Predicate theory having a model, what are the consequences?

Question

While solving some predicate logic exercises I hit a bit of a wall where I can't say if I understood the implications of a theory having a model well enough.

Let $T$ be a predicate theory and $E$ be a statement, if $A\vDash T$ and $A\vDash E$ where $A$ is a structure, what can I say about some of the following?

1. $T\vdash E$, this should be false, let's say that $A$ is an abelian group, $T$ is the group theory and $E$ the abelian statement, this would falsify this.

2. $T \not\vdash \lnot E$, can I use Gödel's completeness theorem to say that $T$ is consistent (since it has a model) and thus it can never prove a contradiction (we can assume that there exist an E such that $E\in T$ right?)

3. $T \cup E$, is consistent (again using Gödel'is completeness theorem)

4. if $ T\not\vdash E$ then T is incomplete, if 2. is true then it can neither prove $E$ nor can T prove $\lnot E$ so T is incomplete.

5. $T$ is consistent, this is an assumption i made for point 2. but does it hold?

6. if every model of $T$ is isomorphic to $A$ then $ T\vDash E$. If this is true then for every model of $T$, $E$ is also satisfied, so $T \vDash E$

Have I made any mistakes in my reasoning?

Answer 1

We have that $T \vdash E \iff T \models E$, from soundness and Gödel's completeness theorems for first-order logic:

1. You have it right, we cannot conclude $T \vdash E$ without further hypotheses, since $T \vdash E \iff T \models E \iff$ for all $A \models T$ we have $A \models E$, for which you provide a counterexample.
2. $T \not\vdash \neg E$, this is true: assume $T \vdash \neg E \implies T \models \neg E \implies A \models \neg E$, contradiction.
3. $A \models T \cup \{E\}$, so $T \cup \{E\}$ is consistent, see here.
4. If $T \not \vdash E$ then $T$ is incomplete, since, given 2, this is the very definition of incompleteness of a theory.
5. $T$ is consistent since $A \models T$, see 3.
6. If every model of $T$ is isomorphic to $A$ then $T \models E$. Indeed, then every model of $T$ would be elementarily equivalent to $A$ (note that the converse does not hold), $T$ would be complete (see here) and combining with 2 we arrive at $T \vdash E$, which implies $T \models E$.