model then $T \nvdash M$ but this is not correct. Because I thought that if we have $T \vdash \bot$ the last rule used in our proof must be rule 2, but this is incorrect, for example we may choose $T = \lbrace \bot \rbrace$ we can very easily prove that $T \vdash \bot$ by using the axiom rule, thus not using rule 2. Thus the proof of my rule 1 is also incorrect because I was using this fact.
Here is how one can prove rule 1:
If our proof ends by rule 1, and $M$ is a model for $T$, by contradiction we can suppose that we have $M \vDash A$, since $M$ is a model we also have $M \vDash T$ thus we have $M \vDash T \cup \lbrace A \rbrace$. This means that $M$ is a model for $T \cup \lbrace A \rbrace$. And since to use rule 1 one needs to already have $T, A \vdash \bot$ thus a shorter proof of $T, A \vdash \bot$ which means that we can use our induction hypothesis to obtain that one has $M \vDash \bot$ but this is not possible thus we must have $M \vDash \neg A$.
(rule 3 can be proved in the same way.)
Thank you for all of your comments they helped me finding this solution.